ANSWER KEY sample final

ANSWER KEY sample final - ANSWER KEY: Sample Final (1) In...

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Unformatted text preview: ANSWER KEY: Sample Final (1) In combination 1) Be is the symbol for beryllium and in 4) the (IV) symbol should not be used to indicate the oxidation state of S. (2) 4 Au + 8 KCN +O2 + 2 H2O --> 4 KAu(CN)2 + 4 KOH 4 + 8 + 1 + 2 + 4 + 4 = 23 (3) 79 + %Mg-25 + %Mg-26 = 100% so .79 + .21 = 1.00. If x = frac. Mg-25 then .21 - x = frac. Mg-26 .79 * 23.99 + x * 24.99 + (.21-x) * 25.98 = 24.31 ( MW of Mg ) 18.95 + 24.99x + 5.46 - 25.98x = .99x = .10 x = .10 or 10% (4) 1) and 3) are properties of the noble gases (main group 8) (5) .6446 * molecular mass = 10 * 12.01(MW of C) molecular mass = 120.1/.6446 = 186.3 (6) 0.0500L * 0.10M = 0.0050 moles of AgNO3 0.0250L * 0.050M = 0.00125 moles of NaCl Since there is more Ag+ (.005 moles) than Cl- (.00125 moles) only .00125 moles of AgCl will precipitate leaving .00500 - .00125 =.00375 moles of Ag+ left in solution. .00375 moles/(.050 + .025 = .075) = .050M (7)175g/91.22(MW of Zr) = 1.918 moles of Zr required Thus 1.918 moles of ZrCl4 must be used to make it. 1 mole ZrCl4 requires 4 moles of Cl2 so 4 * 1.918 = 7.672 moles of Cl2 are required. 7.672*70.9(MW of Cl2) = 544g (8) moles of CO2 = PV/RT = 1.0 * 4.29 / (.082 * 573) = .0913 Since moles CO2 = moles CaCO3 in rock, .0913 * 100.1(MW of CaCO3) = 9.14g of CaCO3 in the rock. (21.8 - 9.1)/21.8 = .583 or 58.3% (9) .025L * 2.50M = .0625 moles of NH3 PV=nRT --> V = nRT/P = .0625 * .082 * 273 / 1 = 1.40L (10) Statement 1) is wrong because PV=nRT fails at both high pressure and low temperature. (11)If x = pressure(PCl3) = pressure(Cl2) then 1.35-x = pressure(PCl5) Thus total pressure = 1.35 + x. x = .24 * 1.35 = .324 Total pressure = 1.35 + .32 = 1.67. (12) For the reaction given: StandardDeltaH(rxn) = 2*(-393.5) + 4*(-241.8) - [ 2*(-238.7) + 3*0 ] = -1276.8kJ Thus [ 1 mol O2 / 3 mol O2 per rxn ] * -1276.8 = -426 (13) First break 1 Cl-Cl bond and 1 Br-Br bond. This costs 243 + 193 = 436 kJ/mol. Then make 2 Br-Cl bonds, which releases 2 * 218 = 436 kJ/mol. The difference 436 - 436 = 0 is StandardDeltaH(rxn) (14)StandardDeltaH(rxn) = sum StandardDeltaH(products) - sum StandardDeltaH(rxn) (reactants) 926 = 435 (for 2 H) + 498/2 (for 1 O) - [ -241.8 (for H2O) ]= 435 (for 2 H) + 498/2 (for 1 O) - [ -241....
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This homework help was uploaded on 04/17/2008 for the course CHE 101 taught by Professor Churchhill during the Spring '08 term at SUNY Buffalo.

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ANSWER KEY sample final - ANSWER KEY: Sample Final (1) In...

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