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AAMC MCAT Test 9 Questions and Answers 2024
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AAMC MCAT Test 9 Questions and Answers 2024A 2 kg mass and a 5 kg mass are connected by a massless cord suspended over a
massless and frictionless pulley. If the acceleration due to gravity is g, what will be the
acceleration of the masses after they are released from rest?
A. 2g/7
B. 3g/7
C. 5g/7
D. g - ANSSolution: The correct answer is B.
A. This is smaller than the magnitude of the acceleration of the masses after they are
released from rest by a factor of 1.5. It implies that only the 2-kg mass is moving after
the release.
B. According to Newton's second law, the net force acting on the 5-kg mass is given by
the expression Fnet= 5 kg × a1 = 5 kg × g - T, where a1 is the acceleration after the
release and T is the tension in the cord. The net force acting on the 2-kg mass is given
by the expression Fnet= 2 kg × a2 = 2 kg × g - T. Because the two masses move
simultaneously but in opposite directions after they are released, a1 = -a2 = a.
Substituting the expression T = 5 kg × (g - a) into the equation of motion of the 2-kg
mass yields -2 kg × a = 2 kg × g - 5 kg × (g - a) = -3 kg × g + 5 kg × a. Then 7 kg × a = 3
kg × g, hence a = 3g/7.
C. This is larger than the magnitude of the acceleration of the masses after they are
released from rest by a factor of 1.7. It implies that only the 5-kg mass is moving after
the release.
D. This implies the two masses fall freely and separately without being connected by the
cord.
A major obstacle to obtaining useful energy from a nuclear fusion reactor is containment
of the fuel at the very high temperatures required for fusion. The reason such high
temperatures are required is to:
A. eliminate the strong nuclear force.
B. remove electrical charge from reactants.
C. decrease the density of the fuel.
D. enable reactants to approach within range of the strong nuclear force. - ANSSolution:
The correct answer is D.
A. The strong nuclear force cannot be eliminated by increasing the temperature
because this force manifests whenever nucleons are present and within a range of
picometers.
B. The electrical charge is an intrinsic property of matter that is independent of
temperature.
C. Decreasing the density of the fuel is detrimental because the probability of fusion
increases with the increase in fuel density because the attractive component of the
strong nuclear force acts at short distances indicative of high density for the fuel
material.
D. The probability of fusion increases with the decrease in the average distance
between fuel particles that enables attractive nuclear forces to overcome the repelling
nuclear forces acting at medium and long distances. An increase in the temperature is
equivalent to an increase in the root-mean-square speed of the fuel particles that will
,AAMC MCAT Test 9 Questions and Answers 2024
travel the average distance between fuel reactants in smaller times. The associated
increase in the kinetic energy of the particles relative to the center of mass of the
nuclear fuel system essentially correlates with a decrease in the electrostatic potential
barrier that repels particles of the same electric charge. In turn, this increases the
probability of particles to undergo the tunnel effect by penetrating the electrostatic
barrier. These combined effects enable reactants to approach within range of the strong
nuclear force.
An ester is prepared by the method of direct esterification using an esterase enzyme as
a catalyst. Which of the following modifications will NOT appreciably increase the final
yield of ester?
RCOOH+RCH2OH-->RCOOCH2R +H2O
A. Using 2 times as much enzyme
B. Using 2 moles of RCOOH instead of 1 mole
C. Using 2 moles of RCH2OH instead of 1 mole
D. Removing RCOOCH2R from the reaction mixture as it is formed - ANSSolution: The
correct answer is A.
A. This modification will increase the rate of formation, not the final yield of ester.
B. Doubling the concentration of RCOOH will increase the yield of ester by Le
Châtelier's principle.
C. Doubling the concentration of RCH2OH will increase the yield of ester by Le
Châtelier's principle.
D. By Le Châtelier's principle, removing the ester product as it forms will drive the
reaction to form more ester.
An ice cube at 0°C and 1 atm is heated to form steam at 100°C and 1 atm. Ignoring
heat loss to the surroundings, what part of the process uses the most heat?
(Note: Specific heat of water = 1 cal/g°C. Heat of fusion = 80 cal/g. Heat of vaporization
= 540 cal/g.)
A. Melting the ice cube
B. Heating all the water from 0°C to 50°C
C. Heating all the water from 50°C to 100°C
D. Vaporizing all the water - ANSSolution: The correct answer is D.
A. Melting the ice cube requires only 80 cal/g, which is less than the heat of
vaporization.
B. Heating all the water from 0°C to 50°C requires 50 cal/g, which is less than the heat
of fusion.
C. Heating all the water from 50°C to 100°C requires 50 cal/g, which is less than the
heat of fusion and the heat of vaporization.
D. Vaporizing all the water requires 540 cal/g, which is a greater heat requirement than
specific heat or heat of fusion.
Approximately how many moles of Kr+ are contained in the laser tube at 0°C and 1
atm? V= 11cm^3
A. 3 x 10-7
,AAMC MCAT Test 9 Questions and Answers 2024
B. 2 x 10-6
C. 4 x 10-5
D. 5 x 10-4 - ANSD
11cm^3= 11x10^-3L
1 mole of gas at STP is 22.4 L
11/22.4= 0.5 (x10^-3), therefore 5e-4 is the answer
Based on Figure 2, what is the approximate Ki of the MCS oligomers? inflection point at
0.03uM
A. 12 nM
B. 30 nM
C. 170 nM
D. 800 nM - ANSSolution: The correct answer is B.
A. This is the MCS oligomer concentration at almost the maximum specific activity.
B. In the semi-log plot shown in Figure 2, the Ki is the inflection point of the sigmoidal
curve because, by definition, the Ki is the concentration of the inhibitor at which the
reaction rate is half of the maximum reaction rate. The inflection point is at
approximately 0.030 mM, or 30 nM.
C. This is the MCS oligomer concentration at about 0.05 × specific activity.
D. This is the MCS oligomer concentration at complete enzyme inhibition.
Based on Reaction 1, when 1.0 atm of CO(g) completely reacts to form carbon
suboxide at 550°C in a sealed container, what is the final pressure in the container?
4Co2--> C3O2 +CO2
A. 0.00 atm
B. 0.10 atm
C. 0.25 atm
D. 0.50 atm - ANSSolution: The correct answer is D.
A. There is still gas present in the container.
B. Since pressure is directly proportional to moles of gas at constant V and T, this is a
smaller pressure than the balanced equation allows.
C. This result ignores the pressure contribution of one of the gases.
D. Based on Reaction 1, 4 mol CO(g) forms 2 mol of gases. Because of the direct
relationship between P and n at constant V and T, that means 1.0 atm CO(g) makes
0.50 atm of gases.
Based on the passage, what most likely causes W96 to be accessible to CHCl3 at
75°C?
A. Peptide bonds are broken, releasing W96.
B. Reduction of disulfide bonds occurs.
C. The protein unfolds and exposes W96 to the buffer.
D. CHCl3 extracts W96 from the protein interior. - ANSSolution: The correct answer is
C.
A. As stated in the passage, there is no change in primary structure when carbonic
anhydrase, is modified respectively by CHCl3 at 75°C, so peptide bonds are not broken.
, AAMC MCAT Test 9 Questions and Answers 2024
B. There is no reducing agent present.
C. As stated in the passage, at 75ºC, carbonic anhydrase fully denatures, which means
that the protein is unfolded and exposes W96 to the buffer.
D. At 10 mM, the reactant CHCl3 is not acting as a solvent. The buffer is the solvent,
which is mostly water.
Based on the passage, which of the following is closest to the pressure exerted on the
chest by a 10 × 5 cm rectangular paddle during defibrillation?
(Note: 1 Pa = 1 N/m2.) (F=100N)
A. 5 kPa
B. 10 kPa
C. 15 kPa
D. 20 kPa - ANSSolution: The correct answer is D.
A. This is 1/4 of the actual pressure. Pressure is the ratio of the force applied
perpendicular to a surface and the area of the surface. Because the normal component
of the force is 100 N and the area is 10 cm × 5 cm = 50 cm2 = 0.005 m2, the pressure is
(100 N)/(0.005 m2)=20 kPa.
B. This is 1/2 of the actual pressure. Pressure is the ratio of the force applied
perpendicular to a surface and the area of the surface. Because the normal component
of the force is 100 N and the area is 10 cm × 5 cm = 50 cm2 = 0.005 m2, the pressure is
(100 N)/(0.005 m2) = 20 kPa.
C. This is 3/4 of the actual pressure. Pressure is the ratio of the force applied
perpendicular to a surface and the area of the surface. Because the normal component
of the force is 100 N and the area is 10 cm × 5 cm = 50 cm2 = 0.005 m2, the pressure is
(100 N)/(0.005 m2) = 20 kPa.
D. Pressure is the ratio of the force applied perpendicular to a surface and the area of
the surface. Because the normal component of the force is 100 N and the area is 10 cm
× 5 cm = 50 cm2 = 0.005 m2, the pressure is (100 N)/(0.005 m2) = 20 kPa.
Based on the passage, which type of inhibitor will provide a flavorful onion that does not
cause tearing?
An inhibitor of:
A. Compound 1 synthesis
B. alliinase
C. Reaction 2
D. LFS - ANSSolution: The correct answer is D.
A. Although inhibiting Compound 1 synthesis will stop the onion from causing tearing,
because Compound 1 is the source of flavor components in onions, the onion will not be
flavorful.
B. Although inhibiting allinase will stop the onion from causing tearing, because allinase
catalyzes the formation of flavor component precursors in onions, the onion will not be
flavorful.
C. Inhibiting Reaction 2 will stop the formation of Compound 3, which is one of the main
flavor components of onions.
D. Inhibiting LFS will stop the formation of Compound 4, a potent tearing agent.
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