Heat of neutralisation between `HCI` and `NaOH` is `13.7kcal` and between `HCN` and `NaOH` is `3`kcal at `45^(@)C`. Calculate the heat of ionisation of `HCN`

Heat of neutralization between HCl and NaOH is -13.7 k.cal . If heat of neutralization between CH_(3)COOH and NaOH is -11.7 k.cal . Calculate heat of ionization of CH_(3)COOH .

heat of neutralisation of HCl against NaOH is 13.7 kcal eq^(-1) what will be the ionisation energy of CH_(3)COOH in kcal mol^(-1) if its heat of neutralisation is 11.7 kcal eq^(-1) .

The heat released in neutralisation of HCI and NaOH is 13.7 kcal/mol, the heat released on neutralisation of NaOH with CH_(3)COOH is 3.7 kcal/mol. The Delta H^(@) of ionsiation of CH_(3)COOH is

The heat released in neutralisation of HCI and NaOH is 13.7 kcal/mol, the heat released on neutralisation of NaOH with CH_(3)COOH is 3.7 kcal/mol. The Delta H^(@) of ionsiation of CH_(3)COOH is

Heat of neutralization of HCl by NaOH is 13.7 kcal per equivalent and by NH_4OH is 12.27 kcal. The heat of dissociation of NH_4OH is

Heat of neutralization of HCl by NaOH is 13.7 kcal per equivalent and by NH_4OH is 12.27 kcal. The heat of dissociation of NH_4OH is

The heat of neutralisation of HCI with NaOH is same as that of HNO_(3) with KOH .

The heat of neutralisation of HCI with NaOH is same as that of HNO_(3) with KOH .