In the following reaction,how much enthalpy change in $kJ$ is associated with the formation of $5.67 \ mol$ of $HCl$ gas? $H_{2(g)} + Cl_{2(g)} \to 2HCl_{(g)} \; ; \Delta H = -184.6 \ kJ$

The $\Delta H^o_f$ for $CO_{2(g)}$,$CO_{(g)}$,and $H_2O_{(g)}$ are $-393.5$,$-110.5$,and $-241.8 \ kJ/mol$ respectively. The standard enthalpy change (in $kJ$) for the reaction $CO_{2(g)} + H_{2(g)} \to CO_{(g)} + H_2O_{(g)}$ is:

The calorific value of glucose is ...... $KJ/g$. Given: $C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O ; \Delta H = -2900 \, KJ/mol$.

$C_6H_{6(l)} + 7.5O_{2(g)} \to 6CO_{2(g)} + 3H_2O_{(g)}$; $\Delta H = -3267.7 \ kJ \ mol^{-1}$. Given that the standard enthalpies of formation of $CO_{2(g)}$ and $H_2O_{(g)}$ are $-393.5 \ kJ \ mol^{-1}$ and $-285.85 \ kJ \ mol^{-1}$ respectively,calculate the standard enthalpy of formation of benzene $(C_6H_{6(l)})$.

In the reaction for the transition of carbon in the diamond form to carbon in the graphite form,$\Delta H = -453.5 \ cal$. This points out that

The enthalpy of neutralization of $NH_4OH$ with $HCl$ is $-51.40 \, kJ/equiv$. The enthalpy of dissociation of $NH_4OH$ is ..... $kJ$.