FL 5 AAMC UPDATED ACTUAL EXAM
QUESTIONS AND CORRECT ANSWERS FULL
SOLUTION GRADED A+
◉ 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?
A. [CaO]
B. [CaCO3]
C. [CO2]
D. [CaO] x [CaCO3]. Answer: C.
From the law of mass action, an equilibrium constant expression
involves a ration of products to reactants with exponents determined
from the stoichiometry of the reaction. Furthermore, solids are
excluded from equilibrium constant expressions. CO2 (g), as the only
non-solid material in the reaction, is the only substance that appears in
the equilibrium constant expression.
◉ During Reaction 2, did the oxidation state of N change?
A. Yes; it changed from -3 to -4
B. Yes; it changed from 0 to +1
C. No; it remained at -3
,D. No; it remained at +1. Answer: C.
The part of the Reaction 2 that involves nitrogen is the protonation of
ammonia (NH3 + H--> NH4+). Acid-base reactions doe not involve
oxidation state changes. Furthermore the oxidation states of N in NH3
is -3 not 0.
◉ If all of Gas X (from Step 6) is held in a sealed chamber at STP,
what will be its appropriate volume?
A. 22.4 L
B. 44.8 L
C. 67.2 L
D. 89.6 L. Answer: A.
The quantity of Gas X was given as 1 mole. One mole of gas occupies
22.4 L at STP.
◉ Why was it important that the cuvettes containing the glucose
oxidase and the blood sample were identical in terms of optical
properties?
A. To enable the comparison of the absorption spectra
B. To reduce the absorption in the glass walls
C. To decrease the uncertainty in the wavelength
D. To increase the absorption in the solutions. Answer: A.
,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?
A. 1 eV
B. 2 eV
C. 3 eV
D. 4 eV. Answer: B. ( I chose C)
The photon energy is E=hc/λ = 19.8 x 10^-26 J.m/ (625 x 10^-9) = 3.1
x10^-19 J , so about 2 eV
◉ According to Table 1, what is the concentration of the glucose in
the blood from which the diluted sample was taken?
A. 60 mg/dL
B. 90 mg/dL
C. 120 mg/dL
D. 150 mg/dL. Answer: D.
From Table 1, the glucose concentration in the diluted sample is
(o.20/0.24) x 6.0 mg/dL = 5.0 mg/dL.
, The blood then has a glucose concentration of 30 x 5.0 mg/dL= 150
mg/dL.
◉ 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?
A. Increase the enzyme concentration.
B. Increase the oxygen pressure.
C. Decrease the content of the oxygen acceptor
D. Dilute the sample with additional solvent.. Answer: D.
By adding solvent, the concentration of glucose will be lowered, and
the resulting absorbance will fall within the range of the standards.
This is easily accomplished, and the resulting calculations that
account for the dilution are not difficult.
◉ Which of the following reasons best explains why it is possible to
separate a 1:1 mixture of 1-chlorobutane and 1-butanol with fractional
distillation?
A. Both 1-chlorobutane and 1-butanol are polar
B. Both 1-chlorobutane and 1-butanol are nonpolar
C. The boiling point of 1-chlorobutane is substantially higher than
that of 1-butanol
QUESTIONS AND CORRECT ANSWERS FULL
SOLUTION GRADED A+
◉ 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?
A. [CaO]
B. [CaCO3]
C. [CO2]
D. [CaO] x [CaCO3]. Answer: C.
From the law of mass action, an equilibrium constant expression
involves a ration of products to reactants with exponents determined
from the stoichiometry of the reaction. Furthermore, solids are
excluded from equilibrium constant expressions. CO2 (g), as the only
non-solid material in the reaction, is the only substance that appears in
the equilibrium constant expression.
◉ During Reaction 2, did the oxidation state of N change?
A. Yes; it changed from -3 to -4
B. Yes; it changed from 0 to +1
C. No; it remained at -3
,D. No; it remained at +1. Answer: C.
The part of the Reaction 2 that involves nitrogen is the protonation of
ammonia (NH3 + H--> NH4+). Acid-base reactions doe not involve
oxidation state changes. Furthermore the oxidation states of N in NH3
is -3 not 0.
◉ If all of Gas X (from Step 6) is held in a sealed chamber at STP,
what will be its appropriate volume?
A. 22.4 L
B. 44.8 L
C. 67.2 L
D. 89.6 L. Answer: A.
The quantity of Gas X was given as 1 mole. One mole of gas occupies
22.4 L at STP.
◉ Why was it important that the cuvettes containing the glucose
oxidase and the blood sample were identical in terms of optical
properties?
A. To enable the comparison of the absorption spectra
B. To reduce the absorption in the glass walls
C. To decrease the uncertainty in the wavelength
D. To increase the absorption in the solutions. Answer: A.
,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?
A. 1 eV
B. 2 eV
C. 3 eV
D. 4 eV. Answer: B. ( I chose C)
The photon energy is E=hc/λ = 19.8 x 10^-26 J.m/ (625 x 10^-9) = 3.1
x10^-19 J , so about 2 eV
◉ According to Table 1, what is the concentration of the glucose in
the blood from which the diluted sample was taken?
A. 60 mg/dL
B. 90 mg/dL
C. 120 mg/dL
D. 150 mg/dL. Answer: D.
From Table 1, the glucose concentration in the diluted sample is
(o.20/0.24) x 6.0 mg/dL = 5.0 mg/dL.
, The blood then has a glucose concentration of 30 x 5.0 mg/dL= 150
mg/dL.
◉ 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?
A. Increase the enzyme concentration.
B. Increase the oxygen pressure.
C. Decrease the content of the oxygen acceptor
D. Dilute the sample with additional solvent.. Answer: D.
By adding solvent, the concentration of glucose will be lowered, and
the resulting absorbance will fall within the range of the standards.
This is easily accomplished, and the resulting calculations that
account for the dilution are not difficult.
◉ Which of the following reasons best explains why it is possible to
separate a 1:1 mixture of 1-chlorobutane and 1-butanol with fractional
distillation?
A. Both 1-chlorobutane and 1-butanol are polar
B. Both 1-chlorobutane and 1-butanol are nonpolar
C. The boiling point of 1-chlorobutane is substantially higher than
that of 1-butanol
