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 neutralization between HCl and NaOH is -13.7 kcal "equiv"^(–1) . Heat of neutralization of H_2C_2O_4 (oxalic acid) with NaOH is -26 kcal mol^(-1) . Hence, heat of dissociation of H_(2)C_(2)O_(4) as H_2C_2O_(4) rarr 2H^(+) + C_(2)O_(4)^(2-) , is :

Heat of neutralization between HCl and NaOH is -13.7 kcal "equiv"^(–1) . Heat of neutralization of H_2C_2O_4 (oxalic acid) with NaOH is -26 kcal mol^(-1) . Hence, heat of dissociation of H_(2)C_(2)O_(4) as H_2C_2O_(4) rarr 2H^(+) + C_(2)O_(4)^(2-) , 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

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 neutralisation of CH_3COOH_((aq)) with NaOH_((aq)) is - 55.2kJ. What is heat of ionisation of CH_3COOH ?

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 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

The heat of neutralization of NaOH with HCl is 57.3 KJ and with HCN is 12.1 KJ. The heat of ionization of HCN is