AAMC Exam 4 FL Review with Complete Solutions
Limestone does NOT decompose when heated D.
to 900 K because, at 900 K, ΔH is: No; it remained at +1. - -Missed, C.
A. No; it remained at -3., involves nitrogen is the
positive and less than TΔS. protonation of ammonia (NH3 + H+ → NH4+).
B. Acid-base reactions do not involve oxidation state
positive and greater than TΔS. changes. The oxidation state of N in NH3 is -3.
C. Each H is +1 and is balanced by the -3 of N to
negative and less than TΔS. make a neutral compound. The oxidation state of
D. N does not change when the N is protonated
negative and greater than TΔS. - -B.
positive and greater than TΔS., the reaction does
not occur, ΔG = ΔH - TΔS > 0. From inspection If all of Gas X (from Step 6) is held in a sealed
of the reaction, it can be concluded that ΔS > 0 chamber at STP, what will be its approximate
(a gas evolves). Consequently, ΔH > TΔS volume?
explains why the reaction does not occur A.
22.4 L
B.
When limestone is heated during Step 1, an 44.8 L
equilibrium is established. Which of the following C.
expressions is the equilibrium constant for the 67.2 L
decomposition of limestone? D.
A. 89.6 L - -A.
[CaO] 22.4 L, The quantity of Gas X was given as 1
B. mole. One mole of gas occupies 22.4 L at STP.
[CaCO3]
C.
[CO2] Why was it important that the cuvettes containing
D. the glucose oxidase and the blood sample were
[CaO] × [CaCO3] - -C. identical in terms of optical properties?
[CO2], solids are excluded from equilibrium A.
constant expressions. CO2(g), as the only non- To enable the comparison of the absorption
solid material in the reaction, is the only spectra
substance that appears in the equilibrium B.
constant expression. The exponent for [CO2] is 1 To reduce the absorption in the glass walls
because the stoichiometric coefficient that C.
appears in the presented balanced reaction is 1 To decrease the uncertainty in the wavelength
D.
To increase the absorption in the solutions -
During Reaction 2, did the oxidation state of N -A.
change? To enable the comparison of the absorption
A. spectra, absorbed radiation is due only to if
Yes; it changed from -3 to -4. glucose is in the blood and not due to the
B. difference in the absorption features of the walls
Yes; it changed from 0 to +1.
C.
No; it remained at -3. What is the approximate energy of a photon in
,AAMC Exam 4 FL Review with Complete Solutions
the absorbed radiation that yielded the data in D.
Table 1? Dilute the sample with additional solvent. -
(Note: Use 1 eV = 1.6 × 10-19 J and hc = 19.8 × -D.
10-26 J•m.) Dilute the sample with additional solvent., adding
A. solvent, the concentration of glucose will be
1 eV lowered, and the resulting absorbance will fall
B. within the range of the standards
2 eV
C.
3 eV Which of the following reasons best explains why
D. it is possible to separate a 1:1 mixture of 1-
4 eV - -B. chlorobutane and 1-butanol by fractional
2 eV, The photon energy is E = hc/λ = 19.8 × 10- distillation?
26 J•m/(625 × 10-9 m) = 3.1 × 10-19 J ≅ 2 eV. A.
Both 1-chlorobutane and 1-butanol are polar.
B.
According to Table 1, what is the concentration Both 1-chlorobutane and 1-butanol are nonpolar.
of the glucose in the blood from which the diluted C.
sample was taken? The boiling point of 1-chlorobutane is
A. substantially higher than that of 1-butanol.
60 mg/dL D.
B. The boiling point of 1-chlorobutane is
90 mg/dL substantially lower than that of 1-butanol. -
C. -D.
120 mg/dL The boiling point of 1-chlorobutane is
D. substantially lower than that of 1-butanol., The
150 mg/dL - -D. molecules have similar molecular weights, but 1-
150 mg/dL, From Table 1, the glucose butanol has a hydroxyl functional group that can
concentration in the diluted sample is (0.20/0.24) participate in hydrogen bonding. Hydrogen
× 6.0 mg/dL = 5.0 mg/dL. The blood then has a bonding is a particularly strong force of
glucose concentration of 30 × 5.0 mg/dL = 150 intermolecular attraction, making its boiling point
mg/dL., multiplied by 30 bc 1/30 dilution ratio higher
Suppose a blood sample tested above the range Which of the following oxidative transformations
(6.0 mg/dL) of the standards used in the is unlikely to occur?
experiment. What modification will provide a A.
more precise reading by data interpolation as A primary alcohol to an aldehyde
opposed to extrapolation using the same B.
standards? A tertiary alcohol to a ketone
A. C.
Increase the enzyme concentration. An aldehyde to a carboxylic acid
B. D.
Increase the oxygen pressure. A secondary alcohol to a ketone - -B.
C. A tertiary alcohol to a ketone, Oxidation of tertiary
Decrease the content of oxygen acceptor. alcohols is difficult because it involves C-C bond
,AAMC Exam 4 FL Review with Complete Solutions
breaking. field in the electrical discharge produced in the
excimer laser tube?
A.
Two vectors of magnitudes |A| = 8 units and |B| = 2.0 × 106 V/m
5 units make an angle that can vary from 0° to B.
180°. The magnitude of the resultant vector A + 4.0 × 105 V/m
B CANNOT have the value of: C.
A. 6.0 × 104 V/m
2 units. D.
B. 8.0 × 103 V/m - -A.
5 units. 2.0 × 106 V/m, 8.0 kV across 4.0 mm is 8000
C. V/0.004 m = 2.0 × 106 V/m.
8 units.
D.
12 units. - -A. What is the frequency of the pulses that deliver
2 units, The magnitude of A + B is as small as 3 laser radiation to the cornea?
units (when A and B are anti-parallel and make A.
an angle of 180°) and as large as 13 units (when 0.4 Hz
A and B are parallel and make an angle of 0°). B.
The magnitude of 2 units is smaller than the 4.0 Hz
smallest possible magnitude of vector A + B. C.
25 Hz
D.
What is the effect produced by the PRK 250 Hz - -B.
technique designed to correct nearsightedness? 4.0 Hz, The frequency is 1/(250 ms) = 4 Hz.
A. 1/period
The density of the cornea is increased.
B.
The radius of curvature of the cornea is The use of pulsed laser radiation in the PRK
increased. procedure, as opposed to continuous laser
C. radiation, allows the cornea to:
The index of refraction of the cornea is A.
increased. absorb more radiation.
D. B.
The thickness of the cornea at the apex is change its index of refraction.
increased. C.
Solution - -B. increase the area exposed to radiation.
The radius of curvature of the cornea is D.
increased., According to the passage, to correct maintain a lower average temperature. - -
nearsightedness, the laser beam is directed onto D.
the central part of the cornea, resulting in a maintain a lower average temperature, pulsed
flattening of the cornea. This means that the laser radiation interacts with the cornea for a
radius of curvature of the cornea is increased. shorter time than a continuous laser radiation,
thus less heat is transferred to the cornea. This
allows the cornea to maintain a lower average
WAS iffy: What is the magnitude of the electric temperature by cooling off between pulses.
, AAMC Exam 4 FL Review with Complete Solutions
Plugging in 10 M for [NH3], which is a good
approximation since very little NH3 ionizes, and
The intermolecular forces that exist among the noting that [NH4+] = [OH-], it is possible to solve
molecules of NH3 gas are: for [NH4+]: 1.76 × 10-5 = [NH4+]2/10 → 1.76 ×
A. 10-4 = [NH4+]2 → 1.32 × 10-2 = [NH4+].
dipole-dipole forces only.
B.
London dispersion forces only. If methane (CH4) were substituted for NH3 in the
C. aluminum can, the crushing of the can would:
both dipole-dipole and London dispersion forces. A.
D. occur because CH4, being polar, would dissolve
neither dipole-dipole nor London dispersion in the water in the tray.
forces. - -C. B.
both dipole-dipole and London dispersion occur because CH4, being nonpolar, would
forces., Since NH3 is a permanent dipole, it will dissolve in the water in the tray.
exhibit dipole-dipole intermolecular forces in C.
addition to the London dispersion forces not occur because CH4, being polar, would not
exhibited by all molecules dissolve in the water in the tray.
D.
not occur because CH4, being nonpolar, would
dipole-dipole forces - -attractions between not dissolve in the water in the tray. - -D.
oppositely charged regions of polar molecules not occur because CH4, being nonpolar, would
not dissolve in the water in the tray., CH4 is a
nonpolar molecule and will not dissolve
London dispersion forces - -the appreciably in water. Since the CH4 will remain in
intermolecular attraction resulting from the the gas phase, the can will not be crushed by the
uneven distribution of electrons and the creation formation of a vacuum inside the can.
of temporary dipoles
Ca(OH)2 and CaCl2 in the reaction shown in
NH3 acts as a weak base in water with Kb = 1.76 Equation 1 are best described as compounds of:
× 10-5 at 25°C. The corresponding equilibrium is A.
shown below. an alkali metal.
NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq) B.
At 25°C, the equilibrium concentration of the an alkaline earth metal.
NH4+ ion in a 10 M aqueous solution of NH3 C.
would be closest to which of the following? a metalloid.
A. D.
0.001 M a transition metal.
B. Solution - -B.
0.01 M an alkaline earth metal., Group 2 of the Periodic
C. Table. Calcium, Ca, is a member of this family
0.1 M found in period 4
D.
1M- -C.
0.1 M, Kb = 1.76 × 10-5 = [NH4+][OH-]/[NH3]. Which conclusion can be drawn from the
Limestone does NOT decompose when heated D.
to 900 K because, at 900 K, ΔH is: No; it remained at +1. - -Missed, C.
A. No; it remained at -3., involves nitrogen is the
positive and less than TΔS. protonation of ammonia (NH3 + H+ → NH4+).
B. Acid-base reactions do not involve oxidation state
positive and greater than TΔS. changes. The oxidation state of N in NH3 is -3.
C. Each H is +1 and is balanced by the -3 of N to
negative and less than TΔS. make a neutral compound. The oxidation state of
D. N does not change when the N is protonated
negative and greater than TΔS. - -B.
positive and greater than TΔS., the reaction does
not occur, ΔG = ΔH - TΔS > 0. From inspection If all of Gas X (from Step 6) is held in a sealed
of the reaction, it can be concluded that ΔS > 0 chamber at STP, what will be its approximate
(a gas evolves). Consequently, ΔH > TΔS volume?
explains why the reaction does not occur A.
22.4 L
B.
When limestone is heated during Step 1, an 44.8 L
equilibrium is established. Which of the following C.
expressions is the equilibrium constant for the 67.2 L
decomposition of limestone? D.
A. 89.6 L - -A.
[CaO] 22.4 L, The quantity of Gas X was given as 1
B. mole. One mole of gas occupies 22.4 L at STP.
[CaCO3]
C.
[CO2] Why was it important that the cuvettes containing
D. the glucose oxidase and the blood sample were
[CaO] × [CaCO3] - -C. identical in terms of optical properties?
[CO2], solids are excluded from equilibrium A.
constant expressions. CO2(g), as the only non- To enable the comparison of the absorption
solid material in the reaction, is the only spectra
substance that appears in the equilibrium B.
constant expression. The exponent for [CO2] is 1 To reduce the absorption in the glass walls
because the stoichiometric coefficient that C.
appears in the presented balanced reaction is 1 To decrease the uncertainty in the wavelength
D.
To increase the absorption in the solutions -
During Reaction 2, did the oxidation state of N -A.
change? To enable the comparison of the absorption
A. spectra, absorbed radiation is due only to if
Yes; it changed from -3 to -4. glucose is in the blood and not due to the
B. difference in the absorption features of the walls
Yes; it changed from 0 to +1.
C.
No; it remained at -3. What is the approximate energy of a photon in
,AAMC Exam 4 FL Review with Complete Solutions
the absorbed radiation that yielded the data in D.
Table 1? Dilute the sample with additional solvent. -
(Note: Use 1 eV = 1.6 × 10-19 J and hc = 19.8 × -D.
10-26 J•m.) Dilute the sample with additional solvent., adding
A. solvent, the concentration of glucose will be
1 eV lowered, and the resulting absorbance will fall
B. within the range of the standards
2 eV
C.
3 eV Which of the following reasons best explains why
D. it is possible to separate a 1:1 mixture of 1-
4 eV - -B. chlorobutane and 1-butanol by fractional
2 eV, The photon energy is E = hc/λ = 19.8 × 10- distillation?
26 J•m/(625 × 10-9 m) = 3.1 × 10-19 J ≅ 2 eV. A.
Both 1-chlorobutane and 1-butanol are polar.
B.
According to Table 1, what is the concentration Both 1-chlorobutane and 1-butanol are nonpolar.
of the glucose in the blood from which the diluted C.
sample was taken? The boiling point of 1-chlorobutane is
A. substantially higher than that of 1-butanol.
60 mg/dL D.
B. The boiling point of 1-chlorobutane is
90 mg/dL substantially lower than that of 1-butanol. -
C. -D.
120 mg/dL The boiling point of 1-chlorobutane is
D. substantially lower than that of 1-butanol., The
150 mg/dL - -D. molecules have similar molecular weights, but 1-
150 mg/dL, From Table 1, the glucose butanol has a hydroxyl functional group that can
concentration in the diluted sample is (0.20/0.24) participate in hydrogen bonding. Hydrogen
× 6.0 mg/dL = 5.0 mg/dL. The blood then has a bonding is a particularly strong force of
glucose concentration of 30 × 5.0 mg/dL = 150 intermolecular attraction, making its boiling point
mg/dL., multiplied by 30 bc 1/30 dilution ratio higher
Suppose a blood sample tested above the range Which of the following oxidative transformations
(6.0 mg/dL) of the standards used in the is unlikely to occur?
experiment. What modification will provide a A.
more precise reading by data interpolation as A primary alcohol to an aldehyde
opposed to extrapolation using the same B.
standards? A tertiary alcohol to a ketone
A. C.
Increase the enzyme concentration. An aldehyde to a carboxylic acid
B. D.
Increase the oxygen pressure. A secondary alcohol to a ketone - -B.
C. A tertiary alcohol to a ketone, Oxidation of tertiary
Decrease the content of oxygen acceptor. alcohols is difficult because it involves C-C bond
,AAMC Exam 4 FL Review with Complete Solutions
breaking. field in the electrical discharge produced in the
excimer laser tube?
A.
Two vectors of magnitudes |A| = 8 units and |B| = 2.0 × 106 V/m
5 units make an angle that can vary from 0° to B.
180°. The magnitude of the resultant vector A + 4.0 × 105 V/m
B CANNOT have the value of: C.
A. 6.0 × 104 V/m
2 units. D.
B. 8.0 × 103 V/m - -A.
5 units. 2.0 × 106 V/m, 8.0 kV across 4.0 mm is 8000
C. V/0.004 m = 2.0 × 106 V/m.
8 units.
D.
12 units. - -A. What is the frequency of the pulses that deliver
2 units, The magnitude of A + B is as small as 3 laser radiation to the cornea?
units (when A and B are anti-parallel and make A.
an angle of 180°) and as large as 13 units (when 0.4 Hz
A and B are parallel and make an angle of 0°). B.
The magnitude of 2 units is smaller than the 4.0 Hz
smallest possible magnitude of vector A + B. C.
25 Hz
D.
What is the effect produced by the PRK 250 Hz - -B.
technique designed to correct nearsightedness? 4.0 Hz, The frequency is 1/(250 ms) = 4 Hz.
A. 1/period
The density of the cornea is increased.
B.
The radius of curvature of the cornea is The use of pulsed laser radiation in the PRK
increased. procedure, as opposed to continuous laser
C. radiation, allows the cornea to:
The index of refraction of the cornea is A.
increased. absorb more radiation.
D. B.
The thickness of the cornea at the apex is change its index of refraction.
increased. C.
Solution - -B. increase the area exposed to radiation.
The radius of curvature of the cornea is D.
increased., According to the passage, to correct maintain a lower average temperature. - -
nearsightedness, the laser beam is directed onto D.
the central part of the cornea, resulting in a maintain a lower average temperature, pulsed
flattening of the cornea. This means that the laser radiation interacts with the cornea for a
radius of curvature of the cornea is increased. shorter time than a continuous laser radiation,
thus less heat is transferred to the cornea. This
allows the cornea to maintain a lower average
WAS iffy: What is the magnitude of the electric temperature by cooling off between pulses.
, AAMC Exam 4 FL Review with Complete Solutions
Plugging in 10 M for [NH3], which is a good
approximation since very little NH3 ionizes, and
The intermolecular forces that exist among the noting that [NH4+] = [OH-], it is possible to solve
molecules of NH3 gas are: for [NH4+]: 1.76 × 10-5 = [NH4+]2/10 → 1.76 ×
A. 10-4 = [NH4+]2 → 1.32 × 10-2 = [NH4+].
dipole-dipole forces only.
B.
London dispersion forces only. If methane (CH4) were substituted for NH3 in the
C. aluminum can, the crushing of the can would:
both dipole-dipole and London dispersion forces. A.
D. occur because CH4, being polar, would dissolve
neither dipole-dipole nor London dispersion in the water in the tray.
forces. - -C. B.
both dipole-dipole and London dispersion occur because CH4, being nonpolar, would
forces., Since NH3 is a permanent dipole, it will dissolve in the water in the tray.
exhibit dipole-dipole intermolecular forces in C.
addition to the London dispersion forces not occur because CH4, being polar, would not
exhibited by all molecules dissolve in the water in the tray.
D.
not occur because CH4, being nonpolar, would
dipole-dipole forces - -attractions between not dissolve in the water in the tray. - -D.
oppositely charged regions of polar molecules not occur because CH4, being nonpolar, would
not dissolve in the water in the tray., CH4 is a
nonpolar molecule and will not dissolve
London dispersion forces - -the appreciably in water. Since the CH4 will remain in
intermolecular attraction resulting from the the gas phase, the can will not be crushed by the
uneven distribution of electrons and the creation formation of a vacuum inside the can.
of temporary dipoles
Ca(OH)2 and CaCl2 in the reaction shown in
NH3 acts as a weak base in water with Kb = 1.76 Equation 1 are best described as compounds of:
× 10-5 at 25°C. The corresponding equilibrium is A.
shown below. an alkali metal.
NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq) B.
At 25°C, the equilibrium concentration of the an alkaline earth metal.
NH4+ ion in a 10 M aqueous solution of NH3 C.
would be closest to which of the following? a metalloid.
A. D.
0.001 M a transition metal.
B. Solution - -B.
0.01 M an alkaline earth metal., Group 2 of the Periodic
C. Table. Calcium, Ca, is a member of this family
0.1 M found in period 4
D.
1M- -C.
0.1 M, Kb = 1.76 × 10-5 = [NH4+][OH-]/[NH3]. Which conclusion can be drawn from the
