Question

When a 19.2-g sample of KCN dissolves in 65.0 g of water in a calorimeter, the temperature drops from 28.1 (C) to 15.4 (C). Calculate DH for the process

Answer 1

The answer is "
`\bold{\ 15.4 (kj)/(mol)} \\` ".

Explanation:

Equation:

`KCN(s) \rightarrow K^+(aq)+CN^-(aq) \ \ \ \` ΔH =?

`\ Q = (84.2) (4.184)(-12.7) \\\\\ Q = -4474\\\\\ Q_(r * r) = (+ 4.474 kj)/(0.29mol) \\\\\ Q= 15.4 (kj)/(mol)`