`Delta_(f)H(H_(2)O) =- 68 kcal mol^(-1)` and `DeltaH` of neutralisation is `-13.7 kcal mol^(-1)`, then the heat of formation of `overset(Theta)OH` is

The enthalpy of the reaction : H_(2)O_(2)(l)toH_(2)O(l)+(1)/(2)O_(2)(g) is -23.5 kcal mol^(-1) and the enthalpy of formation of H_(2)O(l) is -68.3 kcal mol^(-1) . The enthalpy of formation of H_(2)O_(2)(l) is

The enthalpy of the reaction H_(2)O_(2) ("?") to H_(2) O ("?") + 1//2 O_(2) (g) is -23.5 kcal mol^(-1) and the enthalpy of formation of H_(2) O (?) is -68.3 kcal mol^(-1) . The enthalpy of formation of H_(2) O_(2) (?) is -

If heat of neutralisation is -13.7 k cal and H_f^(@)H_2O =-68 k cal , then enthalpy of OH^- would be:

The heat of formation of CO_(2) and H_(2)O is -97 kcal and -68 kcal. The heat of combustion of benzene is -783 kcal. Calculate the heat of formation of benzene ?

The heat of formation of CO_(2) and H_(2)O is -97 kcal and -68 kcal. The heat of combustion of benzene is -783 kcal. Calculate the heat of formation of benzene ?