If heat of neutralization is $-13.7 \, KCal$ at $25 \, ^oC$ and $\Delta H_f^o (H_2O) = -68 \, KCal$,then the standard enthalpy of formation of $OH^{-}$ will be.....$KCal$. (in $.3$)

$C_{(diamond)} + O_2 \to CO_2; \Delta H = -395.3 \ kJ/mole$
$C_{(graphite)} + O_2 \to CO_2; \Delta H = -393.4 \ kJ/mole$
$C_{(graphite)} \to C_{(diamond)}; \Delta H = ?$

Given the thermochemical reactions: $C(\text{graphite}) + \frac{1}{2}O_2 \rightarrow CO; \Delta H = -110.5 \, kJ$ and $CO + \frac{1}{2}O_2 \rightarrow CO_2; \Delta H = -283.2 \, kJ$. Calculate the heat of reaction for $C(\text{graphite}) + O_2 \rightarrow CO_2$ in $kJ$.

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 $..... \, kJ$.

When $10 \ g$ of methane is completely burnt in oxygen, the heat evolved is $560 \ kJ$. What is the heat of combustion (in $kJ \ mol^{-1}$) of methane?

The enthalpy changes for the following reactions are given:
$Cl_{2(g)} = 2Cl_{(g)}, 242.3 \, kJ \, mol^{-1}$; $I_{2(g)} = 2I_{(g)}, 151.0 \, kJ \, mol^{-1}$
$ICl_{(g)} = I_{(g)} + Cl_{(g)}, 211.3 \, kJ \, mol^{-1}$; $I_{2(s)} = I_{2(g)}, 62.76 \, kJ \, mol^{-1}$
Given that the standard states of iodine and chlorine are $I_{2(s)}$ and $Cl_{2(g)}$,the standard enthalpy of formation for $ICl_{(g)}$ is $...... \, kJ \, mol^{-1}$.