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Medical College Admission Test (MCAT) Physics Tutor.

Medical College Admission Test (MCAT) Physics Tutor:

I tutor college students in the subjects of MCAT Physics.

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My Name: Dr. Vasiliy S. Znamenskiy

MCAT Sample Questions : Physical Sciences

Passage I

A series of chemical reactions was carried out to study the chemistry of lead.

Reaction 1

Initially, 15.0 mL of 0.300 M Pb(NO3)2(aq) was mixed with 15.0 mL of 0.300 M Na2SO4(aq).All the Pb(NO3)2 reacted to form Compound A, a white precipitate.Compound A was removed by filtration.

Reaction 2

Next, 15.0 mL of 0.300 M KI(aq) was added to Compound A. The mixture was agitated and some of Compound A dissolved. In addition, a yellow precipitate of PbI2(s) was formed.

Reaction 3

The PbI2(s) was separated and mixed with 15.0 mL of 0.300 M Na2CO3(aq). A white precipitate of PbCO3(s) formed. All of the PbI2(s) was converted into PbCO3(s).

Reaction 4

The PbCO3(s) was removed by filtration and a small sample gave off a gas when treated with dilute HCl.

Some sample questions on this passage are as follows:

• Which of the following reactions depicts the formation of the gas in Reaction 4?

1. PbCO3(s) + 2 HCl(aq) = PbCl2(aq) + CO2(g) + H2O(l)

2. Na2CO3(aq) + 2 HCl(aq) = 2 NaCl(aq) + CO2(g) + H2O(l)

3. PbCO3(s) + 2 HCl(aq) = PbC2(s) + Cl2(g) + H2O(l)

4. PbI 2(s) + HCl(aq) = PbCl2(aq) + HI(g)

Explanation: Reaction 4 is shown in the following equation, which is answer choice A.

PbCO3(s) + 2 HCl(aq) = PbCl2(aq) + CO2(g) + H2O(l)

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.

• The identity of Compound A is:

1. Pb(NO3)2.

2. PbI2.

3. NaNO3.

4. PbSO4.

Explanation: Reaction 1 is shown in the following equation.

Pb(NO3)2(aq) + Na2SO4(aq) = PbSO4(s) + 2 NaNO3(aq)

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.

• Pb(OH)2(s) is slightly soluble in water. How would the amount of Pb(OH)2(s) that normally dissolves in 1 L of water be affected if the pH were 9.0?

1. Less would dissolve.

2. The same amount would dissolve.

3. More would dissolve.

4. There is no way to predict the effect of the change in pH of the water.

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

Pb(OH)2(s)=Pb2(aq) + 2 OH-(aq)

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.

• A soluble form of Pb2+can be carefully added to a solution tosequentially precipitate and separate anions present in the solution. When Pb2+is added, in what order will the following anions be precipitated?

1. SO42- then I-

2. CO32- then I-

3. SO42- then CO32-

4. I- then CO32-

Explanation: 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.

• How many moles of Na+ ions are there in the initial Na2SO4(aq) solution used in Reaction 1?

1. 0.0018 mole

2. 0.009 mole

3. 0.045 mole

4. 0.090 mole

Explanation: The initial Na2SO4(aq) solution in Reaction 1 is 15 mL of 0.300 MNa2SO4(aq).

(15.0 mL) ( 1 L ) (0.300 mol Na2SO4) (2 mol Na+)

(1000 mL) (1 L Na2SO4(aq) ) (1 mol Na2SO4)

= 0.00900 mol = Answer B

MCAT Sample Questions : Physical Sciences

Passage II

At the critical point, the density of liquid CO2 is equal to the density of gaseous CO2.This occurs at specific conditions of temperature and pressure.At temperatures and pressures above the critical point values, CO2 is deemed supercritical.For a supercritical fluid, the density and ability to dissolve other substances are similar to values expected for liquids.The following figures give phase data for CO2.

Figure 1 Phase diagram for CO2

Figure 2 Isothermal curves for CO2 near the critical point.V (x 104) is relative to V = 1.000 at 0oC and 1.000 atm.

Some sample questions on this passage are as follows:

• In Figure 2, which of the points (A-D) is the critical point for CO2?

1. A

2. B

3. C

4. D

Explanation: The liquid and vapor phases coalesce at point D of Figure 2, where the densities of liquid and gaseous CO2are equal. Thus, answer choice D is the best answer

• In an extraction of an organic oil, which of the following is an advantage of using supercritical CO2?

1. It reacts with most organic compounds.

2. It is easily handled at room temperature.

3. It crystallizes easily.

4. It is easily removed by evaporation

Explanation: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

• Water is a liquid at room temperature, yet CO2 at room temperature is liquid only at high pressures.Which of the following best explains this?

1. CO2 is polar and has strong intermolecular forces.

2. CO2 is nonpolar and has strong intermolecular forces.

3. CO2 is polar and has weak intermolecular forces.

4. CO2 is nonpolar and has weak intermolecular forces

Explanation: 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.

• According to Figure 1, what is the critical temperature and pressure of CO2?

1. –56.1oC and 6.0 atm

2. –54.1oC and 119.0 atm

3. 31.1oC and 75.3 atm

4. 25.0oC and 1.0 atm

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

• Which of the following compounds is most soluble in supercritical CO2?

1. NaCl

2. C2H5OC2H5

3. NH4NO3

4. KOH

Explanation: 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.

MCAT Sample Questions : Physical Sciences

Passage III

When X-rays are produced in an X-ray tube, two types of X-ray spectra are observed: continuous spectra and line spectra.

A continuous spectrum is produced by bremsstrahlung, the electromagnetic radiation produced when free electrons are accelerated during collisions with ions.

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A line spectrum results when an electron having sufficient energy collides with a heavy atom, and an electron in an inner energy level is ejected from the atom.An electron from an outer energy level then fills the vacant inner energy level, resulting in emission of an X-ray photon.For example, if an electron in the n = 1 energy level is ejected from an atom, an electron in the n = 2 level of the atom can fill the vacancy created in the n = 1 level, and a photon with an energy equal to the energy difference between the two levels will be emitted.

A scientist produced both types of spectra using the X-ray tube shown in Figure 1 below.

Figure 1 Heated cathode X-ray tube

The tube contains a heated filament cathode (C), which emits electrons.A power supply (LV) regulates the filament temperature, the electrical current in the tube, and the number of X-rays produced at the anode (A).Another power supply (HV) regulates electron acceleration.

The scientist used an X-ray tube to determine the relationship between X-ray wavelength, ?, and X-ray intensity, I, which is proportional to the number of X-ray photons emitted at ?.The scientist then graphed the results of the experiment, as shown in Figure 2.

Figure 2 X-ray intensity versus wavelength

Some sample questions on this passage are as follows:

• In Figure 2, which of the following represents the source of emission peaks P1 and P2?

1. Bremsstrahlung

2. Absorption of X-ray photons resulting in electronic excitations in atoms

3. Emission of X-ray photons as a result of electronic transitions in atoms

4. Acceleration of electrons in a magnetic field

Explanation: 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.

• Based on the tube in Figure 1, to maintain an electron current of 0.005 A and a potential drop of 105 V between the anode and the cathode, approximately how much power must the tube consume?

1. 5 x 102 W

2. 1 x 103 W

3. 2 x 105 W

4. 2 x 107 W

Explanation: 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. Therefore P = (5 x 10-3 A) x (105 V) = 5 x 102 W. Therefore, answer choice A is the best answer.

• The ionization potentials for electrons in the n = 1, 2, and 3 energy levels of Pb are 1,400 x 10-17 J, 240 x 10-17 J, and 48 x 10-17 J, respectively.When an electron in the n = 2 level fills a vacancy in the n = 1 level, what is the energy of the X ray that is emitted?

1. 1.92 x 10-15 J

2. 2.40 x 10-15 J

3. 1.16 x 10-14 J

4. 1.40 x 10-14 J

Explanation: 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 x 10-17 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 will NOT be produced by collisions between electrons and:

1. He

2. He2+

3. Li1+

4. protons

Explanation: 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.

• n order to increase the maximum kinetic energy of electrons colliding with the anode, the scientist made which of the following changes?

1. The voltage of HV was increased.

2. The voltage of HV was decreased.

3. The voltage of LV was increased.

4. The voltage of LV was decreased.

Explanation: 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.

• n Figure 2, peaks P1 and P2were produced by events that occurred with unequal probabilities.Which peak was produced by the more probable event?

1. P1, because the peak has the longer wavelength

2. P1, because the peak has the lower intensity

3. P2, because the peak has the longer wavelength

4. P2, because the peak has the higher intensity

Explanation: 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

MCAT Sample Questions : Physical Sciences

Passage IV

Many chemical bonds are not purely ionic or covalent, but polar covalent.For example, in an HCl bond, chlorine has a greater attraction for electrons than hydrogen does and therefore develops a partial negative charge with respect to the hydrogen atom.

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If the partial charges are separated by a known distance, the dipole moment, a measure of the charge separation in a bond or molecule, can be calculated by the following equation.

dipole moment = charge x separation distance

Dipole moments are usually measured in debyes (D), where 1 D = 3.34 x 10-30 coulomb · meter.

The molecular geometry of some simple molecules can be determined based on the presence or absence of a net dipole moment in the molecule.The observed molecular dipole moments of various compounds are reported in Table 1.

Table 1 Dipole Moments for Various Compounds

 Molecule Dipole moment (D) SnBr4 0 HgBr2 0 SO2 1.47 HF 1.82 HNO3 2.17 BaO 7.95 KCl 10.27

Some sample questions on this passage are as follows:

• Which of the following best explains the observed molecular dipole moment of SnBr4?

1. Sn and Br have the same effective nuclear charge.

2. Sn and Br are the same size.

3. sSn attracts electrons more strongly than predicted by its electronegativity.

4. The geometry of the molecule causes the bond moments to cancel.

Explanation: 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.

• Based on the observed dipole moment of HF in Table 1, what is the observed dipole moment for HCl?

1. 0.02 D

2. 1.08 D

3. 4.22 D

4. 8.97 D

Explanation: 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.

• Which of the following best explains why HCl bonds are polar covalent?

1. H atoms are smaller than Cl atoms.

2. H atoms are more electronegative than Cl atoms.

3. Cl and H atoms have equal electronegativities.

4. Cl atoms have a greater effective nuclear charge than H atoms do.

Explanation: 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.

• If an O atom is removed from a CO2 molecule, the observed molecular dipole moment will:

1. decrease, because a lone pair of electrons will be formed on the carbon.

2. decrease, because the net charge will increase.

3. remain constant, because the geometry will not change.

4. increase, because a charge separation will develop.

Explanation: 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

• Which of the two compounds, PCl3 or PCl5, can be expected to have a larger dipole moment?

1. PCl3, because its geometry is trigonal planar

2. PCl3, because its geometry is pyramidal

3. PCl5, because its geometry is octahedral

4. PCl5, because its geometry is trigonal bipyramidal

Explanation: 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.

MCAT Sample Questions : Physical Sciences

• How much work is done when a constant horizontal 20-N force pushes a 50-kg block a distance of 10 m on a horizontal surface?

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1. 50 J

2. 100 J

3. 200 J

4. 500 J

Explanation: 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.

• Evaporation occurs when molecules at the surface of a liquid overcome the attractive forces of the liquid.This occurs when molecules within the liquid attain a sufficient amount of:

1. resonance.

2. kinetic energy.

3. surface tension.

4. potential energy.

Explanation: 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.

• How long will it take a runner, starting from rest and accelerating uniformly at 1.5 m/s2, to travel 3.0 m?

1. 21/2 sec

2. 1.5 sec

3. 2.0 sec

4. 3.0 sec

Explanation: The relation between distance, acceleration, and time is: d = (1/2)a · t2. To solve for the time it takes the runner to use t = (2d/a)1/2 = (2×3/1.5)1/2 = 2 s. Therefore, answer choice C is the best answer.

• A rectangular sheet of material has a width of 3 m and a length of 4 m.Forces with magnitudes of 3 N and 4 N, respectively, are applied parallel to two edges of the sheet, as shown in the figure below.

A third force, F, is applied to the center of the sheet, along a line in the plane of the sheet, at an angle ? = arctan 0.75 with respect to the horizontal direction.The sheet will be in translational equilibrium when F has what value?

1. F = 3 N

2. F = 4 N

3. F = 5 N

4. F = 7 N

Explanation: A body is in transitional equilibrium when the components of all external forces cancel. For the sheet: F cos ? = 4 N, F sin? = 3 N. The magnitude of F is found by adding the squares of the components: F2cos2 ? + F2 sin2 ?= F2 = 42 + 32 = 25 N2. Therefore F = 5 N. The F vector points in the proper direction since tan ? = 0.75 = 3/4. Thus, answer choice C is the best answer.

• Which one of the following general characteristics is shared by all catalysts?

1. They induce more collisions among reactant molecules.

2. They transfer kinetic energy to the reactant molecules.

3. They increase the reaction rate but do not change the Keq of a reversible reaction.

4. They increase both the reaction rate and the Keq of a reversible reaction.

Explanation: 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.

• Radium, 226Ra, spontaneously decays to radon with the emission of an a-particle and a ? ray. If the speed of the a particle upon emission from an initially stationary radium nucleus is 1.5 x 107 m/s, what is the recoil speed of the resultant radon nucleus? Assume the momentum of the ? ray is negligible compared to that of the a particle.

1. 2.0 x 105 m/s

2. 2.7 x 105 m/s

3. 3.5 x 105 m/s

4. 1.5 x 107 m/s

Explanation: Conservation of linear momentum requires: mradonvradon= mheliumvhelium 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.

• An object is placed upright on the axis of a thin convex lens at a distance of four focal lengths (4 f) from the center of the lens. An inverted image appears at a distance of 4/3 f on the other side of the lens. What is the ratio of the height of the image to the height of the object?

1. 1/3

2. 3/4

3. 4/3

4. 3/1

Explanation: 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.

MCAT Sample Questions : Physical Sciences

Passage V

A chemist performed the following experiments to investigate the melting and freezing behaviors of acetamide.

Experiment 1: Melting

A large beaker of water was heated to a slow boil. A thermometer was placed in a test tube and 10 g of acetamide crystals was added. The test tube was then lowered into the boiling water (100o C). The temperature was immediately read, and was reread every 15 sec. The acetamide was stirred before each reading. When the temperature reached 80o C, the acetamide started melting. After a period of time, when all the acetamide had melted, the temperature began to increase again. Results are shown in Figure 1.

Experiment 2: Freezing

Trial 1

The test tube from Experiment 1 was removed from the hot water and left to cool in air at 20o C. The temperature readings and stirring were continued every 30 sec. The temperature dropped to 80o C, where it remained constant. The acetamide slowly began freezing and was completely solid after 23 min. After this, the temperature again decreased. The time for freezing was considered to be excessive, so another trial was completed.

Trial 2

The same test tube was placed in boiling water until the acetamide was completely melted. For this trial, however, the test tube was then placed in a beaker of water at 20o C. The results are shown in Figure 1.

Figure 1 Melting and freezing behavior of acetamide

Some sample questions on this passage are as follows:

• In Experiment 2, which of the following is the most important difference in the procedures used for Trials 1 and 2?

1. The amounts of acetamide used in each test tube

2. The surroundings that were used to cool the acetamide

3. The temperatures at which the trials were started

4. The lengths of time allowed for the acetamide to melt

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

• During Experiment 1, which of the following would most likely have occurred if the water had only reached 90o C before the test tube was placed into it?

1. More water would have been needed to melt the acetamide.

2. Less water would have been needed to melt the acetamide.

3. The acetamide would not have melted.

4. The acetamide would have taken longer to completely melt.

Explanation: 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.

• How are the designs of the two experiments important for producing useful results?

1. Both processes, melting and freezing, take place under controlled conditions.

2. Both processes, melting and freezing, take place without being controlled or monitored.

3. The amounts of acetamide are shown to control the temperatures of melting and freezing.

4. The amounts of acetamide are shown to control the times needed for melting and freezing.

Explanation: 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.

• During Trial 1 of Experiment 2, if the temperature readings were taken at 1-min intervals instead of 30-sec intervals, the acetamide would most likely have become completely frozen at:

1. 11 min, 30 sec.

2. 23 min.

3. 46 min.

4. a lower temperature.

Explanation: 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.

• In Experiment 2, why was it necessary to place the test tube in hot water for Trial 2, in view of the fact that this was NOT done in Trial 1?

1. The water was boiling for Trial 1, but it needed to be cold for Trial 2.

2. The acetamide was cooled by air in Trial 1, but by water in Trial 2.

3. The temperature lowered more quickly for Trial 2 than it did for Trial 1.

4. The acetamide was liquid before Trial 1, but it was solid before Trial 2.

Explanation: 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

• If the data for Trial 1 were plotted in Figure 1, compared to the data for Trial 2, they would:

1. slope less steeply downward, and not all of the data could be shown.

2. slope more steeply downward, and all of the data could be shown.

3. slope upward, and all of the data could be shown.

4. slope upward, and not all of the data could be shown.

Explanation: 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.

MCAT Sample Questions : Physical Sciences

Passage VI

The timbre, or quality, of a musical tone depends on the number and relative strengths of the harmonics including the fundamental frequency of the note. Figure 1a illustrates the first three harmonics of a tone. The addition of the first two harmonics is pictured in Figure 1b, and the addition of the first 3 harmonics is shown in Figure 1c.

Figure 1 Elements of a complex tone

The graphs in Figure 2 illustrate the characteristics of two adjacent tones from a bassoon. Figure 2a shows the pressure variations and the amplitudes of the harmonics for one of the tones, and Figure 2b shows the same information for the other tone

Figure 2 Pressure variations and amplitudes of harmonics for adjacent bassoon tones.

Following are some sample questions on this passage:

• Which of the waveforms shown in Figure 1 has the shortest period?

1. First harmonic

2. Second harmonic

3. Third harmonic

4. The waveform in Figure 1c

Explanation: 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.

• At the second position where the three curves intersect in Figure 1a, the curves are all:

1. in phase.

2. out of phase.

3. at zero displacement.

4. at maximum displacement.

Explanation: 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.

• If the frequency of the first harmonic in Figure 2a is 100 Hz, what is the period of the second harmonic?

1. 0.005 sec

2. 0.01 sec

3. 50.0 sec

4. 200.0 sec

Explanation: 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 s-1 = 0.005 s. Thus, answer choice A is the best answer.

• Which of the following graphs best illustrates the relative amplitudes of the harmonics in Figure 1?

Explanation: 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.

• If a fourth harmonic exists for the tone graphed in Figure 1, then, compared to the third harmonic, the fourth harmonic will have:

1. lower amplitude.

2. higher amplitude.

3. lower frequency.

4. higher frequency.

Explanation: 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.

• The period of the waveform shown in Figure 1c is the:

1. same as the period of the first harmonic.

2. same as the period of the second harmonic.

3. same as the period of the third harmonic.

4. sum of the periods of the first, second, and third harmonics.

Explanation: 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.

MCAT Sample Questions : Physical Sciences

Passage VII

A student was asked to determine the identity of an unknown acid that was liquid at room temperature (20o C). The student was told that the acid was one of those listed in Table 1.

 Acid Structure Molecular weight (g/mole) Melting point (oC) pKa Propionic CH3CH2COOH 74.08 –21.5 4.88 Crotonic CH3CH=CHCOOH 86.09 71.6 4.69 Butyric CH3CH2CH2COOH 88.10 –7.9 4.82 Oxalic HOOCCOOH 90.04 101 3.14 4.77

Table 1 Characteristics of Several Acids

The student added 0.22 g of the acid to 30.0 mL of H2O(l). The student then titrated the solution with 0.10 M NaOH(aq) while monitoring the pH with a pH meter. The results are summarized in Figure 1.

Figure 1 Titration of the acid with 0.1 M NaOH(aq)

Based on the titration curve, the student proposed that the unknown acid had 1 –COOH group and a molecular weight between 85 and 92.

Following are some sample questions on this passage:

• A comparison of which two compounds from Table 1 best shows the effect of molecular weight alone on melting point?

1. Propionic acid and crotonic acid

2. Propionic acid and oxalic acid

3. Propionic acid and butyric acid

4. Butyric acid and crotonic acid

Explanation: 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.

• Before titrating with NaOH(aq), what was the approximate H3O+(aq) concentration of the solution containing the unknown acid?

1. 0.001 M

2. 0.01 M

3. 0.03 M

4. 0.3 M

Explanation: 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

• The student prepared a 0.1 M aqueous solution of crotonic acid and a 0.1 M aqueous solution of oxalic acid, then adjusted the pH of each to 4.7 by adding NaOH. Which solution has a lower freezing point?

1. The crotonic acid solution, because it contains a lower molar concentration of solute particles

2. The crotonic acid solution, because it contains a greater percent mass of solute

3. The oxalic acid solution, because it contains a greater molar concentration of solute particles

4. The oxalic acid solution, because it contains a smaller percent mass of solute

Explanation: 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.

• During the titration summarized in Figure 1, the concentration of R–COOH equalled the concentration of R–COO- when the pH approximately equalled which of the following? (Note: R is a hydrocarbon.)

1. 4.8

2. 6.2

3. 7.0

4. 9.2

Explanation: 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.

• The student rejected crotonic acid as a possible identity of the unknown acid because crotonic acid:

1. is a strong acid.

2. is insoluble in H2O.

3. is solid at room temperature.

4. has a molecular weight of 86.09

Explanation: 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.

MCAT Sample Questions : Physical Sciences

Passage VIII

Several features of sulfuric acid are given below.

Preparation of Sulfuric Acid

Sulfuric acid is commonly prepared by the combustion of elemental sulfur to sulfur dioxide, followed by the catalytic oxidation of sulfur dioxide to sulfur trioxide. Sulfur trioxide is then absorbed into a 98% aqueous solution of H2SO4, and water is added to maintain a 98% concentration. SO3 reacts with the water in the aqueous solution according to Reaction 1.

SO3(g) + H2O(l) = H2SO4(l)

Reaction 1

Properties

Concentrated sulfuric acid is 98% H2SO4 and 2% water by mass. It has a density of 1.84 g/mL and a boiling point of 338oC.

Preparation of Other Acids

HCl(g) and HNO3(l) may be prepared by the reaction between sulfuric acid and the sodium salt of the corresponding conjugate base (Cl- or NO3-, respectively).

formation of SO2

Sulfuric acid forms SO2 gas when it reacts with several compounds. For example, I2 and SO2 are formed when I- reacts with concentrated H2SO4; Br2 and SO2 are formed when Br- reacts with concentrated H2SO4. Cu+ and SO2 are formed in hot solutions of Cu(s) in H2SO4. This last reaction is unusual, because most metals react with solutions of H2SO4 to form hydrogen gas and a metal sulfate.

Following are some sample questions on this passage:

• When sulfuric acid reacts with copper, how does the oxidation number of the sulfur change?

1. From +4 to +6

2. From +6 to +4

3. From +6 to +8

4. From +8 to +6

Explanation: 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.

• The apparatus shown below can be used to prepare HNO3(boiling point = 86oC)

The yield of HNO3 collected in the tube can be maximized by maintaining the temperatures of the flask and tube, respectively, at:

1. 0oC and 100oC.

2. 100oC and 0oC.

3. 350oC and 150oC.

4. 350

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