From the following data,the enthalpy of vaporization of liquid water in $KJ \, mol^{-1}$ will be:
$H_2(g) + 1/2 O_2(g) \rightarrow H_2O(l); \Delta H = -285.77 \, KJ \, mol^{-1}$
$H_2(g) + 1/2 O_2(g) \rightarrow H_2O(g); \Delta H = -241.84 \, KJ \, mol^{-1}$

What is the enthalpy change when $6.80 \, g$ of $NH_3$ is passed over hot $CuO$ (in $, kJ$)? The standard enthalpies of formation for $NH_3(g)$,$CuO(s)$,and $H_2O(l)$ are $-46.0$,$-155.0$,and $-285.0 \, kJ \, mol^{-1}$ respectively. The reaction is: $NH_3(g) + \frac{3}{2}CuO(s) \to \frac{1}{2}N_2(g) + \frac{3}{2}H_2O(l) + \frac{3}{2}Cu(s)$

The heat of reaction does not depend upon

The enthalpy change for the reaction $2CO_{(g)} + O_{2(g)} \rightarrow 2CO_{2(g)}$ is known as:

If enthalpies of formation of $C_2H_{4(g)}$,$CO_{2(g)}$ and $H_2O_{(l)}$ at $25 \ ^\circ C$ and $1 \ atm$ pressure are $52$,$-394$ and $-286 \ kJ \ mol^{-1}$ respectively,the enthalpy of combustion of $C_2H_{4(g)}$ will be.....$kJ \ mol^{-1}$.

State Hess's Law of Constant Heat Summation.