MCAT Question of the Day

Correct Answer: A. I only
3-bromo-1,1-dimethylcyclobutane structure
A saturated molecule is one where all the valences of the atoms are satisfied by single covalent bonds. In other words, there are no double bonds. Statement I is true.

A planar molecule is a flat one. In this case, we have two opposite corners in a tetrahedral shape. II is not true.

Ring strain is minimized in a cyclohexane and maximized in a cyclopropane, but it is always present in a cyclic molecule. III is not true. Thus the answer is A.

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Correct Answer: B. E(cell) = 0.15 V, galvanic

The standard electrode potential E(cell) for a cell is found using the equation:
E(cell) = E(oxidation) + E(reduction)
In this case, lead will be reduced because it has a higher reduction potential than cobalt. This means that cobalt is being oxidized. Remember that on the MCAT, you will always be given standard reduction potentials, so if we need to find an oxidation potential, we simply switch to the negative reduction potential. Our equations should now look like this:
Co -> Co2+ + 2e- ; E(oxidation)= +0.28 V
Pb2+ + 2e- -> Pb ; E(reduction)= -0.13 V
Thus, using the first equation, we arrive at
E(cell) = (+0.28 V) + (-0.13 V) = 0.15 V
Remember that if E(cell) > 0, the reaction is spontaneous and happens in a galvanic cell. (The opposite is a non-spontaneous, electrolytic cell). Also, lead cobalt batteries were used in some of the first electric cars!

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Correct answer: D. (3.5)(2.0)(60)/(96,500)(3)

Rearrange Faraday’s Law: moles = I x t/n x F. Note that we have to convert min to s, and that 3 electrons are transferred in the redox reaction.

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Correct Answer: B. Activation energy.

A catalyst lowers the activation energy, Ea, of a reaction.

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Correct Answer:C. cyclopentanone

Beta-keto carboxylic acids undergo decarboxylation through ring formations. The carbonyl bond of the ketone will react with the alcohol group of the carboxylic acid to form a ring and release the carboxylic acid as CO2. The remaining enol will tautomerize into a carbonyl group again, leaving cyclopentanone.

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Correct Answer: C. 33.3 s

You can test each of the answers: for 33.3 s intervals, the first 33.3 s finds the concentration dropping by 1/2, the next down to 1/4, the third down to 1/8, for a total of 100 s.

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Correct Answer: A. Decrease.

The addition of a solute to a liquid always lowers the vapor pressure of the liquid.

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Correct Answer: D. 3.7 x 10^-2 M

The dissociation of HF is given by
HF -> (H+) + (F-)
Let [H+] = [F-] = x. Although the actual equilbrium concentration of HF is (2-x) M due to dissociation, we assume that x is very small and ignore the difference. That is, we take [HF] = 2 M at equilibrium. Since the equilbrium expression for this reaction is Ka = [H+][F-]/[HF], we have Ka = x²/2, so

x² = 2Ka = 2(6.8 x 10^-4) = 13.6 x 10^-4, so x = 3.7 x 10^-2.

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Correct Answer: D. Ne and Kr

Two gases having the same temperature have the same average molecular kinetic energy. Therefore, (1/2)m1v1² = (1/2)m2v2², which gives m2/m1 = (v1/v2)². Since v1/v2 = 2, we have m2/m1 = 2² = 4. Among the available choices, the only gases with a mass ratio close to 4 to 1 are krypton (83.8 g/mol) and neon (20.2 g/mol).

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Correct Answer: A. [Ar]

A calcium atom normally contains 20 electrons, so if it acquires a charge of +2, it must have lost 2 electrons, leaving just 18. Only choice A has 18 electrons.

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