The conversion of oxygen to ozone represented by the equation $3O_2 \to 2O_3$ is an endothermic reaction. The enthalpy change $\Delta H$ accompanying the reaction:

$CuSO_{4(s)} + 5 H_2O_{(\ell)} \rightarrow CuSO_4 \cdot 5 H_2O_{(s)} ; \Delta H = -x \ kJ$. The value of $\Delta H$ represents:

The enthalpies of dissolution of $BaCl_2(s)$ and $BaCl_2 \cdot 2H_2O(s)$ are $-20.6 \ kJ \ mol^{-1}$ and $8.8 \ kJ \ mol^{-1}$ respectively. Calculate the enthalpy of hydration for the given reaction: $BaCl_2(s) + 2H_2O(l) \to BaCl_2 \cdot 2H_2O(s)$

Heats of combustion $(\Delta H^o)$ for $C_{(s)}$,$H_{2(g)}$ and $CH_{4(g)}$ are $-94$,$-68$ and $-213 \ kcal/mol$ respectively. The value of $\Delta H^o$ for the reaction,$C_{(s)} + 2H_{2(g)} \to CH_{4(g)}$ is $..... \ kcal$.

On the basis of the thermochemical equations:
$H_{2}O_{(g)} + C_{(s)} \to CO_{(g)} + H_{2(g)} \quad \Delta H = 131 \ kJ$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)} \quad \Delta H = -282 \ kJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_{2}O_{(g)} \quad \Delta H = -242 \ kJ$
$C_{(s)} + O_{2(g)} \to CO_{2(g)} \quad \Delta H = X \ kJ$
The value of $X$ will be $.... \ kJ$.

Nitroglycerine $(MW = 227.1)$ detonates according to the following equation:
$2 C_{3}H_{5}(NO_{3})_{3(l)} \longrightarrow 3 N_{2(g)} + \frac{1}{2} O_{2(g)} + 6 CO_{2(g)} + 5 H_{2}O_{(g)}$
The standard molar enthalpies of formation,$\Delta H_{f}^{\circ}$ for all the compounds are given below:
$\Delta H_{f}^{\circ}[C_{3}H_{5}(NO_{3})_{3}] = -364 \, kJ/mol$
$\Delta H_{f}^{\circ}[CO_{2(g)}] = -393.5 \, kJ/mol$
$\Delta H_{f}^{\circ}[H_{2}O_{(g)}] = -241.8 \, kJ/mol$
$\Delta H_{f}^{\circ}[N_{2(g)}] = 0 \, kJ/mol$
$\Delta H_{f}^{\circ}[O_{2(g)}] = 0 \, kJ/mol$
The enthalpy change when $10 \, g$ of nitroglycerine is detonated is $..... \, kJ$.