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MCAT Content / AAMC MCAT Practice Exam 3 Cp Solutions

AAMC FL3 CP [Web]

Exam 3 Chemistry/Physics Section Passage 1

1) The best way to approach this question is to take what the passage says about the corresponding steps and try to visualize them. It mentions that Compound 3 undergoes an aldol condensation to cyclize, another word for forming a ring. It then goes on to say that the ring undergoes a dehydration reaction to form the two products 4a and 4b. Let’s follow the numbered carbons and determine which one corresponds to the carbon with the asterisk.

  1. C1 is directly attached to C4 and is the carbon to the right of the asterisk.
  2. C2 is indeed the carbon with the asterisk. C3 serves as the nucleophile in the aldol condensation reaction and the carbon with the asterisk is two carbons away (up and to the right) from the nucleophilic alpha carbon.
  3. C3 is the alpha carbon that serves as a nucleophile during the intramolecular cyclization. It should attach to the original five-carbon ring and is below and to the right of the carbon with the asterisk.
  4. C4 was part of the initial five-carbon ring and remains a part of the fused rings structure. The correct answer is C2 as it is the carbon represented by the asterisk.

2) The passage states that upon dehydration of the cyclized product, a thermodynamic mixture of products is formed. Recall that there are two major factors influencing which products of a reaction are preferentially formed: thermodynamics and kinetics. Thermodynamic control will favor energetically stable products even if they require a higher activation energy and more time to produce. Kinetic control will favor products that form more quickly; these products will often be less energetically stable but are formed faster due to lower activation energies.

According to Scheme A, 60% of the product is compound 4a and 40% is 4b.

  1. If both products were equally stable, thermodynamic control would produce roughly equal amounts of 4a and 4b. There is more 4a produced than 4b, so 4a should be more stable.
  2. For this to be true, the amount of 4a produced would need to be less than the amount of 4b produced. This is the opposite of what we see in Scheme A.
  3. This is true; compound 4a must be more stable than 4b because more 4a is produced (60% > 40%) and we are told that a thermodynamic mixture is produced indicating thermodynamic control. Again, thermodynamic control of a reaction will favor the energetically stable product.
  4. This is incorrect. As noted above, relative stability can be inferred because we are told in the passage that the products produced are a thermodynamic mixture. Answer choice C is the best answer.

3) To answer this question, we can look back at a modified version of the schematic we drew for the first question. We know that step 2 involves the aldol condensation followed by a dehydration reaction. If the dehydration reaction does not follow the aldol condensation, the cyclized product will be the major product.

Note that being able to follow and push electrons for common reactions is an important skill to have when test day comes along. Organic chemistry can be really difficult for many of us, but understanding the basic mechanisms and why a reaction occurs allows you to apply that knowledge more broadly. For instance: nucleophiles attack at electrophilic sites, and carbonyl carbons are electrophilic because the oxygen pulls on the carbon’s electrons. The alpha carbon of a carbonyl on the other hand, can serve as a nucleophile when it is made to carry a double bond. To form a ring from compound 2, have the alpha-carbon nucleophile on the first carbonyl attack the second electrophilic carbonyl (see the second image of the general aldol condensation).

  1. This structure looks like a neater version of what we drew above as the product of the aldol condensation reaction prior to dehydration. Looking good!
  2. If you were tempted by this answer choice, it might be worth counting how many bonds the carbon with the alcohol attached has. That carbon has five bonds, one too many! Don’t let the MCAT test writers trick you with a sneaky number of bonds.
  3. This answer choice has a few suspicious alterations. First, the carbonyl that should have been preserved during the nucleophilic attack has been reduced to an alcohol. Also, the second oxygen from the electrophilic carbonyl is missing! This answer choice is too different from the intermediate we should see.
  4. This is almost the final product, but like answer choice C, it has a reduced oxygen on the lower left corner where there should be a carbonyl. It also has a double bond where the alcohol should be located. This means that it has undergone a dehydration reaction, the opposite of what the question stem asks for. We’ll stick with the structure in answer A.

4) Before we draw out the new reaction, let’s work through this question verbally, which some of you may do on test day. Compound X has an additional methyl group attached to the “free end” of the double bond. Wherever this end is located on the original product, there should be an extra carbon/methyl group.

  1. This correctly depicts the “new” methyl group (shown in red above) at the end of the original double bond (shown in blue).
  2. The “extra” methyl group here would correspond to the other carbon on the double bond, the one noted by the green dot above.
  3. There is an extra carbon between the original ring and the methyl group on compound 1. Compound X is replacing compound 2, not compound 1. There should not be any changes to the original ring/compound 1.
  4. This answer choice is incorrect because it has the extra methyl group on the wrong side of the carbonyl, indicating that it would have been attached to the nucleophilic alpha carbon of compound 2. Compound X shows the methyl group on the beta carbon, not the nucleophilic alpha carbon. Answer choice A has the correct configuration.

 

Exam 3 Chemistry/Physics Section Passage 2

5) This question is meant to be relatively straightforward and highlights the importance of having chemistry and especially physics equations memorized and ready to go on test day. If you have not already started working on a master equations sheet, go ahead and add that to your “To Do” list.

Q=VC where Q is the charge stored, and both the voltage and capacitance are given at the end of the first paragraph:

Q= VC = 12V * 100mF = 12V * 100*10-3F = 1,200*10-3 C=1.2 C

U=½ CV2 where U is the energy stored, and both the voltage and capacitance are given at the end of the first paragraph

U=½ CV2 = ½ * 100mF * 12V2 = 50*10-3F *144V2=7,200*10-3 J = 7.2 J

  1. The charge in this answer choice is off by a single decimal place and the energy calculated in this answer is the result of not multiplying the ½ in the equation for energy.
  2. This is the correct pairing; see above for the math.
  3. This answer choice reverses the values for charge and energy, fails to halve the energy and is a decimal place off in each answer. Make sure to thoroughly perform all calculations and be very careful with exponents and moving decimals.
  4. This answer choice has the correct numerical values paired with the wrong measure. Answer choice B is the correct answer.

6) To answer this question, we’ll need to understand the Doppler effect and the effect of relative motion on perceived frequency. Most of us can conceptualize this by imagining an ambulance as the source of the sound/horn.

A. As you get closer to the horn/the ambulance, the frequency you hear is higher than the actual frequency, and once you’re moving away from the horn/the ambulance, the frequency you hear is lower than the true frequency. This answer choice corresponds to what we actually experience and is therefore the correct answer.

B. This is the opposite of what we see above and is incorrect. As we approach the stationary source of the sound, perceived frequency increases and is greater than the actual frequency. As we move away from the stationary source of the sound, perceived frequency drops below the true frequency.

C. The perceived frequency is equal to the actual frequency when there is no relative motion.

D. While f should be greater than f before passing the horn, it should not remain greater after passing the horn. Answer choice A is the most appropriate application of the Doppler effect.

7) P=Force*velocity=Work/time

In the absence of braking, the only force opposing motion is the one mentioned in the last sentence; there is a continuous decelerating force that is present any time the railcar is moving. This decelerating force is 1,000N.

P=Force*velocity= 1,000N * 40m/s = 40,000 W = 40kW

  1. This answer choice is too small and likely results from an arithmetic error.
  2. As shown in the calculation above, this is the correct answer. P=Force*velocity= 1,000N * 40m/s = 40,000 W = 40kW
  3. This answer is too large and is an order of magnitude off. You might have gotten this answer by also using the braking force of 14,000N instead of only using the continuous decelerating force. Recall that the question stem specifies the lack of braking.
  4. This answer choice is too large; you likely reached this answer by combining the force of the decelerating generator force, the maximum braking force and the continuous decelerating force. The question stem specifies that no braking occurs so the force should only be the continuous decelerating force. Answer B is the correct answer.

8) Recall from paragraph two that the generator braking system exerts a decelerating force that declines linearly with speed.

  1. This answer choice is super tempting because the passage mentioned that the force declines linearly with speed. However, it is the force that declines linearly, not the velocity. If force=mass*acceleration and the force decreases linearly, then in the absence of a changing mass it is the acceleration that is changing linearly. Now, if the acceleration is dropping in a linear fashion, the velocity must be dropping exponentially, not linearly. The slope of the velocity-time curve is acceleration, and a linear velocity curve would correspond to a constant deceleration, not one that keeps dropping.
  2. The railcar is slowing down when the brakes are engaged, not speeding up as shown with the increasing velocity in this curve.
  3. As in answer choice B, the railcar should be slowing down but this graph shows it speeding up.
  4. As noted in the explanation for answer choice A, it is the acceleration that is decreasing in a linear fashion, while the velocity decreases in an exponential fashion. The slope of the velocity-time graph is the acceleration and an exponential velocity-time slope corresponds to a linear acceleration-time slope; a linear velocity-time slope corresponds to a constant acceleration-time slope. This is the best answer choice and velocity-time curve.

 

Exam 3 Chemistry/Physics Section Discretes

9)

  1. At first glance, this is a good answer choice because boiling points are a measure of energy required to disrupt a system. However, boiling points are a measure of the strength of INTERmolecular forces that hold molecules together. The thermodynamic stability of a compound is determined by the INTRAmolecular forces. This answer choice addresses the incorrect molecular forces.
  2. UV-vis spectroscopy is all about ground state electron excitation via the absorption of light and is related to degree of conjugation and HOMO-LUMO gaps (highest occupied molecular orbital and lowest unoccupied molecular orbital respectively). While energy is involved in the excitation of electrons, this technique will not give the thermodynamic stability of a compound as a whole. For more on UV-Vis spectroscopy, head over to https://jackwestin.com/resources/mcat-content/molecular-structure-and-absorption-spectra/ultraviolet-region.
  3. Mass spectrometry is used for the identification of a compound via fragmentation and analysis of its mass-to-charge ratio. It is not appropriate for the understanding of thermodynamic stability. Here is a review of how a mass spectrometer works: https://jackwestin.com/resources/mcat-content/molecular-structure-and-absorption-spectra/ultraviolet-region
  4. Heats of combustion measures the heat produced when compounds are combusted, or burned in oxygen. If less heat is produced, the compound is more stable; when more heat is produced, the compound is less stable to begin with. This is a great way to determine the thermodynamic stability of a compound and is the best answer.

10) Because the solvent is water, whatever site we choose should not interact with water for proton exchange. If we were to label the guanine with the radioactive tritium at a site that exchanges its hydrogens for the protons in the solvent (interacts with water), the guanine would lose its radiolabel.

  1. Site I is a double bonded carbon. It is highly unlikely that a carbon at this site will interact with water to exchange its radiolabeled tritium for an unlabeled solvent proton, meaning the guanine will remain radiolabeled. This is promising and correct.
  2. Nitrogen could easily pick up a solvent proton and exchange it with its own radiolabeled tritium. This would make for a poor experiment so it is not the best site and is incorrect.
  3. Like the nitrogen at site II, the nitrogen at site III would readily lose its radiolabel. Nitrogen has a lone pair of electrons that can easily pick up a proton from the solvent.
  4. As with sites II and III, site IV is susceptible to losing its radiolabeled tritium because the nitrogen’s lone pair will readily pick up the protons from solution and could lose its radiolabel when it becomes deprotonated. Answer choice A is the most Appropriate answer.

11) The best way to answer this question is to determine the number of moles of gas particles and then use Avogadro’s number to convert the number of moles to the number of particles.

  1. This answer choice is the number of moles of gas, not the number or particles. This answer needs to be multiplied by Avogadro’s number, ~6 * 1023 molecules/mol.
  2. This answer choice results from the failure to convert degrees Celsius to Kelvin, the SI unit for temperature. Keep track of your units and convert to SI units as needed.
  3. This answer choice is incorrect because the temperature is in Celsius (wrong unit, SI unit is Kelvin) and does not account for the pressure, volume and part of Avogadro’s number.
  4. This is the best answer choice. See above for the full set of calculations.

12) When using Ka and Kb to determine whether a substance is an acid or a base, it’s important to remember that Ka*Kb=Kw so when Ka>Kb, the substance is more acidic than basic and vice versa. The table gives Ka directly, and the second column is equivalent to Kb. Note that we will need to use Ka of HCO3 but the Kw/Ka of H2CO3 because the conjugate base of H2CO3 is HCO3, but the conjugate base of HCO3 is CO32–.

Review the following content pages for a review of constants and conjugate acids and bases respectively: https://jackwestin.com/resources/mcat-content/acid-base-equilibria/equilibrium-constants-ka-and-kb-pka-pkb and https://jackwestin.com/resources/mcat-content/acid-base-equilibria/conjugate-acids-and-bases

  1. a base since Ka > Kb for this ion. – If Ka were greater than Kb, the substance would be an acid, not a base.
  2. a base since Kb > Ka for this ion. – Kb of the H2CO3 (2.3 × 10–8) is indeed greater than Ka for bicarbonate (5.6 × 10–11), meaning that HCO3 is a base.
  3. an acid since Ka > Kb for this ion. – Ka (5.6 × 10–11) is less than Kb (2.3 × 10–8), not greater. Bicarbonate is a base and not an acid.
  4. an acid since Kb > Ka for this ion. – If Kb were greater than Ka, the substance would be a base, not an acid.

 

Exam 3 Chemistry/Physics Section Passage 3

13) Before beginning calculations, we need to imagine the stoichiometry of the reaction. Concerning the zinc and iodine, the passage states that there is a 2:1 ratio of iodine to zinc. For every zinc atom reacted, two atoms of iodine reacted. Methanol is the solvent; it is used to dissolve the iodine and is evaporated at the end. It is not consumed in the reaction.

  1. 2.0g of iodine is 2.0 g / 127 g/mol = 1/64 mol or 0.015 mol of iodine. 2.0g of zinc is 2.0 g / 65 g/mol = 1/32 mol or 0.031 mol of zinc. According to the passage, there should be twice as much iodine as zinc, whereas what was used was twice as much zinc as iodine. There isn’t enough iodine to react with all of the zinc available making the iodine the limiting reagent.
  2. As noted in the explanation above, there is an excess of zinc relative to iodine making iodine the limiting reagent. If there wasn’t enough zinc to react with all of the iodine, then zinc would be the limiting reagent.
  3. As noted above, methanol is the solvent and is not involved in the reaction itself and is therefore not a limiting reagent.
  4. The passage mentions the reaction of iodiNe with zinc and not iodiDe with zinc. Iodide is the I anion and it is iodine, not iodide, that is the limiting reagent.

14) See below for a visual representation of the energy changes involved in a gene reaction.

  1. This would be the Gibbs free energy for the reverse reaction.
  2. This is the Gibbs free energy of the reaction, not the activation energy of the reaction. It represents the thermodynamic favorability of the reaction.
  3. This combines the activation energy and the Gibbs free energy whereas the question stem asks for just the activation energy.
  4. As shown above, the activation energy or energy of activation is the energy needed for the reaction to proceed towards the creation of the products and this energetic “hurdle” is represented by the energy of the activated complex minus the energy of the reactants. This is the best answer.

15) According to the periodic table, zinc has an atomic number of 30, is in the fourth period, is the last element in its respective d-block and has 12 valence electrons.

  1. The 3d orbital follows the 4s orbital according to subshell filling convention. The s orbital holds two electrons and the d orbital holds ten electrons. Zinc has 12 valence electrons which should completely fill both the 4s and 3d orbitals making this the correct answer.
  2. This could apply if zinc had 7 valence electrons instead of 12 valence electrons.
  3. This could apply if zinc had 11 valence electrons instead of 12 valence electrons.
  4. This could apply if zinc had 6 valence electrons instead of 12 valence electrons. Only answer choice A accounts for all 12 valence electrons.

16)

  1. Neither iodine nor zinc are acting as acids or bases to neutralize one another. Table 1 in the passage is also pointing to the importance of reduction potentials for the reaction, indicating that the movement of electrons is the key to answering this question.
  2. Two solids reactants were dissolved and heated to produce a pale yellow solution per the first paragraph. This is not in line with precipitate formation.
  3. This is correct. The iodide of zinc produced is the result of the movement of electrons. Both zinc and iodine change oxidation states as evidenced by the conversion of iodine to iodide, and the table included in the passage with the standard reduction potentials was a big hint that a redox reaction was taking place.
  4. Chelate formation would involve a ligand binding to a central atom with more than one coordinate covalent bond. This is not applicable, making C the best answer.

 

Exam 3 Chemistry/Physics Section Passage 4

17) The first paragraph states that cold plasmas have temperatures of about 35°C. Converting °C to Kelvin is done by adding 273 to the °C. 273+35= 308K

  1. This range is much higher than 308 K and would not correspond to the temperature of cold plasma.
  2. While closer to 308 K than answer choice A, this range is still too high and does not encompass 308 K.
  3. This temperature range encompasses 308 K making it the correct answer.
  4. This range is too low and excludes the 308 K temperature of cold plasma.

18) The period is the amount of time from the beginning of a wave or signal to the beginning of the next wave or signal.

  1. This is too long. The voltage is on for 500ns and off for 500ns, meaning the time from the beginning of the signal/when the voltage first turns on to the beginning of the next signal/next voltage pulse is 1,000ns, not 2,000ns.
  2. This is too long and would be the time for the voltage to be on, off, and then on again before turning off; this is 1.5x the period.
  3. This is correct. As shown in the figure above, the voltage is on for 500ns and then off for 500ns before the next pulse.
  4. This answer fails to account for the time when the voltage is off before the start of the next pulse which is still a part of the period. Only C correctly accounts for the time the voltage is on and then off for a single pulse.

19)

  1. This answer choice is too large and two orders of magnitude off. Make sure that you plugged in the correct values and carried your exponents accordingly. The “7” in this answer choice also suggests that you might have used 300 nm as the wavelength instead of 200 nm. In order to maximize energy as specified in the question stem, frequency would need to be maximized and wavelength, which is inversely proportional to the energy, would need to be minimized.
  2. While closer than answer A, this is still also too large. Double check your arithmetic.
  3. This too, is larger than the estimated and actual energy of the photons. See above for the correct calculation.
  4. This is the best answer, obtained with some rounding on test day; the corresponding calculation is shown above.

20)

  1. If the electric field is uniform, the field lines should be equidistant throughout, not more closely spaced on one end compared to the other.
  2. The electric field is uniform and the field lines should not intersect, they should be parallel to one another throughout the entire electric field.
  3. As noted with answer choice A, the electric field lines should not change spacing at the electrodes in a uniform electric field. They should be parallel to each other and equidistant.
  4. are equally spaced at both electrodes and between them. – This is correct. In a uniform electric field, the electric field lines should be equidistant throughout the entirety of the electric field.

 

Exam 3 Chemistry/Physics Section Passage 5

21) In the second paragraph, we are told that the D-glucose concentration used in the experiment is 72 g/L.

  1. 3.6 × 10–7 g – Both components of this answer choice are incorrect. Be very careful when converting units and make sure to perform the correct operation. 36 is ½ of 72, not twice as much and the calculation requires multiplication, not division of the concentration by the volume used.
  2. 1.4 × 10–4 g – This is the correct answer as shown above.
  3. 3.6 × 10–4 g – While the power is correct, 3.6 is greater than 1.4. This answer is almost 3x larger than the correct answer and like answer choice A, may have incorrectly involved the division of the concentration by the volume used.
  4. 1.4 × 10–1 g – Close, but not quite! This answer might result from the incorrect conversion of μL to L by using 10-3 instead of 10-6.

22) In the final paragraph, the author says that the enzyme used in the blood glucose meter catalyzes a two-electron oxidation. This means that the enzyme is catalyzing a redox reaction.

  1. A transferase transfers a functional group from its substrate onto another molecule. This is different from an oxidation reaction which is the transfer of electrons.
  2. An isomerase catalyzes structural rearrangements, not the transfer of electrons.
  3. Here we have it! An oxidoreductase catalyzes the transfer of electrons via oxidations and reductions. Glucose is oxidized in the reaction so the enzyme must be an oxidoreductase.
  4. A hydrolase cleaves bonds using water; there is no evidence of that occurring in the passage. We’ll stick with an oxidoreductase.

23) This question can be pretty tough at first glance, but it’s actually not too bad. Hear me out: the question stem wants us to find the mass of glucose in the sample and gives us the mole-mass conversion factor, also known as the molar mass. If we can find out the moles of glucose, we can easily find the mass. Now the passage stated that the reaction is a two-electron oxidation which means that whatever moles of electrons were used is twice the amount of glucose in the sample. By keeping track of the variables we’re given, the conversion factors and the units of whatever variable(s) the question is asking for, we can easily tackle MCAT questions.

A. This mass, 20μg, is too small; did you by chance multiply the denominators of the mole fractions to get 1/9 of 180μg? See below for the correct calculation.

B. This is the correct mass.

C. 90 μg – This answer choice is too large and is likely the result of using ½ μmol. Remember that per the passage, the electron-to-glucose ratio is 2:1 and the number of moles of electrons was ⅔ μmol, making the moles of glucose ⅓ μmol.

D. 270 μg – This answer choice assumes that more than one μmol of glucose is present, when in fact there is only about ⅓ of a μmol of glucose present.

24) Recall that the last paragraph says that reaction 2 is an oxidation reaction. Oxidation is the loss of electrons, so whatever answer choice we select should be an oxidizing agent, willing to accept those electrons and be reduced.

  1. FAD is an oxidizing agent and will be reduced to FADH2 upon reaction. Recall that an oxidizing agent is reduced, and a reducing agent is oxidized. FAD is also known as a cofactor oxidant; this is the correct answer.
  2. NADH is already in its reduced form and would serve as a reducing agent, not an oxidizing agent. A reducing agent will give up, not accept electrons.
  3. Water is a poor reducing and oxidizing agent and would need to be in the presence of a strong reducing agent to serve as an oxidizing agent; this is not the case and answer A is a better answer.
  4. Acetyl-CoA is oxidized when in the TCA cycle, not reduced. It would not make for a good oxidizing agent so answer A remains the best answer.

 

Exam 3 Chemistry/Physics Section Discretes

25) Ah, this type of question is a classic. Steroid hormone receptors bind steroids which are classically cholesterol-derived hormones that have four fused rings. Whichever structure has four fused rings will be the correct answer.

  1. As complex as this structure appears, it does not have four fused rings. Yes, there are four rings on the outer corners, however they are not all fused together.
  2. Good ol’ ATP presenting itself for inspection here. ATP lacks the four fused rings characteristic of steroid hormones so we’ll say goodbye until next time!
  3. What a structure! Don’t let those fatty acid tails fool you, this is not a steroid hormone.
  4. Here we have it–four fused rings that are characteristic of steroids. Looking good, and ready to bind to a steroid hormone receptor.

26)

  1. Adding more of a basic salt to increase phosphate anion concentration is not going to bring the pH down. We need to go from a pH of 7.6 to a pH of 7.2 meaning the buffer needs to become a bit more acidic, the opposite effect of adding a basic salt.
  2. NaOH is a strong base and will increase the pH, not decrease it.
  3. The monosodium species is the acid in the H2PO4 and HPO42- equilibrium so favoring monosodium phosphate instead of disodium phosphate (the conjugate base) will cause the pH to drop and approach 7.2 as desired in the question stem.
  4. Simply diluting a buffer will not cause a change in the pH. This idea used to give me trouble as well, but think of it this way: the pH of this buffer is determined by the relative concentrations of the acid and conjugate base. Distilled water will contribute equal parts protons and hydroxide ions (or hydronium and hydroxide, however you prefer to visualize it). This means that the addition of water will not change the relative amounts of acid or conjugate base. If their concentrations don’t change, then the pH will not change. Answer choice C is the only one that will lower the pH.

27) Peptide bonds are formed via a dehydration reaction. The product of a dehydration reaction is water. What is the atomic mass of water?

Atomic mass H2O = 2(mass of H) + mass of O = 2(1)+16=18 amu

  1. This answer is incorrect and fails to account for the mass of the second hydrogen in water.
  2. This is correct; the atomic mass of water is 18 amu. Atomic mass H2O = 2(mass of H) + mass of O = 2(1)+16=18 amu
  3. This is more aligned with the atomic mass of O2, not the atomic mass of H2O.
  4. This answer is also incorrect and much too large. B is the best answer

28)

  1. This is an important assumption for Michaelis-Menten kinetics. Measuring under steady state conditions ensures that the substrate concentration is not influenced by the formation of the enzyme-substrate complex and is instead true to the measurement.
  2. If the pH were to change, the enzyme may not be able to function. Remember that enzymes are proteins and susceptible to pH changes just the same. A constant pH that optimizes enzyme function is a key experimental condition.
  3. This is necessary and therefore not the correct answer. If the enzyme concentration were greater than that of the substrate, the effect of changing the substrate concentration would be lost to poor experimental design. Not all of the enzymes would be catalyzing reactions and the change in substrate concentration would not be noticed.
  4. Not only would this not be necessary, but this would be bad when obtaining Michaelis-Menten kinetics data. Once the reaction reaches equilibrium, changing the substrate concentration will not affect the rate of the reaction. Also, the researchers would not be able to obtain the initial rate/velocity because the reaction is past that point. The data obtained would essentially be useless for Michaelis-Menten kinetics. The question is a NOT question making this the best answer.

 

Exam 3 Chemistry/Physics Section Passage 6

29) This question requires careful examination of the second main paragraph: the first step has two parts. In the first part of the first step, NADPH and FAD react to make NADP+and FADH. The second part of the first step involves FADH reacting with O2 and H+ to make FADH–OOH. In the nonproductive reaction, the passage tells us that FADH–OOH decomposes into H2O2. This means that the products of the nonproductive reaction are NADP+ and H2O2.

  1. This answer choice has the correct products, NADP+ from the first step and H2O2 from the decomposition of FADH–OOH at the end when lysine is the substrate.
  2. While H2O2 is indeed one of the products, this answer fails to account for the other product of the nonproductive reaction, NADP+.
  3. This answer is incorrect because FADH2 is not one of the net products listed above.
  4. Like answer choice B, this answer does not take NADP+ into account and is therefore incorrect. The two key products of the nonproductive reaction. Answer choice A is the correct answer.

30) I don’t know about you, but I certainly don’t know the structure of hydroxamic acid off the top of my head! Luckily, there is a very brief description in the passage that notes the acetylation of the hydroxylamine nitrogen. The key description is the acetylation (carbonyl with a methyl group attached) of the nitrogen in the hydroxylamine (alcohol attached to an amine).

  1. The nitrogen on the left side is lacking an alcohol attachment so it is missing the hydroxyl- part of the hydroxylamine.
  2. There should not be an oxygen atom present between the carbonyl and the nitrogen; the nitrogen needs to have two distinct functional groups attached, the acetyl group and the alcohol.
  3. This is the correct answer. There is a nitrogen with an alcohol attached as well as a carbonyl with an R group attached.
  4. While there is indeed an alcohol attached to an amine, there is an extra carbon that does not belong between the nitrogen and the carbonyl.

31) We will need Fig. 1 to help us select the best answer, but first, let’s rephrase the question stem! The question is asking us to consider how quickly the enzyme catalyzes the second step if you change the substrate. So, at the same pH of 7.5, what is the rate constant (kFAD) for Compound 1 as the substrate relative to or compared to the rate constant for lysine as the substrate? To determine this, we will need to divide the kFAD for Compound 1 by the kFAD for lysine.

  1. The rate with Compound 1 as the substrate is twice that of when lysine is in the active site, not half. You might have placed the rate constant of lysine in the numerator and the rate constant of compound 1 in the denominator but the question stems asks for Compound 1 compared to lysine meaning that the kFAD Compound 1 needs to be in the numerator.
  2. The rate for Compound 1 is greater than the rate for lysine so the answer needs to be greater than 1, not less than 1.
  3. Close, but not quite. At a pH of 7.5, the kFAD for Compound 1 is 2 s-1, and the kFAD for lysine is 1s-1, so the relative rate for Compound 1 is 2s-1/1s-1=2 and not 1.5.
  4. This is the best answer as explained above. At a pH of 7.5, the kFAD for Compound 1 is 2s-1, and the kFAD for lysine is 1s-1, so the relative rate for Compound 1 is 2/1=2.

32) Paragraph 2 describes the reaction mechanism in detail. That said, it can be really difficult to follow each step and each component of the steps from the block of text. This is a great opportunity to practice writing out the reaction from the passage. An example of what you can do on test day is shown below.

  1. There is no indication that NADP+ reacts once generated according to the reaction mechanism in the passage. This means that it does not undergo a reduction.
  2. FAD is regenerated in the reaction and does not undergo any net changes in oxidation number.
  3. Like NADP+, H2O2 does not undergo any further reactions once produced, meaning it does not undergo a reduction.
  4. O2 reacts with FADH and H+ to form FADH-OOH. The FADH attacks the elemental oxygen with its lone pair electron as evidenced by the negative charge it carries. This means that the oxygen is on the receiving end of those electrons and is reduced (gains electrons). This is the only answer choice that contains a substance that was reduced.

33) If we’re not careful, we may end up digging through the passage looking for amino acids in all the wrong places and waste precious time on test day. This is where having at least skimmed the figures and their legends can pay off. The other amino acids are: compound 1 (shown to be L-ornithine in reaction 1), methylornithine and methyllysine.  Note that unlike the amino acids used for protein synthesis, you do not need to have the structure of ornithine memorized for test day; this is why they provide the structure in the passage excerpt. We know that lysine is a basic and positively charged amino acid, and in reaction 1 we see that ornithine has a positively charged amine in its side chain.

  1. Both lysine and ornithine are charged, so they would be hydrophilic and not hydrophobic. They are also positively charged and basic, not acidic. We anticipate that methylornithine and methyllysine will have the same general properties.
  2. This is correct. As noted above, lysine and ornithine are hydrophilic and basic as evidenced by their positive charges.
  3. This is incorrect, they are basic and will be protonated and positively charged at a pH of 7 as shown for ornithine in Reaction 1 and as we know for lysine from our own studies.
  4. This is also incorrect; the amino acids are positively charged and hydrophilic, not neutral and hydrophobic.

34) The highest pH studied was a pH of 10. pH = -log[H3O+] so the H3O+ concentration would be 10-10. However, the question stem asks for the concentration of hydroxide ions. We can use the relationship Kw= Ka*Kb =[H3O+][OH] = 10-14 to conclude that the concentration of hydroxide ions is 10-14/10-10 = 10-4.

  1. This is the concentration of H3O+ , not OH.
  2. This would be correct if the highest pH studied were 8 instead of 10.
  3. This is correct; if the concentration of H3O+ is 10-10 then the concentration of OH must be 10-4.
  4. This would be correct if the highest pH studied were 12 instead of 10.

 

Exam 3 Chemistry/Physics Section Passage 7

35) According to the first sentence of the third paragraph, HRP oxidizes phenols and aromatic amines. This means that the product of the reaction should have either an alcohol or amine substituent attached to the aromatic ring as the site of oxidation.

  1. HRP will oxidize aromatic amines and phenols. This answer choice has a phenol (the alcohol attached to the aromatic ring) and this is the site where the free radical formed and was oxidized; this is the correct answer.
  2. This is tempting because it shows a nitrogen with a radical, however an aromatic amine is an aromatic ring attached to an amine. Compound 1 from Fig. 1 is a better example of an amine substituent that would be oxidized by HRP.
  3. An amine or alcohol substituent will be oxidized, not a carbon on the ring itself.
  4. The alcohol, and not the sulfonate will be oxidized.

36) Like the previous question, we need to remember that HRP oxidizes phenols and aromatic amines. Whatever amino acid we select needs to have either a phenol or aromatic amine as its side chain.

  1. Lysine has an amine group at the end of its side chain, but it does not have an aromatic ring.
  2. Leucine has a branched carbon side chain, not an aromatic amine or a phenol.
  3. Tyrosine is the only amino acid with a phenol in its side chain, and none of the amino acids have an aromatic amine side chain. This is the best answer.
  4. Like lysine, glutamine has an amine in its side chain but no aromatic ring! Tyrosine is the only amino acid with a side chain that could be recognized as an HRP substrate.

37) In order to coordinate with the positively charged calcium ions, the atom will need to act as a Lewis base and carry at least a partially negative charge so as to attract and not repel the calcium ions.

  1. The hydrogens in the backbone and side chains do not carry a negative charge and would make a poor Lewis base.
  2. Like hydrogen, the carbons in the side chains and backbone are uncharged. The correct answer will be an atom with at least a partially negative charge.
  3. According to the passage, calcium ions coordinate with amino acids that only have nitrogens in their backbones: isoleucine, glycine, valine, aspartate, serine and threonine. While nitrogen can serve as a Lewis base and an electron pair donor, the nitrogens in the amino acid backbones are positive and not negative. These would repel the positively charged calcium ions.
  4. Like nitrogen, oxygen can serve as a Lewis base. Unlike the nitrogens in the above amino acids, the oxygens carry a negative charge which would allow them to coordinate with the positively calcium ions. The oxygens in the backbone carry a partially negative charge and a full negative charge in aspartate’s side chain.

38) The second paragraph begins by telling us that HRP contains a heme cofactor. We may not have the exact structure of heme memorized on test day, but we should have a general idea of what it looks like, drawn below. Note that the rings are five-membered nitrogen-containing rings and that heme can be categorized as a porphyrin consisting of 4 pyrroles.

  1. This answer choice corresponds to the pyrroles in heme, one of which is highlighted for ease of identification. This is the answer we’re looking for!
  2. While this is a five-member ring, it does not contain nitrogen. This can’t be the right answer.
  3. This is indeed a heterocycle with a nitrogen, but is six-membered and not five. Answer choice A remains the best answer.
  4. This ring has the wrong number of atoms and no nitrogen here! Answer choice A it is.

39) The passage wraps up with the maximum conductivity of the best-performing PANI: 5.0 × 10–3 (Ω∙cm)–1. Resistivity is the inverse of conductivity so this is a shorter math question:

Resistivity = 1/conductivity = 1/5.0 × 10–3 (Ω∙cm)–1 = (⅕)(103) Ω∙cm = 0.2 * 103 Ω∙cm = 200 Ω∙cm

  1. This is almost the correct answer, but not quite! If you reached this answer, you likely forgot to change the sign of the exponent when you brought it over the line of division. Remember, when you flip to the top, you flip the sign of the exponent.
  2. It is important to keep track of the fact that we need to take the inverse of 5, or ⅕ which is equal to 0.2. There is no “5” in the answer.
  3. This is the correct value as calculated above. Resistivity = 1/conductivity = 1/5.0 × 10–3 (Ω∙cm)–1 = (⅕)(103) Ω∙cm = 0.2 * 103 Ω∙cm = 200 Ω∙cm
  4. Like answer choice B, this answer choice is incorrect because it fails to calculate the multiplicative inverse of 5. Answer choice C is the correct answer.

 

Exam 3 Chemistry/Physics Section Passage 8

40) According to the passage, the conversion of cytosine to 5hmC requires methylation and hydroxymethylation. The enzyme for the first step needs to be able to add a methyl group and the enzyme of the second step needs to be able to hydroxylate that methyl group.

  1. The first step requires a methylation, and a transferase is able to transfer the functional group (such as a methyl group) from one substrate to another which would account for the methylation enzyme. A hydroxylation, the addition of an alcohol into another compound, is a type of reduction-oxidation reaction. An oxidoreductase catalyzes reduction and oxidation reactions. This is the best answer.
  2. Neither of the two steps mentioned above require cleavage of a bond using water which is the function of a hydrolase, so this answer is incorrect.
  3. For the same reason noted in answer choice B, this answer is also incorrect. Neither a methylation nor the hydroxylation of the methyl group require water-mediated bond cleavage.
  4. Transferase and ligase – As in answer choice A, a transferase would be a great way to ensure that a methyl group is transferred from one molecule to the molecule of interest. However, a ligase catalyzes the bonding of two larger molecules, such as a DNA ligase joining two fragments of DNA. However, a hydroxylation is a redox reaction that adds an alcohol to the molecule. Adding an alcohol is not the same as joining two molecules together making A the best answer.

41)

  1. The Gibbs free energy equation includes the natural log “ln,” not its inverse “e.”
  2. The Gibbs free energy equation includes the natural log “ln,” not its inverse “e.”
  3. The passage describes a “denaturation” so the unfolded DNA is the product, and the folded DNA is the reactant. The equilibrium constant Keq is the product relative to the reactant, so equal to [unfolded]/[folded]. Finally, the Gibbs free energy equation is ΔG′° = –RTlnKeq = –RTln([unfolded]/[folded]). This answer choice is incorrect because the Keq is the inverse of what it should be.
  4. This is correct. As described above, Keq for the denaturation described in the passage is equal to [unfolded]/[folded] and when plugged into the Gibbs free energy equation gives: ΔG′° = –RTlnKeq = –RTln([unfolded]/[folded]).

42) A variation of Fig. 1 is shown below for reference.

Before we determine whether the statements below are true or false, we need to figure out how to understand pK and cooperativity in the context of Fig. 1. Here, pK is analogous to pKa and will be represented by the pH at which half of the sequences are unfolded and when the normalized CD signal on the y-axis is equal to 0.5. As for cooperativity, remember that cooperative enzymes display sigmoidal curves which have steeper slopes so we will use steepness as a proxy for cooperativity.

  1. The oligonucleotide with the highest pK (highest pH when the y-axis is 0.5) is 5mC-WT. The highest unfolding cooperativity (steepest slope and most sigmoidal shape) is shown by 5hmC-WT so this statement is incorrect.
  2. The oligonucleotide with the lowest pK (lowest pH when y-axis is 0.5) is 5hmC-WT. The oligonucleotide with the highest unfolding cooperativity is also 5hmC-WT making this statement true and the correct answer.
  3. The oligonucleotide with the second highest pK is the WT while the highest unfolding cooperativity is for the 5hmC-WT oligonucleotide so this statement is false and incorrect.
  4. The oligonucleotide with the second highest pK is the WT while the oligonucleotide with the lowest unfolding cooperativity is 5mC-WT so this statement is false and incorrect. Answer choice B is correct.

43) A variation of Fig. 1 is shown below for our reference. Since all of the answer choices discuss a decrease in stability, let’s pause and use the data in the figure to evaluate relative stability. We discussed pK in the previous question, represented by the pH at which half of the oligonucleotide is denatured. If the oligonucleotide denatures at a lower pH as we increase the pH, then the oligonucleotide is less stable. If it is able to tolerate higher pHs before denaturing, it is more stable. The oligonucleotide with the lowest pK and therefore the least stable oligonucleotide is 5hmC-WT, the one with 5-hydroxymethylcytosine. The 5mC-WT and WT oligonucleotides have very similar pKs (they almost overlap) so they have comparable stabilities.

Figure 1 The pH-dependence of iM denaturation for each oligonucleotide

  1. Methylation, noted by the 5mC-WT curve, did not significantly decrease stability of the oligonucleotide compared to the wild type. This is incorrect.
  2. The pK of the 5mC-WT curve is really close to that of the WT. It would be inappropriate to call it a significant decrease. If anything, the pK for 5mC-WT is actually a bit higher than that of the WT, suggesting increased stability.
  3. The first half of this statement is correct; let’s evaluate the second half. As pH increases, unfolding increases. When the pH is higher, there are less protons in solution and acidic functional groups will lose their protons. The alcohol in the hydroxymethylated cytosines will lose their protons as pH increases. Thus, it’s their deprotonation that decreases their stability making this the correct answer.
  4. While the first half of this statement is correct, the second half is incorrect. The degree of unfolding increases as pH increases meaning that the functional group of interest is losing its proton, not being protonated. Answer choice C is the best answer.

44) Let’s take a quick peek at the wild type sequence: 5′–TTCCCTACCCTCCCCACCCTAA–3′. Well, that sequence doesn’t have any “G”s meaning it does not contain deoxyguanosine. According to the base pairing rules, if G is not present on the original strand, its C complement should not be present on the complement sequence. No G to start, no C to pair.

  1. Deoxythymidine is present in the wild type sequence so deoxyadenosine should be present in the complement.
  2. Deoxyguanosine is not present in the wild type sequence so deoxycytidine should not be present in the complement sequence making this answer choice correct.
  3. Deoxyguanosine is not present in the wild type sequence but should be present in the complementary sequence because its complementary base pair deoxycytidine is present in the WT sequence.
  4. Like its base pair answer choice A, deoxythymidine should indeed be present in the complementary sequence because its deoxyadenosine is in the WT sequence. Only deoxycytidine is not present in the complementary sequence.

 

Exam 3 Chemistry/Physics Section Discretes

45) In order to bind to a cation-exchange column, the protein will need to be positively charged. Remember that the ion exchange column tells us the charge that it wants to hold on to such that cation-exchange columns will elute negatively charged species and retain positively charged species. As for the pI, recall that the isoelectric point tells us the pH at which these proteins are net neutral. Above the pI, they will be negatively charged and below it they will be positively charged. In order to bind the cation-exchange column, the proteins need to be positively charged and the pH has to be less than the pI.

  1. At a pH of 3, all four of the proteins will have a net positive charge. The question asks for two of the four adhering so this is incorrect.
  2. A pH of 7 is less than the pI of exactly two proteins, A and B which have pIs of8.8 and 10.2 respectively, meaning that these two proteins will adhere to the cation exchange column. This is the correct answer.
  3. A pH of 9.5 is only less than one of the pIs given, specifically the pI of protein B. The question stem asks for two proteins to adhere to the column, not just one.
  4. None of the proteins will have a net positive charge at a pH of 11; this pH is greater than all of their pIs so none will adhere to the cation exchange column. The correct buffer pH for this question is pH = 7.0.

46)

  1. Glucose is not aromatic so we have no reason to believe that a polysaccharide made up of glucose will have aromatic properties.
  2. Glucose, and therefore the polysaccharide dextran, is hydrophilic and not hydrophobic. It has a lot of hydroxyl groups or alcohols which make it polar rather than nonpolar.
  3. Salt bridges are used in electrochemistry to prevent the buildup of charges and are a combination of hydrogen and ionic bonding. This does not apply to glucose or dextran.
  4. The oxygens in the hydroxyl functional groups in glucose are capable of hydrogen bonding to the side chains that may be exposed on different proteins. It is the only answer choice with interactions that are applicable to glucose and therefore dextran, making it the correct answer.

47) The first ionization energy is the energy required to remove the first valence electron from an atom. The energy required to remove the valence electron is the result of both the pull of the nucleus (number of protons in the nucleus) and the number of electrons shielding the valence electrons from the pull of the nucleus. As the number of shielding electrons increases, the pull of the nucleus is buffered, so the smallest atom with the fewest number of shielding electrons is the atom that will hold onto its valence electrons the tightest and will have the highest first ionization energy.

  1. Iodine is the halogen with the largest radius and the most shielding electrons on this list. It will have the lowest first ionization energy, not the highest.
  2. Bromine has more shielding electrons than chlorine but fewer than iodine. It will not have the highest first ionization energy.
  3. Chlorine has more shielding electrons than fluorine and fewer than bromine so it will have a lower first ionization energy than fluorine and does not have the highest first ionization energy.
  4. Fluorine is the halogen with the smallest radius and the fewest shielding electrons. It will have the highest first ionization energy and is the correct answer.

48)

  1. The question stem asks for the functional relationship between distance and time. The way this answer choice is written suggests that time is proportional to or responding to distance, specifically the inverse of distance. Time is on the x-axis and distance changes in response to time; the distance does not determine the time. Also, the inverse suggests a negative relationship, whereas here the relationship is positive.
  2. Like the answer choice above, this answer choice is a poor fit because it suggests the time changing in response to the distance. Also, it is inversely proportional to the second power whereas the relationship shown above is positively proportional to the second power.
  3. This represents a linear relationship; as we can see above, the relationship between distance and time is exponential and not linear.
  4. This is the best answer. The distance is roughly equal to the time squared. For instance, at 6s, the distance is about 62 or 36, at 8s just around 60 (82=64) and at 10s about 100 distance units.

 

Exam 3 Chemistry/Physics Section Passage 9

49) The numerical value for the frequency provided in the question stem is irrelevant to this pseudodiscrete question. Rephrasing the question stem, we need to determine what wave property stays the same as the wave changes mediums.

  1. The frequency is determined by the source, not the medium through which the wave travels. It remains unchanged when a wave changes mediums and is the correct answer.
  2. The speed at which a wave travels is influenced by the medium through which it travels so it will change, making this answer incorrect. For a review of the how the medium affects the speed of sound, hop over to https://jackwestin.com/resources/mcat-content/sound/relative-speed-of-sound-in-solids-liquids-and-gases
  3. The amplitude, proportional to what we refer to as the “loudness” of a sound, can change when a sound wave changes mediums.
  4. Wavelength – Like the wave speed, the wavelength will change when the wave changes mediums. Answer choice A is the best answer.

50) The final sentence of the passage gives the size of the opening as ~2.5 × 104 μm2. The question stem asks for the volume flow rate, so we need to multiply the microtubule opening by the speed at which the fluid flows.

Volume flow rate = area * velocity = 2.5 × 104 μm2 * 0.30 mm/s =2.5 × 104 μm2 * 0.30× 103 μm/s  = 0.75 × 107 μm3/s = 7.5 × 106 μm3/s

  1. This answer choice is lower than the correct answer; 2.5*3=7.5, not 4.5.
  2. This is the correct answer as calculated above. Volume flow rate = area * velocity = 2.5 × 104 μm2 * 0.30 mm/s =2.5 × 104 μm2 * 0.30× 103 μm/s = 0.75 × 107 μm3/s = 7.5 × 106 μm3/s
  3. This answer choice incorrectly divides the 0.3 by 2.5 but the two need to be multiplied.
  4. Both components of this answer choice are incorrect. See the calculations above to work through the math and remember to convert units as appropriate.

51) The question stem gives the speed as 1,500 m/s and the passage gives the frequency as 2.3 MHz so we can use the wave equation to calculate the wavelength.

  1. This answer is the result of multiplying the 1500 by 2.3 instead of dividing. The wavelength is inversely proportional to the frequency so the speed needs to be divided by the frequency.
  2. Double check your powers and conversion factors if you obtained this answer; you’re almost there.
  3. Don’t forget to divide by 2.3!
  4. As shown and calculated above, the wavelength of the ultrasound wave is 0.65 mm.

52) 

  1. The cross-sectional areas are inverted here. The new total cross-sectional area in the capillary should be in the denominator and the cross-sectional area of the blood vessel should be in the numerator.
  2. This is the correct expression as shown above.
  3. This answer fails to account for the number of capillaries in the capillary bed. The blood will not all travel through a single capillary so the total cross-sectional area of the capillary bed needs to be plugged into the continuity equation.
  4. The number of capillaries should be in the denominator with the cross-sectional area of a single capillary, not the numerator with the cross-sectional area of the larger blood vessel. We’ll stick with answer choice B.

 

Exam 3 Chemistry/Physics Section Passage 10

53) The solubility product Ksp can be found by multiplying the ions that make up CaCO3 which are Ca2+ and CO32–, according to Equation 1.

  1. The solid product CaCO3 should not be included in the Ksp expression. Solids and pure liquids are not included in any of the equilibrium expressions.
  2. This is correct. The solid CaCO3 dissolves into and is composed of one calcium ion and one carbonate ion. The product of their respective concentrations gives us the solubility product.
  3. The solid product CaCO3 should not be included in the Ksp expression.
  4. This would be correct if the exponents accurately reflected the number of each ion required to form the product. However CaCO3 only requires one calcium ion and one carbonate ion so answer choice B is the correct answer.

54) This is a pseudodiscrete question, and we can use the often forgotten periodic table when we come across this question on test day as needed.

  1. Both calcium and magnesium are in the second group of the periodic table. The alkali metals are the first group or family, not the second.
  2. The metalloids form the “staircase” between the post-transition metals and the reactive nonmetals on the periodic table. None of these are a part of the second family on the periodic table.
  3. Here we have it; the alkaline earth metals are the second family of the periodic table and contain calcium and magnesium.
  4. The halogens are the second to last group on the periodic table, not the second family, and are on the opposite side of the periodic table compared to calcium and magnesium.

55) This is another peudodiscrete question. To answer this question we need to know the relationship between the equilibrium constant, the Gibbs free energy and spontaneity. We’ve gone ahead and summarized the relationships in the table below. For another quick review, hop over to https://jackwestin.com/resources/mcat-content/rate-processes-in-chemical-reactions-kinetics-and-equilibrium/relationship-of-the-equilibrium-constant-and-delta-g

  1. The question stem specifies that the reaction is spontaneous. For the reaction to be spontaneous, the forward reaction and the products should be favored over the reverse reaction and the reactants, meaning that the Keq > 1. This is the correct answer.
  2. This would only be mathematically possible if the RT component of the equation were equal to 1 which would not happen on test day. Answer A was a great answer so we’re not going to let the MCAT test writers distract us with this answer choice.
  3. If Keq < 1, the reaction would favor the reactants and reverse reaction; it would be nonspontaneous.
  4. If Keq = 0, the reaction would be at equilibrium and would not be spontaneous.

56) This question is testing our understanding of Le Chatelier’s principle via Equation 3.

  1. The resin binds both the calcium and sodium ions. Adding more resin will not shift the reaction in either direction.
  2. Adding calcium ions increases the concentration of the reactants, and the reaction will shift to the right to restore equilibrium, not to the left.
  3. Adding sodium ions increases the concentration of one of the products which will produce a leftward shift to restore equilibrium. This is the correct answer.
  4. Like answer choice A, the anions are capable of binding to both the Ca2+ and Na+ which means increasing the concentration will not cause the desired leftward shift.

The bioenergetics content page walks through Le Chatelier’s principle: https://jackwestin.com/resources/mcat-content/principles-of-bioenergetics/bioenergetics-thermodynamics-1d

 

Exam 3 Chemistry/Physics Section Discretes

57) According to the VSEPR theory, an octahedral geometry occurs with an AX6 molecule.

  1. This fits the AX6 requirement for an octahedral geometry where sulfur is the central atom and the six fluorine atoms are attached to the sulfur.
  2. The carbon central atom only has four bromines attached to it. To form an octahedral geometry, the molecule should follow the AX6 template.
  3. Like answer choice B, the central atom only has four other atoms attached, not six.
  4. Like the other answer choices above, this molecule has the wrong number of atoms attached to the central atom for an octahedral geometry.

The sigma and pi bonds content page also reviews the VSEPR theory: https://jackwestin.com/resources/mcat-content/covalent-bonds/sigma-and-pi-bonds

58)

  1. Close, but not quite. Double check the arithmetic on your exponents, this answer is off by a factor of just 10.
  2. We only worked with orders of 10 in our calculations so our answer should be an order of 10.
  3. This is the correct answer. When the stopcock is opened, the fluid will rush into and up the tube until the pressure inside the tube, contributed by the fluid, is equal to the atmospheric pressure outside of the tube. This will occur when the fluid reaches a height of 10 m.
  4. As with answer choice B, we only worked with orders of 10 in our calculations so our answer should be an order of 10. The correct answer is 10 m.

59) This question is a mouthful; let’s break it down. The first requirement is that the element is in the second period, or the second row on the periodic table. The second requirement is that it should be a solid made from covalent bonds when in its standard state. Putting it all together, the question is asking for an element in the second row of the periodic table that is a solid with covalent bonds.

  1. Carbon is indeed in the second period and also happens to be a solid in its standard form, also known as graphite. We’re off to a great start on this last question.
  2. Phosphorus is in the third period, not the second.
  3. While oxygen is in the second period, it is a gas and not a solid in its standard state.
  4. Iodine is in the fifth period, not the second period. Only carbon meets the question stem requirements.

 



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