AAMC FL 4 CP Exam Questions and Answers

AAMC FL 4 CP Exam Questions and Answers Limestone does NOT decompose when heated to 900 K because, at 900 K, ΔH is: -Answer- positive and greater than TΔS. The reaction does not occur (is not spontaneous). This indicates that ΔG = ΔH - TΔS > 0. From inspection of the reaction, it can be concluded that ΔS > 0 (a gas evolves). Consequently, ΔH > TΔS explains why the reaction does not occur. Entropy -Answer-Increases if we go from solid reactant to gas product = positive delta S When limestone is heated during Step 1, an equilibrium is established. Which of the following expressions is the equilibrium constant for the decomposition of limestone? -Answer-[CO2] From the Law of Mass Action, an equilibrium constant expression involves a ratio of products to reactants with exponents determined from the stoichiometry of the reaction. solids are excluded from equilibrium constant expressions. During Reaction 2, did the oxidation state of N change? -Answer-No; it remained at -3. The part of Reaction 2 that involves nitrogen is the protonation of ammonia (NH3 + H+ → NH4+). Acid-base reactions do not involve oxidation state changes. The oxidation state of N in NH3 is -3. Each H is +1 and is balanced by the -3 of N to make a neutral compound. The oxidation state of N does not change when the N is protonated If all of Gas X (from Step 6) is held in a sealed chamber at STP, what will be its approximate volume? -Answer-22.4 L Volume of gas at STP -Answer-22.4 L Why was it important that the cuvettes containing the glucose oxidase and the blood sample were identical in terms of optical properties? -Answer-To enable the comparison of the absorption spectra. The identical optical properties of the cuvettes ensure that the absorbed radiation is due only to the presence of glucose in the blood and not due to the difference in the absorption features of the walls. What is the approximate energy of a photon in the absorbed radiation that yielded the data in Table 1? (Note: Use 1 eV = 1.6 × 10-19 J and hc = 19.8 × 10-26 J•m.) -Answer-2 eV The photon energy is E = hc/λ = 19.8 × 10^-26 /(625 × 10^-9) = 3.1 × 10^-19 J Convert J to eV using the constant in the question. (3.2 x 10^-19) / (1.6 x 10^-19) = 2 eV. According to Table 1, what is the concentration of the glucose in the blood from which the diluted sample was taken? -Answer-150 mg/dL From Table 1, the glucose concentration in the diluted sample is (0.20/0.24) × 6.0 mg/dL = 5.0 mg/dL. The blood then has a glucose concentration of 30 × 5.0 mg/dL = 150 mg/dL. Looking at the table 1- you see that the glucose concentration is between 4.5 and 6 mg/dL and 6 mg/dL. The passage tells you the dilution is 1/30. This means that the concentration of the original sample is between 135 mg/dL and 180 mg/dL. Of these, D is the only feasible answer. Suppose a blood sample tested above the range (6.0 mg/dL) of the standards used in the experiment. What modification will provide a more precise reading by data interpolation as opposed to extrapolation using the same standards? -Answer-Dilute the sample with additional solvent. By adding solvent, the concentration of glucose will be lowered, and the resulting absorbance will fall within the range of the standards. Which of the following reasons best explains why it is possible to separate a 1:1 mixture of 1- chlorobutane and 1-butanol by fractional distillation? -Answer-The boiling point of 1- chlorobutane is substantially lower than that of 1-butanol. The molecules have similar molecular weights, but 1-butanol has a hydroxyl functional group that can participate in hydrogen bonding. Hydrogen bonding is a particularly strong force of intermolecular attractio