$CuSO_{4(s)} + 5 H_2O_{(\ell)} \rightarrow CuSO_4 \cdot 5 H_2O_{(s)} ; \Delta H = -x \ kJ$. The value of $\Delta H$ represents:

Calculate $\Delta H \ (kJ/mol)$ for the reaction:
$2FeO_{(s)} + \frac{1}{2} O_{2_{(g)}} \to Fe_2O_{3_{(s)}}$
Given $\Delta H$ values:
$(i)$ $Fe_2O_{3_{(s)}} + 3C_{(graphite)} \to 2Fe_{(s)} + 3CO_{(g)}$ : $492 \ kJ/mol$
$(ii)$ $FeO_{(s)} + C_{(graphite)} \to Fe_{(s)} + CO_{(g)}$ : $156 \ kJ/mol$
$(iii)$ $C_{(graphite)} + O_{2_{(g)}} \to CO_{2_{(g)}}$ : $-393 \ kJ/mol$
$(iv)$ $CO_{(g)} + \frac{1}{2} O_{2_{(g)}} \to CO_{2_{(g)}}$ : $-283 \ kJ/mol$

Calculate the enthalpy of solution of potassium chloride $(KCl)$ if its lattice enthalpy $\Delta_{L} H = 700 \ kJ \ mol^{-1}$ and hydration enthalpy $\Delta_{hyd} H = -680 \ kJ \ mol^{-1}$.

Calculate the heat of formation of $HCl$ gas from the following reaction: $H_{2(g)} + Cl_{2(g)} \rightarrow 2 HCl_{(g)} ; \Delta H = -194 \ kJ$

The enthalpy of neutralization of $NaOH$ by $HCl$ is $-55.84 \, kJ/mol$ and that of $NH_4OH$ by $HCl$ is $-51.34 \, kJ/mol$. The enthalpy of ionization of $NH_4OH$ is ..... $kJ/mol$.

The heat change $\Delta H$ for the reaction $2CO + O_2 \to 2CO_2; \Delta H = -135 \ kcal$ is called