The enthalpy of neutralization of $NH_4OH$ with $HCl$ is $-51.40 \, kJ/equiv$. The enthalpy of dissociation of $NH_4OH$ is ..... $kJ$.

Using the given reaction enthalpies,find the enthalpy of formation of $H_2O_2(l)$ in $kJ/mol$.
$(i) N_2H_4(l) + 2H_2O_2(l) \rightarrow N_2(g) + 4H_2O(l); \Delta_r H_1^\circ = -818 \, kJ/mol$
$(ii) N_2H_4(l) + O_2(g) \rightarrow N_2(g) + 2H_2O(l); \Delta_r H_2^\circ = -622 \, kJ/mol$
$(iii) H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(l); \Delta_r H_3^\circ = -285 \, kJ/mol$

"The resultant heat change in a reaction is the same whether it takes place in one or several stages." This statement is called

Calculate $\Delta H^{\circ}$ for the reaction,$Na_2O_{(s)} + SO_{3(g)} \longrightarrow Na_2SO_{4(s)}$ given the following:
$(A) \ Na_{(s)} + H_2O_{(l)} \longrightarrow NaOH_{(s)} + \frac{1}{2} H_{2(g)} \quad \Delta H^{\circ} = -146 \ kJ$
$(B) \ Na_2SO_{4(s)} + H_2O_{(l)} \longrightarrow 2NaOH_{(s)} + SO_{3(g)} \quad \Delta H^{\circ} = +418 \ kJ$
$(C) \ 2Na_2O_{(s)} + 2H_{2(g)} \longrightarrow 4Na_{(s)} + 2H_2O_{(l)} \quad \Delta H^{\circ} = +259 \ kJ$

The combustion of carbon produces two oxides,$CO$ and $CO_2$,respectively. Their enthalpies of formation are $26 \ kcal$ and $94.3 \ kcal$ respectively. What is the enthalpy of combustion of carbon in $kcal$?

The enthalpy of solution of $NaOH$ (solid) in water is $-41.6 \ kJ \ mol^{-1}$. When $NaOH$ is dissolved in water,the temperature of the water