AAMC MCAT test 3r a

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AAMC MCAT test 3r a

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Detailed answers to practice test 3R by AAMC Physical Science Part 1-77 Reaction is shown in the following equation, which is answer choice A Answers B and D not show a reaction involving PbCO3 (s), as required by Reaction Answer C shows an implausible and unbalanced equation Thus, answer choice A is the best answer Reaction 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 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 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 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 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 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 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 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 Therefore P = (5 x 10 A) x (10 V) = 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 = 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 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 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=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 vs Trial 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 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 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 Therefore, the sample had to be reheated in a water bath above its melting point to start Trial as a liquid Thus, answer D is the best answer 32 If the data for Trial were plotted, the temperature would drop to 80oC and remain at this melting temperature for 23 (or 23 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 as λ =1.06 x 10-6m The frequency of this light is 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 shows the pH of the solution to be about before any NaOH(aq) is added pH=- log[H3O+] =-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 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 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 not Alternatively, Figure 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 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 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 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.,

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