Heat of solution of `BaCl_(2).2H_(2)O=200 kJ mol^(-1)`
Heat of hydration of `BaCl_(2)=-150 kJ mol^(-1)`
Hence heat of solution of `BaCl_(2)` is

Lattice energy of NaCl_((s)) is -788kJ mol^(-1) and enthalpy of hydration is -784kJ mol^(-1) . Calculate the heat of solution of NaCl_((s)) .

Lattice energy of NaCl_((s)) is -788kJ mol^(-1) and enthalpy of hydration is -784kJ mol^(-1) . Calculate the heat of solution of NaCl_((s)) .

Calculate the enthalpy of hydration of BaCl_(2) to BaCl_(2).2H_(2)O (s) given that the enthalpy of solution of BaCl_(2) (s) is - 20.6 kJ mol^(-1) and that of BaCl_(2).2H_(2)O(s) is + 8.8 KJ mol^(-1) .

Enthalpy of solution for BaCl_(2). 2H_(2)O and BaCl_(2) are 18.8 and -30.6 kL mol^(-1) respectively. Calculate the enthalpy of hydration of BaCl_(2) to BaCl_(2). 2H_(2)O .

Enthalpies of solution of BaCI_(2). 2H_(2)O and BaCI_(2) are 8.8 and -20.6 kJ mol^(-1) respectively.Calculate the heat of hydration of BaCI_(2). 2H_(2)O

The heat of solution of BaCl_(2) .2H_(2) O_((s)) and anhydrous BaCl_(2(s)) are 8.37" kJ mol"^(–1) and –20.62" kJ mol"^( –1) respectively then heat of hydration of BaCl_(2(s)) will be :-

Enthalpies of solution of BaCl_(2)(s) and BaCl_(2).2H_(2)O(s) are -20 kJ//"mole" and 8.0kJ//"mole" respectively. Calculate heat of hydration of BaCl_(2)(s) .

The enthalpy of solution of BaCl_2(s) and BaCl_2.2H_2O(s) are -20.6 kJ mol^-1 and 8.8kJ mol^-1 . The enthalpy change for the formation of hydrated salt is

If the heats of formation of C_(2)H_(2) and C_(6)H_(6) are 230 kJ mol^(-1) and 85 kJ mol^(-1) respectively, the DeltaH value for the trimerisation of C_(2)H_(2) is