The enthalpy of neutralisation of $NH_4OH$ and $HCl$ is numerically:

$C_{(s)} + O_{2(g)} \to CO_{2(g)}; \Delta H = -94 \ kcal$
$CO_{(g)} + \frac{1}{2}O_{2(g)} \to CO_{2(g)}; \Delta H = -67.7 \ kcal$
Then the heat of formation of $CO_{(g)}$ is $..... \ kcal$

The enthalpy of neutralization is about $57.3 \ kJ$ for the pair:

The bond energies of $H-H$ and $Cl-Cl$ are $430 \ kJ \ mol^{-1}$ and $240 \ kJ \ mol^{-1}$ respectively. If the $\Delta H_f$ (enthalpy of formation) of $HCl$ is $-90 \ kJ \ mol^{-1}$,what is the $H-Cl$ bond energy in $kJ \ mol^{-1}$?

Average $C-H$ bond energy is $416 \ kJ \ mol^{-1}$. Which of the following is correct?

On the basis of the following equations,the heat of dimerisation of $NO_2$ will be:
$(i) \ N_2 + 2O_2 \to 2NO_2, \Delta H = 67.9 \ kJ$
$(ii) \ N_2 + 2O_2 \to N_2O_4, \Delta H = 9.3 \ kJ$