Answer:
The answer is "[tex]\bold{\ 15.4 \frac{kj}{mol}} \\[/tex] ".
Step-by-step explanation:
Equation:
[tex]KCN(s) \rightarrow K^+(aq)+CN^-(aq) \ \ \ \[/tex] ΔH =?
[tex]\ Q = (84.2) (4.184)(-12.7) \\\\\ Q = -4474\\\\\ Q_(r \times r) = \frac{+ 4.474 kj}{0.29mol} \\\\\ Q= 15.4 \frac{kj}{mol}[/tex]
