The standard enthalpies of formation of $CO$ and $CO_2$ are $-110 \, kJ \, mol^{-1}$ and $-394 \, kJ \, mol^{-1}$ respectively. What will be the heat of combustion of $1 \, mol$ of graphite in $kJ$?

The heat of neutralization of a strong acid and a weak base is:

The heat of neutralisation of one equivalent of an acid by one equivalent of a base is minimum when:

Enthalpy of formation of $CO_{(g)}$ and $CO_{2(g)}$ are $-110 \ kJ \ mol^{-1}$ and $-393 \ kJ \ mol^{-1}$ respectively. The enthalpy of combustion of $CO$ (in $kJ \ mol^{-1}$) is:

Given that the bond energies of $N \equiv N$ is $946 \ kJ \ mol^{-1}$,$H-H$ is $435 \ kJ \ mol^{-1}$,$N-N$ is $159 \ kJ \ mol^{-1}$,and $N-H$ is $389 \ kJ \ mol^{-1}$,calculate the enthalpy of formation for the gas phase reaction $N_2 + 2H_2 \rightarrow N_2H_4$ in $kJ \ mol^{-1}$.

Given: $S + \frac{3}{2} O_2 \to SO_3 + 2x \ \text{kcal}$,$\Delta H = -2x \ \text{kcal}$ and $SO_2 + \frac{1}{2} O_2 \to SO_3 + y \ \text{kcal}$,$\Delta H = -y \ \text{kcal}$. Find the heat of formation of $SO_2$.