Jack Westin's Cyber Month Extended - Valid through 12/05 11:59PM PST Learn More

Jack Westin's Cyber Month Extended - Valid through 12/05 11:59PM PST
Learn More

MCAT Content / AAMC Chemistry Question Pack Solutions

AAMC QPack Chemistry [Web]

Chemistry Question Pack: Passage 1

1) This question asks about the possible bonds formed by each of the molecules in the answer choices. Additionally, going through our choices, we see four distinct molecules: Methane, glycine, carbon dioxide, methanol. We have to determine which molecules can form extensive networks of intermolecular hydrogen bonds with both participating. What do we know about hydrogen bonds? Hydrogen bonds are formed when a hydrogen atom bonds with a very electronegative atom. In a hydrogen bond, hydrogen atoms gain a partial positive charge, and the electronegative atom gains a partial negative charge. Let’s also pull up and reference Table 1.

We can also draw our four options here.

Carbon dioxide does not have any hydrogen atoms, while methane has 4 hydrogen atoms, all bonded to the same carbon. Alternatively, glycine and methanol have hydrogens bonded to different atoms, and are able to accept and donate hydrogens, so an extensive network of hydrogen bonds can be formed.

  1. Methane and methanol. We already went through and mentioned that while methanol is a viable option, methane does not form a hydrogen bond between carbon and hydrogen. On the MCAT we always look for our FON (fluorine, oxygen, nitrogen) atoms when we’re looking at hydrogen bonding. Let’s try and find a better answer. 
  2. Methane and glycine. Similar to answer choice A, this answer also contains methane. Glycine is a viable option, however, so this answer is only half correct, just like answer choice A.
  3. Glycine and methanol. This answer choice supports what I said in the breakdown: glycine and methanol have hydrogens bonded to different atoms, and are able to accept and donate hydrogens, so an extensive network of hydrogen bonds can be formed. Answer choice C is now our best option.
  4. Methanol and carbon dioxide. While we like methanol as a viable option, carbon dioxide doesn’t have any hydrogen atoms, and therefore we don’t expect hydrogen bonding. We can eliminate answer choice D as well. We’re left with our correct answer, answer choice C. 

2) HCl is one of the common strong acids, and this makes sense because archaebacteria thrive in extreme pH. We’re given the molarity of a pool and asked to solve for pH. After reading the question, you can tell right away that it is almost like a standalone question. It can be solved without using the passage. The only information needed to solve for pH is the molarity provided, and our general knowledge. 

pH can be solved as the negative log of hydrogen ion concentration. Because we know our HCl is going to dissociate completely, the hydrogen ion concentration ends up being the same as our molarity. Our molarity is 0.01 molar which can be rewritten using scientific notation. (If you don’t remember which way to move the decimal point, go back to our mnemonic LARS, which is Left add, right subtract). We start with 0.01 x 10^0. We move the decimal point two places to the right to get a whole number, meaning we subtract two from our exponent to give us 1.0 x 10^-2 molar.

Let’s plug in our numbers. We know ph is the negative log of our hydrogen ion concentration. Our hydrogen ion concentration, again, is the same as our molarity. So plug in 1x 10^-2 as our hydrogen ion concentration. So when we take our negative log, we find the pH is 2.

For the MCAT, pH is measured as the negative log of hydrogen ion concentration. We’re going to see values between 0 and 14. We know acids will have a pH below 7 on the scale, and bases will have a pH above 7 on the scale. So after solving for our pH of 2, we can go back to our answer choices to see which one of them make sense. Remember we said pH below 7 means an acid, a pH above 7 means a base.

  1. 12. If you did the calculation and got a pH of 12, that means you incorrectly took the negative log of the hydroxide ion concentration, instead of the negative log of hydrogen ion concentration
  2. 6. This represents a weak acid. Answer choice B is still better than answer choice A which is actually a base. 
  3. 2. This matches our breakdown of the question and the calculated pH. 
  4. 0.01. While this represents a very acidic pH, it does not match the calculated pH the way answer choice C does. We’re left with answer choice C as our best answer.

3) This is going to be similar to our last question. It’s tangentially related to what we read in the passage, but the passage isn’t completely necessary to answer the question. This is like a standalone question that most likely can be answered without the use of the passage. First and foremost, to be able to identify the compound with the same geometry as methane, we have to know the geometry of methane itself, then compare with the additional answer choices.

Methane has a central carbon atom bonded to four hydrogen atoms. The four hydrogens are organized as far apart as possible from one another at around 109-degree bond angles. 

We use VSEPR theory and know the atoms in the molecules achieve a geometry that minimizes repulsion between the electrons in the valence shell of the atom. We know methane has a tetrahedral molecular geometry, because it has four electron dense areas, and no lone pairs.

Keep in mind we used VSEPR theory to determine methane has a tetrahedral molecular geometry. It is imperative to know the concepts on the MCAT content outline, and this is no different. When given the number of electron dense areas and lone pairs, you should be able to determine the molecular geometry of a compound. 

  1. H2S. Answer choice A is hydrogen sulfide-Hydrogen sulfide has four electron dense areas, and has two lone pairs of electrons. That gives it a bent molecular geometry
  2. CO2 Carbon dioxide has two electron dense areas and no lone pairs. So that gives it a linear molecular geometry
  3. XeF4 Xenon tetrafluoride has six electron dense areas, with two lone pairs. Giving it a square planar molecular geometry
  4. SiCl4 Silicon tetrachloride has four electron-dense areas, with no lone pairs. It has a central silicon atom that is bonded to four chlorine atoms-it has the same molecular geometry as methane. It has a tetrahedral molecular geometry. Answer choice D is our correct answer. 

4) The question is ultimately asking us “which answer choice has similar chemical properties to oxygen?” Elements with similar chemical properties to oxygen will most likely accept electrons in a manner similar to oxygen. Elements that are found in the same group on the periodic table tend to have similar characteristics and chemical properties.

We can consult with the periodic table of elements, and it’s unlikely we have to revisit our passage to answer the question.

The best electron acceptors are those which have the highest electron affinity. Electron affinity increases to the right, and up the periodic table.

Looking at our periodic table, we see sulfur is found in the same group as oxygen, meaning it will have similar chemical properties, and also have a high electron affinity compared to our other answer choices. Elements that are found in the same group on the periodic table tend to have similar chemical properties. 

This periodic table is going to be readily available to you on test day, so make sure you get used to using it as necessary while you are practicing.

  1. S We have answer A which is sulfur, which is found in the same group as oxygen and near the top right of the periodic table. That’s where electron affinity is highest. This is a good option right away.
  2. He Next we have answer choice B, helium. Helium is a noble gas, and what did we say about noble gases? We said they’re unreactive and have very low electron affinity
  3. H2 Next we have hydrogen gas which is already a stable molecule, so it’s not going to want to accept any additional electrons
  4. Fe Answer choice D, we have iron. Iron is more likely to donate electrons than be an electron acceptor. That should make sulfur the best answer choice. We can stick with answer choice A with our best option.

 

Chemistry Question Pack: Passage 2

5) This question is asking us to decide whether limonene or positive carvone will distill first. Why did I phrase it this way? Because even though the test-maker doesn’t mention the two components by name, we recall the components of caraway seed oil. 

Here we have our two structures from the passage. Like I mentione, we’ll look at the structure of both carvone and limonene to compare. Carvone has a hexane ring, just like limonene. The biggest difference is the additional carbonyl group, which I’ve circled in red. How does that affect the properties of the two compounds, and finally the boiling point? We’ll have to rely a bit on our content. Oxygen is more polar than carbon, which makes carvone more polar than limonene. More polar compounds have higher boiling points. Said differently, limonene has a lower boiling point than carvone. The additional double bond is also an indicator that carvone has a higher boiling point. Now what are we doing with this information? 

Let’s discuss the experiment as well. During distillation, temperature is slowly increased until one of the components begins to vaporize out of the solution. This vapor is then condensed into a different container. The component that vaporizes first will do so at a lower temperature than the second. Therefore, the component that has the lower boiling point (limonene) will vaporize first. So before even looking at our answer choices, we have a predicted answer, with some solid reasoning behind the prediction. We can go back to our answer choices and see if this matches any of our answers.  

  1. Limonene, because it has a lower boiling point. Right away, this sounds like a good answer choice for two reasons. First of all, we said the component with the lower boiling point would distill first. Even if we didn’t know if this were limonene or positive carvone, we would still like this answer choice because the reasoning is sound. But the second reason we like this answer choice, is because it matches that prediction we came up with. We said limonene has a lower boiling point and will vaporize first. We like answer choice A. 
  2. Limonene, because it has a higher boiling point. This answer choice also lists limonene, but does the reasoning work here? We said because of the carbonyl group on carvone, it has the higher boiling point. Even though the specific component matches our breakdown of the question, the reasoning does not. This is going to be an incorrect answer choice, and answer choice A is superior.
  3. (+)-Carvone, because it has a lower boiling point. This would be an attractive answer choice, if it were factually correct. Positive carvone has that carbonyl group, and it has a higher boiling point. The component with the lower boiling point is going to distill first, but that’s limonene, not carvone. This is also an incorrect answer.
  4. (+)-Carvone, because it has a higher boiling point. The reasoning here is factually correct. Carvone does have a higher boiling point. But we’re looking for the component that will most likely distill first. That’s limonene because of the lower boiling point. That makes this another incorrect answer. We’re left with our correct answer, answer choice A: Limonene, because it has a lower boiling point.

6) Here we have an excerpt from our passage, as well as the figure. When I pull up these excerpts from the passage, it’s not so you go back and re-read every single detail, but rather so I can demonstrate my thought process. I show you exactly what I’m referencing from the passage so you can follow along at the same time. When you’re actually practicing, and when you’re taking the exam, I don’t recommend going back to the passage unless you’re looking for details in the form of data or figures. 

The passage says an ebulliator was lowered into the distillation flask to introduce small air bubbles into the system. The author gives us the explicit function of the ebulliator. That ebulliator creates an area of imperfections for gases to collect and start boiling. So essentially it’s a boiling chip. In perfectly clean, smooth glassware there’s nowhere for gas bubbles to start forming at boiling temperatures. 

Without the ebulliator, you might super-heat your liquid, or there may be bumping or splashing. The ebulliator is also helping the solution begin boiling at the correct temperature.

  1. keep the condensed vapors cool in the receiving flask. Does this answer choice sound like what we broke down about the ebulliator? We didn’t mention keeping the vapors cool in the receiving flask. But that’s not always a reason to get rid of an answer. There’s a very small chance you’ll be able to predict what every correct answer will be on an exam, so let’s dive a bit deeper. The author says the ebulliator was lowered into the distillation flask. That’s near the bottom left of our figure. Is this ebulliator going to be affecting the vapors in the receiving flask? That receiving flask is on the other side of the apparatus, so not likely. Let’s keep going through our answer choices and we’ll see if we can find a superior answer choice.
  2. promote the establishment of a high vacuum in a system. If the ebulliator did this, that could certainly be helpful. Remember we mentioned using a vacuum allows the experiment to work at a lower pressure and for the compounds to boil at a lower temperature. But the ebulliator isn’t the one establishing this vacuum-instead it’s the vacuum source on the right side of the apparatus. Just like answer choice A, not a great option, so let’s keep comparing.
  3. prevent superheating of the liquid to be distilled. This answer choice matches something we said in our readthrough. For those of you that may have taken organic chemistry lab, or any general science labs, think of a boiling chip – the entire purpose of the chip is to stabilize the boiling and prevent the solution from bumping / splashing everywhere. When it comes down to it, the ebulliator is just a chip that you throw into your solution to help it boil. We explicitly mentioned it prevents superheating during our breakdown, but what if you didn’t make that connection? This answer choice is still superior to answer choices A and B, because it directly relates to the ebulliators function and location in the experiment. We finally have a superior answer, so we can cross off answer choices A and B.
  4. provide an outlet when the pressure inside the system becomes too high. This is essentially saying that chip allows for ventilation, or even the same function as the vacuum. This is going to be similar reasoning to answer choices A and B. The author tells us the function of the chip and we tied that function to the figure in the passage. The ebulliator isn’t functioning to provide ventilation when pressure is too high. This is another incorrect answer. We’re left with our correct answer, answer choice c: prevent superheating of the liquid to be distilled.

7) We know limonene and positive carvone are separated based on their boiling points, but we can revisit the passage to see what the author says specifically about the purpose of the experiment. 

Here we have an excerpt from the passage. It says Because the two compounds have very different boiling points, the chemist decided to separate them by vacuum fractional distillation. The entire experiment revolves around this boiling point difference, and separating the two components of caraway seed oil by vacuum fractional distillation.

We’re looking for the answer choice that would most likely improve the separation between the limonene and positive carvone. Let’s go through our answer choices and see which would most improve the experiment.

  1. Heating the distillation flask at a slower rate. What would be the effect of heating the distillation flask at a slower rate? The experiment would run the exact same way, only the temperature of the caraway seed oil would raise more slowly. If that happens, could that improve the separation between the limonene and positive carvone? Of course! The entire purpose of the experiment is to separate the components of caraway seed oil by taking advantage of boiling point differences. 
  2. Using a vacuum source that can achieve a lower pressure inside the distillation apparatus. Using a vacuum source to lower pressure will affect the boiling point of both limonene and carvone. We said during our readthrough, that lower pressure as a result of the vacuum source will also cause lower boiling points. Although both boiling points are lower, they will be relatively closer together than when they were at higher pressures. This will make less separation, not more. That means A remains superior.
  3. Cooling the condenser with ice water. Cooling the condenser with ice happens after separation, and not during the separation portion of the experiment. This modification would not improve the separation more than answer choice A. We can eliminate this answer choice.
  4. Using a shorter fractionating column. Shortening the column will actually make the separation less. We can visualize this. There’s less room for the vaporization to occur, so we don’t improve the separation between limonene and carvone. That means we’re left with answer choice A, which takes advantage of our boiling point differences. And we can improve the experiment by Heating the distillation flask at a slower rate

8) To answer this question, we’re going to look at our answer choices and see which carbons we want to look at specifically. We know right away by glancing at the answer choices that our efforts should be focused on carbons 2, 5 and 7 because those are the only carbons listed in the answers. We have to remember the distinction between our positive and negative carvone enantiomers. The difference is the positive carvone is the enantiomer that rotates plane-polarized light in the positive direction, and the negative carvone is the enantiomer that rotates plane-polarized light in the counterclockwise, or negative direction. 

Here we have carvone with carbons 2, 5, and 7 labeled. We’re told Carvone exists as two different enantiomers in the passage. Enantiomers are non-superimposable mirror images of each other that differ in configuration. Enantiomers have opposite absolute configurations at every chiral carbon. Chiral carbons have 4 distinct substituents. So, with that in mind, let’s dig into the structure of our carvone, so we can find the ways positive and negative carvone differ. The passage mentions the two are enantiomers, meaning they will have opposite configurations at chiral carbons.

Let’s first look at carbons 2, 5 and 7 to see which have a stereocenter and four distinct substituents.

    –  Carbon 2 has a double bond, so not a stereocenter
    –  Carbon 5 has 4 different substituents
    –  Carbon 7 has a double bond, so not a stereocenter

Ultimately, we’re going to look for an answer choice that has carbon 5

  1. Carbon 2 only We know this is incorrect for two reasons: the orientations differ around carbon 5, and they do not differ around carbon 2. Let’s try and find a better answer choice.
  2. Carbon 5 only Answer choice B is consistent with what we’re looking for in our correct answer. That means answer choice A can be eliminated.
  3. Carbons 2 and 5 only. Again, we know carbon 2 has a double bond and is not a stereocenter. We already said we are going to eliminate any answer choices that include carbon 2.
  4. Carbons 2, 5, and 7 only. We eliminated answer choice C because of carbon 2, so we know we can also eliminate option D here. Even if we didn’t know about carbon 7, we can eliminate this answer choice knowing carbon 2 is not one of our correct options. Also, we actually said carbon 7 has a double bond and is not a stereocenter. We’re left with our correct answer, answer choice B: Carbon 5 only

9) The test-maker is asking the effect of a malfunction in the vacuum distillation in which the vacuum leaks. If the vacuum is not working correctly, it’s not properly lowering the pressure inside our distillation apparatus. We want to know what happens to the boiling points of carvone and limonene at higher pressures. The question is simply asking the correlation between pressure and boiling point. 

Vacuum distillation is distillation performed under reduced pressure, which ultimately decreases the boiling point of compounds. Why would we do this? Because sometimes compounds can have very high boiling points that are hard to reach. These high temperatures could adversely affect the compound itself-we can see denaturing, or the experiment just won’t work. A leak in the apparatus will undo this vacuum effect. That means a higher pressure, and we should see an increase in the boiling point of the compounds.

  1. both increase. This answer choice is consistent with what we came up with as we broke down this question. We said a leak in the apparatus will undo the vacuum effect. That means a higher pressure, and an increase in the boiling point of the compounds. We like this answer choice for now, let’s quickly go through the remaining answers to confirm whether this is our best answer.
  2. both decrease. We know that’s not true because a properly working vacuum is what decreases boiling points. We broke down and visualized the experiment, and that allowed us to determine this is an incorrect answer choice. Answer choice A remains the superior answer choice.
  3. both remain the same. Once again, a leak in the vacuum will alter the boiling points by increasing them. We’re not expecting boiling points to remain the same. That means we’re still sticking with our best answer choice, answer choice A.
  4. become more similar. In reality, the boiling points will likely become further apart, rather than become more similar. Why is that? Because there’s a wider range of boiling points available to these compounds. With the vacuum, both boiling points decreased and were closer to one another. With a malfunctioning vacuum, the boiling points would be increase, and would now be further apart. This is also an incorrect answer. The only answer choice that matches what we said during our breakdown of the question is answer choice A: both boiling points would increase.

 

Chemistry Question Pack: Passage 3

10) The values to calculate the number of sodium ions will come from the passage, but we’ll be solving the problem with general knowledge. One of the most important things to keep in mind for this question or for any math problems in general, is we have to keep our units correct as we work through our question. We can pull up part of the passage here as reference.

The passage mentions there are 7.15 g of Na2CO3 in the colorless solution, and also gives the molar mass. We can find the number of moles of sodium carbonate by dividing the 7.15g in the solution by the molar mass to get 0.0250 moles sodium carbonate. 

Let’s also take a closer look at the formula for sodium carbonate. For every 1 mole of Na2CO3, there are 2 moles of sodium ions (Na+), so the solution contains 0.050 mol Na+.

Avogadro’s number is known as the number of particles (in this case it is ions, but it can be any unit) per mole of a substance. 

(6.022 x 1023 ions per mole)(0.05 mol Na+) = 3.00 x 1022

This corresponds to answer choice B.

I’ve also written everything out below so it’s easier to follow the math:

11) This question is simply testing our ability to know about the acidity or basicity of sodium carbonate. Litmus paper is used to test substances of at various pHs.  Red litmus paper turns blue in base, while blue litmus paper turns red in acid. If red litmus paper is added to an acid, it remains red, and if blue litmus paper is added to a base, it remains blue. 

The first sentence from our passage mentioned the class was experimenting with solubility equilibrium. We expect that in an experiment dealing with solubility equilibrium, the solids are dissolved in water, and the base particles that make up the solids will be form. 

Sodium carbonate is a water-soluble salt, so it’s neutral and wouldn’t change the color of the litmus paper. However, when sodium carbonate is in water, it’s broken down into sodium ions, and we have our carbonate ions. When carbonate ions are in water, they can accept protons, therefore leaving bicarbonate and hydroxide ion products. 

We know bicarbonate from pH buffering in the human body, but we’re focused on the hydroxide ions for now. Hydroxide ions are strongly basic, so we’re expecting a basic pH.

  1. remain red, because carbonate is an acidic salt. If we’re expecting a basic pH, we’re looking for a blue or purple color, not red. We can eliminate answer choice A.
  2. remain red, because sodium carbonate is neutral. Reasoning here is similar to answer choice A. We can group answers as A&B or C&D because only one set of choices indicates the proper color. That means we can eliminate both A and B right away. We’re only deciding between C and D.
  3. turn blue, because carbonate reacts with water to produce OH. This answer choice matches what I said in the breakdown of the question. Hydroxide ions are strongly basic, so we’re expecting a basic pH and the blue color.
  4. turn blue, because sodium ions form sodium hydroxide in water. This is not the case based on our breakdown. Answer choice C remains the best option. The litmus paper will turn blue, because carbonate reacts with water to produce OH.

12) This is a basic math problem. The precipitation reaction forms nickel carbonate as a product-we just need to know exactly how much (maximum number of moles). After we balance the equation discussed in the passage, we find the number of moles of nickel carbonate.

Find the number of moles nickel. Divide grams of sample by molar mass to get moles of NiSO4. There is an equal number of moles of Ni:

6.57 g NiSO4/(262.84 g/mol) = 0.025 mol. 

We can find the number of moles of Na2CO3 by dividing the 7.15g by the molar mass to get 0.0250 moles Na2CO3. For every 1 mole of Na2CO3, there is 1 mole of CO3 2-. There is no limiting reactant between the two as there is an equal number of moles. I’ve worked out the math with the essential information from the passage below:

We can pick our correct answer, answer choice A: 0.025 mol.

13) From the passage, we learned that during experimentation a gas formed, and we need to identify it. We’ll rely on the passage for details about the experiment, but we will use general knowledge to balance chemical equations and quantify the number of moles. 

Second paragraph of our passage is above. Two of the initial solutions are mixed, and then “To one portion is added an excess of 6 M HCl, which results in the disappearance of the precipitate and a rapid evolution of a gas.” We can write down the essential information like we did in our previous problems. Normally, I have no problem typing everything out, but for math problems and chemical equations this tends to be easier to read:

Two possible precipitates form: Na2SOor NiCO3. What kind of metals form green precipitates? Normally chromium, iron, and nickel form green precipitates. This hints at the fact that our precipitate is likely nickel carbonate NiCO3. (Na2SO4 would likely be white)

We add hydrochloric acid to our green precipitate (our nickel carbonate), the precipitate dissolves and a mystery gas is evolved. Wow, so what just happened? Where’s did our precipitate just go? And why is there gas everywhere? Nobody freak out! Let’s break this down by writing out the double replacement reaction 

Solid nickel chloride doesn’t help answer this specific question, but we might recognize carbonic acid from respiratory physiology. What we know about carbonic acid is it exists in chemical equilibrium with carbon dioxide gas and water. In fact, we expect there are only going to be trace amounts of aqueous carbonic acid, and instead predominantly carbon dioxide and water. 

We mentioned earlier in our breakdown that our precipitate is likely nickel carbonate NiCO3, so we can automatically eliminate answer choices A and B. Looking at C and D, only C matches what we expect to see in the experiment (carbon dioxide). 

14) Different colors were formed during the experiment because of one specific ion. We’re just asked to identify this ion, so we’ll reference the passage to find out the components in the colored solutions and details of the solution, but the answer may ultimately come from general knowledge and our knowledge of the colors that different metals ions form. Let’s glance at our answer choices quickly. We have sulfate and nickel as the two recurring ions that may cause the color of the solution. 

I mentioned something about this in our previous question. What kind of metals form green precipitates? Normally chromium, iron, and nickel form green precipitates. What do they have in common? Complex ions containing transition metals are usually colored, and then similar ions from non-transition metals (like sulfur) are not. What does that mean for our question? That suggests that the partly filled d-orbitals (which are characteristic of transition metals) should be involved in generating the color in some way. Answer choice C is going to be our best choice here. Nickel is a transition metal, and we said the reason for the color of the solutions is the unfilled d-orbitals. It does not have to do with charge.

 

Chemistry Question Pack: Questions 15-19

15) This is a standalone question, but we’re going to rely on something presented to us in the question stem to answer, rather than our general knowledge alone. We’ll have to compare the structure of these two compounds and their anions. Alternatively, we can say Compound 1 would be a stronger acid than Compound 2 because Compound 1 has a more stable conjugate base when it loses a hydrogen atom. Phrasing our question differently, why is the conjugate base of Compound 1 more stable than the conjugate base of Compound 2? 

What do we know about about anions? Anions are negatively charged ions. The carboxyl group loses its H and becomes negatively charged carboxylate ion. Why does this happen? Carboxyl groups will ionize often, meaning carboxyl groups in general are acidic functional groups. Ionizing means the carboxyl can break its double bond, which allows it to go to a lower energy state and increase its stability. Then after the hydrogen atom leaves, the remaining 2 oxygen atoms share the remaining negative charge in the carboxylate. 

  1. resonance effect. This answer choice describes delocalized electrons. We see that effect in molecules with double bonds, which we no longer have here.
  2. dehydration. Dehydration occurs following the loss of water from a molecule. That’s not relevant to the question, so we can eliminate answer choice B because it is out of scope of the question
  3. an inductive effect. The inductive effect is going to be the same for both compounds. The inductive effect is based on the vicinity of the compounds. The substituents around the compounds aren’t different, just the orientations of the carboxyl groups.
  4. hydrogen bonding between OH and CO2. Hydrogen bonds are formed when hydrogen atom bonds with a very electronegative atom. For the sake of the MCAT especially, we think of fluorine, oxygen, and nitrogen as the electronegative atoms. The hydrogen gains a partial positive charge, and the electronegative atom gains a partial negative charge. We get hydrogen bonding from the partial positive hydrogen in the hydroxyl group and the negatively charged carboxylate group in Compound 1. In Compound 2, the carboxyl is not in an orientation where it can bond the hydroxyl group, so we don’t get the stabilizing effect from the hydroxyl group. The conjugate base of Compound 1 is more stable than the conjugate base of Compound 2 because of the hydrogen bonding in Compound 1, so our correct answer choice is D

16) We have 4 different substituents around the central carbon. The shaded-in triangles mean a substituent is coming out of the screen. The dotted lines mean the substituents are away from view, into the screen. We can say the grouped substituent orientations should stay the same. What does this mean? The carboxylic acid (COOH) and the hydroxyl group (OH) should be in the same orientation, while the hydrogen atom and the methanol should be in the same orientation. Next thing we want to make sure, is whether the compounds are in an R or S configuration. To do that, go from the highest priority substituent and draw a curved arrow toward the lowest priority substituent. In our question, we would do that from carboxylic acid to hydrogen>the arrow is curving left. Our compound is S-configuration (S stands for sinister, or “left” in latin). 

We’re ensuring the compounds that had identical orientations in compound A continue to have identical orientations in the correct answer choice. We also want the correct answer choice to have S-configuration

a.

Answer choice A is an enantiomer of compound A, it has R-configuration.

b.

Answer choice B is just answer choice A rotated 90 degrees and is also is an enantiomer of compound A, it has R-configuration

c.

Answer choice C is also an enantiomer of compound A. It has R-configuration. It’s the same as answer choice B, but again rotated 90 degrees.

d.

Answer choice D is the exact same structure as compound A, but it is rotated 90 degrees on screen. It’s identical to compound A and has S-configuration. That means we can pick answer choice D as our best option here.

17) We’re breaking a triglyceride up into its base components, and we want to pay close attention to our R groups and our R’, or R prime, groups. What are triglycerides made of? They’re glycerol molecules combined with three fatty acids. There’s a dehydration reaction and the glycerol forms ester bonds to link with the three fatty acids.

Here we’re undoing our triglyceride. Our glycerol is hydrolyzed, meaning our ester bonds are broken, to give 2 fatty acids R, 1 fatty acid R’, and 1 mol glycerol

a.

Answer choice A has three R groups, which contradicts what our initial question said. We’re expecting 2 R groups and 1 R’ group

b.

Answer choice B has three R’ groups. This again contradicts our question stem.

c.

Answer choice C has two R groups and one R prime group. We like this answer choice of the ones we’ve gone through.

d.

Answer choice D has one R group and two R prime groups which contradicts our breakdown of the question. We stick with our best answer, answer choice C.

18) We’re given compound I and compound II and at first glance it looks like the only difference is a nitrogen atom in compound I and one fewer hydrogen atom, while compound II has a carbon atom and an additional hydrogen where compound I has a nitrogen atom.

This question is about solubility. When we look at different solutes and solvents, remember the mantra that’s always drilled in our heads during chemistry classes “like dissolves like.” A polar substance will dissolve best in a polar solvent; a nonpolar substance will dissolve best in a nonpolar solvent. 

This question is asking which of the two Compounds (I or II) is more similar to water. Nitrogen and oxygen can form hydrogen bonds, while carbon does not form hydrogen bonds. Hydrogen bonding happens between a hydrogen atom and an electronegative atom. For the sake of the MCAT especially, we think of fluorine, oxygen, and nitrogen as the electronegative atoms in hydrogen bonding (FON atoms). We’re expecting the compound with nitrogen (Compound I) is more soluble in water. 

  1. Compound I Answer choice A says compound I is more soluble in water. We said nitrogen can form hydrogen bonds like oxygen, so we’re liking answer choice A.
  2. Compound II We’re expecting compound I to be more soluble than compound II because of the nitrogen atom over the carbon atom, so we’re going to eliminate this answer choice. It contradicts what we said about like-dissolves-like so answer choice A remains superior.
  3. They are about equally soluble. Answer choice C says they are equally soluble, which again contradicts our claim that like-dissolves-like and that the nitrogen-containing Compound I would dissolve better.
  4. Neither has appreciable water solubility. Answer choice D says neither has appreciable water solubility. We said Compound I would be soluble in water, so this contradicts our claim. Answer choice A remains the best answer, Compound I is more soluble in water.

19) Most likely, the student is going to be separating the starting material from the ending material. The initial material has carbonyl group; the ending product has a hydroxyl group. The only additional information we’re given about the reaction is that the initial material reacts with sodium borohydride.

Visualize this thin-layer chromatography experiment. Rf is the distance travelled by the materials, so materials that interact more strongly with the silica plates will have lower Rf (less distance travelled), materials that interact less strongly will have higher Rf values. 

In thin-layer chromatography, glass is coated with a thin layer of adsorbent, often silica gel or aluminum oxide, and the mixture is spotted on this layer, similar to paper chromatography. Different substances are going to ascend the plate at different rates and be separated based on their polarity when using thin-layer chromatography. Both the hydroxyl group and the carbonyl group are polar functional groups, but the hydroxyl group can form hydrogen bonds-and that’s what makes it more polar than carbonyl groups. This means the product with the hydroxyl group binds with the silica gel more strongly and will have a lower Rf

  1. Higher, because the product is more polar than the starting material. Answer choice A says the product’s Rf is higher, because the product is more polar than the starting material. We said the product will have a lower Rf, so the first part of our answer choice contradicts our breakdown. We can likely eliminate answer choice A, but for thoroughness, we can check the 2nd part of the sentence which actually looks correct. The product is more polar than the starting material. Let’s still try and find an answer choice that is completely correct. 
  2. Higher, because the product is less polar than the starting material. This is similar to answer choice A. We said the product will have a lower Rf, so the first part of our answer choice contradicts our breakdown of the question. 
  3. Lower, because the product is less polar than the starting material. Answer choice C says the product’s Rf is lower, which is consistent with our breakdown. Second part of the answer choice says the product is less polar than the starting material. This contradicts our breakdown; the product is more polar than the starting material.
  4. Lower, because the product is more polar than the starting material. Answer choice D says the product’s Rf is lower, which we said was true. The reasoning says because the product is more polar than the starting material, which is again consistent with our breakdown. We keep our correct answer choice- answer choice D.

 

Chemistry Question Pack: Passage 4

20) CDP is the monomer discussed in Passage 4. In this question, we’re given the molar mass and asked about Experiment 1 from the passage specifically, so we’ll likely need to go back for specific details about the experiment. We have to make sure we pay particularly close attention to our units and prefixes. 

First bit of information: from our question we’re told the Molar mass of CDP is 403 g/mol and we need to find the mass CDP (in grams) in 10 mL of the solution in experiment 1.

We’re told in Experiment 1, the chemist dissolved 16 mmol CDP in 1L solution. This gives us a molarity of 16mM, which can be converted into 0.016 M. There are 1000 millimoles in a single mole.

Next, to find the number of grams in a single liter of solution, multiply: 0.016 molar x 403 g/ 1 mol = 6.4 grams CDP per liter of solution

Remember, we need to know the mass of CDP in 10 mL of solution. Let’s adjust our units once again. We need to know how many liters are in 10 mL. There are 1000 mL in 1 L, so this means a volume of 0.01 L. See how important it is to keep track of units?

Multiply: 6.4 grams CDP/L x the volume 0.01L = We’re left with 0.064 g CDP OR 6.4 x 10-2 g

Remember our mnemonic LARS for converting to scientific notation, which is Left add, and in this case, right subtract. I want you to note how important it is to keep track of your units during every math problem. Just knowing the proper units can help you get to the right answer, even if you’re unsure exactly how to solve a problem. Glancing at our choices, the initial digits stay the same in all 4 answers, but the exponent next to the 10 varies. This should really hammer the point home, you have to be careful with your units and prefixes to get the correct answer. 

This question was a math problem, so we can eliminate any incorrect values that don’t match our calculated value. We’re left with our correct answer, answer choice C: the mass of CDP in 10 mL of solution is 6.4 x 10-2 grams. I’ve also handwritten the key points here so it’s easier to follow my math:

21) Recall from the passage and the previous question that Experiment 2 improved the yield of the polymer that was obtained in Experiment 1. In this new scenario, the chemist is going to increase the amount of a component in the experiment to increase the yield of polymer. We said the polymer was a product in Equation 1, so we’re trying to increase the yield of products.

  1. H2O. Answer choice A is H2O, or water. Adding more H2O is simply going to decrease the concentration of both reactants and products. It won’t cause the final yield of the polymer to decrease necessarily, but it won’t improve the yield either
  2. PNP. Answer choice B is the enzyme in our passage. Enzymes are catalysts that lower activation energy. They function to catalyze reactions while not getting used up in the process. In this case, the reaction proceeds at a constant rate until it reaches equilibrium, and more enzyme will not create more product. What does it mean when a reaction reaches equilibrium? It proceeds until it reaches a final product concentration
  3. Mg2+. Answer choice C we remember from our passage is magnesium ion, which ensures our catalyst is active. Answer choices B and C work hand-in-hand.
  4. CDP. Adding more reactant (CDP) is the best way the chemist can improve the yield of the polymer. There’s more reactant going into the reaction, and more product is getting formed. We can stick with answer choice D as our best answer here.

22) We’re solving for the equilibrium constant of Equation 1, which we write out here. 

We had to go back to the passage to find Equation 1, but do we need any additional information to find the equilibrium constant? We should know the equilibrium constant for a general reaction, so we should be able to apply that to Equation 1. That makes this a standalone question. What do we know about equilibrium constants for reactions? 

The equilibrium constant (Keq) for any general reaction is the following: aA+bB⇌cD+dD

The products of the reaction (C and D) are placed in the numerator, and their concentrations are raised to the power of the coefficients from the balanced equation. The reactants (A and B) are placed in the denominator, with their concentrations raised to the power of their coefficients.

The concentrations of both aqueous solutions and gases change during the progress of a reaction. For reactions involving a solid or a liquid: the amounts of the solid or liquid will change during a reaction, but their concentrations won’t change during the reaction. Instead, the values for solids and liquids will remain constant. Because they are constant, their values are not included in the equilibrium constant expression. Given this, we can put Equation 1 in the proper format:

Let’s look at our four choices and decide:

Answer choice A correctly lists the concentration of the polymer in the numerator, and the concentration of hydrogen phosphate raised to the power of its coefficient (n) in the balanced equation. The denominator also correctly lists the concentration of the reactants (which is our CDP) raised to the power of its coefficient (n) as well. This is going to be our best answer choice. Answer choice D is similar, but note we’re only getting the concentration of CP, not the polymer.

Answer choice B does not raise the concentration of hydrogen phosphate to the power of its coefficient in the numerator. Answer choice C does not raise the concentration of CDP to the power of its coefficient in the denominator. Answer choice A is our correct answer.

23) We’re comparing the concentration of our two products. We can write out our Equation 1 that we need from our passage.

In the balanced equation, we can add a “1” in front of a molecule that doesn’t have a coefficient already.

Every 1 molecule of (CP)n that is formed, there are n HPO4 2- that are formed. So in the solution, the ratio of (CP)n to HPO4 is 1/n. Compared to the concentration of HPO4, concentration of CPn is 1/n[ HPO4]

For every 1 molecule of polymer that’s formed, there’s N hydrogen phosphates formed. So in the solution, the ratio of polymer to hydrogen phosphate is 1/n. And compared to the concentration of hydrogen phosphate, the concentration of the polymer is 1/n times the concentration of hydrogen phosphate.

So that’s one way of going through this. We can also look and think about this problem a bit differently as well. We can also say every time you make one polymer particle, another hydrogen phosphate is formed as well. The actual number of polymers formed can be large, meaning “n” can be any number for the sake of this question. The polymer, and the hydrogen phosphate are produced at a 1:n ratio. To find the concentration of the polymer in terms of the concentration of the hydrogen phosphate, you have to divide the concentration of the hydrogen phosphate by N. Let’s go through our 4 answer choices: 

  1. Answer choice A says n times the concentration of our hydrogen phosphate. Why might this be a correct answer choice? If we were instead looking for the concentration of our hydrogen phosphate with respect to the polymer, rather than the other way around. So the answer is close, but not quite what we want. 
  2. Answer choice B says n squared times the concentration of our hydrogen phosphate. Again, this doesn’t match our breakdown, and we weren’t dealing with any n squared when we did our breakdown. We’ll stick with answer A as the superior answer, and can eliminate answer choice B for contradicting our breakdown.
  3. Answer choice C says 1/n times the concentration of our hydrogen phosphate. Is this closer to our breakdown? Yes, it’s the exact same as what we came up with, in fact. So, we can now rule out answer choice A as well for contradicting our breakdown as well. I try and wait until I find a superior answer before eliminating any answer choice.
  4. Answer choice D says 1 over n squared. Again, we didn’t deal with an n squared in any of our calculation. This answer may look better than choices A and B, but answer choice C is the only answer choice that matches our breakdown of the question.

24) In other words, we’re trying to compare the concentration of hydrogen phosphate anion to the concentration of dihydrogen phosphate in buffer solution. There are two big details we want to note: We are given pKa here in the question stem, and the question references the buffer solution from Experiment 1. 

Let’s look back at Experiment 1 in the passage. 16 millimol CDP dissolved in 1 L aqueous solution with PNP and magnesium ion. Then we’re told the solution is buffered at a pH of 8.7. Remember, we’re answering a question about acid/base equilibria. We were given the pKa for dissociation of hydrogen phosphate anion to the concentration of dihydrogen phosphate. 

What is the concept being tested here? Looking closely, we’re comparing an acid and its conjugate base. The conjugate base is the hydrogen phosphate anion, and the acid is dihydrogen phosphate. The chemist buffers the solution at pH 8.7

A little background here, and some key points you want to note:

  • In acid/base reactions, an acid will have a conjugate base, while a base will have a conjugate acid. So, in this dissociation, a conjugate base is formed when dihydrogen phosphate loses a proton. 
  • Also, remember from AAMC’s content outline. a Bronsted-Lowry acid is a species that is capable of donating protons to form its conjugate base. So, our dihydrogen phosphate is a Bronsted-Lowry acid as well.
  • So judging by the information we’re given, we’ll be using the The Henderson-Hasselbalch equation.

pH=pKa + log[A]/[HA]

[A] corresponds with the conjugate base (hydrogen phosphate anion), while [HA] corresponds with the acid (dihydrogen phosphate).

Plug in the values we’ve been given. The passage tells us pH=8.7 and question tells us pKa is 6.7

2.0= log [A]/[HA]

Anti log on both sides means:

10^2 = [A]/[HA] and ratio is 100 : 1

  1. 1:1 Answer choice A says 1:1. This does not match our calculated answer. The ratio would be 1:1 if pH and pKa were equal.
  2. 2:1 Answer choice B says 2:1. Our base 10 logarithm increases tenfold with every integer increase. The antilog of 2 is not 2. It’s 100.
  3. 100:1 Answer choice C says 100:1. This is consistent with our calculations and breakdown. We’ll keep this answer choice and we can eliminate choices A and B for being incorrect.
  4. 200:1 Answer choice D says 200:1. Again, this is an incorrect value based on our calculation so we can eliminate this answer choice. We’re left with the correct answer. Answer choice C: a ratio of 100:1

 

Chemistry Question Pack: Passage 5

25) The question explicitly says “according to the passage.” You may have noticed this doesn’t always mean you have to flip back to the passage to answer the question. For example, we should remember from the passage that the cleavage is catalyzed by a base. We’ll keep that in mind for the time being, but what I want you to realize is, ideally, you only go back to the passage rarely for specific details. However, as I go through these solutions you’ll find I pull up excerpts from the passage so you can follow my thought process, and so I’m thorough. 

Our passage says large-scale synthesis of Olestra starts with a base-catalyzed cleavage. So this confirms what we said previously about it being base catalyzed.

The figure in the passage shows us base-catalyzed cleavage of a triglyceride. So, our correct answer should be a base. That’s all the information we needed from the passage, but we still need to be able to identify a base. 

Strong acids and strong bases are listed on AAMC’s content outline, so make sure you know these most common ones. And we also want to know their properties. Strong bases will dissociate completely in an aqueous solution while yielding hydroxide ions. We expect the compound the worker uses to catalyze the cleavage of the triglyceride could be one of the 8 bases listed here.

Strong Acids Strong Bases
HCl- Hydrochloric acid LiOH- Lithium hydroxide
HBr- Hydrobromic acid NaOH- Sodium hydroxide
HI- Hydroiodic acid KOH- Potassium hydroxide
HNO3– Nitric acid RbOH- Rubidium hydroxide
HClO3– Chloric acid CsOH- Cesium hydroxide
HClO4– Perchloric acid Ca(OH)2– Calcium hydroxide
H2SO4– Sulfuric acid Sr(OH)2– Strontium hydroxide
  Ba(OH)2– Barium hydroxide
  1. HCl(aq) Hydrochloric acid, we already established was a strong acid. That contradicts our basic criteria of being basic, so we eliminate this answer choice.
  2. NaCl(aq) Sodium chloride consists of a sodium cation and the conjugate base anion of hydrochloric acid. We mentioned hydrochloric acid is a strong acid. Sodium cations and the conjugate base anions of acids form salts. This would contradict what we know from the passage about the base-catalyzed cleavage.
  3. NaOH(aq) Sodium hydroxide is a strong base. It is one of the common strong bases on our list, so this sounds like our best option so far. Let’s check answer choice D to be thorough.
  4. Na2SO4(aq). This is like answer choice B. This also consists of a sodium cation and the conjugate base anion of an acid, which means it is a salt, not a base. Correct answer is answer choice C, sodium hydroxide.

26) In other words, we’re given stats for the combustion of 10 peanuts and asked about the stats of a single peanut. We need the specific heat of water, which should be general knowledge: 1 cal/g*oC. That should be the only information we may need from the passage, so this is almost like a standalone question. 

We’re going to have to use dimensional analysis, so we’ll write our numbers down and keep our units consistent. Our final answer is in kilocalories.

We’re dealing with 1 kg of water in a calorimeter. So to use the specific heat for water, we can convert the units on the 1 kilogram of water, to grams. Dimensional analysis, multiply by a factor (1000 grams/1kg) which gives us 1000 grams water. 

Our temperature is already in Celsius. So last unit, 1 kilocalorie is 1000 calories, and we raised temperature 50 degrees Celsius with 10 peanuts. So solving for the energy content of 10 peanuts, we get 50,000 calories, or 50 kilocalories because our answers are in kilocalories. Lastly, we want the kilocalories for a single peanut, so we can divide the 50 kilocalories in 10 peanuts by 10. We get an energy content of 5 kilocalories per peanut. I’ve handwritten my work below so it’s easier for you to follow along:

Our calculation corresponds to answer choice C: 5 kcal.

27) Macronutrients provide a certain number of dietary calories per gram. We have to figure out how many are in a one gram sample of olestra. We know the calories per gram of other major macronutrients (that includes fat), but not of Olestra. 

Micronutrient Calories/gram of micronutrient
Carbohydrate 4
Protein 4
Dietary Fat 9

However, the passage said the Olestra is not metabolized. What does that mean? There are no dietary calories contributed to human consumers. We know that any value that isn’t 0 is incorrect. We can pick answer choice A as our correct answer.

28) The structure of glycerine was provided in the passage, but we should know the structure. These structures come in AAMC’s content outline categories 5B and 5D. If you need to review the content outline, we have it on our website with the information we think is most important to know about each topic.

We can compare glycerine’s structure with the structure of isopropyl alcohol:

Glycerine has three carbons, each also bonded to a hydroxyl group. Isopropyl alcohol is our simplest secondary alcohol, and has one hydroxyl group. What are our big differences here? We look at our glycerine, there are three hydroxyl groups (labeled in red). We look at our isopropyl alcohol, we see a single hydroxyl group.

The hydroxyl groups in each molecule, they’re able to hydrogen bond. We know that hydrogen bonding is strong. It takes lots of energy to break them, so an increase in the number of hydrogen bonds increases the boiling point of our molecule. Just looking at the number of potential hydrogen bonds in each molecule, the boiling point of glycerine should be much higher. 

  1. more than 10oC higher. We said the boiling point of glycerine will be higher. Glycerine has three times as many hydrogen bonds, meaning the boiling point should be signicantly higher.
  2. less than 10oC higher. We do expect glycerine to have a higher boiling point, but we expect it to be much higher than just 10 degrees or less. Answer choice A is still better than choice B. And just a reminder, we’re picking the best answer choice, not the perfect answer choice.
  3. less than 10oC lower. We drew out the structures and all signs pointed to glycerine having a higher boiling point. We can eliminate this answer choice. Answer choice A remains the best option.
  4. more than 10oC lower. Similar to answer choice C. We said glycerine has a higher boiling point than isopropyl alcohol. We stick with our correct answer, answer choice A – the boiling point of glycerine is more than 10 degrees higher than the boiling point of isopropyl alcohol

 

Chemistry Question Pack: Passage 6

This passage and the related questions relied heavily on knowing periodic table trends. Make sure to review if you’re not comfortable with any of the topics: https://jackwestin.com/resources/mcat-content/the-periodic-table-variations-of-chemical-properties-with-group-and-row

29) This is almost like a standalone question. It’s tangentially related to the passage we just read, but we don’t need the passage to answer this question. Our passage was more of a history lesson, and didn’t provide any specific details about the elements in the periodic table. Instead, we need to pull up the periodic table:

We can locate the four answer choices on the table. Our atomic radius is highest all the way to the left, and at the bottom of the table, so we can immediately pick the leftmost and downmost element as our answer choice, which is sodium. We were able to compare all 4 answer choices at once and pick the best answer: Answer choice A. 

The MCAT knows you most likely won’t know every specific value for elements on test day.  The radius of two different elements could be nearly identical, but the testmaker knows you can’t possibly memorize such minute details before taking the exam. That’s why it’s important to know periodic trends.

30) Similar to our last question, we don’t need the passage to answer this question. We’re given 4 atoms, and we’re picking the answer choice that has the largest first ionization energy. Our passage didn’t actually provide specific details about the elements in the periodic table. Where is ionization energy the highest? To the top and to the right of the periodic table. 

The first ionization energy is the energy that is needed to knock off the first valence electron of an atom. In general, ionization energy increases to the right across a period. And it also increases up a group because of decreasing radii. So, a noble gas like krypton will have the highest nuclear charge and is also the least willing to give up a valence electron. It already has a noble gas valence electron configuration, so it’s not going to want to change that. An alkali metal like potassium has a much lower ionization energy. Potassium can give up a valence electron and achieve noble gas valence electron configuration, so it’s willing to part with the valence electron. We’re left with our best answer, answer choice D: Krypton.

31) We have a strontium isotope, and the mass number we’re given is 90. We’re solving for the combined number of protons, neutrons, and electrons in the isotope. This is our third straight question where we’re going to rely more on the periodic table than we are our passage to answer the question. 

Find Strontium on the periodic table (element 38). We know an uncharged element has an equal number of positively charged protons and negatively charged electrons. We also know the mass number is the sum of all nucleons. I mentioned strontium is element 38 on the periodic table. So, the number of protons and electrons both should be 38. We’re given a mass of 90. So, solving for neutrons is the total mass of 90 minus the atomic number 38, or the number of protons. This gives us 52 neutrons. We can add up 38 protons, 52 neutrons, and 38 electrons giving us a total sum of 128. 

Quick recap: We know an isotope’s mass number is going to be the sum of the number of protons and neutrons, so those two combined is 90. We just have to add the number of electrons, and we have our combined sum of 128 in a matter of seconds.

  1. 90. This corresponds to the mass number
  2. 126. If we had a mass of 88 (the number we see on the periodic table) this would be the correct answer. Alternatively, if strontium had 36 electrons, or a full valence shell, this could be correct. However, we have a mass of 90 and 38 electrons. This is close, but not the answer we’re looking for.
  3. 128. We went through the calculation and found this is the best answer.
  4. 218. We would get this answer by taking our 128 sum and adding it to the atomic mass of 90. We can stick with the correct answer, answer choice C: 128. 

32) In other words, we’re given 4 sets of elements and asked to identify which pairs of atoms is most likely to form an ionic bond. The question stem explicitly says we’re focused on electronegativity.  Electronegativity trends allow us to determine which types of bonds are formed between elements. It’s the highest to the right, and toward the top of the periodic table. Electronegativity differences are measure on something called the Pauling Scale, but they are very precise. For the sake of our trends, large electronegativity differences mean an ionic bond is formed. A small electronegativity difference means a nonpolar covalent bond. A bond in between is a polar covalent bond. 

We’re trying to find the pair that is most likely to form an ionic bond. We said electronegativity increases from left to right across a period, and also increases from the bottom to the top of the periodic table. Generally, when a metal and nonmetal from opposite ends of the periodic table bond, an ionic bond is formed because of the huge electronegativity difference. 

In this case, we can pick the pair that matches our description-Calcium and iodine.

Quick recap: when elements from opposite ends of the periodic table bond, an ionic bond is formed. A bond is considered an ionic bond because there is a complete transfer of electrons. And that’s exactly what happened in answer choice C. 

 

Chemistry Question Pack: Questions 33-37

33) We’re give the planar structure and the chair confirmation of cholic acid. There’s a circled hydroxyl group in the figure; it’s in the planar structure. We’re trying to identify the same hydroxyl in the identical molecule, but in its chair confirmation.

This is an organic chemistry question from AAMC’s content category 5B which talks about intermolecular interactions, and in this question, we’re trying to visualize a molecule in two different views. The chair confirmation in particular minimizes strain, so it’s standard practice to switch from the planar structure to the chair confirmation. 

The circled hydroxyl group is found on a secondary carbon. What does a secondary carbon mean? It means the hydroxyl group is bonded to a carbon, that’s bound to two other carbon atoms.

What else can we recognize about the hydroxyl group? It looks like it’s two carbons removed from the fusion of the third hexose ring and the pentose ring. The hydroxyl group is also five carbons removed from each of the other hydroxyl groups. 

  1. I Answer choice A corresponds to labeled position I in the bottom figure. If the hydroxyl were on position 1, it would be located on a secondary carbon like the hydroxyl group in the top figure. So far so good. But the hydroxyl at position 1 is 4 carbons removed from one hydroxyl group and 6 carbons removed from the other. It also looks like the hydroxyl group in the top figure is closer to the fusion of the 2nd and 3rd rings, while labeled position 1 in the bottom figure is closer to the fusion of the 1st and 2nd ring. Let’s continue to compare to the other options.
  2. II. Answer choice B corresponds to labeled position II and would be bound to a secondary carbon again. The hydroxyl group would also be 5 carbons away from each of the other hydroxyl groups, the same as it is in the top figure. Position 2 is also closer to the fusion of the 2nd and 3rd ring, so our basic criteria are met. We’re keeping answer choice B for now, and we can eliminate answer choice A because it contradicted some of our criteria
  3. III. Answer choice C corresponds to position III and is bound to a tertiary carbon. Same with position I, it’s 4 carbons removed from one hydroxyl group and 6 carbons removed from the other. It’s also closer to the fusion of rings 2 and 3 than the fusion of rings 1 and 2
  4. IV. Answer choice D corresponds to position 4, which is again bound to a tertiary carbon. It’s 3 positions removed from the closest hydroxyl group and bonded to the carbon along the fusion of rings 2 and 3. Answer choices A, C, and D all contradict information we were able to gather about the circled hydroxyl, so we’re left with the correct answer, answer choice B (position 2).

34) We’re covering a nucleophilic substitution. Alcohols have hydroxyl groups, and we know hydroxide ions are poor leaving groups. Weak bases are the best leaving groups, but hydroxide ion is strong base. But this question specifically focuses on the mechanism by which the acid catalyst enhances the reaction. Adding an acid allows for the conversion of our alcohol to its protonated state (its oxonium ion). This allows water to be the leaving group instead of the more difficult hydroxide ion. We’re assuming our answer will at least be tangentially related to this process.

  1. increasing the solvent polarity. This doesn’t relate much to the process of getting the hydroxide ion to leave in a substitution reaction.
  2. creating a better leaving group. Right away, this answer sounds better than answer choice A. We said that ultimately what we want to do is have water be our leaving group instead of the more difficult hydroxide ion. That’s exactly what is being described in the answer choice, so we like answer choice B for now
  3. neutralizing basic impurities. This again does not describe what is going on in the reaction. An acid catalyst could theoretically neutralize basic impurities, but does that actually relate to what is being done in making water the leaving group over the hydroxide ion? The answer choice isn’t necessarily wrong, but it’s not answering the question being asked. We’ll stick with answer choice B over answer choice C
  4. protecting the alcohol group. Again, the mechanism we described is the conversion of the alcohol to its protonated state and ultimately allowing water to be the new leaving group. If we protect the alcohol group, this is the exact opposite of what we anticipate is happening in the enhanced reaction. This contradicts what we said about the mechanism by which the acid catalyst enhances the reaction. We eliminate answer choice D. The best answer is going to be answer choice B: The acid catalyst enhances the reaction by creating a better leaving group.

35) What exactly can we use the IR spectrum for? To tell us which functional groups are found in our starting material and products.

First and foremost, we want to know about what occurred in the reaction in the question stem before we make observations about the infrared spectrum of the reaction mixture. This is as simple as knowing what occurred in this reaction. The aldehyde (our CHO) is replaced with a primary alcohol (CH2OH). The single OH group is present in both the reactants and products. Our OCH3 is also present in both the reactants and products. Our answer should include something about the aldehyde group being replaced with a primary alcohol. 

  1. The disappearance of an O–H stretchWe said there is a new O-H in our product. There should be the appearance of an OH stretch, not the disappearance of an O-H stretch. This contradicts what we said in our breakdown of the question.
  2. The disappearance of a C=O stretch. The -OH stretch that appeared is preceded by the disappearance of the aldehyde. Aldehyde groups (CHO groups) have a double bonded carbon and oxygen. This is the only carbonyl group in the reactants, or the products. In fact, our OCH3 and our hydroxyl groups (which we said don’t change from reactants to products), but both only contain single bonds. We can eliminate answer choice A now, because it contradicted our breakdown, and answer choice B is a better answer.
  3. The appearance of an aliphatic C–H stretch. Aliphatic hydrocarbons contain carbons and hydrogens. In this case, the reactants and products both would have C-H stretches, so the appearance of the primary alcohol doesn’t cause the appearance of a C-H stretch. That would have already been found in the IR spectrum. This is a contradiction and answer choice B is still superior.
  4. The appearance of an aldehyde C–H stretch. Like we mentioned in answer choices A and B, the reactant had an aldehyde group, and the product has a primary alcohol. The aldehyde functional group is not present in the product, so there would be no appearance of an aldehyde C-H stretch. Once again, this is a contradiction, so we can eliminate answer choice D and we are left with our correct answer choice, answer choice B. If the reaction shown in the question stem occurs, we would observe the disappearance of a C=O stretch.

36) The name Proton Sponge could imply the compound is basic if it is a good proton acceptor. What makes a good nucleophile, and what do we know about the structure of the compound? The compound is aromatic and has tertiary amino groups that are located on the same side of the compound. We already hypothesized that the compound was basic based on its name and its structure-it’s a good proton acceptor.

Nucleophilicity depends on how readily a species will donate electrons. So basic compounds are typically good nucleophiles, but a big factor in determining nucleophilicity is the structure of the compound itself. Substitution reactions don’t proceed when there is steric hindrance like we’re seeing in our compound here. We mentioned both tertiary amino groups are on the same side of the compound. 

  1. Yes, because it is a strong base. This sounds like a decent answer. Strong bases will often make strong nucleophiles. But this answer doesn’t consider the steric hindrance from the two tertiary amino groups that we said will make the compound a poor nucleophile.
  2. Yes, because the nitrogens have high electron density. Again, we have an answer that says “yes” even though our breakdown was leaning “no.” The lone pair on nitrogen normally mean amine is a good nucleophile. Again, in this case the answer choice doesn’t consider the steric hindrance.
  3. No, because it is aromatic. We already established the compound is aromatic when we went through step 3. But does this actually influence the nucleophilicity? Aromatic rings can be nucleophiles, but also electrophiles. This answer choice is partially consistent with our breakdown.
  4. No, because the nitrogens are sterically hindered. Looking at our compound, the nitrogens are sterically hindered, and we established this doesn’t make the compound a good nucleophile. What this question really demonstrates is the importance of understanding the scope of the question. All of the other answer choices list true statements, but they don’t answer the question being asked about whether the compound is a good nucleophile. We stick with our correct answer, answer choice D.

37) The compound shown contains 3 amino acids. We’re trying to use these 3 same amino acids to form different tripeptides. A tripeptide means a peptide derived from three amino acids, the same as the compound in the question stem. Do we have to know the identity of the amino acids in the compound? In general, we do, but for this question? Not at all. 

We have to know the possible number of peptides given we have one of each of our three amino acids. The formula for the number of peptides if we have one of each of three amino acids is 3 factorial, or 3*2*1, or 6.

What’s the general formula? N! (N factorial) if we’re trying to find the possible number of different peptides that contain each of N amino acids. So, if n=4, then 4! Is 24. 

  1. 1. We know this is not the case because that is in incorrect value compared to our prediction. Just switching two of the amino acids in the compound in the question stem shows we can get more than one different tripeptide. This would be a correct answer if we were trying to find the possible number of different peptides that contain one single amino acid. 1 factorial is 1.
  2. 3. This also contradicts our calculated value of 6, but it’s closer than answer choice A
  3. 6. This is consistent with our calculation. If we want to, we can actually number, then write out the different combinations to double check: 123, 132, 213, 231, 312, 312. Answer choice C is going to be our best answer.
  4. 9. This would also be an incorrect value. How might we get answer choice D? If we incorrectly calculated the number of possible peptides as n squared instead of n factorial. We stick with answer choice C as our correct answer.

 

Chemistry Question Pack: Passage 7

38) We were told different properties of elements and bonds in the passage. This question focuses sulfur hexafluoride, and we have to describe its bonding and electrical conductivity. That means we can use the information given in the passage, likely about covalent and ionic compounds.

The passage says nonmetallic elements make up covalent compounds. Ionic compounds are a metal and nonmetal. Sulfur and fluoride are both nonmetals, they’re found on the right side of the periodic table. We can start forming our prediction-the bonding should be covalent, and we’ll want to kill any ionic answer choices. 

The passage says many covalent compounds are liquids or gases (so they are aqueous) and they will not conduct electricity, meaning our compound will be a nonconductor. We’re looking for an answer that mentions covalent bonding and a nonconductor in terms of electrical conductivity.

  1. covalent and a nonconductor. This corresponds to our breakdown. Sulfur and fluoride are both nonmetals and we’re saying it’s a covalent compound. We also said covalent compounds don’t conduct electricity, so answer A looks good. 
  2. ionic and a nonconductor. We said ionic compounds are a metal and nonmetal, which is not the case here. This contradicts what we’re told in the passage. Answer choice A is still our best answer.
  3. covalent and a conductor. This matches half of our breakdown, but we said we’re looking for answers that say nonconductors.
  4. ionic and a conductor. This is the exact opposite of our prediction and the passage. We can eliminate this answer choice for being a contradiction. Our correct answer is A, covalent and a nonconductor.

39) The periodic table is based on trends, so all this is asking is about a specific trend, where nonmetals are typically found. Metals are generally found on the left side and middle of the periodic table, the nonmetals are found on the right side, Metalloids separate the metals and nonmetals and share characteristics of metals and nonmetals. We’re looking for an answer that mentions the right side of the periodic table.

  1. Right side. The right side of the periodic table has nonmetals like the noble gases and halogens. This is consistent with our breakdown. Let’s check out the additional answers just to be thorough.
  2. Left side. This is where metals are found, so we eliminate this answer choice.
  3. Top half. The metals and nonmetals are split between left and right. The top and bottom halves don’t have anything to do with the split
  4. Bottom half. This is similar to answer choice C. We can eliminate answer choice D. We’re left with the correct answer, answer choice A. Nonmetals are found on the right side of the periodic table.

40) We’re told trends in the passage, and now we’re asked to pick examples of both ionic and covalent compounds. Correct answer will have an ionic compound first, and a covalent bond 2nd. The definition of ionic and covalent will come from the passage, but we’ll need to use the periodic table as well.

The passage said that covalent compounds are comprised of nonmetallic elements. Ionic compounds are made up of a metal and nonmetal. So, let’s find our compounds on the periodic table, and see which pair lists an ionic compound first, and a covalent bond 2nd. We can start going through our answer choices and compare. 

  1. HBr(g) and NaCl(s) Hydrogen and bromine are both nonmetals. Sodium is a metal, chlorine is a nonmetal. So answer choice A lists a covalent compound and an ionic compound. This is the opposite order of the correct answer according to the question stem
  2. NaCl(s) and NaI(s) NaCl we said is ionic. NaI is a metal and nonmetal, which means it is also ionic. This doesn’t match our question stem
  3. NaI(s) and NaCl(s) Answer choice C is the same as answer choice B in reverse order. Both are ionic
  4. NaCl(s) and HBr(g) Answer choice D is the opposite of answer choice A. We have an example of ionic bonding with NaCl and an example of covalent bonding with HBr. This matches the correct answer according to the question stem. That means answer choice D is the best example of ionic bonding and covalent bonding listed in order.

41) We’re given four solid compounds in our answer choices. We have to decide which compound has the most ionic character. The passage tells us what makes compounds ionic and covalent. That’s what we’re going to use to pick the correct answer here. We’ll also need the periodic table again to see where each element is found.

Once again, the passage said that covalent compounds are comprised of nonmetallic elements. Ionic compounds are made up of a metal and nonmetal. So, let’s find the elements that make up our compounds on the periodic table. We’re looking for a metal and nonmetal. And we want the compound with the most ionic character. That usually means an element near the bottom left of the periodic table where electronegativity is the least. And the top right of the periodic table where electronegativity is greatest.

  1. KBr(s) Find potassium and bromine in row 4. We have a metal and a nonmetal, so this is an ionic compound, we’ll hold on to answer choice A for the time being.
  2. CsCl(s) Cesium is near the very bottom left of the periodic table, while chlorine is near the top right. This is the prototypical ionic compound. The electronegativity difference doesn’t get much greater. We like this answer choice better than answer choice A. Remember we’re looking for the most ionic character.
  3. NaI(s) Sodium is near the top left of the periodic table, while I is near the bottom left. This is a metal and nonmetal, but the electronegativity difference isn’t as much as answer B based on our trends.
  4. RbBr(s) Rubidium is near the bottom left of the periodic table, and bromine is on right right side of the table. This is likely our second-best answer choice, but notice that cesium is going to be less electronegative than rubidium and chlorine will be more electronegative than bromine. This means answer choice B is still the best answer because it has the highest electronegativity difference. Always keep in mind how important the trends are for periodic table questions. All four answer choices list ionic compounds. Answer choice B is our best answer choice because it has the most ionic character.

42) We have to explain why HCl is not an ionic compound. The passage went into details about ionic compounds, and we have to explain why HCl matches these criteria, even though it is not an ionic compound. Ionic compounds are solid at room temperature, made of a metal and nonmetal, and aqueous solutions conduct electricity. Covalent compounds are liquids, gases, or solids, mostly nonmetals, and aqueous solutions do not conduct electricity.

HCl is a covalent compound. But one of the answers lists a property of an ionic compound.

  1. It is a gas at room temperature. We just said covalent compounds can be gases at room temperature, and ionic compounds are solid at room temperature. Choice A does not imply HCl is an ionic compound. Let’s keep comparing our answer choices
  2. A 1 M solution freezes below 0°C. Once again, ionic compounds are solid at room temperature, so they aren’t likely to freeze below 0 degrees Celsius. This isn’t consistent with HCl being an ionic compound
  3. A 1 M solution conducts electricity. We said ionic compounds conduct electricity, and covalent compounds do not. This would be consistent with HCl being an ionic compound
  4. It is composed of two nonmetals. Ionic compounds are made of a metal and a nonmetal. Hydrogen and chlorine are both nonmetals. That is consistent with HCl being a covalent compound. Answer choice C is the only property of ionic compounds. Even though HCl is a covalent compound, answer choice C is the only statement that would be consistent with being an ionic compound.

 

Chemistry Question Pack: Passage 8

43) Diamond is formed from graphite at extreme pressures and it’s formed from carbon. This question is asking about the structure and bonding of diamond and which elements its similar to. Elements found in the same group will often have similar chemical and physical properties, which is why we can group them together and give the columns names. So, we likely want to pick elements that are in the same group as carbon. We’ll be using the periodic table to find the elements in the same period as carbon.

The passage says that elements subjected to high pressures will assume the structure and bonding characteristics of a heavier element in the same column of the periodic table. So which pair of elements is found in the same group as carbon? Choice A is aluminum and gallium. Choice B is silicon and germanium. Choice C is phosphorus and arsenic. Choice D is sulfur and selenium. Only silicon and germanium are both in the same column as carbon. The correct answer is answer choice B.

44) This is a multi-level question. First, we have to know the type of bonding that’s present in solid nitrogen. Next, we have to know about the bonding in methanol. We’re ultimately trying to explain the increase in energy density in the solid nitrogen that was discussed in the passage. Energy density is how much energy the nitrogen has, compared to its volume.  The only information we’re given in the passage related to this question is nitrogen having an energy density that’s equal to or greater than hydrocarbon fuels. The actual reasoning is going to come from general knowledge. We’re comparing solid nitrogen with methanol, which is a liquid. We can visualize here-a solid is much more compact than a liquid. 

Next, we know diatomic nitrogen contains triple bonds. For homonuclear, diatomic molecules, there’s no electronegativity difference. Normally we have to use our periodic trends, but in this case both nitrogens have equal electronegativity values. The bonding is going to be covalent. Methanol has the -OH group, so there is hydrogen bonding taking place. We classify hydrogen bonding as intermolecular forces. Chemical bonds are much stronger than inter-molecular forces. 

Last thing we want to mention, the nitrogen molecules are well packed in small volume, so their energy density is so much higher than that of the methanol liquid. Solid molecules are much closer to each other, compared to liquids. 

  1. Solid nitrogen contains covalent and ionic bonds; methanol contains only weak ionic bonds. We said solid nitrogen will only contain covalent bonds. Ionic bonds occur when elements bond and there are large electronegativity differences. Methanol has polar covalent bonds, and also has hydrogen bonds.
  2. Solid nitrogen contains covalent and ionic bonds; methanol has covalent bonds within each molecule and weak van der Waal’s interactions between molecules. Again, nitrogen will have the covalent bonds only, not ionic bonds. We’re looking for something that also mentions the hydrogen bonding in methanol
  3. Solid nitrogen contains only covalent bonds; methanol contains only weak ionic bonds. First part of the answer looks good. Solid nitrogen has covalent bonds. It says methanol contains weak ionic bonds. We said methanol has polar covalent bonds and hydrogen bonds.
  4. Solid nitrogen contains only covalent bonds; methanol has covalent bonds within each molecule and weak intermolecular interactions. First part of the answer looks good. Solid nitrogen contains covalent bonds. Second part of the answer says methanol has covalent bonds, which is the case between the carbon and oxygen. It also mentions weak intermolecular interactions-that’s our hydrogen bonding because of the hydroxyl group. This sounds like the best answer choice and it doesn’t contradict our breakdown like answer choices A-C, so we stick with our correct answer, answer choice D.

45) The passage tells us about the general background of solid nitrogen, but doesn’t go into specific characteristics. We’re using general knowledge to explain the entropy change that accompanies a phase change to a solid. 

Entropy decreases following a phase change from a gas to a liquid to a solid. Solids have the fewest microstates. They’re by far the most ordered state. Glancing at the answer choices as well, we’ll have to explain delta S as well. ΔS represents the change in entropy of the system and it is expected to be negative to correspond to a decrease in entropy. 

Entropy will increase if there are additional microstates in a system (solid>liquid>gas). That’s when delta S would be positive.

We’re looking for an answer along the lines of: delta S is negative and entropy will decrease.

  1. ∆S is positive, entropy decreases. We do expect entropy to decrease, but that corresponds to a negative delta S. This contradicts our breakdown and general knowledge.
  2. ∆S is positive, entropy increases. This is the exact opposite of what we’re looking for. We’re still keeping answer choice A as our best answer. 
  3. ∆S is negative, entropy decreases. A negative delta S corresponds to a decrease in entropy. We said a solid nitrogen will mean a decrease in entropy as well. This is consistent with our breakdown.
  4. ∆S is negative, entropy increases. The negative delta S is consistent with the synthesis of solid nitrogen, but would correspond to a decrease in entropy. Not an increase. Best answer choice here is C. ΔS is negative, and entropy decreases.

46) The passage mentions solid form of nitrogen is stable at above 65 GigaPascals, but we’re not given much additional information. We’re going to have to know how gases behave under various conditions. This is general knowledge, and we’re not just using the passage to answer our questions. 

As we mentioned, the passage mentions solid form of nitrogen is stable at above 65 GPa. 65 Gigapascals is 65 x 10^9 pascals, or a BIG number. Continuing with this, at 180 GPa, we can’t assume the ideal gas law is in play, just because the solid is stable. Gases deviate from the ideal gas law at low temperatures and at high pressures. 180 gigapascals is an extremely high pressure. So, we would expect deviations due to the pressure difference, but not necessarily because of the room temperature.

  1. No, because both temperature and pressure must increase before such deviations are observed. We mentioned deviations from the ideal gas law happen at very low temperatures and at high pressures. In our breakdown we said that we would see a deviation from the ideal gas law.
  2. No, because nitrogen molecules are symmetrical and do not interact with each other. Again, our prediction is that we see a deviation from the ideal gas law.
  3. Yes, because at this pressure, molecular volumes and intermolecular forces become significant. We expect gases deviate from the ideal gas law at low temperatures and extremely high pressure. At a pressure this high, and where solid nitrogen is expected to be stable, molecular volumes and intermolecular forces do become significant. Normally the ideal gas law expects them to be insignificant.
  4. Yes, because at room temperature, molecular volumes and intermolecular forces become significant. Room temperature is not an extremely low temperature. We expect a deviation at extremely low temperatures and high pressures. Room temperature would not mean molecular volumes and intermolecular forces become significant. Our correct answer choice is C. We expect deviations because at this pressure, molecular volumes and intermolecular forces become significant.

 

Chemistry Question Pack: Passage 9

47) We’re given four different mixtures here and told specific details about the components in each set of mixtures. Each component is present at a concentration of 0.1 molar. To answer this question, we want to know which mixture has a pH closest to 7. We’re going to need the passage to get specific details, but we need to know how to analyze the details about the pH and Kas using general knowledge.

Quick overview of our answer choices. First component of each mixture listed is one of the four acids from the passage. Second component of each mixture is the conjugate base that reacted to sodium ion from sodium hydroxide. We’re using the details from the passage.

We have a table that gives us Ka values. 

What general knowledge do we need to know? When concentrations are equal like we have in this question. pH of our buffer is equal to pKa of our acid, so in this case we want a pKa of close to 7. We’ll have to convert our Ka to pKa. pKa can be found by taking the negative logarithm of Ka. We can approximate this using the following formula:

-log (y-10^-x) =x-0.y formula.

The only option here that will give us a value of close to 7 is hypochlorous acid, which gives an approximate value of 7.7. 

We can eliminate the other answer choices for being incorrect values-we already compared all of the answers. Keep our correct answer choice, answer choice A: HClO(aq) and NaClO(aq)

48) What is dissolution? Another word for dissolving. So we’re given the same reactant in every answer choice, our ammonia nitrate, and we have to show the equation for what exactly is taking place. There wasn’t anything in the passage that would hint at the reactants in this dissolution, so we’ll use general knowledge to pick the correct equation. 

We’re going to need to use solubility rules. First and foremost, we know ammonium ion (NH4+) and nitrates (NO3-) are soluble in water. We have an ionic salt being dissolved in water, so we’re predicting there to be ionization that occurs with our two ions being ammonium ion and nitrate ion. Let’s consider our four options:

  1. Answer choice A goes against our breakdown. We said ionization occurs when ionic salts are dissolved in water. We’ll compare this answer choice to the other choices before eliminating
  2. Answer choice B, shows ions, but doesn’t have the correct charge on our ammonium ion or our nitrate ion. Choice B is better than A for the time being, we can eliminate answer choice A.
  3. Answer choice C shows an unbalanced equation. We should still be able to see the transition from reactants to products. We eliminate answer choice C for being unreasonable.
  4. Answer choice D, shows ionization occurring, and the formation of ammonium ion and nitrate ion, just like our breakdown. Answer choice D matches what we said in the breakdown of the question exactly, so we’ll keep it as our correct answer. We can eliminate answer choice B for contradicting that breakdown.

49) We’re noting a change in the pH of a solution when strong base is added. Specifically, we have 2 mL of 0.1 molar sodium hydroxide being added to our solution. The question wants to know how much of change in pH we will see.

We’re told we’re adding the strong acid to a solution containing 0.1 molar hypochlorous acid and 0.1 molar sodium hypochlorite. This was something we needed to know for an earlier question. When an acid and conjugate base are found in equal concentration, then the pH of a buffer equals the pKa of the acid. 

Said differently, we’re at the half-equivalence point of the titration curve. The half-equivalence point is the point at which pH=pKa. And the concentration of acid in solution is equal to the concentration of the conjugate base in solution. At this point when a strong base like sodium hydroxide is added, pH would still become more basic, but change very slowly. 

  1. A slight (<0.1 pH unit) increase. This sounds exactly like our breakdown. Adding a strong base is going to increase pH. But we expect pH will only change slightly, and would change very slowly
  2. A slight (<0.1 pH unit) decrease. We do expect there to be minimal change, but adding a strong base is going to make pH increase, not decrease. We can eliminate answer choice B for being unreasonable.
  3. A significant (>1.0 pH unit) increase. It’s true pH is going in the right direction here, but we’re expecting only a small increase in pH. Answer choice A remains superior.
  4. A significant (>1.0 pH unit) decrease This is the opposite of our breakdown. We said that pH will increase if we add strong base. We’re expecting the change in pH will be minimal; we’re not expecting a significant change. Answer choices C and D contradict what we said in our breakdown. We’re left with our correct answer choice. Answer choice A: A slight pH increase.

50) This question is going to rely on knowing our solubility rules, and properties of sodium nitrite. Sodium nitrite is a salt. Salts containing group I elements (such as Na+) are soluble. They’re usually listed as solubility rule #1 because they are completely soluble. The nitrite ion is basic, it can pull out a H+ from water, leaving hydroxide ion OH-. What do we know about hydroxide ions? They increase pH. 

  1. NO2 is hydrolyzed with the formation of OH (aq) ions. The nitrite ion is hydrolyzed, and following the reaction with water, there are hydroxide ions remaining. Hydroxide ions increase pH. Good answer choice so far.
  2. Na+ is hydrolyzed with the formation of OH (aq) ions. The sodium ions won’t react in the water like the nitrite ions. Alkali metals cations are soluble and inert in water.
  3. NaNO2(aq) decreases the Ka of HNO2(aq). The Ka of the reaction isn’t going to change by adding more reactant. This answer choice is unreasonable
  4. NaNO2(aq) increases the Ka of HNO2(aq). This is similar to answer choice C. Both mentioned a change in Ka via the reactants. Answer choice A is going to be our best answer here.

 

Chemistry Question Pack: Questions 51-55

51) We’re given an equilibrium equation above and we have to give a name for the transition from the compound on the left to the compound on the right.

First thing we want to do is break down what’s going on in the figure. A hydrogen atom from this CH2 moves to the oxygen atom on this carbonyl group. The formula on both sides of the equation is the same, the only difference is the connectivity. We essentially are dealing with constitutional isomers. Again, the connectivity is different, but if we count the atoms on both sides, they are the same.

  1. reduction. In reduction, there’s an increase in the number of covalent bonds between an atom and atom(s) that are less electronegative. Usually this means an increase in the number of bonds between carbon and hydrogen. We’re doing the opposite here. Let’s still compare with our other chances.
  2. resonance. Resonance structures are two forms of a molecule where the chemical connectivity is the same. In this case, we said the connectivity differs-the hydrogen moved. Let’s keep comparing and see if we find a superior answer. 
  3. tautomerism. Constitutional isomers are tautomers. They have the same molecular formula, just different connectivity. We’ll keep this answer choice and eliminate A and B for being the opposite of our breakdown.
  4. mutarotation. Mutarotation is the change in the optical rotation because of the change in the equilibrium between two anomers. The corresponding stereocenters interconvert. We see this a lot in chemistry with sugars, but in this case, that’s not what’s happening. We stick with answer choice C. The phenomenon exemplified in the equation is tautomerism.

52) Glycine is an amino acid, and polarity is the result of uneven distribution of electron density. It has a specific property that makes it a dipolar ion that we’re looking to identify. We’re working with amino acids, and dipolar ions. How can we either link these using our content outline? Or what do we know about each individual topic? 

Amino acids can exist as zwitterions, or dipolar ions, at neutral pH. So first things first, what is a zwitterion? Zwitterions are molecules with a positive charge and a negative charge in different functional groups. Amino acids are amphoteric-they have both an acidic carboxyl group, and a basic amino group. So why are amino acids so polar? Because of the uneven distribution of electron density.

  1. High dipole moment. This isn’t exactly what we predicted, but it’s at least consistent with our prediction. Amino acids are polar, like we mentioned. They’ll have high dipole moments as zwitterions. We’ll hold on to answer choice A, and let’s compare to our other answers.
  2. High molecular weight. The molecular weight of glycine is not what makes it a dipolar ion in aqueous solution.
  3. Low dielectric constant. That would mean glycine has low ability to store electrical energy. That’s not expected for the amino acid.
  4. Low solubility in water. Amino acids are generally soluble in water, particularly glycine. This contradicts our breakdown. We’re left with our correct answer choice, and our best answer choice in this case: Answer choice A, the high dipole moment.

53) This is a standalone question, so we’re using the figure in the question stem, and using external knowledge to answer this question. First things, we want to number our carbons. We have several substituents and functional groups attached to our carbons around the ring. However, only one carbon is attached to multiple substituents-that’s going to be carbon 1 (I’ve labeled it on the right side of the figure below).

If we look at the straight-chain structure, this is the carbon that was part of the carbonyl group, so in this case, we can also call it the anomeric carbon. 

The difference between our alpha and beta-D-glucoronides is going to be the side on which the -OR group is found. In the compound shown here in our question stem, the C-1 -OR group is on the same side of the pyranose ring as the carboxyl group. In alpha-D-glucuronide, the C-1 -OR is on the opposite side. So, our difference comes at the C-1 carbon.

  1. C-1.
  2. C-2.
  3. C-3.
  4. C-4.

Comparing answer choices here is going to be easy. We already broke down the structure and compared all of the carbons. We know the C-1 carbon is going to be where the configuration difference comes in. Answer choice A is the correct answer.

54) To answer this question, we’re going to be using stoichiometry and dimensional analysis. That means it’s going to be crucial to keep our units straight. Note our final answer is given in milliliters, and all answers are given in whole numbers. 

Stoichiometry’s going to help us balance the equation given. And we’ll more clearly see the relationship between the substances in our reactants and products. 

If we rewrite our equation, we need an additional H on the left side, so we add a 2 coefficient in front on HCl. We now need another chlorine on the right side and an extra sodium. We add a 2 coefficient in front of our sodium chloride as well.

We’re also told conditions are STP. That means standard temperature and pressure. 273 Kelvin and a pressure equal to 1 atm. We’re dealing with carbon dioxide, which is a gas. Gases occupy a volume of 22.4 L for 1 mole of ideal gas. I’ve handwritten my work here, and I’ll break down the process right below.

We need to know the mL of a 2 molar solution of sodium carbonate to produce 11.2 L of CO2. We can quickly convert to moles of CO2 using our STP information. If 1 mole occupies 22.4 L of volume, ½ a mole will occupy 11.2 L.

For every ½ mole of CO2, there are also ½ mole Na2CO3. Both have the same coefficient.

Next, divide the ½ moles of sodium carbonate by molarity of 2.0 moles/liter.

We get 0.25 liters. But we said our final answer should be in mL. So we use dimensional analysis once again. Multiply by a factor (1000 mL/1 L) and get 250 mL of the 2 molar sodium chloride will produce 11.2 L of carbon dioxide at STP.

This is a math problem, and we did minimal rounding, so the calculated value can be used as our predicted value. We said our prediction is 250 mL. Comparing our answer choices. Answer choice B is the correct answer.

What are the key takeaways here? Always keep your units and conversions straight. If that means writing them out like I did here, then please make sure you do so for every dimensional analysis question. We want to build these good habits. The next thing, we want to make sure we start making the connections to our content outline as you work through these problems. We were told the reaction was at STP, and we knew right away what that meant and how to use that information. The better you know the content outline, and the more you practice using it, the better off you’ll be on exam day.

55) We’re shown a reaction that at first glance looks balanced. We’re going to analyze the reaction to see which species is oxidized, and which is reduced. That’s going to help us find the oxidizing and reducing agent in the reaction. What we want to be careful about the most, is the verbiage. This can get tricky. In the reaction, a molecule, atom, or ion in the oxidizing agent is reduced-meaning it gains electrons, and we’ll see a more negative oxidation number. A molecule, atom, or ion in the reducing agent is oxidized-meaning it loses electrons and we see a more positive oxidation number. Let’s look at our specific reaction:

We’ll go one by one through the reactants first, and use our rules for oxidation numbers. 

  • We have a neutral hydrochloric acid. Oxidation number of hydrogen is +1, and group 17 elements is -1. 
  • We have hydrogen peroxide. Again, oxidation number of hydrogen is +1, and is normally -2 for oxygen, but -1 in peroxide. 
  • Lastly, we have manganese dioxide. Oxidation number of oxygen is -2, so oxidation number of manganese is +4.

Products: 

  • First we have diatomic oxygen. Oxidation number of an atom is zero in a neutral substance with only atoms of one element. So oxidation number for oxygen is now zero.
  • Next we have manganese chloride. Group 17 elements have oxidation number of -1. Manganese now has oxidation number of +2. Last we have H2O. Hydrogen is +1, oxygen is -2.

What was reduced? Manganese was reduced from +4 to +2. What was oxidized? Oxygen from -1 to 0. 

The manganese dioxide is the oxidizing agent. It is an electron acceptor-it gains electrons. Hydrogen peroxide is the reducing agent. Reducing agent will lose electrons.

  1. H2O2; HCl. Answer choice A says hydrogen peroxide is the oxidizing agent and hydrochloric acid is the reducing agent. We went through our reactants one-by-one. HCl isn’t the oxidizing OR reducing agent in this situation. Also, we said hydrogen peroxide is the reducing agent.
  2. H2O2; MnO2. Answer choice B says hydrogen peroxide is the oxidizing agent, and manganese dioxide is the reducing agent. This is the opposite of our prediction and what we found going through our oxidation numbers. This answer choice isn’t much better and answer choice A.
  3. MnO2; HCl. Answer choice C says manganese dioxide is the oxidizing agent, and hydrochloric acid is the reducing agent. We said manganese dioxide is our oxidizing agent, but hydrochloric acid isn’t the oxidizing OR reducing agent. This is our best answer choice thus far.
  4. MnO2; H2O2. Answer choice D says manganese dioxide is the oxidizing agent, and hydrogen peroxide is the reducing agent. This sounds like a keeper. It matches our breakdown exactly. We said manganese was reduced from +4 to +2 and manganese dioxide is the oxidizing agent. Oxygen is oxidized from -1 to 0, and hydrogen peroxide is the reducing agent. We’ll stick with answer choice D as our correct answer. 

 

Chemistry Question Pack: Passage 10

56) Table 1 is essentially a summary of all of the various conditions in the experiment, and how they affected the rate of Reaction 1. Which of the options can be added in excess without influencing the rate? We can pull up Table 1 and see how Solution A and B are affected. We can determine the role of each of our four answer choices to answer correctly.

As I mentioned, we need to pull up Table 1 to answer our question. Answer choices are near the top right of the figure, and the breakdown of the two solutions is right below the table.

When we have an experimental passage, it’s not uncommon to be asked about specific components of an experiment. The test-maker wants to see if you understand the experiment and the various components that go into the experiment.

The four answer choices listed correspond to the 4 compounds in Solution A and Solution B. All of the aqueous compounds are directly related to Reaction 1 and Reaction 2 in the passage. Starch on the other hand, was not found in the reactants or products. It was used as an indicator. Adding excess starch doesn’t affect the rate of the reactions, because starch is simply an indicator. Answer choice D is going to be our best answer choice because starch doesn’t participate in the reactions.

57) In the passage we learned that “The students added 1 drop of 0.1 M CuSO4(aq) to Tube 6.” To answer this question, we simply have to explain the function of this drop. We can go back to the passage to check the effect of CuSO4 on the rate of Reaction 1. 

Copper sulfate is added to Tube 6. It was one of the lowest times in the rightmost column. This is despite all of the other conditions being identical to Tube 1. Looking at Tube 6, we see a time that is ~33% faster than Tube 1, with the other conditions being identical. Seemingly, the copper sulfate increases the rate of the reaction. That sounds like an enzyme or catalyst of sorts. 

The AAMC outline emphasizes that we need to be able to demonstrate the understanding of important components of scientific research, meaning you have to be able to read and understand experiments. 

In this specific case, we have to know what a catalyst is, and how it is used in experiments. Enzymes are catalysts that are able to lower activation energy without being used up or being destroyed in the reaction. Catalysts work to increase the rate of a reaction by lowering activation energy.

  1. Reactant. Adding more reactant wouldn’t cause the rate of the reaction to increase the way it did. We didn’t actually change the reaction taking place; we just make the reaction proceed more quickly. This goes against what we saw in the experiment. Let’s look for a better answer choice.
  2. Indicator. We already had an indicator in the passage. We used starch as an indicator, and indicators won’t affect the rate of the reaction. Copper sulfate actually affected the rate, so we’re still looking for a different answer.
  3. Inhibitor. The rate of the reaction actually increased, while an inhibitor would have the opposite effect. This answer choice is unreasonable and it goes against Table 1.
  4. Catalyst. This matches our breakdown exactly. The increased rate of reaction in Tube 6 was because the copper sulfate acted as a catalyst.

58) We’re comparing Tube 1 and Tube 4. We’re also explaining why the rate of the reaction decreases in Tube 4. We’ll go back to Table 1 once again. We should be able to see a decreased rate of reaction in Tube 4. We want to justify why that is.

Starch is used as an indicator that turns solution dark blue. We recall that from the passage. The time column of the table is measuring the time it takes for the solution to turn dark blue. Once again, we’re explaining why Tube 1 shows less time in that column that Tube 4. Why did Tube 1 turn more rapidly than did Tube 4?

Tubes 1 and 4 have the same volume conditions, but the only difference we see is with temperature. Temperature in Tube 1 is 10 degrees higher. The lower temperature in Tube 4 correlates to a decreased rate of reaction. 

  1. rate of Reaction 2 was slower in Tube 1 than in Tube 4. We’re comparing the rate of Reaction 1 in the tubes, not the rate Reaction 2. There’s a decreased rate of reaction in Tube 4, so this contradicts Table 1. 
  2. average kinetic energies of I(aq) and S2O8(aq) were greater in Tube 1 than in Tube 4. Tubes 1 and 4 have the same volume conditions, but the only difference we see is with temperature. Temperature in Tube 1 is 10 degrees higher. Increased temperature means higher average kinetic energy. This is better than answer choice A. 
  3. concentrations of I(aq) and S2O8(aq) were greater in Tube 1 than in Tube 4. The concentrations don’t change and are the same in both tubes. We have to make sure to know the experimental conditions well. Answer choice B is still the best option.
  4. concentration of starch was greater in Tube 1 than in Tube 4. The concentration of starch will not affect the reaction beyond being an indicator. The increase of 10 degrees Celsius will increase the average kinetic energy of the components in Tube 4. We’re left with our correct answer, answer choice B: average kinetic energies of I(aq) and S2O8(aq) were greater in Tube 1 than in Tube 4

59) We’re using the variables given in the passage and looking at them in a different format: a line graph. We can go back to Table 1 once again. 

We want a graph that shows moles of S4O62–(aq) in Tube 6. First thing we want to note, is the reaction turns dark blue in 19 seconds. Dark blue is the result of the starch indicator telling us thiosulfate was used up in Reaction 2, and the maximum amount of tetrathionate has been formed at this point.

What do we expect the graph to look like? We expect the product to steadily increase from 0-19 seconds (19 seconds is the max number of moles), then we should see the graph plateau at that point.

a.

Graph A implies the moles continue rising, even though the reaction is complete in 19 seconds. We expect the line will continue to rise until 19 seconds, then plateau

b.

Graph B does exactly what we expect, rises for 19 second, then plateaus. This is now our best option.

c.

Graph C says the moles increase after 19 seconds, but the reaction is complete at that point, not starting

d.

Graph D says the number of moles increases, but then subsequently decreases. Not the case. The moles do not disappear. We’re going to stick with our best option, answer choice B. 

 

Chemistry Question Pack: Passage 11

60) We had a reaction take place at 300 Kelvin, and the question is asking us to write out the net reaction. To answer this question, we need to see the figure in the passage to see what reaction actually took place.

We have our figure above. Look at the figure to see that for every X in reactants (or before) side of Figure 1, two Ys bind to form the ultimate Y2X product. So, the ratio of X reactants to Y reactants in the net reaction will be 1:2.

  1. X + Y2 → Y2X. Are there any diatomic Y reactants in our figure? No, there were only single X and single Y reactants. This is close to what our breakdown said and we still want to compare with our other answers, but keep in mind we didn’t have diatomic Y reactants
  2. X +2Y → Y2X. Our ratio we came up with is consistent with answer choice B. And our product, the Y2X, is consistent with figure 1. We like answer choice B better than answer choice A
  3. 2X + Y → YX2. This is the opposite of our breakdown, and the product in the figure. It contradicts both. So, we still like choice B as the best answer so far.
  4. X2 + Y → YX2. Answer choice D is similar to answer choices A and C. There is no diatomic X reactant in our figure, and our product is Y2X, not YX2. Our best answer is B.

61) This question is very straightforward if you’ve taken the time to read the passage. To answer this question, we can use the results of the experiment to predict the reaction rate at 400 K. What was the biggest thing we should’ve picked up from reading this experimental passage? As temperature decreases, reaction rate decreases. As temperature increases, reaction rate increases. This ties the whole experimental passage together. The purpose of the experiment was to explain the relationship between temperature and the amount of product formed, and reaction rate. So, what do we expect to happen here? Answer choice A: Reaction rate should increase. 

62) We already know the reaction in the passage is exothermic, so there’s no real reason to go back to the passage. Because it’s exothermic, the potential energy for our product will be lower than for our reactants. 

Energy (that’s shown on our y-axis in our answer choices) is going to be lower at the end of our reaction. Next thing we want to note, do we have to overcome an activation energy barrier? We absolutely do, and we can do that simply by increasing temperature.

a.

Answer choice A shows a reaction where we overcome the activation energy threshold. But energy ends up in the same location on the y-axis. We’ll still hold on to this answer choice for now to compare with our additional answers

b.

Answer choice B shows an endothermic reaction. That contradicts what we’re told in the question stem. That’s no better than choice A

c.

Answer choice C shows the activation energy as a threshold of sorts. But we know it’s not a dip in energy as shown. In the experiment we were able to overcome the activation energy barrier with an increase in temperature. The final energy is lower than our initial energy here though, so that part matches our breakdown at least

d.

Answer choice D shows an exothermic reaction, and the proper activation energy barrier that matches our breakdown. Answer choice D is going to be our best answer. We can eliminate the remaining answer choices that we already invalidated

63) To answer this question, we want to visualize the product molecule, our Y2X, and determine the molecular shape. We can go back to the passage, but the issue is, the passage doesn’t tell us about the identity of our reactants. There’s no way to know how many electrons are found in our product molecules. So even though we know our central X atom is attached to two electron dense areas, our Y atoms, we can have any number of valence electrons on our X atom. There are two possible options: The molecular geometry can either be linear or bent, depending on the identity of the individual reactants, and the number of electron dense areas in our final molecule.

  1. Linear. This is possible. The simplest linear molecule would be a central X attached to two Y atoms with no lone pairs. 
  2. Bent. Again, we said this is possible. If the central X attached to two Y atoms with one lone pair or two lone pairs, the molecular geometry would be bent. We’ll still hold on to this answer choice as well
  3. Trigonal planar. This molecular geometry would only be possible if our central X was attached to three Y atoms. This contradicts our breakdown so we can eliminate this answer choice.
  4. The shape of the product molecules cannot be determined without more information. This seems to be the correct choice in this situation. Answers A and B both give possible right answers, but we know every question only has one correct answer. That means we’re left with answer choice D.

 

Chemistry Question Pack: Passage 12

64) Triacylglyerols are formed through the joining of three fatty acids to a glycerol backbone by a dehydration reaction. We want to know specifically about the triacylglycerols and its number of stereogenic centers. We have Reaction 1 in the passage that showed us the structure of triacylglycerol. We’re going to be using general knowledge to identify the number of stereogenic centers. We can draw out the reaction of our glycerol with the three fatty acids to form triglyceride. 

We know the structure of glycerol. We’re reacting with three fatty acids. Ultimately, we form this triacylglycerol. Any carbons attached to four different groups will be a stereogenic center. The top carbon and the bottom carbon along the glycerol backbone are bonded to two hydrogens each. Carbon 2 is attached to four different groups, so we have a single stereogenic center. 

Looking at our answer choices, we’re given numbers from zero to three. We already predicted there is only a single sterogenic center on carbon 2. Our correct answer is going to be B: there is 1 stereogenic center in the product triacylglycerol.

65) In the passage we were shown reaction 1 that produced fatty acid salts. This question wants us to pick the formula that represents the general structure of the fatty acid salts produced in the reaction. We have to know the general formula of a fatty acid, and then the general formula for fatty acid salts that were produced in Reaction 1. We can use the reaction in the passage as a reference point, but the answer here is going to come from external knowledge.

We have reaction 1 listed here. Again, first step, let’s draw out the general formula for a fatty acid. General formula here is going to be our R group attached to a carboxyl group.

Our fatty acid salt was formed through saponification with sodium hydroxide, so we’re expecting our fatty acid salt to have a corresponding sodium ion attached. The general formula for the fatty acid salt is going to be our R group. At the end, we have CO2 minus and a positive sodium. This is going to be the general formula for the sodium salt of the fatty acids.

  1. Rn—CH2 Na+ This answer completely gets rid of both oxygens that were found on the fatty acid carboxyl group. We said that’s not the case. We’re losing a single hydrogen on the hydroxyl, and we’ll have the negative charge on that remaining oxygen. Answer choice A contradicts our breakdown, but we’ll compare to our other options
  2. Rn—CH2O Na+ This answer similarly gets rid of one of our oxygen atoms. We’re left with two hydrogens on our last carbon, but compare to our prediction. We still have the double-bonded oxygen, plus the negatively-charged oxygen bonded to the carbon in our breakdown. We don’t see that in answer choice B. This choice is better than A because it’s closer to our breakdown, we can eliminate answer choice A
  3. Rn—C(O) Na+ This answer has our double bonded oxygen atom on the carbon, but the other oxygen from the carboxyl group is no longer a part of this answer choice. This answer choice is similar to answer choice B. One aspect of our breakdown is present, but it still does not fully match our expected answer.
  4. Rn—CO2 Na+ This answer matches our breakdown completely. We had a carboxyl group at the end of our fatty acid. The fatty acid salt is going to have the sodium ion that’s present in every answer choice, and the general formula is going to match our breakdown. We stick with answer choice D for matching our criteria.

66) We want an explanation for the four different fatty acid salts that were the result of the saponification of the triacylglycerol. Normally we’d expect three, but in this case, there were four, and why is that? The passage actually mentioned there were four fatty acid salts produced, but we didn’t get any specific reasons.

Saponification is the formation of a salt following the mixture of a long-chain carboxylic acid with a strong base. In this case, the strong base was sodium hydroxide, and there were four fatty acid salts instead of three. The two most likely outcomes would be if the original molecule was broken down into 4 acyl groups that turned into fatty acid salts, or one of the fatty acid salts was counted twice for some reason. Is it possible there were 4 acyl groups? Not likely. We broke down a triacylglycerol in basic conditions. There’re only going to be 3 fatty acid units. In fact, the passage explicitly says it was pure triacylglycerol. Now why might one of the fatty acid salts get counted twice? One of the fatty acids must have reacted. Unsaturated fatty acids can undergo addition and elimination under basic conditions, meaning cis/trans isomers were produced. This is a possibility that would explain why there were four different fatty acid salts.

  1. The triacylglycerol molecule consisted of four different fatty acid units. We actually discussed this in our breakdown of the question. A triacylglycerol contains 3 fatty acid units. We can’t have 4 acyl groups from triacylglycerol. The passage also says the triacylglycerol was pure.
  2. Glycerol was transformed into a fatty acid salt under the reaction conditions. Think about this logically, if glycerol can be treated under saponification conditions to form a fatty acid salt, then wouldn’t every saponification reaction yield 4 fatty acid salts? This answer is unreasonable
  3. One of the fatty acid salts was unsaturated, and it completely isomerized under the reaction conditions. We did mention that one of the fatty acids could have isomerized, but what would happen if it completely isomerized? We’d still only see three fatty acids. It’s not likely the fatty acid completely isomerizes, and even if it did, we won’t see a fourth fatty acid. This answer is also unreasonable
  4. One of the fatty acid salts was unsaturated, and a small percentage isomerized under the reaction conditions. This is similar to answer choice C; we said one of the fatty acids could’ve isomerized. In this case only a small percentage isomerize. What does that mean? We can have both the cis and trans isomers, for example. When we’re analyzing our results, because of the addition and elimination that took place, we have 4 different fatty acid salts. Answer choice D is going to be our best answer here.

67) The example reaction in the passage used sodium hydroxide to saponify a triacylglycerol and formed glycerol and fatty acid salts. We want to know how much. And a quick glance at the answers shows this is not going to be a simple quantitative answer. To answer this question, we’re going to need to know about triacylglycerols and the breaking of the ester bonds during saponification.

Triacylglycerols consist of glycerol attached to three fatty acid chains. The strong base that is typically used in the saponification process is an alkali hydroxide. Just like in our passage, that alkali hydroxide is commonly sodium hydroxide. In the presence of the strong base, there’s cleavage of the ester bond which releases glycerol and fatty acid salts known as soap. Additionally, the triacylglycerols contain three ester linkages. So, our strong base, our sodium hydroxide, has to be present to cleave these three linkages. 

  1. A catalytic amount, because OH– is continuously being regenerated during saponification This doesn’t match our breakdown of the question, or the method of action of our sodium hydroxide. We need each hydroxide to be able to hydrolyze one of three ester linkages. There’s no regeneration taking place
  2. One-third of an equivalent, because each OH– ion reacts to form three fatty acid salts. One third of an equivalent wouldn’t be able to hydrolyze one of the ester linkages, let alone all three that are necessary. This answer choice is unreasonable.
  3. One equivalent, because each OH– ion reacts to produce one molecule of glycerol. We’re going to need three equivalents. Even if we’re only producing one molecule of glycerol, there are three ester linkages in a triacylglycerol. We’ll need three equivalents of strong base. Note, the reasoning here isn’t inherently incorrect, but the number of equivalents is.
  4. Three equivalents, because one OH– ion is required to saponify each of the three fatty acid groups. This sounds very close to our breakdown. First and foremost, we know we’ll need three equivalents. What’s the reasoning for that? Because there are three ester linkages. The strong base has to saponify each of the three fatty acid groups. That means we stick with our correct answer choice: Answer choice D: Three equivalents, because one OH– ion is required to saponify each of the three fatty acid groups.

68) We want to know about the fatty acid salts that were formed in Reaction 1 in the passage. Are the solubility properties going to come from the passage, or external knowledge? Likely from external knowledge. We weren’t given much information about the fatty acid salts that we shouldn’t have known coming into the passage. We still need to break down the structure of the fatty acid salts, then we can explore the solubility properties. I’ve handwritten below, and explained exactly how to break this question down.

We have our triacylglycerol and the 3 equivalents of strong base. Our products are our glycerol and the sodium salt of our fatty acid. We’re focused on the fatty acid salt to answer our question. We have a hydrocarbon chain, and we also have this charged CO2- and Na+.

What’s the golden rule for solubility? Like dissolves like. So first, Lipids are hydrophobic and nonpolar and are not very soluble in water. While structures may vary based on the type of lipid, all lipids have low solubility in water, and higher solubility in nonpolar organic solvents.

But, we have the charged group at the end of our salt. That’s going to be soluble in polar solvents. For example, table salt can dissolve in water because both are polar. Again, like dissolves like. So, we’re expecting our fatty acid salts to be able to dissolve partly in polar solvents and partly in nonpolar solvents.

  1. They are soluble in polar media only. We said the charged group at the end of the salt is going to cause the fatty acid salts to partially dissolve in polar media, but we also expect the hydrocarbon portion to be soluble in nonpolar media. This answer choice is partially correct, but the “ONLY” at the end makes this an extreme answer choice.
  2. They are soluble in nonpolar media only. This is similar to answer choice A. The hydrocarbon portion allows for solubility in nonpolar media. But the charged group at the end of the salt means it’s also soluble in polar solvents. This answer choice is also extreme
  3. They can partially dissolve in both polar and nonpolar media. This is similar to our prediction. We have the hydrocarbon chain that’s soluble in nonpolar solvent. The charged group is soluble in polar solvents. That means we can eliminate answer choices A and B for being too extreme
  4. They are completely insoluble in both polar and nonpolar media. This answer goes completely against our prediction. Part of our fatty acid salts are soluble in polar media, while the other part is soluble in nonpolar media. They aren’t insoluble in both. We stick with our correct answer, answer choice C: They can partially dissolve in both polar and nonpolar media. 

 

Chemistry Question Pack: Questions 69-73

69) We’re given the mass of each individual component in a compound, so we have to find the corresponding empirical formula. We’re going to need to know the molar mass of each individual component of the compound, so we actually need our periodic table.

Empirical formula is the smallest ratio of numbers to represent the proportions of the atoms. That means we’re going to need to divide the grams of each individual component in the compound to find the number of moles of each. For example, 12.0 grams of carbon in the compound divided by the mass on the periodic of 12.0 gives 1.0.

After dividing all three components, we have a ratio of 1:2:1. Carbon:hydrogen:oxygen. Can this ratio be simplified any further? No, it cannot. Before we go through each answer choice individually, let’s be careful here. We mentioned empirical formulas can’t be simplified. If we look at answer choices C and D, can those be simplified? Yes, they can, so we’re skeptical about those answer choices right away since they aren’t proper empirical formulas.

  1. CH2O. That actually has the components of the compound in the exact same ratio as our breakdown. There’s not much subjectivity here, so it’s likely this will be our correct answer. Let’s still consider the other options.
  2. C6HO8. This would be the correct answer if the molar mass of each individual element was 2.0, but we know that’s not the case because we just checked our periodic table. Answer choice A remains superior.
  3. C6H12O6. Answer choice C lists a fairly common hexose that we see in our science passages, but we already mentioned it’s not an empirical formula. In fact, if we simplify answer choice C, it would give us the same ratio as answer choice A.
  4. C12H2O16. Answer choice D simply puts the grams of each individual component into the subscript next to the elemental symbol. This is the ratio in terms of mass of each component, but not the correct empirical formula. We can stick with answer choice A as our correct answer.

70) The way we’re going to approach this question is to pay attention to subscripts and the equivalents of different components in the prepared solution. First thing we want to note, is we’re focused on Na+ in each of our initial solutions. By looking at the subscripts, we can see there is 1 equivalent of Na+ in the NaHCO3 solution (no subscript), and 2 equivalents in the Na+ Na2CO3 solution (subscript 2 on the sodium).

Given that information and also noting the molarity of each solution, there is also 0.010 M Na+ in the NaHCO3 solution and 0.020 M Na+ in the Na2CO3 solution. However, we’re not done yet. We’re mixing these two solutions to form a third solution. The test-maker makes this easy for us and tells us there are equal volumes of the two solutions. What does that mean? The final molar concentration of Na+ is going to be the average of the two solutions being mixed:

(0.010 M + 0.020 M) / 2 = 0.015 M Na+(aq). You might be looking at the question stem and this calculated answer and wondering about units. To find the molar concentration, we divide the number of moles of a substance by the volume of the solution. We are already given molar concentrations, so we simply find the relative amount contributed by both. We have the proper units.

  1. 0.010 mole/L This represents the molarity of the two initial solutions being mixed and the molar concentration of Na+ in NaHCO3. This answer is not giving us our desired value.
  2. 0.015 mole/L. This answer choice matches our calculation exactly. We had identical volumes of both solutions, so we found the molar concentration by finding the average molar concentration of Na+ in both.
  3. 0.020 mole/L. This answer choice represents the molar concentration of Na+ in the Na2CO3 solution which is not our desired value.
  4. 0.030 mole/L. We might incorrectly get this answer choice by adding the molar concentrations of Na+ rather than taking the average of the two. We can stick with our correct answer, answer choice B.

71) This question is asking which of the two Compounds (1 or 2) can form hydrogen bonds water. This boils down to a difference in electronegativity between elements determining the type of bond that will form between the elements. Hydrogen bonds are formed when hydrogen bonds with a very electronegative atom. The hydrogen gains a partial positive charge, and the atom gains a partial negative charge. Nitrogen is an electronegative element relative to hydrogen, but carbon is not. Like I mentioned, hydrogen bonding happens between a hydrogen atom and an electronegative atom. For the sake of the MCAT especially, we think of fluorine, oxygen, and nitrogen as the electronegative atoms in hydrogen bonding (FON atoms). We’re expecting the compound with nitrogen (Compound 2) is more soluble in water. 

We answered an objective question here, so we can actually compare all of our answers at once here. Answer choice B is the only answer choice that contains Compound 2, and does not contain Compound 1.

72) To answer this question, we need to know the structure of the molecules in the four answer choices. Our answers have a central carbon surrounded by one to four bromines. 

We’re more focused on the carbon here. Look at the figure above and you see it’s tetrahedral and bond angles are all roughly 109o when all substituents are identical. Electrons are all shared equally between carbon and the 4 bromines, meaning the vector sum of the bond polarities is zero. Where do we see this happening? In answer choice A. Carbon is sp3 hybridized in all of our four answer choices, but there’s only one answer choice in which the vector sum of the bond polarities is zero. 4 of the same substituents, all contributing the same bond polarity. We can stick with our best answer, answer choice A.

73) To answer this question, we have to know the difference between chlorine’s normal electron configuration, and its electron configuration in NaCl. An ionic bond is formed, meaning the sodium atoms are cations while chlorines are anions. Chlorine has an extra electron, giving it a full octet in this case with 18 electrons, or the same electron configuration as argon.

A neutral chlorine atom will have 7 valence electrons and a total of 17 electrons, with two being found in the first shell, eight being found in the second shell, and seven remaining electrons that are found in the outermost shell as valence electrons. Chlorine is very reactive and instead is typically found as a diatomic molecule or part of a compound (like in NaCl). With 18 electrons, chlorine has a full third shell with electrons found in the s-subshell and p-subshell.

  1. 1s22s22p63s23p4. This answer choice incorrectly includes 16 electrons around chlorine. We know chlorine will have an extra electron compared to the 17 we see in the chlorine atom on the periodic table.
  2. 1s22s22p63s23p5. This is what we expect if chlorine is not part of a compound. As I mentioned, we’re dealing with 18 electrons in this case, not 17.
  3. 1s22s22p63s23p6. This answer choice matches our breakdown exactly. Chlorine has 18 electrons like we mentioned, with two being found in the first shell, eight being found in the second shell, and the eight remaining electrons that are found in the outermost shell as valence electrons. The outermost electrons are found in the 3s and 3p subshells. This is our best answer choice so far.
  4. 1s22s22p63s23p44s2. This answer choice correctly lists 18 electrons, but incorrectly fills the subshells. We don’t expect the final 2 electrons to fill the 4s subshell, but rather they would fill up the 3p subshell so there are 6 total electrons. We can eliminate this answer choice as well. We’re left with answer choice C as our best answer. 

 

Chemistry Question Pack: Passage 13

74) During the passage we were shown different fertilizers and corresponding equations. This question is asking specifically about the fertilizer shown in Equation 2. We want to know the effect of moist soil conditions (additional water in the reactants). We’re going to need the equation from the passage primarily. And we need to know the effect of adding more reactant to a chemical equation. Glancing at our answer choices, we want to answer in terms of ionization, and the effect on hydroxide ion.

We have Equation 2, and we want to know the effect of additional moisture, or more water reactant. A quick look at our equation shows that there is ionization taking place. There’s the release of hydroxide ions. 

What happens when we add reactants to a chemical equation? This is Le Chatelier’s principle. The reaction equilibrium should shift toward the products. We’re expecting more hydroxide ion to be released. We know hydroxide is charged (hence the ion). What do we expect in terms of our degree of ionization? We’re going from neutral water and ammonium monohydrogen phosphate reactants, to ionized products like the hydroxide ion. That means our degree of ionization increases

  1. The degree of ionization will be greater, releasing more OH. This is consistent with both parts of our breakdown of the question. We said as more reactants are added, Le Chatelier’s principle is in effect. More hydroxide ion is released. Going from neutral reactants to ionized products increases the degree of ionization
  2. The degree of ionization will be greater, consuming more OH. We do expect the degree of ionization to be greater, but adding more water reactant should not mean consuming more of the hydroxide ion product. That’s impossible, so we can call eliminate this answer choice, it’s unreasonable
  3. The degree of ionization will be reduced, releasing more OH. The degree of ionization increases as we have more ionized reactants. We do expect more hydroxide ion released, but this answer is only partially correct. We can also eliminate answer choice C, it contradicts our breakdown of the question. Answer choice A remains the best option.
  4. The degree of ionization will be reduced, consuming more OH. This is the opposite of our breakdown. If the degree of ionization were reduced, more hydroxide ion would be consumed. That correlation is technically true, but not to answer our question. We stick with our correct answer: answer choice A.

75) We were told common fertilizers contain ionic salts of nitrogen. So why can’t plants use the nitrogen in the atmosphere instead as a sort of fertilizer? To answer this question, we have to know properties of nitrogen to explain why plants can’t simply use nitrogen from the atmosphere as fertilizer. 

Nitrogen from the atmosphere is going to be diatomic nitrogen. The two nitrogen atoms are held together by a triple bond, and it’s going to be found in a gaseous state. If need be, we can have our periodic table and periodic trends ready.

  1. N2 is present in very low concentrations in the atmosphere. We know nitrogen is the most abundant element in the air. It makes up roughly 4/5 of earth’s atmosphere, so it’s plentiful in high concentrations. This answer is unreasonable, but we’ll still compare to our additional answer choices
  2. N2 is too polar. Diatomic nitrogen is made of two nitrogen atoms. Two of the same elements in a molecule can’t be a polar molecule. This is unreasonable, but it’s not much better than answer choice A. That means we’ll still have to use choices A and B to compare and see if there are better options.
  3. N2 is very unreactive because it is a noble gas. This would be an interesting answer choice, but there’s a big problem with the reasoning. Nitrogen gas isn’t a noble gas. It’s one of the seven common diatomic molecules, but not a noble gas. Make sure to not get this confused.
  4. N2 is very unreactive because of the great strength of the N≡N triple bond. As we mentioned in the breakdown of the question, nitrogen atoms are held together by a triple bond. These are very hard to break. Triple bonds are extremely strong. Plants can only utilize nitrogen in reduced form, and not directly from the air. Answer choice D looks like the most likely answer choice here.

76) We’re given a component of equation 1, one of the products. We have to explain whether it is a conjugate acid or base of different compounds. We only need the definition of conjugate acids and bases to answer. Even though the question references Equation 1, that doesn’t mean we need to go back. Oftentimes the testmaker will reference something in the passage, but you can use external knowledge to answer. That’s what we’re doing here. We save so much time here just by knowing the definition of conjugate acids and bases.

Step 3: A conjugate base is the compound that remains after an acid has donated a proton during a chemical reaction. A conjugate base is formed when an acid loses a proton. A conjugate acid is formed when a base gains a proton. Hydrogen phosphate is the conjugate base of dihydrogen phosphate ion. The dihydrogen phosphate ion loses a proton to form the hydrogen phosphate. Hydrogen phosphate is the conjugate acid of phosphate ion. Phosphate ion gains a proton to form hydrogen phosphate.

  1. acid of NH4+. This answer choice contains nitrogen. There’s no nitrogen in hydrogen phosphate, so this answer choice looks very unreasonable. It does not match our explanation of our conjugate acid, nor does it match our conjugate base. 
  2. base of NH4+. Reasoning here is going to be the same as answer choice A. This answer contains nitrogen so it’s unreasonable. 
  3. acid of H2PO4. Hydrogen phosphate is the conjugate acid of phosphate ion. It’s the conjugate base of dihydrogen phosphate. 
  4. base of H2PO4. This is the perfect answer choice here. Conjugate of dihydrogen phosphate matches our breakdown. We can stick with this answer choice as our correct answer. 

77) To answer this question, we’re referencing Equation 2 from the passage and we’re asked about the equilibrium constant. We need to explain two things here: why the concentration of H2O is omitted, and how it relates to the salt in Equation 2. We’ll need the passage to reference equation 2, and we’re going to need to know about acid-base equilibria.

We have equation 3 given above. First and foremost, our water is a liquid. For reactions involving a solid or a liquid. The amounts of the solid or liquid will change during a reaction, their concentrations and densities won’t change. Those values remain virtually constant. Because they’re constant, the values are not included in the equilibrium constant expression.

Next order of business, the fertilizer is our salt here. The ammonium hydrogen phosphate. When we read through the passage, we should’ve noticed the formation of hydroxide ions in the product. Does that make the salt acidic or basic? It’s basic.

  1. only weakly basic, and [H2O] is nearly constant. This matches our breakdown. Again, we said the concentration of H2O stays virtually the same, so it doesn’t need to be included in the equilibrium constant expression.
  2. strongly basic, and [H2O] is nearly zero. We said the salt is basic, but the concentration of H2O is not nearly zero. It’s constant. The answer is half right, but doesn’t match our breakdown. Answer choice A remains the superior option at this point.
  3. only weakly acidic, and [H2O] is nearly constant. The concentration of H2O being nearly constant matches our prediction, but the salt is not acidic, we said basic. This contradicts our breakdown and the equation given in the passage.
  4. strongly acidic, and [H2O] is nearly zero. We said the salt is slightly basic, and the concentration of H2O is constant, and not necessarily zero. We can eliminate answer choice D as well. We’re left with our correct answer choice, answer choice A.

 

Chemistry Question Pack: Passage 14

78) Experiment 2 involved electrolysis cells, and we want to know what is taking place at the cathode. To answer this question, we need to know how electrolytic cells work. Electrical energy is used to drive a nonspontaneous redox reaction. The oxidation half reaction occurs at the anode, and the reduction half reaction takes place at the cathode. Electrons flow from the positive anode to negative cathode. 

  1. oxidation by a loss of electrons. This answer choice already goes against our breakdown because it starts with “oxidation.” We said the reduction half reaction takes place at the cathode. Oxidation takes place at the anode.
  2. oxidation by a gain of electrons. This is going to be similar reasoning to answer choice A. We want an answer choice that mentions reduction, not oxidation.
  3. reduction by a loss of electrons. Electrons flow from the positive anode to the negative cathode. That means there is a gain of electrons at the cathode. Answer choice C is better than answer choices A and B for simply starting with “reduction.” The reasoning here does not, however, match what we’d expect.
  4. reduction by a gain of electrons. Once again, an answer that starts with “reduction” which is what we want. However, it also says there is a gain of electrons. Is that what we expect at the cathode? Electrons flow from the positive anode to the negative cathode, so there is a gain of electrons. This answer choice sounds like a keeper! We can eliminate answer choice C. Answer choice D is our best answer. 

79) To answer this question, we can figure out the electron configuration without using the passage. We just need to know about the reaction of calcium and water. Then, we can find the electron configuration of the ion. Another thing we want to note is we’re dealing with ions, so there’s a strong chance we’ll be using the periodic table as well.

What do we know about the reaction? We know the resulting solution is basic, meaning hydroxide in the product. Let’s write out the equation:

Ca(s) + 2H2O(g) → Ca(OH)2(aq) + H2(g)

We have solid calcium reacting with liquid water. That yields calcium ion, hydroxide ion like we just mentioned, and hydrogen gas. We can balance the equation by adding a 2 as the coefficient in front of water, and in front of our hydroxide ions. The charge on our calcium ion is positive 2. Quickly reference the periodic table to see how many electrons calcium normally has. 

Elemental calcium is element number 20, meaning it has 20 protons and 20 total electrons typically. In this case, we’re dealing with calcium ion. The calcium loses 2 electrons to get a +2 charge and giving it a full octet, or the same electron configuration as argon, a noble gas. Argon’s electron configuration is 1s22s22p63s23p6. First energy level holds 2 electrons, 2nd energy level holds 8 electrons, and 3rd energy level will hold an additional 8 electrons.

  1. 1s22s22p63s23p64s2 Answer choice A has an extra 4s2. That is the electron configuration of elemental calcium, not the calcium ion. Let’s keep comparing to our other choices
  2. 1s22s22p63s23p64s1 Answer choice B has an extra 4s1. That’s the electron configuration for potassium, or a +1 calcium ion, not +2.
  3. 1s22s22p63s23p6 Answer choice C matches the electron configuration from our breakdown. We can stick with this answer as our best option. 
  4. 1s22s22p63s23p64s23d2 Answer choice D is the electron configuration if the charge on calcium were -2 instead of +2. Be careful as always with these details because they can change the entire meaning of an answer choice! We’re left with our correct answer choice, answer choice C.

80) Experiment 1 gave us brief observations about the reaction between the metals and water. This question is asking to pick a trend to explain the reactivities of the different metals. We’re going to need to use Table 1 from the passage to find the observations made about the different metal ions, then we’re going to use the periodic table and periodic trends to help answer this question.

I mentioned we’re going to need the passage to answer our question, so reference Table 1 above. Magnesium is an outlier. There’s no obvious reaction and the resulting solution is neutral. This could mean no magnesium ion even formed. Next thing we want to note, is the sodium and potassium reacted the strongest. Those ions formed fairly readily it looks like. Let’s locate all five of our elements on the periodic table next to see if we can see any trends. 

Our five elements were magnesium, calcium, lithium, sodium, and potassium. The magnesium was an outlier because there was no obvious reaction, meaning there likely wasn’t an ion formed. The sodium and potassium reacted the strongest, meaning they ionized more readily. Lithium, sodium and potassium are all group 1 elements. Calcium and magnesium are group 2 elements. So, it seems like the further right and up the periodic table we go, the less likely we are to form an ion. That corresponds to what are referred to as E trends-electronegativity, electron affinity, and ionization energy.

  1. Electronegativity. Electronegativity has to do with attracting electrons in a bond. We did say the general trend is consistent with electronegativity, but that is not consistent with our justification.
  2. Ionization potential. Ionization energy is the energy that is necessary to remove an electron from an atom. The atoms that are most strongly attracted to an atom’s nucleus are most difficult to detach. Potassium is in group 1 and the lowest element in group 1, meaning it has the smallest ionization potential. It makes sense why it reacted the most violently. Magnesium would have the greatest ionization potential, which explains why no reaction took place. Answer choice B is consistent with the general trend, and our reasoning is consistent with the experimental results.
  3. Electron affinity. Electron affinity is the ability of an atom to accept an electron. In this case, our metals are ionizing by losing electrons, not accepting more electrons. This contradicts what is happening in the experiment.
  4. Polarizability. Polarizability decreases across a period, and up the columns of the periodic table. This contradicts our general trend. Answer choice B is the best answer choice listed.

81) In this case, we’re redoing Experiment 1 with calcium and trying to identify the gas that evolved from the experiment and its volume. Note, we’re given pressure and temperature, and we shouldn’t need the passage to answer this question. 

We’re going to use the ideal gas law to solve for the volume of the gas, and the volume is given in mL. That means we’ll have to solve for moles and we’ll have all the information necessary to find volume. 

Next thing we’re going to do is write out the reaction taking place. We actually used this reaction two questions ago:

Ca(s) + 2H2O(g) → Ca(OH)2(aq) + H2(g)

We have solid calcium reacting with liquid water. That yields calcium ion, hydroxide ion, and hydrogen gas. Make sure to balance the equation. We needed to identify the gas produced to answer this question, and we can see it’s hydrogen gas. Let’s move to the rest of our question:

We have 0.4 grams calcium, let’s solve for moles of calcium so we can subsequently solve for moles of hydrogen gas. Molar mass of calcium is 40.1 g/mol, so we can use dimensional analysis and solve for moles. We have 0.01 mol calcium. Looking at our balanced equation above, for every mole of solid calcium, we have 1 mole of hydrogen gas. That means we also have 0.01 moles hydrogen gas.

Now is when our ideal gas law comes in. PV=nRT.

We can plug in our values and solve for our unknown volume. 

Isolating volume, we get 0.25 mL. We said our answers are in mL, so we can use dimensional analysis to get 250 mL. I’ve handwritten everything here to allow you to follow more closely:

  1. H2, 250 mL. This answer choice matches our breakdown exactly. We’ll still go through our other answer choices.
  2. H2, 500 mL. Answer choice B also says hydrogen gas, but this volume is not what we solved for.
  3. O2, 250 mL. Right away we don’t like this answer choice because it mentions oxygen gas instead of hydrogen gas. Even though the volume matches our calculation, we still like answer choice A better because of the hydrogen gas.
  4. O2, 500 mL. Similar reasoning to answer choice C, but this also has the incorrect volume of gas as well.

 

Chemistry Question Pack: Passage 15

82) To answer this question, we’re going to be comparing formulas for compounds 1 and 2.

Compound 1 is B2P2H2N2R4. Compound 2 is B3P3H3N3R6. The molecular formula here is not the same for the 2 compounds, but we can check the empirical formula. The empirical formula is the smallest ratio of numbers to represent the proportions of the atoms. 

Empirical formula for Compound 1 is B1P1H1N1R2. Empirical formula for Compound 2 is B1P1H1N1R2. So, in this case, even though the molecular formula of Compound 2 represented a larger compound, its empirical formula contained a smaller ratio of whole numbers, which was the same as that of Compound 1.

Let’s jump into our answers and remember that both compounds have the same empirical formula, but the molecular formula for Compound 2 represents a larger compound.

  1. Valence-bond. This is an interesting answer. The verbiage is not something AAMC uses on their content outline, even though it sounds like it could be a type of formula. This answer choice is possibly made up, and therefore unreasonable. We don’t deal with valence bond formulas on the MCAT. Let’s compare with our additional answers
  2. Empirical. We said in our breakdown that both compounds had identical empirical formulas. That means we like this answer choice better than answer choice A. 
  3. Molecular. We mentioned the molecular formula of Compound 2 represented a larger compound. Answer choice B remains the best option so far.
  4. Structural. This is essentially the same as answer choice C. Compound 2 is a larger compound with an additional number of atoms, so the structural formula can’t be the same. We’re left with our correct answer choice: answer choice B. Compound 1 and 2 had the same empirical formula

83) We’re given specific quantitative values and conditions here. We have to solve for the maximum volume of PH3 gas (one of the products in Equation 1a) we can get given the conditions. We’ll need to pull up Equation 1a, and we’ll need the molar ratios of the reactants and products. We can use these molar ratios to find the volume of phosphine gas. We’ll use external knowledge to manipulate our ratios, and to determine the volume of gas per mole. Answers are given in mL and liters.

We have Equation 1a above, and the conditions we were given right below. Our equation is already balanced. Equation 1a shows that there is a ratio of 2 moles R2NBCl2 to 5 moles of LiPH2 to produce 3 moles phosphine gas. 

Conveniently, the volumes given are in the same ratios as our molar ratios. That means we have 0.003 moles phosphine according to our molar ratio. 

Look at pressure and temperature. This reaction takes place at STP, meaning 1 mole of a gas will occupy a volume of 22.4L. Pay attention to every aspect of the question and what each number could potentially mean. 

Using the volume 22.4 liters/1 mole, multiply by the 0.003 moles phosphine to find the volume to be 0.0672L or we can convert using dimensional analysis to 67.2 mL. I’ve handwritten the math to solve this question below:

  1. 0.672 mL. The decimal point is off by two places here.
  2. 6.72 mL. We have the decimal point off by a single place again here
  3. 67.2 mL. This matches the calculation above. The value is the same and the units are the same. Remember to always double check units! That’s especially true with questions like this one where answer choices are given in multiple units.
  4. 67.2 L. We know this isn’t the correct answer because we kept track of our units. But notice how easy it could be to pick this answer choice. The numerical values are the same, and the only thing different between answers C and D are one letter. We stick with our correct answer choice, answer choice C.

84) We can draw out the Lewis structure, but we should check the periodic table of the elements first.

We need to know the number of valence electrons on phosphorus, and the phosphorus is bonded to three hydrogen atoms. So, we expect three of the valence electrons will be part of the bonds with hydrogens. That should leave two as a lone pair. That gives 8 electrons around phosphorus, and each hydrogen is accounted for.

The central phosphorous in this molecule has five valence electrons and only binds with three hydrogen atoms. The two remaining valence electrons will be unbonded, so that lone pair would be the only one in the molecule. Our final predicted answer is a central phosphorus with 5 valence electrons. The one lone pair, and three will be a part of bonds with hydrogens. Total of 8 electrons around phosphorus.

a.

Answer choice A looks exactly like our drawn-out Lewis structure above. Let’s still compare to our additional answer choices

b.

Answer choice B only has 6 total electrons around the central phosphorus. There should be an additional lone pair. This is incorrect.

c.

Answer choice C has too many electrons around the central phosphorus. The phosphorus should only have 8 electrons around it. This answer choice is incorrect as well.

d.

Answer choice D is similar to answer B. We only have 6 electrons around the central phosphorus. This is also incorrect. The central atom here should have 8 electrons around it. The correct answer is answer choice A. 

85) In this question, we’re given a line graph showing how Compound 1’s volume varies with temperature. We want to compare this with what the graph of Compound 2 would look like. We’re going to need to know the details about Compound 1 and Compound 2 from the passage. We’ll also use external knowledge to properly compare the two compounds. 

Let’s take a look at both of our compounds. Then we’ll analyze our variables; we want to know if our variables are directly related or inversely related.

We’re told both compounds are in the gaseous state and at constant pressure. Volume is the dependent variable on the line graph. When temperature increases, volume will also increase. 

Next, we write out our ideal gas law: 

Pv=nrt where p and r will be the same for both compounds. Pressure is constant. The number of moles in Compound 2 is fewer than Compound 1 because Compound 2 has a greater molecular weight. So, we’re expecting that at the same temperature, volume will be less in Compound 2. I’ve demonstrated below:

  1. Below the plot for Compound 1. We said we’re expecting that at the same temperature, the volume for Compound 2 will be less. That means it would be below the plot for compound 1. This is consistent with our breakdown.
  2. Above the plot for Compound 1. This would be accurate if the mass of Compound 2 were less than that of Compound 1. Then the volume for Compound 2 would be greater than that of Compound 1 at the same temperature. That’s not the case here, and this also contradicts our breakdown.
  3. Precisely on top of the plot for Compound 1. The difference in mass between the compounds makes this answer choice unreasonable. The plot for the two compounds is not identical.
  4. Intersecting the plot for Compound at its midpoint with an opposite slope. The general direction of both plots is going to be the same. The slopes will not be opposites. The plot for Compound 2 should be below the plot of Compound 1. We can eliminate answer choice D for going against our breakdown. That means our correct answer choice is A. The plot of the 1-gram sample of Compound 2 would appear below the plot for Compound 1.

 

Chemistry Question Pack: Passage 16

86) Silver sulfide was formed following the reaction of hydrogen sulfide and silver. We want to say what type of reaction forms the silver sulfide. To answer this question, we’re going to use the passage to identify the reaction in the question stem, then we’ll be using external knowledge to explain the reaction type. 

Let’s take a look at Equation 2 from the passage. The silver from the coins replaces the hydrogen in the hydrogen sulfide compound. This looks like a single replacement reaction. 

General formula for a single replacement, or substitution reaction is A + BY > AY + B. We can match up Equation 2 with this general formula. The element silver replaced the hydrogen in the compound in the reactants. The hydrogen is then found alone in the products.

  1. A combination reaction. A + B > AB is the formula for a combination reaction. This is not what’s going on in equation 2. We’re seeing a replacement, like the name suggests instead. This answer contradicts what we see in Equation 2.  
  2. A decomposition reaction. AB > A + B is the formula for a decomposition reaction. Again, not consistent with our breakdown, or what we see in Equation 2. 
  3. A single replacement reaction. General formula for a single replacement, or substitution reaction is A + BY > AY + B. We said we can match up Equation 2 with this general formula, just like we’ve done at the bottom of the screen with our breakdown.
  4. A double replacement reaction. The general formula for a double-displacement reaction is given by AX + BY > AY + BX. A double-displacement reaction can also be known as a metathesis reaction or double replacement reaction. Double replacement reactions occur when bonds are exchanged in two different species. We only have a single replacement in Equation 2. We can eliminate answer choice D for contradicting Equation 2 and our breakdown.

87) In other words, if 2 moles of sulfate ion are consumed, how much hydrogen sulfide is produced (in grams)? We’re going to need Equation 1 from the passage, then we’re going to use dimensional analysis to convert between moles and molar masses. Notice our answers are given in grams, and are rounded to whole numbers.

We have our balanced equation from the passage. Look at the coefficients on the sulfate ion and the hydrogen sulfide. Both are the same. So, for every mole of sulfate ion consumed, there is one mole of hydrogen sulfide produced. 

We’re told 2 moles of sulfate ion are consumed in the question stem, so we also have 2 moles of hydrogen sulfide produced. Now we can use dimensional analysis. Molar mass of hydrogen is 1 gram/mol. Molar mass of sulfur is 32 grams/mol. So two hydrogen atoms and one sulfur atom together in hydrogen sulfide have a molar mass of 34 grams/mole. Multiply this molar mass by the number of moles-we have 2 moles. This gives us a maximum number of grams produced of 68 grams. I’ve handwritten the solution below so it’s easier to follow:

  1. Answer choice A says 68 grams. This matches our calculated answer. We have a molar mass of 34 grams/mole and two total moles. That equals 68 maximum grams of hydrogen sulfide. 
  2. Answer choice B says 34 grams. This is the molar mass and would be the correct answer if we were looking for molar mass, or the mass in one mole of hydrogen sulfide. That’s not the case here, so we can eliminate answer choice B for being an incorrect value. 
  3. Answer choice C says 96 grams. This would be the correct answer if we were looking for the mass in three moles hydrogen sulfide. Again, we can eliminate this answer choice for being an incorrect value.
  4. Answer choice D says 192 grams. The only way we can get this answer choice is if we multiply the molar mass of the sulfate ion by 3 moles to get 192 grams. That is not the value we’re looking for. We can eliminate answer choice D as well. We’re left with our correct answer choice, answer choice A. 

88) What is the ionization constant? It’s essentially telling us the strength of an acid. Remember strong acids dissociate well, but weak acids do not. So, in essence, we’re asked if the hydrogen sulfide is a weak or strong acid. While the passage told us about hydrogen sulfide in particular, we’re going to have to know about the ionization constant coming into the exam. The passage explicitly told us hydrogen sulfide is a weak acid. It was a small detail that the author added to the end of a sentence, but luckily we read slowly and we included that bit in our mental summary. 

Some quick background here related to the ionization constant for an acid. The acid dissociation constant, Ka, is a quantitative measure of the strength of an acid. A strong acid will dissociate completely in solution. So a high value of Ka will correspond with a stronger acid that dissociates well. A low value of Ka will correspond with a weaker acid that dissociates only slightly. So, we expect our hydrogen sulfide to have a small ionization constant. 

  1. near zero. We do expect a small ionization constant, but we’re still dealing with a weak acid. The acid is still going to dissociate in water, so we don’t expect our constant to be near zero. Let’s compare with our other answers
  2. much less than 1. This is a standard value for any weak acid. Remember, we’re still dissociating, just not completely. This is a better answer choice than answer choice A which was too extreme.
  3. about 1. We expect weak acids to have ionization constants several times smaller than 1. This answer choice contradicts our breakdown. 
  4. much more than 1. This would be the correct answer for a strong acid. Strong acids can have ionization constants in the thousands, but in this case, we can eliminate this answer choice for contradicting our prediction about weak acids. We’re left with our correct answer choice, answer choice B.

89) This topic is always tricky for students because of the verbiage, so make sure to pay close attention as you’re working through it. We need to know the reducing agent in equation 2. We’re not asked the species that is reduced. We want to know the reducing agent from the equation in the passage. We’re going to need to analyze Equation 2 and we’re going to use external knowledge to actually identify the reducing agent.

We have Equation 2 from the passage and we need to identify the reducing agent. The reducing agent is the species being oxidized. The oxidizing agent is the species that is reduced. A species that’s oxidized loses electrons. A species that’s reduced, gains electrons. 

In Equation 2, silver reacts with hydrogen sulfide to produce silver(I) sulfide and hydrogen. 

Silver goes from 0 charge in the reactants, to +1 in the products. It donates electrons and is the reducing agent/the species being oxidized.

  1. S2– The sulfide ion maintains its -2 charge in both the reactant and products. It doesn’t donate electrons, it’s not oxidized. This contradicts our breakdown.
  2. H2S. Our silver reacted with hydrogen sulfide to produce our products, but the hydrogen sulfide was not the reducing agent, it didn’t lose electrons
  3. H+. The hydrogen did go from a +1 charge to zero in the products. This is the opposite of our breakdown and what we’re looking for in an answer
  4. Ag. This matches what we came up with in our breakdown. We said silver donates electrons and is the reducing agent, and the species being oxidized. This is our correct answer.

 

Chemistry Question Pack: Questions 90-94

90) In the question stem we’re shown the phase diagram for water and asked the phase shift between two points. This is a standalone question and we’re going to have to read the phase diagram given. We’re going to answer the question using the diagram itself and our external knowledge about the phase diagram of water.

Our phase diagram of water has two variables on its axis, pressure and temperature. So as these conditions vary, so do the phases of water. The phases vary from solid, to liquid, or gas based on the variation in pressure and temperature.

The phase all the way to the left is the solid phase. The rightmost phase is the gaseous phase. The topmost and middle phase is the liquid phase. This should inherently make sense. Ice is present at the lowest temperatures, steam is present at the highest temperatures, and liquid is present in between these two phases. Point A is liquid, and Point B is gas. 

Evaporation, or vaporization, is when molecules are heated and have enough energy to transition from the liquid to the gaseous phase. That means Point A to Point B is vaporization.

  1. melting. Melting occurs when a solid transitions into a liquid. This is inconsistent with what’s happening in the question stem.
  2. sublimation.  When a solid goes directly into the gas phase, this process is referred to as sublimation. Note that’s much easier at lower pressure. This is also inconsistent with what’s happening between Point A and Point B. 
  3. condensation. Condensation is the process of a gas being converted to a liquid (that would be Point B to Point A). This is the opposite of what we saw in our breakdown
  4. vaporization. We said vaporization describes going from liquid to a gas. That’s exactly what’s going on in our diagram. Answer choice D is going to be our best answer here.

91) To answer this question, we’re going to use external knowledge to explain the Bohr model. Niels Bohr proposed an update to Ernest Rutherford’s model. The Bohr model focuses more on electrons and their movement around the central nucleus. That’s consistent with what we’re being asked in the question stem. 

In Bohr’s model, electrons move at a constant speed and in fixed orbits. Bohr’s model explained that the energy of an electron depends on the size of an orbit. Smaller orbits correspond to lower energy. Moving away from the nucleus will mean an increase in energy level with each orbit of electrons. Electrons that move from energy levels further away from the nucleus to an energy level closer to the nucleus will emit energy.

  1. change orbits. This is partially true. We said when electrons move from energy levels further away from the nucleus to energy levels closer to the nucleus, energy is emitted. We can hold on to answer choice A as a good option for the time being.
  2. undergo acceleration. The electrons typically move at a constant speed and in a fixed orbit. But radiation is only emitted when electrons move to energy levels closer to the nucleus. This contradicts our prediction. We can eliminate this answer choice; answer choice A is still the best answer.
  3. move to orbits of lower energy. This is exactly what we said in our breakdown. Electrons move closer to the nucleus and radiation is emitted. Even though answer choice A was partially true, answer choice C is fully true, and is our best option at this point.
  4. move to orbits of larger radius. This sounds like electrons are moving away from the nucleus, not moving closer to the nucleus. That’s the opposite of our breakdown. We can stick with our correct answer, answer C. Radiation is emitted whenever electrons move to orbits of lower energy.

92) In this question, we’re given a hypothetical soluble metal hydroxide and given a corresponding molar solubility. We have to solve for the Ksp value for this hypothetical metal hydroxide. 

First thing we’re going to do is write out the general formula for the solubility product constant, our Ksp. Remember with Ksp, we raise the product to that coefficient power (and also multiply the concentration by that coefficient. A lot of students will forget this step or think we’re counting it twice. Make sure to not skip this).

Write out the Ksp: M(OH)2 = [M][OH-]2

We can plug in our molar solubility, S, and multiply the concentration by 2 for our hydroxide:

We have [s]*[2s]^2

When we multiply numbers raised to exponents, we can add the exponents. 

Multiplying this out, we get 4s^3. I’ve hand written this out below to make it easier to follow: 

  1. S2. This would be the case if there was no 2 on the hydroxide ion and we just multiplied the molar solubility with itself twice. 
  2. 2S2. That would be the case if we only had the hydroxide portion of the compound and there was no M
  3. 2S3. This is also incorrect. We would get this value if we did not also raise our 2 to the 2nd power.
  4. 4S3. This answer choice is correct. It’s consistent with the calculation we did in our breakdown. We stick with answer choice D as our best answer.

93) All of the 4 molecules in the question stem look like they have similar formulas, but only one is liquid at room temperature. We have to explain why water is an outlier, while the other three are gases. 

Water is unusual. It has high boiling and freezing points, it has a high specific heat, high surface tension, and high polarity. We would expect it to be gas, rather than liquid considering its molecular weight, but it’s not. We expect water to behave in a certain way given its molecular weight, but it doesn’t. Why is that? It’s because water molecules can form hydrogen bonds. Hydrogen bonding is a type of dipole-dipole interaction in which a hydrogen is covalently bound to a very electronegative element. The big three are fluorine, oxygen, and nitrogen. 

Hydrogen (2.1) is a relatively electronegative element. Oxygen (3.5) is more electronegative than sulfur (2.5), selenium (2.6) and Telluriium (2.1). Water is a polar covalent molecule. The rest are closer to nonpolar covalent.

  1. Hydrogen bonds form between H2O molecules. This reflects exactly what we said in our prediction. Water is unusual, and a lot of its unique characteristics come from water being able to form hydrogen bonds. The other three molecules, these other gases-cannot
  2. Oxygen lacks d orbitals. This is a true statement, but it doesn’t answer the question being asked. Another thing we should point out, selenium and tellurium have full D orbitals, but sulfur also lacks d orbitals.
  3. H2O has a lower molecular weight. Again, another true statement, but water having the lowest molecular weight would actually suggest to us that it should be a gas, not a liquid.
  4. H2O is more volatile. Substances with higher volatility are more likely to exist as gases. Water being liquid at room temperature would correspond to it being less volatile. Answer choice A remains our best answer.

94) To answer this question, we’re going to use external knowledge to compare partial pressures of two components of an equation. 

First thing we’re going to do before we jump into our partial pressures is to write out balanced equation to visualize better:

CH4(g) + 2O2 (g) > CO2 (g) + 2H2O

First thing we want to notice, is there are 2 moles water to 1 mole of carbon dioxide, meaning the effect on total pressure is a ratio of 2:1.

We know the total pressure is 1.2 torr, so let’s use our ratios. For every 3 total parts of our gases, 2 parts are water, and 1 part is carbon dioxide.

2/3 of our total pressure is 0.8 torr water, while 1/3 of the total pressure is 0.4 torr carbon dioxide.

  1. 0.4 torr. This is the partial pressure of carbon dioxide, not of the water vapor.
  2. 0.6 torr. This would be the correct answer if we had an equal number of moles of both products. But in this case, it’s also an incorrect value. 
  3. 0.8 torr. This matches what we came up with in our breakdown and our calculated value. This is the best answer choice so far. 
  4. 1.2 torr. This is the total pressure, not the partial pressure of our water. Answer choice C is the best answer.

 

Chemistry Question Pack: Passage 17

95) We’re asked specifically about solution A and given the cation portion. We want to know the corresponding anion component, or negatively charged component of the solution. We’re going to need to use the passage. The two tables in our passage provided good summaries of the experimental results. At the very least, we’re going to be the results of mixing silver with the 4 anions.

We have Table 1 here, and the information about Solution A. We were told Solution A contains silver ion. And the passage also says “Solution A was colorless.” What does that mean? It’s completely soluble in water.

Look at the table. We’re looking at the anion component of the solution. Anions are negatively charged ions that’ve gained one or more electrons. So, we’re looking at our Ag+ column. Going down the column, there’s only one anion that does not have a reaction with silver ion. That’s our fluorine ion.

  1. CrO42–. chromate ion forms a red precipitate according to our table.
  2. Cl. chlorine ion forms a white precipitate with silver according to our table.
  3. F. There’s no reaction with silver according to our table. That’s exactly what we want according to our breakdown of the question.
  4. S2–. sulfur ion forms a black precipitate with silver according to our table. We can eliminate this answer choice. We’re left with our correct answer choice: The anion component of solution A is C) fluorine ion if solution A contained silver ion.

96) We’re being asked why we had the results we did during experimentation. We got results in Table 1. We mixed anions and cations, but there were only certain situations in which precipitates formed. This question is asking why precipitates formed between anions and cations in the situations when they did.

We want to identify why precipitates form in general. There are two ways to do this, we can look at solubility tables, and we can determine the concentration of the ions to see if their reaction quotient is greater than the solubility product constant: the Ksp value. We get this value when a solution is saturated. So how can we read Ksp? A lower Ksp equals lower solubility of compounds in water. Once the Ksp is exceeded, precipitates would form.

  1. few aqueous solutions can contain more than one cation or anion. This is not true. Aqueous solutions can often contain more than one cation/anion.
  2. the anions precipitated as solid metals. Precipitates are insoluble, ionic solid products. They form when cations and anions combine in solution. Often the anions will be nonmetals as well, not solid metals.
  3. the solubilities of cations were decreased by the other cations. Cations and anions were mixed one at a time. The solubility of cations wasn’t decreased by the other cations.
  4. the solubility product of a compound was exceeded. This sound exactly like our breakdown. Once our Ksp, the solubility product constant, is exceeded, then precipitate forms. Answer choice D is our correct answer here. 

97) The four cations that were listed around Table 1 are shown as possible answer choices for this question. We want to know which of the four options had the fewest visible reactions with anions. Looking at Table 1, we either had colored precipitate formed, or we had no reaction, meaning we have a soluble compound. We’re going to need to use the passage to see which mixtures were soluble and which are insoluble. We also have to reference our external information to know the definition of soluble and insoluble.

Take a look at Table 1 once more. We have the results of mixing aqueous solutions of cations and anions here. We have to determine which cation allowed for the greatest number of soluble compounds. Cations are positively charged ions, while anions are negatively charged ions. Table 1 shows cations along the top part of the table. Soluble compounds are going to be the mixture results that correspond to no reactions. So, we can go through our cations one by one to compare. Silver ion forms one soluble compound. Calcium ion forms three soluble compounds (only one precipitate). Copper and iron ions both form two soluble compounds each. So which cation allowed for the greatest number of soluble products? Our calcium ion.

That means we’ve analyzed all 4 of our answer choices listed here. We compared the number of soluble compounds allowed by each cation. We’re left with our correct answer choice: Answer choice B, calcium ion.

98) In other words, barium cation is toxic, but mixing or dissolving it in water can possibly minimize this toxicity. We want to know which of the compounds listed as answer choices with corresponding Ksp values would be the safest for mammals to consume. 

Precipitates form following Ksp being exceeded. Lower Ksp means lower solubility of compounds in water. A lower Ksp means the most amount of solid precipitate and least amount of toxic, aqueous barium ion.

Do we have to know the specifics about each compound in our answer choices? Not necessarily. All we’re focused on here is finding the most favorable Ksp value. The lowest Ksp value is the safest, and also our correct answer.

  1. BaSO4, Ksp = 1.1 × 10–10. Note the Ksp value in this answer choice because we’re comparing with the other answers. 
  2. BaCO3, Ksp = 8.1 × 10–9 How does this compare with answer choice A? It’s a larger number, so answer choice A remains the best option. 
  3. BaSO3, Ksp = 8.0 × 10–7 This is larger than both answer choices A and B. Answer choice A is still the best. 
  4. BaF2, Ksp = 1.7 × 10–6 Again, this Ksp is larger than all the other answer choices. We stick with our correct answer: Answer choice A.

99) We have the same cation in both solutions, but iron cation is bonded to chlorine ion in one solution, fluorine ion in another. We have to distinguish between the two in the question stem using one of the answer choices listed. We’re going to need to use the passage to see which mixtures were soluble and which are insoluble. We should be able to answer the question using just Table 1. 

Looking at our table, what’s the first thing we notice? There is no reaction for both iron chloride and iron fluoride. These compounds are soluble. We need to find a way to differentiate between the two. So, we have to get creative with the rest of our chart. Our question stem said we can add a cation or anion to differentiate between the solutions. So which ions can help us differentiate? Let’s keep looking through our table from right to left. If we add copper to the solutions, would we have able to differentiate between iron chloride and iron fluoride? Well, both fluorine ion and chlorine ion don’t react with copper ion, so we’d have soluble solutions. Next, check the calcium ion. Calcium fluoride forms a precipitate, but calcium chloride is water soluble. So FeCl3 and FeF3 can be differentiated by adding Ca2+. 

Next, can silver ion differentiate the two? AgF is soluble, and AgCl forms a white precipitate. So FeCl3 and FeF3 can be differentiated by adding either calcium ion or silver ion. 

  1. CrO42– We have a negatively charged anion. We would expect the iron in our two solutions to bind to any anions. So there would be no way to differentiate between the two solutions.
  2. Ca2+ calcium ion matches our breakdown. Calcium fluoride forms a precipitate, but calcium chloride is water soluble.
  3. S2– Again, we’re not looking for a negatively charged anion. That would mean our iron binds the anion, and there would be no way to differentiate between the two solutions.
  4. Cu2+ Both fluorine ion and chlorine ion don’t react with copper ion, so we have a soluble solution. We stick with our answer choice B: calcium ion.

 

Chemistry Question Pack: Passage 18

100) From the passage we recall that KHP was used while the strong base (sodium hydroxide) was standardized. We want to know specifically the number of moles KHP in student A’s sample. We’re going to need to use information from the passage about student A’s KHP sample; we’ll have to manipulate the numbers we get from the passage to solve for the number of moles.

Looking back at some information from our passage above, we have Equation 2 that shows us how KHP is used in the standardization. We’re also told the molar mass of 204.2 here. 

We have a bit of Table 2 right below, we’re told the mass of KHP in student A’s sample is 0.5500 grams. If we have the grams of KHP, we can use the molar mass and solve for the number of moles. 

The molar mass of KHP can also be written as 204.2 grams/mole. When we’re using dimensional analysis, we want to cancel out units that aren’t our final units. In this case, we want our final answer in moles. So, we can divide the 0.5500 grams by molar mass. The grams unit cancels, and we can estimate our answer here. 0.5500 grams divided by 2 x 10^2 grams per mole, gives us 0.275 x 10^-2 moles. Or in proper scientific notation, approximately 2.75 x 10^-3 moles. We move the decimal point one place to the right, so we subtract 1 from the exponent. Review the handwritten solution right below as that can be easier to visualize:

Keep in mind that we rounded our answer, just in case we have to be more accurate with our calculation. Glancing at the four answers, we have a coefficient in each one, following by 10 raised to different exponents. Which exponent is consistent with our calculation? Answer choice C. We did some rounding, so our answer here isn’t exact, but AAMC doesn’t expect it to be. This question is done this way on purpose. The other answer choices are so far away from one another so we can be confident in answer choice C.

101) We recall students prepared the sodium hydroxide solution in Equation 1, and we have to explain whether any of the three changes in state functions occurred. Recall from the passage the temperature of the solution rose during the mixing process in Equation 1 and that the process is exothermic. We can pull up part of the passage, but we’ll need external knowledge to explain the changes in state functions-we’re dealing with change in Gibbs free energy.

ΔG=ΔH–TΔS where ΔG is the change in Gibbs free energy, ΔH is the change in enthalpy, T is temperature (kelvin), and ΔS is the change in entropy. We said heat is given off, so the process is exothermic. That means ΔH is negative.

Entropy also increases as the sodium hydroxide is dissolved and moves from solid to aqueous. The more positional probabilities that are available for an atom, the more entropy it has. Aqueous solutions have far more positional options than solids, so they have more entropy. That means ΔS is positive. 

If we plug in a negative value for change in enthalpy, and a positive value for change in entropy, ΔG, or the change in Gibbs free energy, is always negative because absolute temperature has to be positive also.

We have a negative ΔH, we have a positive ΔS, and we have a negative ΔG. 

That means options I and III are our correct options. Every answer choice ends in “only” so we’re looking for an answer choice that matches our breakdown and predicted answer exactly in this case. Our correct answer is going to be answer choice C, options I and III ONLY.

102) Recall the instructor prepared a solution of sodium hydroxide at the beginning of the experiment. We have to calculate the approximate molarity of the solution. We’re going to use the passage to find the specific information we need about sodium hydroxide. We’re also going to use dimensional analysis to solve for molarity. 

The excerpt from the passage above says 8 grams of NaOH dissolved in 2L of H2O.

Divide 8 grams of NaOH by the given molar mass to find number of moles of sodium hydroxide (40.0 grams/mol) = 0.2 mol NaOH.

Molarity is moles of solute per liter of solution, so it can be found through dividing 0.2 mol NaOH by the 2 liters of H2O to get a molarity of 0.1M. See my handwritten solution below which may be clearer:

  1. 0.1 M. This matches our calculated answer above so we’re liking this answer choice.
  2. 0.2 M. How might we have gotten this answer? If we never divided by 2 liters of H2O. Answer choice A remains the best answer. 
  3. 0.4 M How might we have gotten this answer? Maybe if we multiplied by 2 liters of H2O by accident. Luckily that would also mean our units would not check out, so we should never make this mistake!
  4. 4.0 M This is an incorrect value as well so we can stick with answer choice A as our best answer.

103) We’re going to use external knowledge about equivalence pH and equivalence points to answer this question. Coming in, we should know that a weak acid and a strong base, like this situation, would yield an equivalence point at a pH of above 7. But how do we know this?

The pH at the equivalence point will be the same as the pH of the sodium benzoate and water formed. The question stem mentions that benzoic acid is a weak acid. And we know NaOH is a strong base. How do we know that? 

Strong Acids Strong Bases
HCl- Hydrochloric acid LiOH- Lithium hydroxide
HBr- Hydrobromic acid NaOH- Sodium hydroxide
HI- Hydroiodic acid KOH- Potassium hydroxide
HNO3– Nitric acid RbOH- Rubidium hydroxide
HClO3– Chloric acid CsOH- Cesium hydroxide
HClO4– Perchloric acid Ca(OH)2– Calcium hydroxide
H2SO4– Sulfuric acid Sr(OH)2– Strontium hydroxide
  Ba(OH)2– Barium hydroxide

We should have our common strong bases memorized (above), but also group I hydroxides are typically strong bases. But, because benzoic acid is a weak acid, benzoic acid’s conjugate base can react with water to produce hydroxide ion. Hydroxide ions cause pH to increase, or be more basic. 

We can check our work here, and we should note this general trend. If we only know the equivalence point occurs at a pH above 7, we could conclude the acid is a weak acid, because sodium hydroxide is a strong base. If benzoic acid was a strong acid, the equivalence point would be at pH 7. 

  1. < 4. Answer choice A is strongly acidic, that is unreasonable given our breakdown.
  2. > 4 and < 7. Answer choice B is weakly acidic which is not what we’re looking for.
  3. = 7. Answer choice C is neutral. We said this would be the case if benzoic acid was a strong acid.
  4. > 7. Answer choice D says basic. This is our correct answer choice, and it matches our breakdown. We can stick with answer choice D as our best answer.

104) This is almost like a Standalone question. No need to go back to the passage to answer, except if you didn’t know chlorobenzoic acid’s formula. Chlorobenzoic acid’s formula is actually given in Table 1 as HC7H4ClO2

A conjugate base is the compound that remains after an acid has donated a proton during a chemical reaction. The conjugate base is C7H4ClO2

  1. OH Answer choice A is hydroxide ion. This is not close to the conjugate base we came up with.
  2. H2O Answer choice B is water. Once again, not consistent with the conjugate base we came up with. 
  3. C7H4ClO Answer choice C is closer. We’ve donated a proton, but we also lost an oxygen in this answer choice. That’s not necessary, but this is still a better answer than answer choices A and B. 
  4. C7H4ClO2 Answer choice D matches our breakdown exactly. Chlorobenzoic acid donated a proton, and this is the result. We stick with answer choice D as our best answer.

105) To answer this question, we need information from Table 1, Table 2, and information about equivalence weights. 

Equivalence weight of an acid is the quantity of the acid that yields one mole of hydrogen ions. For succinic acid, find the molar mass using Table 1. We have H2C4H4O4. We can use the periodic table and the corresponding subscripts to solve for molar mass.

2+48+4+64= 118 g/mol

Student E’s equivalent wt. for succinic acid is found to be 59.1. This means that 59.1 grams of succinic acid, when dissociated, will yield one mole of hydrogen ions. Think about what that means.

Because this is only one half of the molar mass of succinic acid, the titration needs another mole of NaOH per mole of succinic acid. There are two titratable hydrogen atoms per molecule.

  1. diprotic and requires one-half the number of moles of NaOH expected for a monoprotic acid. The first part of answer choice A matches our breakdown. We said two titratable hydrogens. However, the second part of answer choice A contradicts our breakdown. We said the titration needs another mole of NaOH per mole of succinic acid.
  2. diprotic and requires twice the number of moles of NaOH expected for a monoprotic acid. The first part of answer choice B, again matches our breakdown. Second part of answer choice B also matches our breakdown. The titration needs another mole of NaOH per mole of succinic acid, or twice the number of moles. Answer choice B is the superior answer choice at this point. 
  3. triprotic and requires one-third the number of moles of NaOH expected for a monoprotic acid. We said only two titratable hydrogen atoms per molecule, and twice the number of moles of sodium hydroxide. Answer choice C contradicts our breakdown. Answer choice B remains the best option.
  4. triprotic and requires three times the number of moles of NaOH expected for a monoprotic acid. I said only two titratable hydrogen atoms per molecule, and twice the number of moles of sodium hydroxide. Answer choice D contradicts our breakdown so we’ll stick with our correct answer choice: answer choice B.

 

Chemistry Question Pack: Passage 19

106) To answer this question, we’re locating and identifying the block of the periodic table where we’ll find aluminum, and we’re picking from the 4 blocks: S, P, D, and F. 

This is almost like a Standalone question so there’s no need to go back to the passage to answer. Our passage was focused on aluminum, but we can answer this question using just the periodic table:

We are dealing with the properties of a specific element, aluminum, so it’s expected we use our periodic table. Identify our element of interest, aluminum. It’s in group 13 and atomic number is 13 as well.

How do we classify blocks of the periodic table? Blocks are split and named based on the valence electrons or valence shells of elements. In general, the first two groups are S-block, the transition metals are D-block, the right side of the table is P-block, and the F-block elements are the lanthanides and actinides (shown at the bottom of the periodic table). Where does aluminum land? P-block

  1. S-block. Those are the first two groups, on the left side of the periodic table. This contradicts our breakdown of the question. 
  2. P-block. These are the elements on the right side of the table. Aluminum is in group 13, so it blocks to the p-block elements. This is the best option so far.
  3. D-block. Those are the transition metals near the middle of the table. This contradicts our correct answer. 
  4. F-block. F-block elements are the lanthanides and actinides. This is another incorrect answer. Aluminum is a P-block element, or answer choice B.

107) We want to know the oxidation number of a specific atom here, the aluminum in sodium aluminate. This is like a Standalone question just like our last question. We can answer this question without using the passage. The passage was about aluminum, but we don’t need to know specifics about aluminum from the passage to answer this. We’ll go one by one through the components of our compound, and use our rules for oxidation numbers. Keep in mind the sum of all oxidation numbers in our compound is going to be zero because our compound is uncharged.

Oxidation number of alkali metals is +1 so we can write that above our sodium. 

Oxidation number of an ion like hydroxide ion is equal to the charge on the ion, so we have -1. We have four hydroxides, so a total of -4. 

Now we can solve for the oxidation number of aluminum. Remember, we want the compound to be neutral. That means aluminum has an oxidation number of +3.

We have a numerical value, so we can quickly go through our answer choices. We’ll stick with our correct answer choice C, the oxidation number of aluminum in the compound is +3.

108) Just like the previous questions in this question set, we won’t need the passage to answer this question. We’re going to use VSEPR theory.

We need to break down the compound one step at a time. Aluminum has a charge of +3. Every fluorine has a charge of -1. So, we have a central aluminum, surrounded by 6 ligands. The molecular geometry and electron geometry is going to be octahedral because of the 6 electron-dense areas according to VSEPR theory. So main takeaway here? Know your content! This question can be answered in a matter of seconds because we know VSEPR theory.

  1. Octahedral. This matches our breakdown, but we still want to compare with the additional answer choices.
  2. Tetrahedral. This would be the case if our central aluminum had 4 electron dense areas. Answer choice A remains the best option.
  3. Trigonal bipyramidal. This would be the case if the central aluminum had 5 electron dense areas.
  4. Hexagonal. This is a nonsense answer that is trying to trick you to pick it because of the hex prefix. There is no hexagonal geometry, so this is unreasonable. Correct answer choice is A-the geometry is octahedral.

109) We’re going to use the information in the passage, and then use dimensional analysis to solve for the aluminum oxide required. Note our units are in kilograms, and the question stem says “approximately.”

I’ll write out the process here, then I’ll include handwritten notes that are easier to follow visually. 

2 moles of liquid aluminum oxide (Al2O3) forms 4 moles of pure, solid aluminum.

Next step, 100 kg of Al is how many moles?

Molar mass of aluminum is 27 g/mol, so we’re going to round to 25 g/mol. 

Next: dimensional analysis: 

100 kg is 100,000 g. Dividing 100,000 by 25 gives 4000 moles of solid aluminum. One thing we want to remember and note from our math lecture. This is an overestimation. We rounded our divisor down. So the actual number of moles should be slightly less than 4,000. 

Next thing we want to do, is multiply by the ratio of moles (2 moles aluminum oxide / 4 moles pure aluminum) = 2,000 moles aluminum oxide.

Multiply by the molar mass of aluminum oxide. Molar mass is 102, we can round that down to 100. Notice again, we rounded a number, but this time we rounded our multiplier down. This helps compensate for our last set of rounding where we said we had an overestimation.

2000 moles aluminum oxide x ~100 g/mol molar mass= 200,000 g aluminum oxide, or approximately 200 kg

  1. 500 kg. That’s much bigger than our predicted answer, that looks like an incorrect value
  2. 200 kg. Answer choice B actually matches our approximated value exactly. But note I said “approximated” so we have to keep looking at the other answer choices. If we have an answer that’s close, we may have to re-do some of our approximated calculations. Remember, our calculated answer isn’t exact.
  3. 80 kg. This is much smaller than our predicted answer. Answer choice B remains the best option.
  4. 50 kg. Same reasoning as answer choice C here. Answer choice B is going to be our correct answer.

Rest assured AAMC knows we’re not doing this math with calculators. They’re not going to make the calculations very easy on purpose, but typically we see questions like this. We’re expected to approximate, and the answers differ by large orders of magnitude. The answer choices they usually give us differ in how they’re obtained, rather than the details of the specific calculations. Another thing we can potentially do is look at the answer choices beforehand and get a sense of how close our approximations need to be. In most cases, the most important thing is to keep our units straight, and do our calculations properly. Rounding is expected of us.

110) We’re identifying the role of aluminum hydroxide in Equation 1. We’re going to use the equation in the passage, then we’re going to analyze aluminum hydroxide using what we know about acids and bases. Glancing at the answer choices, we have to decide between Lewis and Bronsted acids and bases.

Lewis acids and Lewis bases deal with the movement of electrons. Lewis bases donate electrons, and lewis acids are electron acceptors. Bronsted-Lowry acids donate protons and Bronsted-Lowry bases accept protons. Be aware that every Bronsted-Lowry acid and base will also be a Lewis acid or base. Therefore, it’s unlikely our answer will be Bronsted acid or base. We can only have one correct answer choice.

The excerpt from the passage says aluminum hydroxide reacts with sodium hydroxide to produce sodium aluminate. What happens to our aluminum here? Is it donating or accepting electrons? The aluminum in aluminum hydroxide is accepting that electron pair from the hydroxide. Lewis acids are electron acceptors, so our answer is going to be a Lewis acid.

We already compared the 4 choices briefly. We said our predicted answer is that the aluminum in aluminum hydroxide is accepting that electron pair from the hydroxide. That makes it a Lewis acid by definition. That’s answer choice A.

We also said every Bronsted-Lowry acid and base will also be a Lewis acid or base, but not necessarily the other way around. We can eliminate answer choices C and D because it’s not possible for a question to have two correct answers. That means we’re left with our correct answer choice: Lewis Acid, answer choice A.

111) We’re asked about the final step in producing pure aluminum through electrolysis. We’re going to have to use what we know about galvanic and electrolytic cells. 

First step, we need to know how electrolytic cells work. Electrical energy is used to drive nonspontaneous reactions. The oxidation half reaction occurs at the anode, and the reduction half reaction takes place at the cathode. Electrons flow from the positive anode to negative cathode.

Next, in a simple Galvanic cell, chemical energy is converted to electrical energy. The oxidation half reaction occurs at the anode, while the reduction half reaction takes place at the cathode. 

So, the definition of both cathode and anode is the same for both. The reduction takes place at the cathode and oxidation occurs at the anode.

Aluminum ion going from a +3 charge in aluminum oxide, to a 0 charge in solid aluminum is a decrease in the oxidation state of aluminum. This means aluminum is reduced.

  1. at the anode in both cells. This is the opposite of our breakdown. Aluminum is reduced, and reduction takes place at the cathode, not the anode. This contradicts our breakdown so let’s see if we can get a better answer.  
  2. at the cathode in both cells. This is consistent with our breakdown. Aluminum is reduced, and in both galvanic cells and electrolytic cells this takes place at the cathode. Answer choice B is superior to answer choice A. 
  3. at the anode in the galvanic cell and cathode in the electrolytic cell. This is only partially correct. We said reduction takes place at the cathode in both cells, not just the electrolytic cell. This contradicts our external knowledge and our breakdown. Answer choice B remains superior.
  4. At the cathode in the galvanic cell and anode in the electrolytic cell. Answer choice D is essentially the same as answer choice C. We can also eliminate this answer choice for contradicting our breakdown. We’re left with our correct answer, answer choice B: Aluminum is produced at the cathode in both galvanic and electrolytic cells.

 

Chemistry Question Pack: Passage 20

112) We want to know the effect on our experiment if a gas besides oxygen enters into our experimental apparatus. The experimental apparatus is used to determine ultrasmall concentrations of oxygen, so we want to know how adding another gas to be reduced by the apparatus will affect our concentration reading. The passage specifically mentions how the apparatus works in detail, so we should be able to identify the effect of adding another gas. 

We were told coulometry is a means of monitoring gases. Coulometry measures the amount of matter transformed during the reaction, but is not going to specifically find the amount of oxygen. That’s a small distinction, but an important one. 

So, what do we expect happens to the accuracy of our reading? Accuracy will go down-there’s now an increased amount of reduced gas being measured. The experimental setup is not going to discriminate between oxygen and a new gas.

  1. Increased, because the smaller the amount of oxygen in the stream, the more effective is microanalysis. This would be true if the gas in the stream wasn’t also reduced by a reaction analogous to that of oxygen. Our experiment can’t distinguish between the two gases. This contradicts the experimental setup in the passage
  2. Increased, because the larger the sample reduced, the less effect a small variation in measurement will have on results. There’s an increased sample of gas that has been reduced, but the gas isn’t just oxygen. The larger sample won’t make the measurement more accurate.
  3. Decreased, because the partial pressure of oxygen will be decreased. The first part of the answer choice is consistent with our breakdown. We do expect accuracy to go down. But the total pressure of a mixture of gases is the sum of the pressures of the individual gases. The oxygen partial pressure doesn’t necessarily decrease.
  4. Decreased, because the method cannot distinguish oxygen from the added gas. Again, first part of our answer choice is correct as we expect accuracy to go down. We said the reasoning is because coulometry will measure the amount of matter transformed during the reaction. The gas is added into the stream and is reduced in a reaction analogous to that of oxygen. The experiment is not going to specifically find the amount of oxygen. We’re left with our correct answer choice, answer choice D.

113) To answer this question, we’ll be using the diagram in the passage, and we’ll be explaining a specific aspect of the diagram: the cadmium electrode. We’ll be using external knowledge to explain the role of the specific electrode.

The passage says reduction occurs at the silver electrode and the electrochemical reaction is completed at the cadmium electrode. In a Galvanic cell, the oxidation half reaction occurs at the anode, while the reduction half reaction takes place at the cathode. 

In this case, the silver electrode is the cathode where the reduction half reaction takes place; the cadmium electrode is the anode where oxidation takes place. 

In Equation 3, oxidation is taking place, the cadmium is losing two electrons. 

  1. the anode, because oxygen gas is reduced there. First part of our answer choice looks good. But oxygen is reduced at the silver electrode according to the passage.
  2. the anode, because the silver electrode is where reduction occurs. Again, first part of the passage looks consistent with our breakdown. Second part is also consistent with what we know from the passage. Answer choice B is going to be superior to answer choice A. 
  3. the cathode, because the silver electrode is where reduction occurs. The cadmium electrode is the anode, not the cathode. The second part of the passage is consistent with our breakdown. We can still eliminate answer choice C for contradicting our breakdown.
  4. the cathode, because the silver electrode is where oxidation occurs. Again, cadmium electrode is the anode. And our silver electrode is where reduction happens. This answer choice contradicts our breakdown. We’ll stick with answer choice B as our best answer choice. 

114) PPM is parts per million. We’re finding a numerator, that when divided by 1,000,000, will give us 1%. So the “part” in “parts per million” to make it equal to 1%. We can solve this using the question stem alone; we just need to use dimensional analysis.

I’ve handwritten the process above so it’s easier to follow along. 

First, write out the ratios. 1 part per 100 is equal to 1%. An unknown part per 1,000,000 is also equal to 1%. We’re simply solving for X here. Isolating X gives us 10,000 parts per million. Dimensional analysis to the rescue! It’s not uncommon to get an extra two or three questions right per section just by keeping your units straight and using dimensional analysis.

We did a simple math conversion without any rounding. Answer choice C matches our calculated 10,000.

115) In other words, can methane gas replace oxygen in the experiment to be able to find its concentration? We’re going to need to know specifics about the experiment from the passage, but we’ll have to know about the properties of oxygen. Additionally, we’ll have to know specifically about methane to see if it’s a candidate for determination by coulometry described in the passage. 

The experiment requires a gas to be reduced at the anode, exactly how oxygen is reduced in the passage. We would have to reduce methane gas at the cathode. Remember, the oxidation half reaction occurs at the anode, while the reduction half reaction takes place at the cathode.

  1. Yes, because carbon, like oxygen, is a nonmetal. Carbon and oxygen are both nonmetals, but this doesn’t affect the procedure.
  2. Yes, because carbon, like oxygen, is in the 2nd period of the periodic table. The period of the periodic table will not affect the procedure. Group number affects the experiment and chemical properties. Group number determines how we can solve for the oxidation state of carbon.
  3. No, because hydrogen is already at its lowest oxidation state in methane. Hydrogen could be reduced to form hydrogen gas. That would be an oxidation state of 0.
  4. No, because carbon is already at its lowest oxidation state in methane. Carbon has an oxidation state of -4 in CH4. It can’t be reduced any further. That means it can’t undergo the same method as the oxygen in the passage. Answer choice D is going to be the best answer choice. 

 

Chemistry Question Pack: Questions 116-120

116) To answer this question, we can quickly go through what we know about catalysts. Catalysts are chemical compounds that increase the rate of a reaction by lowering the activation energy required to reach the transition state. Unlike reactants, a catalyst is not consumed as part of the reaction process. Catalysts do not change the equilibrium constant of the reaction. Biggest example of a catalyst we’ll see on the MCAT is undoubtedly enzymes. 

When an enzyme binds its substrate, it forms an enzyme-substrate complex. This complex lowers the activation energy of the reaction and promotes its rapid progression by providing certain ions or chemical groups that actually form covalent bonds with molecules as a necessary step of the reaction process. Enzymes also promote chemical reactions by bringing substrates together in an optimal orientation and lining up the atoms and bonds of one molecule with the atoms and bonds of the other molecule. This can contort the substrate molecules and facilitate bond-breaking. The active site of an enzyme also creates an ideal environment, such as a slightly acidic or non-polar environment, for the reaction to occur.

  1. They induce more collisions among reactant molecules. As was mentioned in the breakdown, catalysts (enzymes) promote reactions by promoting an optimal orientation, but this does not necessarily mean more collisions among reactant molecules. That typically happens with increased temperature. 
  2. They transfer kinetic energy to the reactant molecules. This is similar to answer choice A. All catalysts do not transfer kinetic energy to reactant molecules. Rather they work to promote chemical reactions by making the ideal environment for the reaction. 
  3. They increase the reaction rate but do not change the Keq of a reversible reaction. This answer choice is consistent with our breakdown. I mentioned catalysts do not change the equilibrium constant of the reaction. Additionally, catalysts/enzymes work to lower activation energy and increase reaction rate. This is going to be the best answer choice so far.
  4. They increase both the reaction rate and the Keq of a reversible reaction. I mentioned catalysts do not change the equilibrium constant of the reaction. This answer choice is only half correct, so answer choice C remains the best option. 

117) Let’s think about the definition of conjugate bases. A conjugate base is the compound that remains after an acid has donated a proton during a chemical reaction. In other words, a conjugate base is formed when an acid loses a proton. A conjugate acid is formed when a base gains a proton.

When we’re looking at the bisulfate ion, we can remove a proton from HSO4 which gets us SO42-. This is a straightforward and objective answer. If you accidentally find the conjugate acid instead, you might pick answer choice D. However, we solved for our correct conjugate base: SO42- which is answer choice C. 

118) This question just boils down to reading the phase diagram carefully. Our phase diagram of water has two variables on its axes, pressure and temperature. As these conditions vary, so do the phases of water. The phases vary from solid, to liquid, or vapor based on the variation in pressure and temperature.

The phase all the way to the left is the solid phase. The rightmost phase is the gaseous (vapor) phase. The topmost and middle phase is the liquid phase. This should inherently make sense. Ice is present at the lowest temperatures, steam is present at the highest temperatures, and liquid is present in between these two phases. Note, in this question specifically, we’re looking at negative 0.1oC, while the x-axis shows -3 oC and 0.01 oC. That means we’re looking at somewhere between the two labeled points along the x-axis. We can start at 1.0 torr along the y-axis which intersects our phase diagram at -3.0 oC. As we move slightly right (toward -0.1 oC) we’re in the vapor phase. We’re told pressure increases from 1.0 torr to 200 atm. 200 atm is roughly 152,000 Torr, so there’s quite a bit of pressure increase. As we trace up from the x-axis from -0.1 oC, we note we go from vapor to solid at around 2 to 3 torr. However, as we keep increasing in pressure, we will once again cross into another phase. This time we cross into the liquid phase. So, we expect to go from vapor to solid to liquid. 

  1. vapor will become a solid and then a liquid. This is exactly what we said in our breakdown of the question. Find -0.1 oC along the x-axis and trace your finger up. That’s exactly what this answer choice is laying out for us.
  2. vapor will become a liquid and then a solid. This is not the case. From -0.1 oC going to higher pressure, vapor goes to solid and then to liquid. Vapor does not go to liquid first.
  3. vapor will become and remain a solid. This might be the case at a much lower temperature, but at -0.1 oC this is not the case. Answer choice A remains superior.
  4. solid will become a liquid. This answer choice is incorrect. We know the sample starts as vapor first. Eventually it goes to a solid and a liquid, but it does not start as solid.

119) First thing we want to note is, what do we know about sodium hydroxide? It’s a strong base: 

Strong Acids Strong Bases
HCl- Hydrochloric acid LiOH- Lithium hydroxide
HBr- Hydrobromic acid NaOH- Sodium hydroxide
HI- Hydroiodic acid KOH- Potassium hydroxide
HNO3– Nitric acid RbOH- Rubidium hydroxide
HClO3– Chloric acid CsOH- Cesium hydroxide
HClO4– Perchloric acid Ca(OH)2– Calcium hydroxide
H2SO4– Sulfuric acid Sr(OH)2– Strontium hydroxide
  Ba(OH)2– Barium hydroxide

Strong bases will dissociate completely in an aqueous solution while yielding hydroxide ions.

pOH can be solved as the negative log of hydroxide ion concentration. The sodium hydroxide will dissociate completely, and hydroxide ion concentration ends up being the same as our molarity. Our molarity is 0.001 molar which can be rewritten using scientific notation. (If you don’t remember which way to move the decimal point, go back to the mnemonic LARS, which is Left Add, Right Subtract). We start with 0.001 x 10^0. We move the decimal point three places to the right to get a whole number, meaning we subtract three from our exponent to give us 1.0 x 10^-3 molar.

Let’s plug in our numbers. We know pOH is the negative log of our hydroxide ion concentration. Our hydroxide ion concentration, again, is the same as our molarity. So plug in 1x 10^-3 as our hydroxide ion concentration. So when we take our negative log, we find the pOH is 3. However, we’re looking for pH. To find pH, we can simply subtract pOH from 14:

14-3 = 11.

  1. .001. This is a very acidic pH which is not what we’re looking for. This answer choice incorrectly lists the molarity or hydroxide ion concentration as the answer.
  2. 3. This is giving you the pOH of the solution. You’re dealing with a strong base, so it doesn’t make sense to have a very acidic pH. 
  3. 7. This is a neutral pH.
  4. 11. This is the correct answer and corresponds to our calculated value. 

120) A gas is considered ideal if its particles are so far apart that they do not exert any attractive forces upon one another. In real life, there is no such thing as a truly ideal gas, but at high temperatures and low pressures (conditions in which individual particles will be moving very quickly and be very far apart from one another so that their interaction is almost zero), gases behave close to ideally. For ideal gases, at Standard Temperature (273K or 0 oC) and Pressure (1 atm), 1 mole of any gas will occupy a volume of 22.4 L. 

  1. The law PV = nRT2 is strictly obeyed. The ideal gas law is PV=nRT, not T2. This answer choice is close to the correct law, but it is incorrect.
  2. Intermolecular forces are infinitely large. Intermolecular forces include forces of attraction or repulsion, but I mentioned gases do not exert any attractive forces upon one another.
  3. Individual molecular volume and intermolecular forces are negligible. This answer choice is consistent with ideal gases. The ideal gas law works when intermolecular forces are negligible and when molecules do not occupy a significant volume. We like this answer choice better than options A and B. 
  4. One mole occupies a volume of 22.4 L at 25°C and 1 atm pressure. This answer choice is also factually incorrect. We talked about STP, but STP is 0 oC not 25 oC. That leaves answer choice C as our best answer. 


Billing Information
We had trouble validating your card. It's possible your card provider is preventing us from charging the card. Please contact your card provider or customer support.
{{ cardForm.errors.get('number') }}
{{ registerForm.errors.get('zip') }}
{{ registerForm.errors.get('coupon') }}
Tax: {{ taxAmount(selectedPlan) | currency spark.currencySymbol }}

Total Price Including Tax: {{ priceWithTax(selectedPlan) | currency spark.currencySymbol }} / {{ selectedPlan.interval | capitalize }}