MCAT AAMC practice exam Chem Phys Question already solved 2023
MCAT AAMC practice exam Chem Phys Question already solved 2023What quantity of Compound 1 (483.5g) must be provided to prepare 100.00 mL of solution with a concentration equal to Ki (60.3uM)? M=mol/v 1 umol=0. moles (1x10^-6) 2.92mg If only [I] is increased, then [ESI] or [EI] increases. This is an example of: A. the Bose-Einstein Principle. B. the Heisenberg Uncertainty Principle. C. the Le Châtelier's Principle. D. the Pauli Exclusion Principle. Solution: The correct answer is C. A. The Bose-Einstein Principle states that a collection of atoms cooled close to absolute zero will coalesce into a single quantum state. B. The Heisenberg Uncertainty Principle states that one cannot know both the momentum and position of an object with absolute certainty. C. Le Châtelier's Principle states that in a reversible process, the application of stress to the system will cause the system to respond in a way that will relieve this stress. In this case, the reversible process is E + I forming EI or ES + I forming ESI. In either case, increasing [I] induces stress of the system, and the system relieves that stress by converting I to either more EI or ESI. D. The Pauli Exclusion Principle states that two or more identical fermions, e.g. electrons, cannot occupy the same quantum state. Which structural change to Compound 1 would make it more water soluble? A. Replacing benzene CH with N in the ring B. Replacing C=O with C=CH2 C. Replacing N-N=N with CH-CH=CH D. Replacing NH with NCH3 Solution: The correct answer is A. A. Nitrogen in the benzene ring would have a lone pair that could accept a hydrogen bond from water, thus increasing the solubility of the compound. B. Replacing C=O with C=CH2 would decrease the polarity of the compound and make it less water soluble. C. Replacing N−N=N with CH−CH=CH would decrease the polarity of the compound and make it less water soluble. D. Replacing NH with NCH3 removes a hydrogen bond donor, thus decreasing water solubility. In μM•s-1 and μM, what should the approximate values of kcat/KM and Ki be, respectively, when [I] = 180 μM? Neither is affected by a change in I, therefore they will stay the same (150 and 60.3). Ki is the equilibrium constant and is never affected by increasing I. What functional group transformation occurs in the product of the reaction catalyzed by Na+-NQR (NADH and ubiquinone)? A. RC(=O)R → RCH(OH)R B. ROPO32- → ROH + Pi C. RC(=O)NHR'→ RCOOH + R'NH2 D. RC(=O)OR'→ RCOOH + R'OH Solution: The correct answer is A. A. This is two-electron reduction of a ketone to an alcohol, which is the reaction catalyzed by Na+-NQR. A. This is the reaction catalyzed by a phosphatase. A. This is the reaction catalyzed by a protease or amidase. A. This is the reaction catalyzed by an esterase. What is the chemical structure of a component found in four of the five cofactors (flavins) used by Na+-NQR? Solution: The correct answer is B. A. This is the structure of adenine. It is only found in FAD. B. This is the structure of flavin, found in four of the five cofactors used by Na+-NQR. C. This is the structure of ubiquinone. It is a substrate, but not a cofactor. D. This is the structure of histidine. It is an amino acid, not a cofactor. What is the ratio of cation (0.150M) to enzyme (0.75mM) in the spectroelectrochemical experiments described in the passage? A. 1:2 B. 2:1 C. 20:1 D. 200:1 0.150M=150mM 150/0.75= 200: 1 (D) The reaction between NADH and ubiquinone is exergonic, but the reaction, when catalyzed by Na+-NQR, does not generate much heat in vivo. What factor accounts for this difference? The reaction catalyzed by Na+-NQR in vivo: A. is more exothermic as a result of the lower activation energy. B. occurs sequentially in several small steps. C. maintains a large separation between the reacting centers. D. is coupled to the movement of a charged particle against a concentration gradient. Solution: The correct answer is D. A. This is impossible. Even if it were true, this would make the heat generation larger, not smaller, for the catalyzed reaction. Catalysis does not change thermodynamics. B. By Hess's Law, the heat of reaction will sum and be the same. The fact that the reaction can be broken down into steps will not change the overall thermodynamics. C. This is also impossible. The reactants ultimately must be close together to react. D. The movement of a charged particle against its concentration gradient is energetically costly. Coupling the two processes: the redox reaction between NADH and ubiquinone and the movement of Na+ up a concentration gradient makes the overall process less exothermic. Two open flasks I and II contain different volumes of the same liquid. Suppose that the pressure is measured at a point 10 cm below the surface of the liquid in each container. How will the pressures compare? A. The pressures will be equal. B. Pressure in flask I will be less. C. Pressure in flask II will be less. D. The pressures cannot be compared from the information given. Solution: The correct answer is A. A. The pressure at a point 10 cm below the surface of the liquid is the same in both flasks because the pressure is equal to the liquid density multiplied by the gravitational acceleration multiplied by 10 cm. B. The pressure at a point 10 cm below the surface of the liquid in flask I is the same as the pressure in flask II at 10 cm below the surface because the pressure is equal to the liquid density multiplied by the gravitational acceleration multiplied by 10 cm. C. The pressure at a point 10 cm below the surface of the liquid in flask II is the same as the pressure in flask I at 10 cm below the surface because the pressure is equal to the liquid density multiplied by the gravitational acceleration multiplied by 10 cm. D. The pressures can be compared because both pressures are calculated according to the hydrostatic pressure formula p = ρgd, where ρ is the liquid density, g is the gravitational acceleration, and d is the depth where pressure is measured. What is the identity of an atom that contains six protons and eight neutrons? A. Nitrogen B. Carbon C. Oxygen D. Silicon Solution: The correct answer is B. A. The atom cannot be nitrogen because nitrogen contains seven protons. B. Carbon contains six protons because it also contains six electrons as a neutral atom. C. Oxygen contains eight protons because as a neutral atom it contains eight electrons. D. The silicon atom contains 14 protons. Which of the following substances is polar? A. NF3 B. CCl4 C. CO2 D. Li2 Solution: The correct answer is A. A. The geometry of trifluoroamine is impacted by the lone pair on nitrogen, making it trigonal pyramidal. No bond dipoles cancel; this results in a polar molecule. B. While each C-Cl bond in carbon tetrachloride is polar, the sum of the dipole moments cancel as a result of its tetrahedral molecular geometry. C. Carbon dioxide is a linear molecule. The bond dipole moments of each C=O bond cancel as they are in opposite directions. D. This molecule is necessarily non-polar as it is comprised of two identical atoms. If all else is held constant, which of the following changes would NOT double the volume of a gas? A. Doubling the pressure B. Doubling the absolute temperature C. Halving the pressure D. Doubling the number of gas molecules Solution: The correct answer is A. A. Based on Boyle's Law, P is inversely proportional to volume, thus, doubling the pressure of a gas sample will decrease, not increase, the volume. B. Doubling the absolute temperature of a gas sample will double the volume because T is directly proportional to V for an ideal gas according to Charles' Law. C. Halving the pressure of a gas sample will double its volume because P is inversely proportional to V for an ideal gas. D. Doubling the number of gas molecules present will double the volume according to Avogadro's Law. What is the number of neutrons in the nucleus of the atom used to produce laser radiations? A. 48 B. 49 C. 50 D. 51 86Kr+ 36 protons, so 86-36=50 (C)
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mcat aamc practice exam chem phys question already solved 2023
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what quantity of compound 1 4835g must be provided to prepare 10000 ml of solution with a concentration equal to ki 603um mmolv
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