AAMC Exam 4 FL Review
Comprehensive Questions with
Verified Answers Graded A+
Limestone does NOT decompose when heated to 900 K because, at 900 K, ΔH is:
A.
positive and less than TΔS.
B.
positive and greater than TΔS.
C.
negative and less than TΔS.
D.
negative and greater than TΔS. - Answer: B.
positive and greater than TΔS., the reaction does not occur, 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
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] × [CaCO3] - Answer: C.
[CO2], 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. The exponent for [CO2] is 1 because the stoichiometric coefficient that appears in
the presented balanced reaction is 1
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: Missed, C.
No; it remained at -3., 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?
A.
22.4 L
B.
44.8 L
,C.
67.2 L
D.
89.6 L - Answer: A.
22.4 L, 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.
To enable the comparison of the absorption spectra, absorbed radiation is due only to if glucose
is 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.)
A.
1 eV
B.
2 eV
C.
, 3 eV
D.
4 eV - Answer: B.
2 eV, The photon energy is E = hc/λ = 19.8 × 10-26 J•m/(625 × 10-9 m) = 3.1 × 10-19 J ≅ 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.
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., multiplied by 30 bc 1/30 dilution ratio
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 oxygen acceptor.
Comprehensive Questions with
Verified Answers Graded A+
Limestone does NOT decompose when heated to 900 K because, at 900 K, ΔH is:
A.
positive and less than TΔS.
B.
positive and greater than TΔS.
C.
negative and less than TΔS.
D.
negative and greater than TΔS. - Answer: B.
positive and greater than TΔS., the reaction does not occur, 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
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] × [CaCO3] - Answer: C.
[CO2], 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. The exponent for [CO2] is 1 because the stoichiometric coefficient that appears in
the presented balanced reaction is 1
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: Missed, C.
No; it remained at -3., 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?
A.
22.4 L
B.
44.8 L
,C.
67.2 L
D.
89.6 L - Answer: A.
22.4 L, 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.
To enable the comparison of the absorption spectra, absorbed radiation is due only to if glucose
is 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.)
A.
1 eV
B.
2 eV
C.
, 3 eV
D.
4 eV - Answer: B.
2 eV, The photon energy is E = hc/λ = 19.8 × 10-26 J•m/(625 × 10-9 m) = 3.1 × 10-19 J ≅ 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.
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., multiplied by 30 bc 1/30 dilution ratio
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 oxygen acceptor.
