The molar neutralization heat for $KOH$ and $HNO_3$ as compared to the molar neutralization heat of $NaOH$ and $HCl$ is:

If $C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)}$,$\Delta H = -X$,and $CO_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{2(g)}$,$\Delta H = -Y$,calculate $\Delta_f H$ for $CO_{(g)}$ formation.

$C + O_2 \to CO_2; \Delta H = X$
$CO + \frac{1}{2} O_2 \to CO_2; \Delta H = Y$
Then the heat of formation of $CO$ is

Hess's law of constant heat summation is based on

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 $..... \ kcal$.

If the combustion of $1 \ g$ of graphite produces $20.7 \ kJ$ of heat,what will be the molar enthalpy change? Give the significance of the sign also.