# PART VIII: PCAT PRACTICE TEST in Visual Studio .NET Generate ECC200 in Visual Studio .NET PART VIII: PCAT PRACTICE TEST

PART VIII: PCAT PRACTICE TEST
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79 10 3 = x, which is the concentration of H3O+ pH = log[H3O+] = log(79 10 3) = 210 Even without a calculator, you can estimate the pH Starting with 63 10 5 = x2, you can rewrite this in a more convenient form: 63 10 6 = x2 This is very close to 64 10 6, which has a square root of 8 10 3 = x You can estimate the pH by using powers of 10 The value of x is between 10 3 and 10 2, but closer to 10 2 The pH is between log(10 3) and log(10 2) or between 30 and 20, but closer to 20 The only value that meets these criteria is choice A 25 Answer: A + The weak acid CH3COOH dissolves to produce an acidic solution NH4Cl contains NH 4 , which is the conjugate acid of a weak base and produces an acidic solution The anion Cl is the conjugate base of a strong acid which has no acidic or basic properties The salt Mg(NO3)2 will not affect the pH because neither ion is acidic or basic The salt NaCH3COO dissolves to form Na+ ions and CH3COO ions Sodium ion has no acid or basic properties, but CH3COO is the conjugate base of the weak acid, CH3COOH This anion will make the solution basic by removing a proton from water according to the following equilibria: CH3COO (aq) + H2O(l) CH3COOH(aq) + OH (aq) 26 Answer: D The pH of buffers is determined by the Henderson-Hasselbalch equation: pH = pKa + log For this problem: pH = 314 + log [(03)] [(30)] [A ] [HA]
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pH = 314 + log(010) pH = 314 + ( 10) = 214 27 Answer: B The solubility equilibrium equation, ICE table, and solubility product constant are: CuSCN(s) Cu+(aq) + SCN (aq) Initial 0 0 Change +x +x Equilibrium x x Ksp = [Cu+][SCN ] = x2 16 10 11 = x2 40 10 6 M = x
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To work this problem without a calculator with a square root key, rewrite the equation as 16 10 12 = x2 The left side is now two perfect squares and the solution is easily obtained as 40 10 6 M = x 28 Answer: A The spontaneity of a reaction is given by the value of the free energy, G , which is de ned by: G = H T S If the value of G is negative, the reaction is spontaneous (product favored), a positive value means the reverse reaction is spontaneous, and a value of 0 means the system is at equilibrium For this reaction, the H term is negative and the term T S is also negative, because S is positive The two terms make the overall value of G negative, which means the reaction is spontaneous as written 29 Answer: C The species that is reduced will have a decrease in the oxidation number The oxidation number of each species is given in the equation below Ca(s)+Sn2+(aq) Ca2+(aq)+Sn(s) 0 +2 +2 0 The oxidation number of Ca changes from 0 to +2, which is an oxidation The Sn changes from +2 to 0, which is a reduction The Sn2+ ion picks up two electrons and is reduced to Sn metal The correct half reaction for the species that is reduced is: Sn2+(aq) + 2 e Sn(s) 30 Answer: C Choice A is a termination step, and choice B is a highly endothermic fragmentation Choice D resembles a propagation step, but it actually consumes product Choice C obeys all the guidelines of a propagation step: it provides a continuation of the radical cycle, it consumes stoichiometric reagent, and it provides one of the observed products 31 Answer: C Since the rst step in bromination is highly endothermic, the transition state energies resemble those of the intermediate carbon radicals This is why bromination is a selective process Therefore, the most highly substituted center is most rapidly brominated 32 Answer: D This step represents bond formation, which is always exothermic; therefore, H is negative Since two species are combining to form one, translational degrees of freedom are lost, and S is also negative 33 Answer: D The Baeyer-Villiger reaction converts ketones to esters; thus choices C and D are the most valid These two choices differ with regard to which alkyl group has migrated
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