Enthalpy of neutralization of `HCl` by `NaOH` is `-57.1 J//mol` and by `NH_(4)OH` is `-51.1 KJ//mol`. Calculate the enthalpy of dissociation of `NH_(4)OH`.
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Enthalpy of neutralization of HCl by NaOH is -55.84 kJ/mol and by NH_(4)OH is 51.34 kJ/mol. The enthalpy of ionization of NH_(4)OH is :
The enthalpy of neutralization of NH_(4)OH and CH_(3)COOH is – 10.5 kcal/mole and enthalpy of neutralization of strong base and CH_(3)COOH is – 12.5 kcal/mole. Calculate the enthalpy of dissociation of NH_(4) OH -
The enthalpy of neutralisation of HCl by NaOH IS -55.9 kJ and that of HCN by NaOH is -12.1 kJ mol^(-1) . The enthalpy of ionisation of HCN 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
Enthalpy of neutralisation of NH_(4)OH and HCl is numerically
Enthalpy of neutralization of H_(3)PO_(3) "with " NaOH " is" -106.68 "kJ"//"mol" . If enthalpy of neutralization of HCL with NaOH is -55.84 "kJ"//"mole" , then calculate enthalpy of ionization of H_(3)PO_(3) in to its ions in kJ.
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