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  1. 1. MCAT Full-Length TestsDear Future Doctor,The following Full-Length Test and explanations are an opportunity to bring it all togetherin simulation. Do not engage in Full-Length practice until you have adequately preparedyour knowledge and critical thinking skills in Subject, Topical, and Section tests. Simplyg the tests is inadequate; a solid understanding of your performance through your ScoreReports and the explanations is necessary to diagnose your specific weaknesses andaddress them before Test Day.All rights are reserved pursuant to the copyright laws and the contract clause in yourenrollment agreement and as printed below. Misdemeanor and felony infractions canseverely limit your ability to be accepted to a medical program and a conviction canresult in the removal of a medical license. We offer this material for your practice in yourown home as a courtesy and privilege. Practice today so that you can perform on testday; this material was designed to give you every advantage on the MCAT and we wishyou the best of luck in your preparation.Sincerely,Albert ChenExecutive Director, Pre-Health Research and DevelopmentKaplan Test Prep 2003 Kaplan, Inc.All rights reserved. No part of this book may be reproduced in any form, by Photostat, microfilm,xerography or any other means, or incorporated into any information retrieval system, electronicor mechanical without the written permission of Kaplan, Inc. This book may not be duplicated,distributed or resold, pursuant to the terms of your Kaplan Enrollment Agreement.
  2. 2. PHYSICAL SCIENCES ANSWER KEY11. A2. D3. B4. C5. D6. D7. A8. C9. C10. A11. B12. D13. B14. D15. C16. C17. C18. D19. D20. C21. C22. D23. A24. D25. C26. A27. D28. B28. A30. B31. D32. C33. A34. B35. D36. C37. C38. C39. A40. D41. B42. B43. C44. C45. C46. B47. A48. D49. A50. A51. A52. B53. D54. C55. B56. B57. B58. C59. B60. C61. D62. A63. D64. A65. D66. D67. C68. A69. A70. B71. A72. C73. B74. B75. A76. D77. C
  3. 3. Passage I (Questions 16)1. AThis question is asking us to apply the definition of momentum, p = mv. Since each bullethas a mass m = 5.010-3 kg (remember to convert to kilograms) and leaves its gun at 250 m/s,its momentum is p = (5.010-3 kg) (250 m/s) = 1.25 kgm/s The correct answer is choice A.2. DThis question asks about the conservation of the total momentum of the system and the con-servationof kinetic energy in Stunt 2. The total momentum of the system is not conserved, sincegravity acts on both bullets. The total momentum of a system of objects is conserved only whenthere is no net force on the system. The kinetic energy of the system is also not conserved, sincethe collision is inelastic. Remember that the kinetic energy of a system is only conserved in acollision that is perfectly elastic. The correct answer is choice D.3. BLets apply Roman Numeral strategy by evaluating the statement that appears in the mostanswer choices: statement II. The targets translational kinetic energy is zero throughout; thetargets motion is all rotation, it never changes position. Statement II must appear in the correctanswer; eliminate answer choice A.What about statement I? The targets potential energy is a function of how high it sits abovethe ground. In fact, the zero level of gravitational potential energy is arbitrary. We usually set thepotential energy at ground level equal to zero; but we could give it any number at all. So thepotential energy of the target doesnt have to be zero; any answer containing statement I iswrong. Only answer choice B remains.We dont have to check statement III, but lets check it just in case. The targets rotationalkinetic energy is not zero; according to the passage, the target rotates at constant angular veloc-itybetween the two shots.4. CThe first bullet causes the target to rotate because it applies a torque. The second bullet willstop the targets rotation by applying an equal torque in the opposite direction. Since the bulletsstrike on opposite sides of the center of the target, we know the torques will be in opposite direc-tions.We need to know the magnitude of the torque from the first bullet. Applying the defini-tionof torque ( = rF sin , but= 90, so= rF), we see that this torque is 1 = (0.5m)(F0) =0.5F0, using the fact that the radius of the target is 0.5 m. The second bullet strikes the target0.25 m away from the center with a force well call F. Setting the two torques equal, we see that0.5F0 = 0.25F; F = 2F0.2
  4. 4. 5. DThe two bullets travel the same horizontal and the same vertical distance before they collide:250 m/s 250 m/sAB250 cos250 cos In this projectile motion problem, it would be easier for us to deal with motion in the hori-zontaldirection, since there is no acceleration horizontally. The horizontal component of thevelocity of each bullet is (250 m/s) cos. Since there is no acceleration horizontally, we canapply d = vt for each bullet. How long does it take each bullet to travel the 2.5 horizontal metersbefore the collision?(2.5 m) = (250 m/s cos) t; t = = s.The correct answer is choice D.6. DHow should the shooter compensate for the wind? The wind blows to the northeast. The east-erlypart of the wind will act to speed up the bullet a little bit, but wont affect its aim. However,the northerly component of the wind will blow the bullet off course. To compensate for that, theshooter will have to aim a bit south of the target. This eliminates answer choices A and C.Answer choice D offers a different wrinkle: does the shooter have to aim high to compensate forgravity? Yes, she does. The acceleration due to gravity will start pulling the bullet downwardonce it leaves the gun. Over a long distance, the bullet could fall far enough to pass under thetarget, so the shooter should also aim a bit high, so that the bullet will arc upwards, and strikethe target on the way down.Passage II (Questions 712)7. AThis question asks us to know the distinction between molecular geometry and electronicgeometry. In XeF2, the central xenon atom is bonded to two fluorine atoms and carries threeelectron lone pairs. The three lone pairs inhabit sp2 hybridized orbitals (technically, Xe is sp3dhybridized). These three orbitals fall into a trigonal planar configuration, with the two fluorineatoms positioned on opposite sides of the xenon atom, perpendicular to the planar electron con-figuration,as shown below:FXeF0.01cos2.5 m(250 m/s) cos3
  5. 5. We could look at this electronic geometry and mistakenly choose answer choice D, trigonalbipyrimidal. But why would this be wrong? Because though the electronic geometry is indeedtrigonal bipyrimidal, the molecular geometry is simply linear. Lone pairs are not considered partof molecular geometry. Choice A is correct.8. CVSEPR Theory allows us to generate the three-dimensional structure of XeF4. Xenontetrafluoride has 36 valence electrons: 7 from each of the four fluorides and 8 from the centralxenon atom. Xenon therefore has 4 fluoride single bonds and two lone pairs. Six regions of elec-trondensity (4 single bonds + 2 lone pairs) correspond to octahedral electronic geometry andsquare planar molecular geometry, as shown below:FXeFF FHowever, the information relevant to answering the question is simply the number of lonepairs on xenon, which again is 2. Choice C is the correct response.9. CTo determine the intermolecular forces in a sample of XeF6, we must again look to themolecular structure. Xenon has 8 valence electrons: 6 fluoride single bonds and 1 lone pair.Since there is but one lone pair on xenon, there can be no electronic symmetry in the molecule.XeF6 must have a dipole moment. Therefore the predominant type of intermolecular interactionis dipole-dipole interaction, choice C. Answer choice D can be immediately discounted sincethere are no hydrogen atoms in XeF6. Choices A and B are wrong since XeF6 is not ionic.10. AAccording to the kinetic theory of gases, the average kinetic energy of a gas particle is pro-portionalto the absolute temperature of the gas, KE=3/2kT, where k is the Boltzmann constant.Thus the average kinetic energy of a xenon gas particle can be calculated as follows:KEXe = (3/2)(1.3810-23 J/K)(400K273K) (3/2)(673)(1.410-23 J) (1000)(1.410-23 J) 1.410-20 JThe correct answer is therefore choice A.11. BTo produce XeF4, xenon reacts with fluorine as shown:Xe(g)2F2(g) XeF4(g)4
  6. 6. In Experiment 2, the chemist uses 19 g, or 0.5 mole, of F2(g). If she adds xenon and fluo-rinein a 1:5 molar ratio (as stated in the passage) , she must then add 0.1 mol Xe(g). Based onthe stoichiometry of the above reaction, xenon is the limiting reagent in Experiment 2. If all 0.1mol of Xe get consumed, then twice as much fluorine, or 0.2 mol to be exact, get consumed aswell. This leaves 0.5 mol 0.2 mol = 0.3 mol of fluorine gas left after the reaction has gone tocompletion. At STP, the volume of 0.3 mol of F2(g) is , choice B.12. DAlthough this question asks about a topic (lasers) with which we may not be familiar, thequestion can be answered using the information provided in the question stem, as well as ourfundamental understanding of light and wave properties. The question stem tells us that xenon-chlorideand argon-fluoride lasers can be used for medical procedures. Why? Because they havesufficient energy (actually, this is in addition to other desirable properties of rare gas-halidelasers but we do not need to know this to answer the question!) So why might xenon-fluoridelasers not be a good choice for medical lasers? Probably because they do not have sufficientenergy. But since we do not know everything about these lasers, we should use this predictionas a good (but not limiting) starting point as we look through the answers.Choice A tells us that the instability of gas-halide molecules precludes the use of xenon-flu-oridelasers. While gas-halide molecules are indeed unstable (again, we do not need to knowthis), we can discount this choice for the simple reason that the passag