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文档下载 免费文档下载 http://doc.guandang.net/ chemistry-subject test5 w. solutions 本文档下载自文档下载网,内容可能不完整,您可以点击以下网址继续阅读或下载: http://doc.guandang.net/bbca35c11081d34250955e480.html MCAT(The Medical College Admission Test)美国医学院入学考试超全模拟试题及全部答案。 MCAT Subject Tests Dear Future Doctor, The following Subject Test and explanations contains questions not in test format and should be used to practice and to assess your mastery of the foundation content necessary for success on the MCAT. Simply memorizing facts is not sufficient to achieve high scores; however, an incomplete understanding of basic science knowledge will limit your ability to think critically. Think of building your content knowledge as learning the vocabulary and practicing MCAT-like questions as actually speaking. All rights are reserved pursuant to the copyright laws and the contract clause in your enrollment agreement and as printed below. Misdemeanor and felony infractions can severely limit your ability to be accepted to a medical program and a conviction can result in the removal of a medical license. We offer this material for your practice in your own http://doc.guandang.net/bbca35c11081d34250955e480.htmlhome as a courtesy and privilege. Practice today so that you can perform on test day; this material was designed to give you every advantage on the MCAT and we wish you the best of luck in your preparation. 文档下载 免费文档下载 http://doc.guandang.net/ Sincerely, Albert Chen Executive Director, Pre-Health Research and Development Kaplan 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, electronic or 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. _________________________________________________________ GENERAL CHEMISTRY SUBJECT TEST 5 General Chemistry Subject Test 5 1. What is the ahttp://doc.guandang.net/bbca35c11081d34250955e480.htmlpproximate F-Xe-F bond angle in the compound XeF4? A. 60° B. 90° C. 109.5° D. 120° E. 160° 文档下载 免费文档下载 http://doc.guandang.net/ 2. For which one of the following mixtures is the ideal gas law most likely to produce an accurate prediction of volume at STP? A. B. C. D. E. 3. HCl and HBr HCl and NH3 HF and HCl HF and CH4 HCl and CH4 6. 5. At STP, a 29.0 g sample of an unknown gas occupies a volume of 11.2 liters. This gas could be which of the following? A. CO B. Ar C. B2H6 D. Kr E. C4H10 A mixture of gases is produced at a total pressure of 234 torr. If the mixture contains 1.35 moles of H2(g) and 2.07 moles of CO(g), what is the partial pressure of CO(g) in the mixture? A. B. C. D. 1.352.07) 234 torr 2.07)234 torr 1.35 2.07 torr 234 2.071.35 ) torr ∞ 234 The equilibrium constant for Reaction 1 is K. What is the equilibrium constant for Reaction 2? A. K B. –K C. 1/K 1http://doc.guandang.net/bbca35c11081d34250955e480.html/K2 2C 7. D A 2B D. K2 E. 文档下载 免费文档下载 http://doc.guandang.net/ 2.07 E. 1.35 2.07)234 torr For a certain chemical reaction, ?H = 128 kJ/mol and ?S = 64 J/mol?K. What is the minimum temperature at which this reaction will be spontaneous? A. 500 K B. 2000 K C. 2 K D. 2000°C E. 2°C 8. Iron reacts with element X to form the compounds FeX and Fe3X2. When copper reacts with element X, the expected compound(s) should be A. CuX and Cu3X2. B. Cu2X and Cu6X. C. Cu3X only. D. CuX2 only. E. Cu3X and Cu3X2. 4. The value of Kc for the reaction below is 2.0. 2 CO(g) C(s) CO2(g What is the concentration of CO at equilibrium if the CO2 concentration is 0.50 M? A. 2.0 M B. 1.5 M C. 1.41 M D. 1.0 M E. 0.50 M KAPLAN__________________________________________________________________________ ______________1 文档下载 免费文档下载 http://doc.guandang.net/ GENERAL CHEMISTRY SUBJECT TEST 5 _______________http://doc.guandang.net/bbca35c11081d34250955e480.html___________ ______________________________________ 9. In which of the following is bromine in a 5 oxidation state? A. NaBr B. BrO4- C. NaBrO D. HBrO3 E. BrO2- 13. For the reaction below, Keq = 96.2 at 450oC. What is the equilibrium concentration of PCl3 at 450oC when the equilibrium concentrations of Cl2 and PCl5 are 3.50 M and 2.50 M, respectively? PCl5(aq) A. B. C. D. E. 14. Which of the following pairs of elements would form the most ionic compound? A. B. C. D. E. 15. Silver hydroxide, AgOH, is a slightly soluble ionic compound whose Ksp = 1.5 x 10-8 at 20oC. What is the minimum pH at which AgOH will precipitate from a solution, 0.010 M in Ag ? A. B. C. D. E. 6 - log(1.5) 8 log(1.5) 8 - log(1.5) 14 - log(1.5) 14 Cl Pt and Cl La and Br La and Cl log(1.5) Au and S Au and 文档下载 免费文档下载 http://doc.guandang.net/ 2 . 5096. 2 × 3 .5096. 2 × 2 . 503 . 502 . 50× 3 .http://doc.guandang.net/bbca35c11081d34250955e480.html50 96. 2 96. 2 2 . 50× 3 .50 96. 2 2 . 50× 3 .50 PCl3(aq) Cl2(g) M M M M M 10. The unbalanced redox reaction shown below occurs in a solution of aqueous base. Which species is the oxidizing agent? K2Cr2O7(aq) KClO(aq) ? Cr(OH)3(s) KClO3(aq) A. Cr in Cr2O72– B. Cl in ClO– C. Cr in Cr(OH)3 D. Cl in ClO3– E. K 11. If 25 ml of 0.75 M NaOH(aq) is required to completely neutralize 150 mL of HCl(aq), what is the molarity of the acidic solution? A. 0.125 M B. 0.15 M C. 0.25 M D. 0.75 M E. 4.5 M 12. The pH of a solution of a weak, monoprotic acid 4.5. The molarity of the acid solution is: A. B. (Ka = 4 x 10-6) 文档下载 免费文档下载 http://doc.guandang.net/ 4 × 10? 6 10( ? 4. 5 2 ) 4 . 5 4 × 10) ( ? 6 2 C. ? log D. E. . 5 ? ? 4 ? ? 4 × 10? 6 ? 16. What is thttp://doc.guandang.net/bbca35c11081d34250955e480.htmlhe oxidation number on the arsenic atom in arsenous acid, H3AsO3? A. -3 B. -2 C. 0 D. ( 3 E. 5 文档下载 免费文档下载 http://doc.guandang.net/ 5 10? 4. ) 2 4 × 10? 6 ? log 4 .52 log 4 × 10? 6 ( ) 2 ________________________________________________________________________________ ________KAPLAN _________________________________________________________ GENERAL CHEMISTRY SUBJECT TEST 5 17. Which of the following pairs of compounds could be mixed to produce a buffer solution? A. CH3NH3I and HI 文档下载 免费文档下载 http://doc.guandang.net/ B. CH3NH2 and CH3NH3I C. CH3NH3I and NH4I D. HI and NH4I E. CH3NH2 and NH3 22. How much 2 M H2SO4 is required to neutralize 200 mL of 2 M NaOH ? A. 75 mL B. 100 mL C. 150 mL D. 200 mL E. 500 mL 23. 20 mL of an ideal gas is placed in a piston with an initial pressure of 700 tohttp://doc.guandang.net/bbca35c11081d34250955e480.htmlrr. If the pressure is decreased to 630 torr, and the temperature is held constant, what is the final volume of the gas? A. 14.4 mL B. 18.8 mL C. 22.2 mL D. 26.4 mL E. 31.8 mL 24. Which of the following could be characteristic of a reaction with a positive ?H and a positive ?S? A. B. C. D. E. 25. Why is it more difficult to compress a liquid to a solid than it is to compress a gas to a liquid? A. B. C. D. 文档下载 免费文档下载 http://doc.guandang.net/ van der Waals forces exist within a liquid. Gas particles are close together. Liquid molecules are close together. Liquid molecules possess more kinetic energy than gas molecules. E. Liquid molecules move at a slower speed than gas particles. 26. Which of the following processes involves an increase in entropy? A. Condensation B. Freezing C. Sublimation D. Cooling E. Solidification The reaction is shttp://doc.guandang.net/bbca35c11081d34250955e480.htmlpontaneous. The reaction is nonspontaneous. The reaction is at equilibrium. The reaction is exothermic. Two of the above 18. A voltaic cell is constructed by connecting a nickel electrode in a Ni2 (aq) solution to a tin electrode in a Sn2 (aq) solution. Given that the standard reduction potentials for Ni2 and Sn2 are -0.28V and -0.14V respectively, what is the cell EMF? A. 0.84V B. 0.42V C. 0.14V D. -0.14V E. -0.42V 19. How many seconds will it take to produce 11.2 liters of 文档下载 免费文档下载 http://doc.guandang.net/ Cl2(g), measured at STP, by electrolysis of molten NaCl with a 12 amp current? (F = 96500 C/mol) A. 96500/12 B. 12/96500 C. (12)(96500) D. (2)(12)/(96500) E 96500/(2)(12) 20. 10.0 mL of 0.40 M CuSO4 solution contains what mass of CuSO4? A. 0.40 mg B. 4.0 mg C. 384 mg D. 400 mg E. 638 mg 21. If 625 ml of 0.75 M NaOH(aq) is diluted to 0.45 M NaOH(aq), the finahttp://doc.guandang.net/bbca35c11081d34250955e480.htmll volume, in mL, will be: A. (625)(0.75/0.45) B. (625)(0.45/0.75) C. (1/625)(0.75/0.45) D. (1/625)(0.45/0.75) E. (625)(0.75 - 0.45) KAPLAN__________________________________________________________________________ ______________3 GENERAL CHEMISTRY SUBJECT TEST 5 ________________________________________________________________ 27. 23 g of ethanol (C2H5OH) is dissolved in 250 g of water. What is the approximate freezing point of this solution? 1.86_C_kg/mol) (kf of water = 文档下载 免费文档下载 http://doc.guandang.net/ A. –9_C B. –1.9_C C. –2.7_C D. –3.7_C E. –7.6_C 28. What is the boiling-point elevation when 60 g of glucose (C6H12O6) is dissolved in 667 g of water? (kb of water = 0.51_C_kg/mol) A. B. C. ( 0 . 005 ) ( 1. 8 × 10-5) 0 .01( 180 ) ( 667 ) 29. Which of the following solutions will have the greatest boiling point? A. 0.5 m NaCl B. 1.5 m C6H12O6 C. http://doc.guandang.net/bbca35c11081d34250955e480.html1.0 m Na3PO4 D. 1.5 m KCl E. 1.0 m K2CO3 30. The concentration of Pb2 is 1.15 ∞ 10–3 M in a saturated solution of PbF2. Calculate the Ksp of PbF2. A. 1.3 ∞ 10–6 B. 5.2 ∞ 10–6 C. 1.5 ∞ 10–9 D. 6.0 ∞ 10–9 E. 1.5 ∞ 10–10 ? ( 0 . 51) 文档下载 免费文档下载 http://doc.guandang.net/ ( 667 ) ( 60) ( 1000) D. 3 - ? ( 0 . 51) ( 667 ) ( 60) ( 1000) E. ? ( 0 .51) ( 180 ) ( 667 ) ( 6 ) ( 1000) ( 180 ) ( 1000) ? ( 0 . 51) STOP! END OF TEST. 4 ________________________________________________________________________________ ________KAPLAN _________________________________________________________ GENERAL CHEMISTRY PRACTICE TEST 1 THE ANSWER KEY AND EXPLANATIONS BEGIN ON THE FOLLOWING PAGE. 文档下载 免费文档下载 http://doc.guandang.net/ KAPLAN__________________________________________________________________________ _____________5 PREPARhttp://doc.guandang.net/bbca35c11081d34250955e480.htmlATION FOR THE DENTAL ADMISSION TEST _____________________________________________________ GENERAL CHEMISTRY SUBJECT TEST 5 ANSWER KEY 1. B 2. E 7. B 13. B 3. C 4. D 8. E 5. E 16. D 17. B 18. C 19. A 20. E 21. A 14. E 6. E 22. B 9. D 15. B 23. C 10. A 24. E 11. A 12. D 文档下载 免费文档下载 http://doc.guandang.net/ 25. C 26. C 27. D 28. E 29. C 30. D 6 ________________________________________________________________________________ ________ KAPLAN _________________________________________________________ GENERAL CHEMISTRY PRACTICE TEST 1 EXPLANATIONS 1. B This is an example where an inert gas forms a compound with another element. The Lewis structure is: Note that Xe has to expand its octet, which it can do because it is behttp://doc.guandang.net/bbca35c11081d34250955e480.htmlyond the second period. According to VSEPR theory, the six pairs of electrons (4 bonding, 2 non-bonding) will arrange themselves in an octagonal structure. The two lone pairs will occupy opposite 文档下载 免费文档下载 http://doc.guandang.net/ positions to be away from each other as far apart as possible, leaving the fluorine atoms in the corners of a square planar molecular geometry: The F—Xe—F bond angle, then, would be about 90°. 2. E The ideal gas law assumes (among other things) that there are no intermolecular forces among the gas particles, which hold the molecules tighter together, leading to a smaller volume than that predicted by the ideal gas law. Intermolecular forces among neutral particles are due to hydrogen bonding, dipole-dipole interactions and dispersion forces (in the order of decreasing strength). For the ideal gas law to give an accurate prediction of the volume, then, we are looking for gases that do not have ahttp://doc.guandang.net/bbca35c11081d34250955e480.html strong dipole moment. Methane, CH4, does not have a dipole moment: What little dipole moments exist from the relatively nonpolar C–H bonds cancel one another because of the tetrahedral arrangement. Between choices D and E, HF can participate in hydrogen bonding while HCl cannot. (Hydrogen bonding occurs when a hydrogen atom is bonded to F, O, or N.) The intermolecular forces will be weaker overall for choice E than for choice D. 3. C Note that reaction 2 is the reverse reaction of reaction 1. If a reaction has an equilibrium constant of K, then the reverse reaction must have an equilibrium constant 文档下载 免费文档下载 http://doc.guandang.net/ of 1/K. 4. D The expression for the equilibrium constant for the given reaction is: [CO]2 Kc = [CO]2 KAPLAN__________________________________________________________________________ _____________7 PREPARATION FOR THE DENTAL ADMISSION TEST _______http://doc.guandang.net/bbca35c11081d34250955e480.html___________________ ___________________________ The concentration of CO is squared because of the stoichiometric coefficient of 2 in the balanced equation. Carbon does not appear in the expression because it is a pure solid. We are told that the value of Kc is 2.0, which enables us to solve for the equilibrium concentration of CO: [CO] = Kc[CO2] = 2.0 ∞1.0 = 1.0 M 5. E 文档下载 免费文档下载 http://doc.guandang.net/ One mole of any gas that behaves ideally will occupy a volume of 22.4 liters at STP. A volume of 11.2 liters thus implies that there is 1/2 mole of gas. The sample has a weight of 29.0 g, which means that the gas has a molecular weight of 29.0/(1/2) = 58.0 g/mol. Choice E, C4H10, has a molecular weight of 4 ∞ 12 1 ∞ 10 = 58 which is what we are looking for. Choice A, CO, has a molecular weight of 28 g/mol. Choice B, Ar, has a molecular weight of 40 g/mol. Choice C, B2H6, has a molecular weight of about 28 g/mol.http://doc.guandang.net/bbca35c11081d34250955e480.html Choice D, Kr, has a molecular weight of about 84 g/mol. 6. E Dalton’s law of partial pressures states that the partial pressure of a gas in a mixture is equal to the total pressure of the mixture times the mole fraction. The mole fraction of gas A is the number of moles of A divided by the total number of moles of gases present (including A itself). If a mixture of 1.35 moles of hydrogen and 2.07 moles of CO exert a total pressure of 234 torr, then the partial pressure of CO is given by: 2.07 PCO= XCOPtot1.35 2.07 7. B Since _H and _S are both positive, the reaction will be spontaneous only at high temperatures. The minimum temperature at which this reaction becomes spontaneous 文档下载 免费文档下载 http://doc.guandang.net/ occurs when _G, the free energy change, is zero. We therefore need to solve for T in the equation: _G = _H – T_S = 0 128000 – 64T = 0 T = 128000/64 = 2000 K 8. E://doc.guandang.net/bbca35c11081d34250955e480.htmlpar Iron forms the 3 or the 2 ion: Fe3 (ferric) and Fe2 (ferrous). The fact that it reacts with X to form FeX and Fe3X2 suggests that X forms an anion with a –3 charge. X3– reacts with the ferric ion to form FeX, since both have a –3 charge, but reacts with the ferrous ion to form Fe3X2 to balance the charge. Copper usually forms either the 1 or the 2 ion. When the former reacts with X, then, there must be three Cu ions to balance the charge, resulting in Cu3X. When the latter reacts with X, it does so in the same way that the ferrous ion does, forming Cu3X2. 8 ________________________________________________________________________________ ________ KAPLAN _________________________________________________________ GENERAL CHEMISTRY PRACTICE TEST 1 9. D [charge][value], e.g. By sticking 3, and ionic charges are written as [value][charge], e.g. 2–. to a system yhttp://doc.guandang.net/bbca35c11081d34250955e480.htmlour you'll chances decrease of careless errors during redox reactions & oxidation state calculations]. causing 文档下载 免费文档下载 http://doc.guandang.net/ The sum of the oxidation numbers of the atoms in a molecule or ion have to equal the charge of the molecule or ion. charge of the polyatomic ion. on the atom. The sum is either zero for a neutral molecule or the For single atoms, the oxidation state equals the charge Oxygen always gets a –2 oxidation number, unless it is in the form of a peroxide (e.g. H2O2), in which case it is –1. Hydrogen always gets a it is in the form of a hydride (e.g. LiAlH4), in which case it is 1. 1, unless The oxidation states of other atoms has to be deduced using the above rules, with a little help from the periodic table (knowing how many electrons an atom wants to gain or lose based on its electron configuration helps determine possible oxidation states). choice A, Na is 1, so Br must be –1. In choice B, each O is –2, so four give a totahttp://doc.guandang.net/bbca35c11081d34250955e480.htmll of –8. the polyatomic ion is 1–, so Br must be 7 (x C, there is a Na is at is a H is at (–8) = –1; x = 1, an O at –2, and so Br must be 1. The charge of 7). In choice In choice D, there 1, three O's (–2 each) giving a total of –6, and so Br must be the correct answer. In 5, In choice E we have two O's giving a total of –4, the charge of the ion is 1–, so Br must be 3. 10. A A reducing agent is the species that gets oxidized in a redox reaction, while an oxidizing agent is one that gets reduced. Using the mnemonic OIL RIG, we remember that reduction is gain of electrons, and a decrease in oxidation number. In choice A, the chromomium goes from agent (it gets reduced). 6 in Cr2O72- to 3 in Cr(OH)3, so it is an oxidizing In choice B, the chlorine goes from ClO3–, so is a reducing agent (it gets oxidized). a d subshell, allowing it 1 in ClO– to [Cl is in period 3, so it has to have oxidhttp://doc.guandang.net/bbca35c11081d34250955e480.htmlation multiple states.] choice C, Cr(OH)3 is a product of the reaction, so cannot be the answer. E, K 5 in In In choice does not change during the reaction, as it is just a spectator ion. 文档下载 免费文档下载 http://doc.guandang.net/ 11. A To solve neutralization questions, we use the formula NAVA = NBVB; eq NALxA, then N = M _ x . For example, a 3 M H3PO4 is a 9 N H3PO4. VA is the volume of acid used; NB is the normality of the base. then y. Bases that bind H For any given base with the general formula B(OH)y, are treated similarly, e.g. 2 M Na2CO3 is 4 N. VB is the volume of base used. Essentially, the formula states that the molar amount of acid equals the molar amount of base, which it must if the solution has been neutralized. Here, both the base and the acid are monoequivalent, so their normalities equal their normalities. Solving for the NV(.75)(25)751 normality ofhttp://doc.guandang.net/bbca35c11081d34250955e480.html NAV1506008A equation, no conversion to liters is necessary. acid, 文档下载 免费文档下载 http://doc.guandang.net/ 12. D The chemical equation for the dissociation of a monoprotic acid is HA The stoichiometry tells us that the concentration of H of A–. By solving for the concentration of H concentrations of both. be ? H A–. must equal the concentration using the pH, we will have found the Since pH = –log [H ], then [H ] = 10-pH (since 10a = b can expressed as log b= KAPLAN__________________________________________________________________________ _____________9 N = M _ PREPARATION FOR THE DENTAL ADMISSION TEST _____________________________________________________ [H ][A–] a[HA]a). Here the [A–] = [H ] = 10–4.5. The acid dissociation constant for the reaction is K[H ][A–](10–4.5)2 concentration of acid, [HA] = [K]4 x 10a 13. B This is a straightforward 文档下载 免费文档下载 http://doc.guandang.net/ queshttp://doc.guandang.net/bbca35c11081d34250955e480.htmltion that asks us to manipulate the formula for Keq and solve for an unknown [PCl][Cl]K[Cl] concentration. For this reaction the Keq[PCl and solving we get [PCl33[PCl5]5]96.2 x 2.5 14. E The two elements that would form the most ionic compound will have the greatest difference in electronegativities. Electronegativity increases from left to right and from bottom to top on the periodic table of the elements. The elements with the greatest difference in electronegativities will be the furthest apart from one another on the periodic table. Lanthanum and chlorine are the furthest apart on the periodic table. 15. B To answer this question, we must first solve for the concentration of OH–, then use that to figure out the pH. = [Ag ][OH–]. The solubility product constant for this reaction is Ksp Solving for the concentration of OH– we get [OH–] = K1.5 x 10–8 –6http://doc.guandang.net/bbca35c11081d34250955e480.html–6 [Ag ]1 x 10–2 1.5 x 10. To calculate the pOH we use the formula pOH = –log[OH] = –log(1.5 x 10) = –(log1.5 log10–6) = –(log1.5 –6) = 6 – log1.5. use the pOH so determine the pH. Since pH Now we can pOH = 14, then pH = 14 – pOH = 14 – 文档下载 免费文档下载 http://doc.guandang.net/ (6 – log 1.5) = 8 log 1.5. It is essential to understand the mathematical rules involving logarithms, as it goes hand in hand with acid/base chemistry. 16. D Using the oxidation rules outlined in question 9, we have to determine the oxidation state of arsenic in the polyatomic ion AsO33–. The oxygens give a total oxidation state of –6 (3 x (–2)), and the oxidation number of arsenic plus that of the oxygens have to sum to the charge of the ion. Therefore arsenic must be 3 (x (–6) = 3–; x = –3). Alternatively, you could solve for the entire molecule, H3AsO3, with the hydrogens giving total oxidation state of 3 (3 x ( 1)), the oxygenhttp://doc.guandang.net/bbca35c11081d34250955e480.htmls –6, and arsenic (x 3 (–6) = 0; x = 3 3). 17. B Buffers are solutions that resist change in pH. acid and its conjugate base. A buffer is a solution of a weak Weak acids and bases must be used, since a strong acid or base would completely dissociate and not have any buffering capacity. If the acid and base in the buffer were not conjugates, they would react with one another, producing salt and water with no buffering capacity. a weak acid and a strong acid, so cannot make a buffer. Choice A is Choice B is the correct answer, 文档下载 免费文档下载 http://doc.guandang.net/ the buffer consists of the weak base methyl amine and the conjugate base methylammoniumiodide. Choices C, D and E are not conjugates. 18. C A voltaic or galvanic cell is a spontaneous cell that has a positive EMF (electromotive force) and a negative _G. When given half reactions you have to determine which species will be oxidized ahttp://doc.guandang.net/bbca35c11081d34250955e480.htmlnd which we be reduced such that a positive electrical cell potential is obtained. The half reactions with the corresponding reduction potentials are as follows: 10 ________________________________________________________________________________ _______ KAPLAN _________________________________________________________ GENERAL CHEMISTRY PRACTICE TEST 1 Ni2 (aq) 2e– ? Ni(s) ? Sn(s) E_ = –0.14 V E_ = –0.28 V Sn2 (aq) 2e– Since the nickel has the more negative reduction potential, it will get oxidized. The reaction for nickel above will therefore proceed to the left. the cell is: The reaction for 文档下载 免费文档下载 http://doc.guandang.net/ Sn2 (aq) Ni(s) ? Sn(s) Ni2 (aq) The EMF of the cell is the reduction potential of the species reduced minus the reduction potential of the species oxidized, as expressed in the formula EMF = Ecathode – Eanode. The purpose of the subtraction is to rhttp://doc.guandang.net/bbca35c11081d34250955e480.htmleverse the sign of the reduction potential since that species is getting oxidized. Be sure not to make the mistake of reversing the sign of reaction occurring at the anode, and then insert it into the preceding equation--you will have reversed the sign twice and gotten the answer wrong! For this reaction, EMF = (–0.14) – (–0.28) = 0.14 V. 19. A This question requires dimensional analysis to solve, as it would take too many equations and way too much time so solve directly. In dimensional analysis, a series of fractions are multiplied together, canceling the units you don't need and eventually giving the units you are looking for. You start with a fraction that has the units you are solving for in the numerator, and proceed to cancel out the denominator and subsequent units until you are left with the answer. This question is asking for seconds, so we must start with a fraction with seconds in the nhttp://doc.guandang.net/bbca35c11081d34250955e480.htmlumerator. proceeds as follows: (1 sec12 C) current The analysis 文档下载 免费文档下载 http://doc.guandang.net/ x 96500 Cmol efaraday's constant x 1 mol Cl2 2 mol e 2Cl- ? 2 Cl2 2e- x (1 mol Cl22.4 Lx (11.2 L) = 1296500 given volume gas law The trick is to recognize that amps are C/sec, and that coulombs can be canceled using Faraday's constant. 20. E This is another question that can be solved using dimensional analysis: 160 g 文档下载 免费文档下载 http://doc.guandang.net/ (1 mol CuSO) 4 x 1 L 0.40 mol x (.01L) = 0.64 g = 640 mg 21. A This is an example of a dilution question, where the volume has been changed but not the molar amount of the substance. in neutralization questions: Mi is The formula used is very similar to that used MiVi = MfVf: the initial molarity of solutionhttp://doc.guandang.net/bbca35c11081d34250955e480.html. initial volume of the solution. Vi the is Mf is the final molarity of the solution. the Vf is the volume volume of the solution. Essentially, since M x V = moles, the formula states ni = nf. This formula can only be used when the molar amount MVhas not changed. This question is looking for the final volume, so solving and plugging in we get VfM2 文档下载 免费文档下载 http://doc.guandang.net/ (0.75)(625) 0.45 KAPLAN__________________________________________________________________________ ____________11 PREPARATION FOR THE DENTAL ADMISSION TEST _____________________________________________________ 22. B This is a neutralization question that is looking for the volume of acid. Using the techniques outlined in question 11, NV(2)(200) we rearrange the equation NAVA = NBVB for acid volume and solve; VAA 23. C To solve this question we have to use Boyle's http://doc.guandang.net/bbca35c11081d34250955e480.htmlP1V1 = P2V2. the equation and solving for the final PV(700)(20)1400 Law, Rearranging 文档下载 免费文档下载 http://doc.guandang.net/ volume we get V2 = P630? 602 24. E The change in enthalpy (?H) and change in entropy (?S) of a reaction are related to the change in free energy (?G) of a reaction by the formula ?G = ?H – T?S. Since ?H is positive and ?S is positive, the reaction will be spontaneous when temperatures are very high, and will be nonspontaneous when temperatures are very low. of this is ice melting: An example ?H is positive because breaking bonds requires energy, and disorder is increasing (solid _ liquid), so ice will only melt when T > 0_C. choice C is incorrect because when a reaction is at equilibrium, ?S = 0. D is incorrect because a Answer Answer choice ?H means heat is being absorbed, and is defined as endothermic. 25. C As the phases of matter change from gas _ liquid mhttp://doc.guandang.net/bbca35c11081d34250955e480.htmlolecules closer together. _ solid, are the becoming Since molecules in the liquid phase are already close together, it is more difficult to compress them further into the solid phase than it is to compress gas phase molecules (which are far apart). Answer choice A is incorrect because van der Waals forces do not effect how easily a substance can be compressed. Choice B is incorrect because gas molecules are far apart from one another, and are the least dense of all the phases of matter. Liquid molecules possess less kinetic energy than gas molecules, since 文档下载 免费文档下载 http://doc.guandang.net/ liquids exist at lower temperatures than gases, making choice D wrong, and although choice E is a correct statement, it does not answer the question, as it does not address a correlaton between solids and liquids. 26. C An increase in entropy means an increase in disorder. become gases, with no liquid Sublimation is when solids phase in between. Sincehttp://doc.guandang.net/bbca35c11081d34250955e480.html gas molecules are more disorder than solids, entropy increases. Condensation is when gases become liquids, a decrease in entropy. Freezing and solidifaction are both processes by which liquids become solids, also a decrease in entropy. Cooling is also a decrease in entropy, as the molecules are moving slower and less randomly. 27. D The freezing point of a liquid decreases when a substance is dissolved in it according to the formula ?Tf = kfm ; where kf is the freezing point depression constant (a constant specific for a given solvent), and m is the molality (mol solute/kg solvent) of the solution. Remember that the formula reports the change–if a question asks for the final temperature, you must subtact ?T from the initial freezing point of the liquid (Tf = Ti – ?Tf). To We are given 23g ethanol, 250 g H2O (= 1/4 kg), and a kf of 1.86_C_kg/mol. calculate the molality we have to first dethttp://doc.guandang.net/bbca35c11081d34250955e480.htmlermine the molar amount 文档下载 免费文档下载 http://doc.guandang.net/ m23 gmol1/1/2 mol. Therefore m?Tf = kfm; = (1.86)(2) ? 4 (2)(2) = 4_C. The freezing point of water is 0_C, so the new freezing point is –4_C. [Notice: calculations are usually made easier when numbers are used in fractional format instead of decimals, and values are rounded to the nearest whole number when possible.] 12 ________________________________________________________________________________ _______ KAPLAN _________________________________________________________ GENERAL CHEMISTRY PRACTICE TEST 1 28. E Similar to freezing point depression is boiling point elevation. The boiling point of a solution increases according to the formula ?Tb= kbm ; where kb is the boiling point elevation constant. asks for the final Remember, the formula reports the change, so if a question boiling point, the ?Tb has to be added to the ihttp://doc.guandang.net/bbca35c11081d34250955e480.htmlnitial boiling point (Tf = Ti ?Tb). Before doing any calculation, you should always look at the answer choices to help determine how the question is to be solved. 文档下载 免费文档下载 http://doc.guandang.net/ mol Here, we are not required to do any math, we need only set up the formula: kbmkgm60 gMW180 /(0.51_C_kg/mol)kg = (0.51)1kg. ?Tb= After dropping the units and rearranging the complex fraction we get: 1000g (60)(1000)(0.51)_(667)(180) 29. C Boiling point elevation and freezing point depression are colligative properties. Colligative properties depend on the amount of the substance, and does not depend on the properties of the substance itself. If a molecule dissociates when it dissolves, the molality has to be multiplied by the number of particles formed upon dissociation, that way the total number of dissolved particles are taken into consideration when calculating the new freezing point or boiling point. We ahttp://doc.guandang.net/bbca35c11081d34250955e480.htmlre of given molalities and asked which will have the greatest boiling point. a list The solution with the greatest boiling point will have the greatest number of dissolved particles in solution. Let's go through the individual answer choices and determine the equivalent molality: A. B. C. D. E. .5 mol NaCl2 mol ions _ mol NaClm ions [1 mol Na and 1 mol Cl– are formed] kg1.5 mol CHO1 mol 文档下载 免费文档下载 http://doc.guandang.net/ _ molm [glucose does not dissociate] kg 1.0 mol NaPO4 mol ions _m ions [3 mol Na _ mol KClm ions and 1 mol PO43- are formed] 341.5 mol KCl2 mol ion [1 mol K and 1 mol Cl- are formed] kg 1.0 mol KCO_ (3 ions,mol K2CO3) = 3.0 m ions [2 mol K and 1 mol CO32- are formed] kg Since Na3PO4 has the greatest amount of particles in solution, it will increase the boiling point the most. 30. D The formula for the dissolution of PbF2 is: PbF Thttp://doc.guandang.net/bbca35c11081d34250955e480.htmlhe stoichiometry tells us that for every mole of PbF2 that dissolves, 1 mol of Pb2 is formed and 2 moles of F- are formed. If we let x = [Pb2 ], then the [F-] = 2x. Plugging this into the formula we get Ksp = [Pb2 ][F-]2 = (x)(2x)2 = (x)(4x2) = 4x3. 1 x 10–3 D. Pb2 2F- The [Pb2 ] = 1.15 x 10–3 ? and so Ksp = 4x3 = 4(1 x 10–3)3 = 4 x 10–9, which is close to choice 文档下载 免费文档下载 http://doc.guandang.net/ Ksp = [Pb2 ][F-]2 KAPLAN__________________________________________________________________________ ____________13 文档下载网是专业的免费文档搜索与下载网站,提供行业资料,考试资料,教 学课件,学术论文,技术资料,研究报告,工作范文,资格考试,word 文档, 专业文献,应用文书,行业论文等文档搜索与文档下载,是您文档写作和查找 参考资料的必备网站。 文档下载 http://doc.guandang.com/ 亿万文档资料,等你来下载