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 home 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 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 General Chemistry Subject Test Which one of the following reactions will be accompanied by an increase in entropy? A Na(s) + H2O(l) ∅ NaOH(aq) + H2(g) B I2(g) ∅ I2(s) C H2SO4(aq) + Ba(OH)2(aq) ∅ BaSO4(s) + H2O(l) D H2(g) + 1/2O2(g) ∅ H2O(l) E None of the above A B C D E 7.9 % 31.6 % 15.8 % 12.7 % 22.3 % 316.1 kcal 12.5 kcal –291.1 kcal –316.1 kcal –337.3 kcal The reaction below is carried out at constant temperature in a 2.0 liter vessel N2 + O2 ∅ 2NO 40.5 kcal 60.3 kcal 75.8 kcal 81 kcal 162 kcal At equilibrium, the vessel is found to contain moles of NO, moles of N2 and moles of O2 What is the value of Keq for this reaction under these conditions? A B C D E What is the percent composition by mass of Al in Al2(SO4)3? A B C D E –68.4 –57.8 –20.2 +12.5 +54.2 –26.4 –94.1 Based on the values contained in the table, what is the heat of combustion of one mole of ethylene at 298K and atm pressure? Na+ K+ Mg++ Al3+ Cl– The heat of combustion of gaseous ammonia, NH3(g), is 81 kcal/mole How much heat is released in the reaction of 34 grams of ammonia with excess oxygen? A B C D E H2O(l) H2O(g) C2H6(g) C2H4(g) C2H2(g) CO(g) CO2(g) Which one of the following has the largest ionic radius? A B C D E _H of formation (kcal/mole, 25 °C, atm) Compound I 0.25 0.5 A slow → B fast II B + C → BC* (activated complex) III BC* → D + E fast Given the above series of reactions leading to the formation of products D and E, which is the ratedetermining step? A B C D E I II III I and II II and III K A P L A N G E N E R A L C H E M I S T R Y S U B J E C T T E S T K A P L A N _ According to VSEPR Theory, the H-Te-H bond angle in H2Te should be closest to A B C D E 90° 120° 109.5° 149.5° 180° A 300 mL flask containing nitrogen at a pressure of 100 torr and a 200 mL flask containing oxygen at a pressure of 200 torr are connected such that the gases are allowed to fill the combined volume of the two flasks What is the partial pressure of nitrogen in the combined volume? A B C D E 60 torr 80 torr 100 torr 150 torr 200 torr GENERAL CHEMISTRY SUBJECT TEST A B C D E F Cl Br K C 12 Which one of the electronic configurations below represents the atom with the highest ionization energy? A B C D E 1s22s22p63s1 1s22s22p63s23p64s23d104p65s14d10 1s22s22p63s23p64s23d104p65s24d105p5 1s22s22p63s23p63s23p5 1s22s22p63s23p6 13 Electron density studies have revealed that X and Y have an equal number of electrons Which of the following could be X and Y respectively? A B C D E 10 Which of the following statements is true with regard to the diagram below? Ca+ and K H+ and He Cl and F O– and S+ None of the above 14 Free Energy U W X Y Z Reaction Coordinate A Going from U to W requires less energy than going form X to Y B Going from U to X occurs more readily than going from X to Z C Although in the forward direction the reaction X to Z will occur more rapidly than from U to X, in the reverse reaction Z to X will occur more slowly than X to U D The reaction U to X is exothermic while the reaction from X to Z would be endothermic E The overall reaction U to Z is endothermic U 1st Ionization Energy (eV) 5.6 2nd Ionization Energy (eV) 5.6 V 1.7 2.9 X 1.1 13.6 Y 12.4 2.8 Z 2.9 1.7 From the information given in the table above, which of the following is most probably a Group IA metal? A B C D E U V X Y Z 11 Which of the elements below is the least electronegative? K A P L A N G E N E R A L C H E M I S T R Y S U B J E C T T E S T 15 HCl + Na2CO3 ∅ NaCl + H2O + CO2(g) (Atomic weights: H = 1, Cl = 35, Na = 23, O=16, C=12) 36 g of HCl are mixed with excess Na2CO3 If the reaction is carried out at STP, how many liters of CO2 are formed? A B C D E 5.60 11.2 16.8 22.4 44.8 16 Which of the following elements would be the most reactive with fluorine? A B C D E Ar Ti Cl Li I Questions 17 -18 refer to the following kinetic rates: Trial [A]o (M) x 10–2 x 10–2 x 10–2 [B]o (M) x 10–3 x 10–3 1.2 x 10–2 rate (M/min) x 10–3 x 10–3 3.2 x 10–2 17 What is the reaction order with respect to A? A B C D E 1.5 18 What would be the initial rate of the reaction in a fourth trial if [A]o = x 10–2 M and [B]o = x 10–3 M? A B C D E x 10–3 M/min x 10–3 M/min x 10–3 M/min x 10 –2 M/min Cannot be determined from the information given 19 A spontaneous reaction is always accompanied by A a negative change in enthalpy B a positive change in enthalpy C a negative change in enthalpy and a positive change in entropy D a positive change in enthalpy and a negative change in entropy E a negative change in Gibbs free energy 20 Combustion of a 50 gram sample of an unknown hydrocarbon yields 132 grams of CO2 and 126 grams of H2O Approximately how many grams of carbon were contained in the original sample? A B C D E 14 g 28 g 36 g 63 g 66 g 21 Under which of the following sets of conditions (temperature and pressure) will the ideal gas law be least likely to apply? A B C D E 250 K, atm 250 °C, 400 atm 25 °C, torr 25 K, 400 torr 25 K, 400 atm 22 The following reaction is first order with respect to W and first order with respect to X If the concentration of W and X are doubled what happens to the rate of the reaction? W+X∅Y+Z A B C D It is doubled It is quadrupled It is squared It is increased but it cannot be determined how much from the information given E It remains the same 23 If atom A has a valence of and atom B has a valence of 2, which compound consisting of atoms A and B would be the most likely to form? A B C D E BA B2A BA2 B2A2 B3A2 K A P L A N _ 24 In moving down column II A of the Periodic Table A B C D E electron affinity increases ionization energy increases electronegativity remains constant the oxidation state remains constant the shielding effect decreases GENERAL CHEMISTRY SUBJECT TEST 28 An ideal gas is placed in a 10 liter container that is free to expand If the temperature (K) is doubled and the pressure is changed from 60 mmHg to 20 mmHg, what is the new volume in liters? A B 25 Which of the following choices lists elements, from left to right, with decreasing atomic radii? A B C D E Li, Be, B, C, N Be, Mg, Ca, Sr, Ba Li, Mg, Sc, Zn, Ta Au, Ag, Cu Two of the above D E 26 Under which of the following conditions will a reaction always be in thermodynamic equilibrium? A B C D E ∆H ∆S ∆H ∆S ∆H ∆S ∆H x x 10 x x 10 x x 29 The concentration of which of the following substances would NOT decrease as a result of an increase in the temperature? A B C D E K+, verifying choice E as the credited choice E This question combines some basic stoichiometry with thermodynamic values for a reaction We are given the ∆Hcomb per mole of ammonia and the mass of ammonia in grams Applying dimensional analysis, we must convert the mass of ammonia into moles, then multiply by kcal/mol to find the total number of kcal Ammonia has a formula weight of 14 + 3(1) = 17 g/mol, therefore: (34 g)/(17 g/mol) = mol (2 mol).(81 kcal/mol) = 162 kcal Note that it is not necessary to come up with a balanced equation for the combustion reaction to this problem K A P L A N _ GENERAL CHEMISTRY SUBJECT TEST C The percent by mass of an element in a compound is calculated by multiplying the atomic mass of that element by the number of atoms of that element in one formula of the compound, then dividing by the total formula weight of the compound and, finally, converting the result to a percentage In this example there are two atoms of aluminum (atomic weight 27) and three sulfate ions (formula weight = 32 + 4(16) = 96) per formula of aluminum sulfate, so the calculation goes: 2(27)/(2(27) + 3(96)) = 54/342 which rounds off and converts to 15.8% Calculation can, as usual, be avoided by noticing that 54/342 will be approximately equal to 50/350, or roughly 15% Choice A is the result of inserting one aluminum atomic weight into the top of the calculation, while choices B, D, and E are the results of other random errors in calculation E This question simultaneously tests the basic definition of combustion and thermodynamic calculations involving heats of formation, along with a bit of basic organic nomenclature To calculate thermodynamic values for a reaction or process from given heats of formation, we must set up a correctly balanced equation for the process and then subtract the sum of the heats of formation, ∆Hf, of the reactants from the sum of the heats of formation of the products To so for this problem, we must first identify ethylene as the C2H4(g) entry in the table and then write a balanced equation for its combustion: C2H4(g) + O2(g) ∅ CO2(g) + H2O(l) Adding the values for the ∆Hf(products) = ∆Hf(CO2(g)) + ∆Hf(H2O(l)) and subtracting the sum of the ∆Hf(reactants) = ∆Hf(C2H4(g)) + ∆Hf(O2(g)), the result is : ∆Hcomb(C2H4(g)) = ∆Hf(products) - ∆Hf(reactants) = [2 ∆Hf(CO2(g)) + ∆Hf(H2O(l))]- [∆Hf(C2H4(g)) + ∆Hf(O2(g)) = 2(-94) + 2(-68.4) -(12.5) -3(0) = -337.3 kcal/mol (Recall that the heat of formation of any standard state element, like O2(g) here, is zero.) As for the wrong choices, notice first that choices A and B could have been immediately eliminated since combustion, by nature, is exothermic; a correct answer to this question must therefore be negative Choice D results from using the heat of formation of gaseous water rather than that of the liquid water that should be expected at 25oC, and choice C is the result obtained by adding 12.5 (when it should have been subtracted) in addition to using the value for gaseous rather than liquid water B We can calculate the equilibrium constant for this reaction by setting up the reaction quotient in terms of the concentrations of the reactants and products, then solving based on the numbers given in the problem Since we are told that the vessel is at equilibrium the reaction quotient is the equilibrium constant For the given reaction, the reaction quotient is: [NO]2 Qc = Keq = [N ][O ] 2 Since the reaction takes place in a 2.0 liter container, we can convert the given numbers of moles into moles per liter by dividing each by two, then plug the results into the expression for Qc: [NO] = mol/2 L = mol/L [N2] = mol/2 L = mol/L [O2] = mol/2 L = mol/l K A P L A N G E N E R A L C H E M I S T R Y S U B J E C T T E S T [NO]2 42 ∴ Keq = [N ][O ] =(1)(2) = 2 A This question can be answered based on fundamental definitions; the rate-determining step is simply the slowest step in a proposed mechanism Of the three steps shown, only step I is labeled as 'slow' and must therefore be the bottleneck in the overall process As such, it must be the rate-determining step C Tellurium, Te, is a Group VIA element, found below oxygen and sulfur in the Periodic Table It should thus have six valence electrons available for bonding If the central Te atom uses two of its valence electrons to form sigma bonds to the two hydrogens in H2Te, it will still have two pairs of nonbonding electrons remaining According to VSEPR theory H2Te fits the general formula AB2U2 and will have tetrahedral electronic geometry and angular molecular geometry, with bond angles of 109.5_ VSEPR theory makes the same prediction for H2O and H2S, and similar predictions for other tetrahedral electronic species such as methane and ammonia (Note: In reality, the bond angles generally vary from those predicted by the theory because of differences in repulsion between bonded and nonbonded electron pairs and because of orientational differences in the orbitals involved in the bonding.) A This question about gas volumes and pressures can best be answered by applying Dalton's law of partial pressures, which states that the pressure exerted by any one gas in a mixture of ideal gases is the same as it would be if the gas were to occupy the same volume alone If we focus on the nitrogen we can approach this problem as if the volume increases from the initial 300 mL to the combined 500 mL and apply Boyle's law: P1V1 = P2V2 (at constant temperature ) Since an increase in volume will lead to a decrease in pressure, the final pressure must be less than 100 torr Thus we can automatically eliminate choices C, D, and E before completing the calculation as follows: P2 = P1V1/V2 = (100 torr)(300 mL)/(500 mL) = 3/5(100 torr) = 60 torr 10 C This question is probably best approached by examining the answer choices in search of the one true statement required Choice A is false; it appears from the reaction coordinate diagram provided that W and Y are at approximately the same level while U is lower in energy than is X It will thus take more energy to go from the lower energy state U to the W/Y level than it will to go from the higher X energy level to that of W/Y In other words, the activation energy for the reaction U ∅ X is larger than the activation energy for the reaction X ∅ Z Choice B is essentially equivalent to choice A, since a reaction "occurs more readily" if and only if its activation energy is smaller Choice C is true In the forward direction, X ∅ Z has a smaller energy of activation (and is therefore a faster reaction) than does U ∅ X Meanwhile, the free energy of activation for Z ∅ X, that is, the energy required to go from Z to Y, is larger than the energy required to move from X to W Therefore, in the reverse direction, the reaction Z ∅ X will be slower than the reaction X ∅ U Choices D and E are both wrong for two reasons First, the diagram shown is a free energy diagram (∆G on the vertical axis) rather than an enthalpy diagram (∆H on the vertical axis); as such we cannot directly determine changes in enthalpy to which the terms exothermic and endothermic apply What we can determine from a free energy diagram is whether a given reaction is spontaneous or not Secondly, even if we did attempt to infer that enthalpy and free energy were following parallel trends, the reaction U ∅ X is an uphill move on the diagram and would thus be 10 _ K A P L A N _ GENERAL CHEMISTRY SUBJECT TEST endothermic (making choice D wrong) while the reaction U ∅ Z results in a net decrease in energy and would thus be exothermic (disqualifying choice E) 11 D Electronegativity increases toward the upper right-hand corner of the Periodic Table, with fluorine, F, the most electronegative element and cesium, Cs, the least Electronegativity is a property of atoms in bonds, i.e., it is actually a molecular property, therefore rare gases like helium, He, which not form compounds, are not included in the electronegativity scale (Heavier elements like francium, Fr, are not included because their half-lives are so short that their compounds can not be isolated.) The question can therefore be translated as "Which of the following is closest to the lower left-hand corner of the Periodic Table?" Choice D, potassium, lies far to the left of the other entries and as such is the least electronegative 12 E Ionization energy is defined as the amount of energy required to remove an electron from a gas phase atom or ion, and increases toward the upper right-hand corner of the Periodic Table, with helium, He, having the largest such energy requirement of any neutral atom This question can therefore be reworded as "Which of the following configurations corresponds to the element which is closest to the upper right-hand corner of the Periodic Table?" We can count the total number of electrons and, given that these are atoms rather than ions, thereby translate the answer choices into their corresponding elemental symbols: A total number of electrons = atomic number = 11 _ Na B total number of electrons = atomic number = 47 _ Ag C total number of electrons = atomic number = 53 _ I D total number of electrons = atomic number = 17 _ Cl E total number of electrons = atomic number = 18 _ Ar We can then see that choice E, argon, lies furthest to the upper right and therefore has the greatest ionization energy It is not necessary to convert the electron configurations into the atoms which they represent if one notices that choice E is the only one with all its subshells filled and would therefore be the hardest to remove an electron from 13 A Two species with the same number of electrons are said to be isoelectronic; this question can thus be reworded to read "Which of the following is an isoelectronic pair?" Calcium has atomic number 20 This means that a neutral atom of calcium has twenty protons in its nucleus and twenty electrons around its nucleus Positive charges result from the removal of outer electrons, thus to acquire a charge of positive one, calcium would need to lose one electron from its outermost, or 4s, orbital, leaving Ca+ with a total of 19 electrons Potassium, meanwhile, has atomic number 19 and thus also has 19 electrons as a neutral atom These two species are then seen to be isoelectronic, verifying choice A as the correct choice The species in each of the other choices not have the same number of electrons; H+ has no electrons while He has two, Cl has 17 while F has 9, and O- has electrons while S+ has 15 These choices, along with choice E, can thus be eliminated 14 C Ionization energy is defined as the amount of energy required to remove an electron from a given species Ionization energy is usually expressed in energy per particle (as it is here) or in energy per mole, and energy, in turn, is usually expressed in electron volts (eV), joules, or kilojoules The first ionization energy of an element is the energy required to remove an electron from a neutral atom of that element, while the second ionization energy is the energy required to remove a second electron, i.e., to remove an electron from the +1 cation A particularly useful piece of logic/knowledge K A P L A N _ 11 G E N E R A L C H E M I S T R Y S U B J E C T T E S T here is that it will be more difficult to remove an electron from a positively-charged species than it will be to remove an electron from a neutral version of the same species In other words, the second ionization energy of an element is always greater than its first ionization energy Based on this fact alone, choices A, D, and E can be eliminated To choose between the remaining two choices, we must reason out that a small value for ionization energy corresponds to relative ease of removal of the electron A Group IA metal atom will lose its first electron with relative ease, but after that it will possess an electronic configuration similar to that of a noble gas It will therefore be difficult to remove another electron, implying a high second ionization energy Looking at the numbers in the table, we see that both remaining choices have relatively small values for first ionization energy, but that X has a much higher second ionization energy It follows that element V is most likely a Group IIA element (e.g., Mg, Ca) while element X is most likely a member of Group IA (e.g., Na, K) Choice C, as indicated above, is thus the best choice 15 B According to the balanced equation, the complete reaction of two moles of HCl should result in the formation of one mole of CO2 We are given that 36 grams, or one mole, of HCl reacts; there should thus be half a mole of CO2 formed Assuming that the gas formed behaves ideally, and recalling that one mole of any ideal gas occupies 22.4 liters at STP, this half a mole of CO2 should occupy a volume of 11.2 liters as stated in choice B 16 D Reading from left to right across the Periodic Table, metallic character decreases as electronegativity increases The most reactive combinations of elements are those with the greatest differences in electronegativity, and this difference can be most easily predicted by observing the distance between the elements on the Periodic Table This question can thus be translated to read: "Which of the following lies furthest from F on the Periodic Table?" Lithium, in Group IA, is therefore the correct answer Titanium, choice B, would be the second most reactive of the options offered, while choices C and E, chlorine and iodine, are Group VIIA elements like fluorine, and thus would be only slightly reactive with it Choice A, argon, is an inert gas and would not be reactive at all 17 E The rate law for any reaction, A + B ∅ C, is given by: rate = k[A]x[B]y where k is the rate constant at the temperature at which the reaction is carried out, [A] and [B] are the concentrations of the two reactants, and x and y are the reaction orders with respect to A and B respectively This question asks us to find x from the experimental rate data provided To simplify the tabulated rate data, we can rewrite the table as small whole number multiples of the smallest entry in each column: line # [A]o [B]o rate 1 1 2 4 16 Looking at lines and 2, where the concentration of reactant A is held constant, we see that a doubling in the concentration of B results in a quadrupling of the rate, therefore the reaction is second order with respect to B We can now look at lines and (or and 3) to determine the required reaction order From line to line 3, the concentration of B is doubled; the reaction rate should thus quadruple due to the influence of the concentration increase of second order reactant B We can divide out this quadrupling effect and create a hypothetical line 4: 12 _ K A P L A N _ GENERAL CHEMISTRY SUBJECT TEST line # [A]o [B]o rate 1 1 2 4 16 2 From this revision of the experimental data it can be seen (lines and 4) that a doubling of [A] results in no change in the rate, i.e., 2x = Therefore the reaction is zero order with respect to A 18 B This question is based on the same reaction, and can be answered quickly based on the results outlined above As determined in the explanation above, the reaction is zero order in A; as a result, changing the concentration of A will have no effect on the reaction rate The original entry labeled as 'trial 1' has the same initial concentration of B, x 10-3 M, as that offered in this question The answer will thus be that the rate is the same as that in trial 1, or x 10-3 19 E This is a question on fundamental concepts of thermodynamics To be classified as spontaneous, in thermodynamic terms, a reaction or process must be accompanied by a decrease in Gibbs free energy, i.e., ∆G < Recall that ∆G = ∆H T∆S; depending on the sign and magnitude of ∆S, ∆H can be either positive or negative and still result in a negative ∆G, and choices A and B can be eliminated Choice C is a bit tricky: while it is true that the conditions listed in choice C will always result in a negative value for ∆G, these conditions are not, conversely, required for a reaction to be spontaneous and thus they will not always be true for any spontaneous reaction In other words, while the statement in C guarantees a spontaneous reaction, a spontaneous reaction does not necessarily guarantee the statement in choice C Finally, the conditions provided in choice D will never allow a process to proceed spontaneously; since T is the absolute temperature and hence always positive, ∆H - T∆S will always be positive for positive ∆H and negative ∆S 20 C This question of stoichiometry can be easier than it appears if we notice that we are only asked for the carbon content of the original sample, rather than for its complete formula We can therefore ignore the sample size and the water produced, focusing instead on the carbon containing product CO2 Since each formula of CO2 contains one carbon atom, it follows that each formula weight of CO2 will contain 12 grams of carbon The calculation can then be completed as follows: (132 g CO2)(12 g C/44 g CO2) = (132)(12)/(44) g C = 36 g C The reactants in a combustion reaction are the substance being combusted and oxygen All the carbon content in the form of carbon dioxide in the product thus has to come from the substance itself originally Therefore 36 g of carbon in the product means 36 g of carbon in the original sample 21 E Despite the numerical nature of the answer choices, this is a qualitative question on fundamental concepts regarding ideal gases Ideal gases are those which obey the ideal gas law, PV = nRT This obedience is most likely to occur for relatively inert, nonpolar gases under conditions of high temperature and low pressure Conversely, it follows that the ideal gas law will be least likely to apply for highly reactive, polar gases at low temperature and high pressure Since the K A P L A N _ 13 G E N E R A L C H E M I S T R Y S U B J E C T T E S T question does not identify the chemical nature of a specific gas, we can thus translate the given question as “Which of the following combinations has the lowest temperature and the highest pressure?” Choice E does Keep in mind that a torr is a much smaller unit of pressure than is an atmosphere (1 atm = 760 torr) 22 B The rate law for any reaction of the type W + X ∅ products is given by: rate = k[W]x[X]y where k is the rate constant at the temperature at which the reaction is carried out, [W] and [X] are the concentrations of the two reactants, and x and y are the reaction orders with respect to W and X respectively This question gives us x and y and asks us to predict the effect on the overall rate resulting from the given changes in concentration Since x and y are each equal to one, as implied by the fact that the reaction is first order with respect to each reactant, we may rewrite the rate law as: rate = k[W][X] If we substitute double concentrations for each of W and X, the new rate becomes k[2W][2X] = 4(k[W][X]), i.e it has quadrupled Note that if we were not given the information about the orders of the reaction, then the equation by itself is insufficient for us to predict the rate 23 C This question is best answered based on the chemical properties of elements as read from the Periodic Table A valence of puts element A in column VIIA along with fluorine, chlorine, etc., while a valence of puts element B in group IIA along with calcium and the rest of the alkaline earth metals Group IIA metals most readily form 2+ cations while the halogens prefer a 1- charge as anions; charge balance then requires that two A combine with each B, giving the empirical formula BA2 (This question could have been answered most conveniently, and quickly, by comparison to a reference compound such as CaCl2.) 24 D Group IIA is the group commonly referred to as the 'alkaline earth metals' Metallic character is greatest toward the lower left-hand corner of the Periodic Table Metallic character includes physical properties such as malleability, ductility, and luster, as well as chemical properties such as low ionization energy, electron affinity, and electronegativity Choices A and B are thus in complete opposition to the properties of metals, and should therefore be disposed of Choice C is also untrue, as electronegativity decreases as you move down the group On the other hand, the low ionization energy of metals allows them to easily give up electrons to a willing acceptor; the metals of Group IIA are easily oxidized to 2+ cations, making choice D a true statement Finally, choice E is incorrect since the effective nuclear charge, Zeff, remains roughly constant as one moves down a column of the Periodic Table: all the elements in column IIA have the electron configuration [G]ns2, where G is the symbol for the previous rare gas and n is the principal quantum number of the valence shell of the metal in question 25 E This is yet another question on atomic properties and periodic trends Atomic radius is greatest toward the lower left-hand corner of the Periodic Table; it generally decreases from left to right across a period, and increases from top to bottom down a group Choice A lists elements in order from left to right across period and thus represents elements in order of decreasing atomic radius Choice D lists three elements in order moving up a group and thus likewise in order of decreasing atomic radius Choice E, two of the above, is therefore the credited response Choice B lists the Group IIA elements in order of increasing atomic radius, while choice C is a somewhat randomized list of elements in (mostly) increasing order by size; choices B and C can thus be eliminated 14 _ K A P L A N _ 26 GENERAL CHEMISTRY SUBJECT TEST D To be classified as spontaneous, in thermodynamic terms, a reaction or process must be accompanied by a decrease in Gibbs free energy, i.e., ∆G < For a system to be at equilibrium, the forward and reverse processes must be equally spontaneous, therefore the net change in Gibbs free energy for the system must be equal to zero, i.e., ∆G = Recall that ∆G = ∆H - T∆S; it follows then that at equilibrium ∆H - T∆S = 0, or, that ∆H = T∆S An algebraic rearrangement of this last expression then gives choice D as the credited response 27 D This question is asking us to determine the minimum additional information necessary to find an ideal gas's volume If we begin with PV = nRT, we can focus on what is given, and figure out what is missing Volume is given as liter and temperature as 350K; we can thus rearrange the ideal gas law as: P = n(RT/V), where all the values inside the parentheses are known To complete the determination of pressure, we need only to find n, the number of moles of gas in the container Molecular weight alone does not tell us how many moles we have, so choice A can be eliminated If we know the number of molecules, as given in choice B, we can divide by Avogadro's number to find the number of moles present, so choice B will be sufficient Likewise, if we know the mass and molecular weight of the gas, as choice C offers, we can calculate the number of moles present and thereby complete the pressure calculation Since choices B and C are each sufficient while choice A is not, choice D is the credited response 28 E This question on gas volumes could be answered quantitatively by applying the ideal gas law, PV = nRT, but the format of the choices indicates that calculation is unnecessary One approach is to use the combined gas law: P1V1/T1 = P2V2/T2, and rearrange it to solve for V2 = T2P1V1/P2T1 = (T2/T1)(P1/P2)V1 = (2)(60/20)(10) = 60/20 x x 10, as written in choice E An alternate, more logical approach is to use the choices and think: if temperature is increased, the gas will expand and the volume therefore increases; we thus need to multiply by rather than divide by 2, and choices A, C, and D can be eliminated Similarly, if the pressure is decreased, the gas will expand, and we must multiply the original volume, 10 liters, by 60/20 rather than by 20/60; this observation alone allows us to eliminate choices A and B Choice E is then the only surviving choice after both of these observation/elimination sequences are completed 29 D This question is on equilibrium and le Châtelier's principle The reaction shown is exothermic, as evidenced by the negative value provided for ∆H Heat can thus be thought of as a product of this reaction, and raising the temperature should favor reactants, i.e., le Châtelier's principle predicts that the equilibrium will shift to the left A shift toward the formation of the reactants would increase the concentrations of X2Y and BD, making choice D the credited choice for this 'NOT' question, while decreasing the concentrations of AB and P 30 C The atomic weight of an element is calculated by multiplying the atomic mass of each isotope of that element by the relative abundance of that isotope In this example, there are only two isotopes of boron so the calculation set-up goes: (x)(10.013) + (1–x)(11.0093) = 10.811, where x = the relative abundance of 10B, and thus 1–x = the relative abundance of 11B To simplify the arithmetic, we can approximate and perform the claculation to just one decimal place: x(10.0) + (1–x)(11.0) = 10.8 10.0x + 11.0 – 11.0x = 10.8 –1.0x = –0.2 x = 0.2 = 20% K A P L A N _ 15 ... _ GENERAL CHEMISTRY SUBJECT TEST General Chemistry Subject Test Which one of the following reactions will be accompanied by an... K A P L A N _ GENERAL CHEMISTRY SUBJECT TEST GENERAL CHEMISTRY SUBJECT TEST ANSWER KEY A A 13 A 19 E 25 E E C 14 C 20 C 26 D E A 15 B 21 E... remains constant the oxidation state remains constant the shielding effect decreases GENERAL CHEMISTRY SUBJECT TEST 28 An ideal gas is placed in a 10 liter container that is free to expand If the