MCAT Question of the Day

Correct Answer: C. Lead (Pb).

First of all, we can immediately disregard answers A, B, and E. Nearly all compounds containing sodium, potassium, or zinc are soluble. Thus, we are left with answers C and D. Of the two, we realize that sulfur does not bind with sulfate. Sulfur sulfate does not exist. We recognize as well that lead compounds are rarely soluble, but lead perchlorate is. Thus, answer C is the correct answer.

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Correct Answer: C. Adding argon to the vessel.

The correct answer is C. Temperature change does increase the solvation ability of gas into liquid, but not significantly. However, we can easily recognize that a warm soda loses air much faster than a cold soda. Thus, it seems that answer B could be correct. However, in a sealed vessel, the factor that will increase the amount of air dissolved most immediately is the increase of pressure. The pressure is increased when another gas in forced in the system, like argon. The increase pressure will drive the equilibrium to force more air into the water.

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Correct Answer: A. A.

The correct answer is A. Looking at the graph for A, we see the pressure at point 1 is 10 kPa, and at point 2 is 25 kPa. The volume at point 1 is 50L. 50 x 10 = 500, and following Boyles Law, the volume at point 2 must be 500/25, or 20l. The other graphs do not show this correctly.

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Correct Answer: D. Cl

Electronegativity increases to the right of the periodic table, and also from bottom to top. We note that chlorine is the closest to the upper right of the periodic table, meaning it is the most electronegative of the available choices. Thus, it is the most likely to react with Mg, which normally carries a 2+ charge.

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Correct Answer: D. CaCl2.

Since Calcium is ionic and has attached to two chlorides ions which are anions, D is the correct answer. Barium does not form ions very well. A and B only have one ionic bond each.

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Correct Answer: B. [Kr] 4d¹⁰

Cadmium in it’s basic state has Configuration [Kr] 5s²4d¹⁰. Positively charged cadmium (II) has lost 2 electrons. It will lose its two 5s-electrons first, making B the most reasonable answer. Answer A is illogical, as it has a higher shell number (5) for the d-electrons than the s-electrons (4). C is the correct configuration for neutral cadmium. Finally, D is incorrect because the d-electrons were removed rather than the s-electrons.

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Correct Answer: A. H3N⁺-CH2-COO^–

Recall that you can use isoelectric point (PI) to predict an amino acid’s charge at any given pH. Specifically, if PI < pH, then this means that pH is high -- which means that [H+] is low, and charge is negative. Likewise, if PI > pH, then this means that pH is lower — which means [H+] is high and charge is positive. In this case, PI = pH, which means that charge is zero (the amino acid is neutral). Thus, the answer is A since this is the amino acid where both charges cancel each other out, leaving a neutral zwitterion glycine.

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Correct Answer: A. (CH3CH2)3CF; SN1

This is an example of a SN1 reaction where first step is Br (good leaving group) dissociates to produces a carbocation (stabilized by the polar protic solvent, water). The tertiary carbocation intermediate is very stable, especially in the highly polar protic solvent water. The carbocation then combines with the strong nucleophile, the the F- ion. (the original substituent (Br) must be a better leaving group than the nucleophile (F-) in order for the reaction to occur). C and D are wrong because this is not an elimination reaction. B is wrong because this is not an SN2 reaction, due to the highly hindered substrate, and the type of solvent used in the reaction (SN2 reactions favor polar aprotic solvents, not polar protic solvents).

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Correct Answer: C. C2H2

In this question, we use the ideal gas law, PV = nRT. First of all, we calculate the number of mols in the volume of 11.2 L. n = PV/RT, or n = (1*11.2)/(0.082*298). This gives us roughly n= 0.5 mol. You can bypass this by remembering that there are 22.4 L in a gaseous mol at STP. Thus, there is 0.5 mol of a substance that we know weighs 13g. The molar mass must be 26 g/mol then. With the knowledge that C = 12g/mol and H = 1g/mol (standard knowledge), we find that the hydrocarbon should be C2H2.

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Correct Answer: B. Octane.

The boiling point of hydrocarbons increases as the number of carbons increases. Therefore, Heptane has a higher boiling point than hexane, which has a higher boiling point than pentane. This is because the intermolecular forces between molecules keep the hydrocarbons in solution, requiring more heat to overcome these forces and become a gas. In the case of branched molecules, the hydrocarbons are not able to “stack” as easily as straight-chain hydrocarbons. This reduces intermolecular interactions and, subsequently, the boiling point. Even though octane, 2,2-dimethylhexane, and 2-methylheptane have eight carbons, octane has the highest boiling point.

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