Ag+(aq)+Cl−(aq)→AgCl(s)ΔH=−65.5kJ
Please explain how to do this as well as..
Calculate ΔH for the formation of 8.00g of AgCl.
and
Calculate ΔH when 9.21×10−4mol of AgCl dissolves in water.
1 Answer

(0.480 mol AgCl) × (−65.5 kJ/mol AgCl) = −31.44 kJ
(8.00 g AgCl) / (143.3212 g AgCl/mol) × (−65.5 kJ/mol AgCl) = −3.66 kJ
(9.21 × 10^−4 mol AgCl) × (+65.5 kJ/mol) = 0.0603255 kJ = +60.3 J
[Note that the sign on ΔH changed because the dissolving of AgCl in water is represented by the backwards form of the given equation.]