AAMC MCAT test 3r a

Detailed answers to practice test 3R by AAMC
Physical Science Part 1-77
1. Reaction 4 is shown in the following equation, which is answer choice A.
Answers B and D do not show a reaction involving PbCO3 (s), as required by Reaction 4. Answer C shows an
implausible and unbalanced equation. Thus, answer choice A is the best answer.

2. Reaction 1 is shown in the following equation.
Compound A, the white solid, is PbSO4 (s ). Neither the reactant Pb(NO3)2 nor the product NaNO3 can precipitate
because all nitrates and sodium salts are water soluble. PbI2 cannot precipitate because iodide is not present. Thus,
answer choice D is the best answer.

3. The dissolution of Pb(OH)2(s ) is represented by the following equation.

At pH 9, the concentration of OH-(aq) is greater than the concentration of OH-(aq) at pH 7. According to Le
Châtelier's principle, the additional common ion, OH-(aq ), will shift the position of equilibrium to the left, and less Pb
(OH)2 will dissolve. Thus, answer choice A is the best answer.

4. The reactions described in the passage show that lead(II) is successively precipitated as PbSO4, PbI2 , and
PbCO3 . This sequence shows (assuming equal anion concentrations, as must be done here) that PbCO3 is
less soluble than PbI2 , and PbI2 is less soluble than PbSO4 . The order in which the anions precipitate Pb2+
is: CO32- then I- then SO42-. When this sequence is applied to the question, answer choice B is in the correct
order, and answers A, C, and D are all in the opposite order. Thus, answer choice B is the best answer.


5.

6. The liquid and vapor phases coalesce at point D of Figure 2, where the densities of liquid and gaseous CO2
are equal. Thus, answer choice D is the best answer.

7. The question does not compare CO2 to a specific solvent, so we are looking for an inherent property of CO2

that makes it a good solvent for an organic oil. Supercritical CO2 is similar to a liquid and can be easily
removed by evaporation because it changes into a gas when the pressure is lowered. Answers A,B, and C are
not true of CO2, and answers A and C are not desirable properties of an extraction solvent. Thus, answer
choice D is the best answer.

8. Polar water molecules are held together by relatively strong hydrogen bonds; whereas, the linear, nonpolar
molecules of CO2 are held together at room temperature by weak London dispersion forces. Thus, answer
choice D is the best answer.

9. The critical point, shown as a dot (·) in Figure 1, is near 30oC and 80 atm. Answer choice C is the best
answer.

10.

According to the principle of “like dissolves like,” the covalent compound CO2 is a better solvent for a
covalent compound than it is for an ionic compound. Diethyl ether, C2H5OC2H5, is a covalent compound and
NaCl, NH4NO3, and KOH are ionic compounds. Thus, answer choice B is the best answer.

11.

The emission peaks P1 and P2 are described in the passage as due to an electron from an outer energy
level filling a vacant inner energy level, resulting in emission of an X-ray photon. These photons have discrete
energies, and therefore discrete wavelengths, so they appear in the spectrum as peaks. Thus, answer choice C
is the best answer.

The power P , supplied by the battery to accelerate the electron beam is given by the formula P = I
·V , where I is the beam current and
V is the potential difference between the cathode and anode.
-3
5

Therefore P = (5 x 10 A) x (10 V) = 5 x 102 W.
Therefore, answer choice A is the best answer.

12.

13.

14.

The emitted X- ray has the positive energy difference between the atomic energy levels as measured by
their ionization potentials. For Pb n = 2 to 1, that is (1,400 x 10-17 J) - (240 x10-17 J) = 1160 x 10-17 J =
1.16 x 10-14 J. Thus, answer choice C is the best answer.
According to the passage, bremsstrahlung is produced when electrons are accelerated during collisions


with ions. All the choices of answers are ions except He, a neutral atom. Therefore, answer choice A is the
correct answer.

15.

To increase the kinetic energy of the electrons, they must be accelerated by a higher voltage between
the cathode and anode, thus the voltage of HV was increased. Thus, answer choice A is the best answer.

16.

The probability of an X-ray emission event at a given wavelength is measured by its intensity in the
spectrum. In Figure 2, P2 has a higher intensity than does P1. Thus, answer choice D is the best answer.

17.

The dipole moment of a molecule is the vector sum of all of the bond moments. According to the data
in Table 1, the dipole moment of SnBr4 is zero; therefore, its bond moments add to zero or cancel. Thus,
answer choice D is the best answer.

18.

The

σ2s electron cloud in NO is in a bonding molecular orbital (MO) that forms by the overlap of the

2s orbital of an oxygen atom with the 2s orbital of a nitrogen atom. Because O is more electronegative than
N, the electron cloud in the resulting σ
MO is larger around the oxygen atom than it is around the nitrogen

2s

atom. Thus, answer choice D is the best answer.

19.

Table 1 gives the dipole moment of HF as 1.82 D. Chlorine is just below fluorine in the periodic table;
therefore, the electronegativity of chlorine, though significant, is less than that of fluorine. Chlorine is less
effective than fluorine in creating a separation of charge when bonded to hydrogen, and the dipole moment of
HCl is slightly less than that of HF. Thus, answer choice B is the best answer.

20.

HCl is polar covalent because H and Cl share a pair of bonded electrons that are more strongly attracted
to the chlorine atom. The higher effective nuclear charge (i.e., the charge of the nucleus minus the shielding
caused by extranuclear electrons) of chlorine accounts for its greater electronegativity. Thus, answer choice D

is the best answer.

21.

Carbon dioxide, O=C=O, is linear. Therefore, the two CO dipoles cancel because they are in opposite
directions. If one of the oxygen atoms is removed, the resulting CO will have a dipole because the species is
linear and comprised of two different atoms. Thus, the dipole moment will change from zero in CO2 to a
positive value in CO. Thus, answer choice D is the best answer.

22.

An analysis of the two structures shows that the bond moments in PCl5 add to zero; whereas, those in
PCl3 do not. As shown in the figure, PCl3 is pyramidal not planar. Thus, answer choice B is the best answer.

23.

Work is the product of the force on an object and the distance the object moves in the direction of the
applied force. In this case, work = 20 N x 10 m = 200 J. Thus, answer choice C is the best answer.

24.

Evaporation occurs when a molecule attains sufficient speed or kinetic energy to overcome the
attractive forces of a liquid. Resonance, surface tension and potential energy all relate to molecules that are
not in motion. Thus, answer choice B is the best answer.

The relation between distance, acceleration, and time is:
.
To solve for the time it takes the runner to use
t=(2 d /a )1/2=(2 * 3/1.5)1/2=2s.
Therefore, answer choice C is the best answer.

25.

d=(1/2)a·t2

26.

A body is in transitional equilibrium when the components of all external forces cancel. For the sheet:
Fcosθ=4N, Fsinθ =3N. The magnitude of F is found by adding the squares of the components: F2cos2θ + F2


sin2θ =F2=4 2 + 3 2=25N2.
Therefore F=5N. The F vector points in the proper direction since tanθ =0.75= 3/4.
Thus, answer choice C is the best answer.

27.

The only experimental difference in Trial 1 vs. Trial 2 is that, in Trial 2, the test tube is placed in water
to cool rather than in air (also 20oC). In other words, only the surroundings were different. Thus,
answer choice B is the best answer.

28.

The melting point of acetamide is 80oC; therefore, acetamide will melt when it is in a test tube that is
placed in a water bath at 90oC. The temperature of the water in the bath, not the amount of water in the bath,
determines whether or not the acetamide will melt. The period of time for acetamide to melt, starting at 90oC,
is more than the corresponding period, starting at 100oC (i.e., the temperature of boiling water). Thus, answer
choice D is the best answer.

29.

Without controlling the temperature (i.e., raising the temperature of the water bath above 80oC), the
experimenter could not have observed melting or freezing. Without monitoring the time, the experimenter
could not have determined the period of time for the samples to melt or freeze. The temperature of melting
(freezing) of a pure substance such as acetamide is independent of the amount melted, and Experiment 2
shows that the surroundings control the period of time for freezing to occur. Thus, answer choice A is the best
answer.

30.

The time period of melting is independent of the time intervals used by the experimenter to record
temperatures. The sample would freeze completely after 23 min regardless of the time interval used by the
experimenter to record temperatures. Thus, answer choice B is the best answer.

31.

After Experiment 1, the sample was removed from a hot water bath as a liquid. Subsequently, the
sample froze during Trial 1. Therefore, the sample had to be reheated in a water bath above its melting point
to start Trial 2 as a liquid. Thus, answer D is the best answer.

32.

If the data for Trial 1 were plotted, the temperature would drop to 80oC and remain at this melting
temperature for 23 min (or 23 min x 60 sec/min = 1380 sec). The line at 80oC would not slope downward at
all in the figure, and it would extend well past 270 sec, the maximum time shown in the figure. Thus, answer
choice A is the best answer.

33.

The tone with the shortest period has the shortest wavelength. In Figure 1a, the period of the third

harmonic (the curve with the smaller dashes) is seen to be shorter than the other two harmonics. Thus, answer
choice C is the best answer.

34.

The three curves in Figure 1a intersect at three points in time. The second intersection occurs in the
middle of the time axis. At that point all three curves have zero displacement. Therefore, answer choice C is
the best answer.

35.

The period T and frequency f of a tone are related by T=1/ f. If the first harmonic has a frequency of
100 Hz, then the second harmonic has a frequency of 200 Hz. The period corresponding to 200 Hz is 1/200 s1=0.005 s. Thus, answer choice A is the best answer.

36.

The amplitudes of the three harmonics can be compared in Figure 1a. The first harmonic is seen to be
largest while the other two have equal amplitudes. Answer choice A best represents these observations.

37.

A fourth harmonic would have a shorter period than the other three. Since T=1/ f, the fourth harmonic
would have a higher frequency than the third harmonic. Therefore, answer choice D is the best answer.

(20oC)


38.

The waveform in Figure 1c begins to repeat at the zero displacement point near the end of the time

axis. This is the same time period as the first harmonic as seen in Figure 1a. Thus, answer choice A is the best
answer.

39.

The relation of wavelength, frequency, and wave velocity is λ ·f=v. For light, v=3 x 108m/s. The
wavelength for Material B at an efficiency of 0.42 is read from Table 1 as λ =1.06 x 10-6m. The frequency of
this light is 3 x 108 m/s divided by 1.06 x 10-6 m, giving 2.8 x 1014Hz. Thus, answer choice C is the best
answer.

40.

The passage states "a coating that maximized the absorption of light." Therefore, the coating also
maximizes the conversion efficiency. Thus answer choice C is the best answer.

41.

Identical voltage sources connected in parallel produce the same output voltage as a single source.
(Whereas if they were connected in series, the source voltages would be added.) Therefore, answer choice D
is the best answer.

42.

The equation K=hf –ø given in the passage implies that the photon energy must be greater than the
work function of the material in order to liberate an electron, i.e. for K to be positive. Thus, answer choice A
is the best choice.

43.

Applying a coating that makes ε independent of the wavelength means ε is the same for all λ . Thus,

ε is constant when plotted versus λ , a horizontal line. Therefore, answer choice C is the best answer.

44.

According to the data in Table 1, both structure and molecular weight (i.e., molar mass) affect the
melting point of a compound. In order to assess the effect of molecular weight or mass alone, any other effects
such as obvious structural differences must be minimized. This is best done by comparing two compounds
that are structurally similar. Because the structures of propionic acid and butyric acid (Answer C) differ by
only a CH2 group, they best show that melting point increases with molar mass. All of the other answer
choices compare two compounds that differ significantly in structure. Therefore, the melting points of these
compounds include both molar mass and structural effects. Thus, answer choice C is the best answer.

45.

Figure 1 shows the pH of the solution to be about 3 before any NaOH(aq) is added.

pH=- log[H3O+]
3 =-log[H3O+]
[H3O+]=10-3 M=0.001 M= Answer A

46.

The freezing point depression of an aqueous solution is a colligative property (i.e., it depends on the
number of solute particles in a given volume of water.) Given two solutions, the one with the greater number
of solute particles per liter of solution freezes at the lower temperature. Answer C is the only answer that
relates a larger number of solute particles directly to a lower freezing point. Oxalic acid is diprotic and ionizes
in accord with the pKa values in Table 1 to a greater extent than does crotonic acid. Subsequently, oxalic acid
requires more NaOH than does crotonic acid to reach a pH of 4.7, and oxalic acid produces a larger number of
particles in solution.Thus, answer choice C is the best answer.

47.

In a titration of R–COOH, the concentrations of R–COOH and R–COO- are equal at the mid-point of
the titration. This is often called the half-equivalence point. From the expression for the equilibrium constant
of a weak acid HA, when [HA]=[A-], then [H3O+]=Ka and pH=pKa. Table 1 shows the pKa value for a


monoprotic acid to be 4.69 – 4.88. Answer choice A (4.8) lies in this range, the other choices do not.
Alternatively, Figure 1 shows the pH at the half equivalence point of a weak acid to be about 4.8. Thus,
answer choice A is the best answer.

48.

The first sentence of the passage states that the unknown “was a liquid at room temperature (20oC).”
Table 1 shows that the melting point of crotonic acid is 71.6oC, which means it is a solid at room temperature
(i.e., it melts 51.6oC above room temperature). Thus, answer choice C is the best answer.

49.

In general, catalysts lower the activation energy of the slowest step in a reaction. Thus, they increase
the rate of the reaction without increasing the number of collisions, the kinetic energy of the reactants, or the
Keq of a reversible reaction. Thus, answer choice C is the best answer.

50.

Conservation of linear momentum requires: mradon vradon=mhelium vhelium with helium identified
as the alpha particle. The nuclear masses can be approximated by their mass numbers (222 and 4). Thus, the
recoil speed of the radon is (4/222) * 1.5 x 107 m/s= 2.7 x 105 m/s. Therefore, answer choice B is the best
answer.

51.

52.

The overall order of a reaction is the sum of the exponents of the concentrations in the rate law. The
exponent of [NO2] is 1 and that of [F2] is 1, and their sum is 1+1=2. Thus, the overall order is two or second
order. Thus, answer choice C is the best answer.
The problem gives ∆Hfo for HCl as -92.5 kJ/mol. This means that the formation of one mole of HCl
from its elements liberates 92.5 kJ of heat, as shown in the following equation.

Therefore, the formation of two moles of HCl liberates twice this amount or –185.0 kJ.

The question asks for the enthalpy change (∆H) for the reverse reaction. When the reaction is reversed, the sign of
∆H is changed from –to +. Thus, the reverse reaction requires +185.0 kJ = Answer D.

53.

The ratio of object to image distance equals the ratio of object to image height. The ratio of image to
object height is found by rearranging the ratios to give 4f/(4/3)f=1/3. The image is demagnified by a factor of
3. Thus, answer choice A is the best answer.

54.

In the artery, the product of blood speed and artery cross-sectional area is everywhere constant because
the volume flow rates are equal. At the point with half the normal area, the speed must double so that the same
volume of blood passes through the constriction as does through the normal part of the artery. Thus, answer
choice C is the best answer.

55.

A stronger B field increases the magnetic force, Fm = q v B┴. The electric force must also increase to
achieve equilibrium. This implies a larger electric field in the artery and a larger voltage across the artery.
Thus, answer choice C is the best answer.

56.

Volume flow rate is the product of blood speed and artery cross-sectional area: (0.20 m/s)·( л /4) ·
(1.0 x 10-2 m)2 = 5л x 10-6 m3/s. Therefore, answer choice B is the best answer.


57.

Since volume flow rate is proportional to blood speed, it doubles when v doubles. Therefore, answer
choice B is the best answer.

58.

A magnetic force acts on a moving charge in a direction that is perpendicular to both the velocity of the
charge and the direction of the magnetic field. This is a basic law of the interaction of electric currents and
magnetic fields. Thus, answer choice D is the best answer.

59.

The passage states that sulfuric acid reacts with Cu(s) to produce Cu+ and SO2. Thus, sulfuric acid is
converted into sulfur dioxide, or H2SO4→SO2. The oxidation number of sulfur in H2SO4 can be found by
assigning oxidation numbers of +1 for hydrogen and -2 for oxygen. For the formula H2SO4 to be neutral, the
sum of the oxidation numbers must be zero. If x is the oxidation number of sulfur in H2SO4, then: 2(1) + 4(2) + x=0, and x=+6. Likewise, for SO2: 2( - 2) + x=0, and x=+4. The change in oxidation number is from +6
to +4, which is answer choice B.

60.

The boiling point of HNO3 is given in the question as 86oC. Because HNO3 must boil out of the flask
and be trapped in the tube, the temperature of the flask must be above the boiling point of HNO3 (i.e., <86oC)
and the temperature of the tube must be less than the boiling point of HNO3 (i.e, <86oC). Answer B meets
these criteria; the other answers do not. Thus, answer choice B is the best answer.

61.

62.

The combustion of elemental sulfur involves a reaction between oxygen (O2) and sulfur (S). [Note:
Though sulfur exists as S8 molecules, its reactions are normally written in terms of its empirical formula S.]
Only Answer D shows such a reaction. Thus, answer choice D is the best answer.
The reaction involves the formation of gaseous SO3 from gaseous O2 and gaseous SO2

According to Le Châtelier's principle, any action that causes the reaction to shift toward the right will cause O2 and
SO2 to react and increase the yield of SO3 Of the four possible actions (answers A-D), the removal of SO3 as it forms
will shift the reaction toward the right and is the most likely action to increase the yield of SO3. Thus, answer C is the
best answer.

63.

64.

The passage states that concentrated H2SO4 is 98% H2SO4 and 2% water by mass and that the density
of concentrated H2SO4 is 1.84 g/mL.

The problem hypothesizes the liberation of hydrogen in accord with the unbalanced equation below.
Fe(s) + H2SO4 → H2(g)

In the conversion, hydrogen goes from an oxidation state of +1 in sulfuric acid to 0 (zero) in H2. Thus, the hydrogen in
H2SO4 is reduced (i.e., it undergoes an algebraic decrease in oxidation state). Fe(s) is the reducing agent (i.e., it causes
the reduction). Thus, answer choice A is the best answer.


65.

The first ionization of sulfuric acid, H2SO4, is normally 100% in water. However, under conditions of
low water content, all of the H2SO4 cannot ionize. Qualitatively, the only source of SO42- is HSO4-. The Ka
for the second ionization step of a parent acid is a few orders of magnitude smaller than that of the first step;
therefore, SO4-2 must be the least abundant species, because it is only formed in the second ionization step.
Thus, answer choice A is the best answer.

Quantitatively, the mass relationship in 100g of 98% H2SO4 is 98g H2SO4 and 2g H2O. The molar mass of H2SO4 is
98 g/mol and of water is 18 g/mol. Thus, in 100 g of 98% H2SO4 there is one mole of H2SO4 and 2/18 or 1/9 mole of
H2O. In excess water, H2SO4 would ionize completely. However, in this case (i.e., very low water content), only about
1/9 of a mole can react stoichiometrically with water to form H3O+ and HSO4-. Of this 1/9 mol, only a small fraction of
the HSO4- further ionizes to H3O+ and SO4-2, because HSO4-2 is a weak acid. Therefore, SO4-2 is the chemical with
the lowest concentration. Thus, answer choice A is the best answer.

66.

The reflected radar signal has a frequency shift due to the Doppler effect. Mercury must have been
moving toward the Earth for the shift to be to a higher frequency. Thus, answer choice A is the best answer.

67.

An equatorial bulge of the Sun would be caused by some process that is not the same in all directions.
Of the four answers, only the Sun's rotation about an axis fulfills this condition. The bulge is related to the
balance between the centripetal force and the gravitational force. Therefore, answer choice A is the best

answer.

68.

The perihelion will have moved through an angle θ=Ω·t after a time t has elapsed. The conversion of
angles is given by the following. An arcsecond is 1/60 of an arcminute. An arcminute is 1/60 of a degree. The
time for 500 arcseconds per century to accumulate to θ=360 degrees is 360/(500/(60 x 60)) = 360 x 60 x
(60/500) centuries. Thus, answer choice C is the correct answer.

69.

The orbit of Mercury around the Sun is, by Kepler's and Newton's laws, an ellipse with the Sun at one
fixed focus. Therefore, answer choice C is the best answer.

70.

The time for the radar signal to travel from the Earth to Venus and return is the total distance divided by
the velocity of light: 2 x 5 x 1010 m / 3 x 108m/s = 3.3 x 102s = 300 s. Thus, answer choice B is the correct
answer.

71.

The ellipse precesses around the Sun at one focus. Any fixed point on the ellipse is a constant distance
from the focus. Therefore it traces out a circle around the focus. Answer choice A is the best answer.

72.

As written, the equation shows the reduction of Zn2+ to Zn. The negative reduction potential for Zn2+
means that Zn has a positive oxidation potential (i.e., Zn is easily oxidized.) Zinc will displace hydrogen,
which has a zero standard reduction or oxidation potential (i.e., hydrogen is the standard against which other

substances are measured.). As given in the problem, ionic zinc is Zn2+. Therefore, zinc metal liberates
hydrogen gas and produces ZnCl2(aq). Answer choice B is the best answer.

73.

Power is defined as the rate of doing work. For the automobile, the power output is the amount of work
done (overcoming friction) divided by the length of time in which the work was done. Therefore, answer
choice D is the best answer.

74.

The bonds labeled C and D in the figure are of equal length but shorter than bond B. This is because
two resonance structures can be drawn: in one resonance structure, bond C is a double bond, and in the second
resonance structure, bond D is double bond. Double bonds (bond order = 2) are shorter than single bonds
(bond order = 1), so the bond order for bonds C and D is about 1.5 and for bond B about 1. Bond B is longer
than bond A because of the small atomic radius of hydrogen compared to nitrogen. Thus, answer choice B is


the best answer.
Alternatively, the figure shows that C = D, so answer choices C and D can be eliminated. Also, A is clearly
shorter than B; therefore B is the longest. Thus, answer choice B is the best answer.

75.

Conservation of energy requires that the 15.0 eV photon energy first provides the ionization energy to
unbind the electron, and then allows any excess energy to become the electron's kinetic energy. The kinetic
energy is this case is 15.0eV - 13.6 eV = 1.4 eV. Thus the correct answer is A.

76.

In radioactive decay, the sum of the mass numbers A and atomic numbers Z, before and after decay,
must balance. The numbers for beryllium undergoing positron decay are: mass (7= 7 + 0) and atomic (4 = 3 +
1). The resulting nucleus is 73Li. Thus, answer choice B is the best answer.

77.

Work is the product of force and distance. The easiest way to calculate the work in this pulley problem
is to multiply the net force on the weight mg by the distance it is raised: 4 kg x 10 m/s2 x 5m = 200J.
Therefore, answer choice D is the best answer.

Verbal Reasoning Section of Practice test 3R by AAMC:
78.


79.

80.


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84.

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92.

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