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

From the following data:
$CH_3OH_{(l)} + \frac{3}{2}O_{2(g)} \longrightarrow CO_{2(g)} + 2H_2O_{(l)}$; $\Delta_rH^{\circ} = -726 \ kJ \ mol^{-1}$
$H_{2(g)} + \frac{1}{2}O_{2(g)} \longrightarrow H_2O_{(l)}$; $\Delta_rH^{\circ} = -286 \ kJ \ mol^{-1}$
$C_{(graphite)} + O_{2(g)} \longrightarrow CO_{2(g)}$; $\Delta_rH^{\circ} = -393 \ kJ \ mol^{-1}$
The standard enthalpy of formation of $CH_3OH_{(l)}$ in $kJ \ mol^{-1}$ is:

Identify the law for the statement "Overall the enthalpy change for a reaction is equal to the sum of enthalpy changes of individual steps in the reaction".

The heat of neutralization for $2 \ moles$ of $\text{LiOH}$ and $1 \ mole$ of $H_2SO_4$ at $25^{\circ} C$ is $-69.6 \ kJ$. The heat of ionisation of $\text{LiOH}$ will be nearly $:-$

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.....$kJ \ mol^{-1}$.

The bond energy (in $kcal \ mol^{-1}$) of a $C-C$ single bond is approximately: