Suppose 23.00 g ammonium chloride (53.491 g/mol)dissolves in water in a calorimeter to form 100.00 g of solution.The temperature decreases by 10.75°C.Calculate the $\Delta$ H of solution of NH₄Cl in kJ/mol.Assume the specific heats of the solution and of the empty calorimeter are 4.18 J/(g . °C)and 0.170 kJ/°C,respectively.
NH₄Cl(s) $\to$ NH₄⁺(aq)+ Cl^-(g) $\quad$ $\Delta$ H = ? kJ/mol

Question

Suppose 23.00 g ammonium chloride (53.491 g/mol)dissolves in water in a calorimeter to form 100.00 g of solution.The temperature decreases by 10.75°C.Calculate the

\Delta

H of solution of NH₄Cl in kJ/mol.Assume the specific heats of the solution and of the empty calorimeter are 4.18 J/(g . °C)and 0.170 kJ/°C,respectively.
NH₄Cl(s)

\to

NH₄⁺(aq)+ Cl^-(g)

\quad

\Delta

H = ? kJ/mol

Quizplus · Accepted Answer

The Answer of Suppose 23.00 g ammonium chloride (53.491 g/mol)dissolves in water in a calorimeter to form 100.00 g of solution.The temperature decreases by 10.75°C.Calculate the $\Delta$H of solution of NH₄Cl in kJ/mol.Assume the specific heats of the solution and of the empty calorimeter are 4.18 J/(g . °C)and 0.170 kJ/°C,respectively. NH₄Cl(s)$\to$ NH₄⁺(aq)+ Cl^-(g) $\quad $$\Delta$H = ? kJ/mol

Suppose 23 Δ\DeltaΔ H of Solution of NH4Cl in KJ/mol →\to→ NH4+(aq)+ Cl-(g)

Suppose 23 $\Delta$ H of Solution of NH₄Cl in KJ/mol $\to$ NH₄⁺(aq)+ Cl^-(g)