If $(i)$ $C + O_2 \to CO_2$,$(ii)$ $C + 1/2 O_2 \to CO$,$(iii)$ $CO + 1/2 O_2 \to CO_2$,the heats of reaction are $Q$,$-12$,and $-10$ respectively. Then $Q =$ ?

Consider the following data:
$\Delta_{f}H^{\Theta}(CH_{4}, g) = -X \ kJ \ mol^{-1}$
Enthalpy of sublimation of graphite = $Y \ kJ \ mol^{-1}$
Dissociation enthalpy of $H_{2} = Z \ kJ \ mol^{-1}$
The bond enthalpy of $C-H$ bond is given by:

What is the heat of atomisation of $P_4O_{6(s)}$?
Given that the heat of sublimation of $P_4O_6$ is $x \ kJ/mol$ and the $P-O$ bond energy is $y \ kJ/mol$.

The heat of solution of $Na_2SO_{4(s)}$ and $Na_2SO_4 \cdot 10H_2O_{(s)}$ are $-2.34 \ kJ \ mol^{-1}$ and $78.87 \ kJ \ mol^{-1}$ respectively. The heat of hydration of $Na_2SO_{4(s)}$ is:

Calculate the enthalpy change in $kJ$ for the reaction: $2C_{(graphite)} + 2H_{2(g)} \to C_2H_{4(g)}$
$C_{(graphite)} + O_{2(g)} \to CO_{2(g)} \quad \Delta H = -393.5 \ kJ$
$C_2H_{4(g)} + 3O_{2(g)} \to 2CO_{2(g)} + 2H_2O_{(l)} \quad \Delta H = -1410.9 \ kJ$
$H_{2(g)} + 1/2O_{2(g)} \to H_2O_{(l)} \quad \Delta H = -285.8 \ kJ$

When $1 \, \text{mol}$ of anhydrous salt $AB$ is dissolved in water,$21.0 \, J \, \text{mol}^{-1}$ of heat is released. The enthalpy of hydration of $AB$ is $-29.4 \, J \, \text{mol}^{-1}$. What is the enthalpy of solution of the hydrated salt $AB \cdot 2H_2O_{(s)}$ in $J \, \text{mol}^{-1}$ (in $.4$)?