MCAT Question of the Day

Correct Answer: A. 1, 4

Rotation of plane polarized light is indicative of a chiral molecule. If not, the compounds must be meso or achiral. Compounds 1 and 4 are meso compounds, which, despite having two or more stereocenters, are not chiral (superposable on its mirror image). Compounds 2 and 3 are chiral molecules.

Subscribe below to receive the MCAT Question of the Day delivered straight to your inbox every morning.

Correct Answer: B. tert-butyl carbocation

The most stable fragment likely formed from 2,2-dimethylpentane fragmentation is t-butyl species.

Subscribe below to receive the MCAT Question of the Day delivered straight to your inbox every morning.

Correct Answer: D. Benzoic acid, an organic compound, is insoluble in water. Extraction into water can be achieved with aqueous sodium hydroxide (deprotonation -> polar sodium benzoate, now soluble in water). Transfer of the benzoate back into organic solution is accomplished by treatment with hydrochloric acid (protonation -> less polar benzoic acid).

Subscribe below to receive the MCAT Question of the Day delivered straight to your inbox every morning.

Correct Answer: B. 1/16

F1 generation is the heterozygous offspring of the homozygous parents. F2 generation is the offspring of the heterozygous offspring generation F1. The ratio is 9:3, 3:1 in this generation, meaning that only one of the offspring will have black fur and a long tail.

Subscribe below to receive the MCAT Question of the Day delivered straight to your inbox every morning.

Correct Answer: B. Atomic radius increases

As you move from left to right across a Period, metallic properties decrease (metals > metalloids > non-metals), atomic radius decreases (the atoms get smaller), electronegativity (the ability to attract electrons in a compound) increases, first ionization energy (the energy required to remove the outermost electron form a neutral atom) increases and electron affinity (the energy released when an electron is bonded to an atom) increases.

Subscribe below to receive the MCAT Question of the Day delivered straight to your inbox every morning.

Correct Answer: A. Zinc.

To provide a sacrificial coating the coating metal must be more reactive than the base metal being protected. Thus gold and copper are not applicable as they are both less reactive than iron and can only provide a passive coating.

Aluminum and zinc are more reactive than iron. Aluminum and zinc both form protective oxide layers that will stop corrosion of the iron, but aluminum is too reactive to be used in exposed weathering conditions. It is zinc that has just the right degree of reactivity that allows it to both protect the iron and not be corroded away, providing a strong and long lasting sacrificial coating.

Subscribe below to receive the MCAT Question of the Day delivered straight to your inbox every morning.

Correct Answer: B. Equilibrium favors the products.

In an acid-base equilibrium such as this, the relative strength of an acid can be determined using the magnitude of Ka (or pKa): as Ka increases, the strength of the acid increases; as pKa (inverse log of Ka) decreases, the strength of the acid increases. The equilibrium lies to the left because t-butanol is the weaker acid compared to water. You can also think this way: hydroxide is the weaker base (more stable) compared to t-butoxide because of e- donation of alkyl groups, making the resultant anion less stable (more reactive).

Subscribe below to receive the MCAT Question of the Day delivered straight to your inbox every morning.

Correct Answer: D. The rate order.

“SN” stands for nucleophilic substitution and the “1″ represents the fact that the rate-determining step is unimolecular.

Subscribe below to receive the MCAT Question of the Day delivered straight to your inbox every morning.

Correct Answer: C. 120^o

An sp²-hybridized carbon has a trigonal planar geometry. This means that the bond angle will be 120^o, so answer choice C is the best choice. An sp-hybridized carbon has a linear geometry, meaning a bond angle of 180^o. Finally, an sp³-hybridized carbon has a tetrahedral geometry with bond angles of 109.5^o.

Subscribe below to receive the MCAT Question of the Day delivered straight to your inbox every morning.

Correct Answer: C. N2(g) 0.75 and H2(g) 2.25 mol.L^-1.s^-1

The rate of consumption of the reactants will be the product of the rate of formation of the product multiplied by the ratio of the (co-efficient of the reactant) / (co-efficient of the product).

Rate N2(g) = 1.50 x 1/2 = 0.75 mole.L^-1.s^-1 and H2(g) 1.50 x 3/2 = 2.25 mole.L^-1.s^-1.

Subscribe below to receive the MCAT Question of the Day delivered straight to your inbox every morning.