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what mass of Pb^2+ ion is left in solution when 50 mL of 0.20 M Pb(NO3)2 is added to 50 mL of 1.5 NaCl? given k sp of PbCl2 is 1.7 x 10^-4
Dear Nancy,
q = mass x specific heat capacity x change in temperature q = [mPb(NO3)2(aq) + mKI(aq)] x cg x (Tf - Ti) q = [50 + 30] x 4.184 x (22.2 - 19.6) = 870.27 J
n = M x V n[Pb(NO3)2(aq)] = 0.20 x 50 x 10-3 = 0.010 mol
H = -q/1000 ÷ n H = -0.870/1000 ÷ 0.010 = -87 kJ mol-1 H is negative because the reaction is exothermic.
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