The enthalpy change for the combustion of `N_(2)H_(4)(l)` (Hydrazine) is `-622.2 kJ mol^(-1)`. The products are `N_(2)(g)` and `H_(2)O(l)`. If `Delta_(f)H^(Theta)` for `H_(2)O(l) is -285.8 kJ mol^(-1)`. The `Delta_(f)H^(Theta)` for hydrazine isA. `-336.4 kJ mol^(-1)`B. `+50.6 kJ mol^(-1)`C. `-622.2kJ mol^(-1)`D. `+336.4 kJ mol^(-1)`

The enthalpy change for the combustion of `N_(2)H_(4)(l)` (Hydrazine)

1.	The enthalpy change for the combustion of `N_(2)H_(4)(l)` (Hydrazine) is `-622.2 kJ mol^(-1)`. The products are `N_(2)(g)` and `H_(2)O(l)`. If `Delta_(f)H^(Theta)` for `H_(2)O(l) is -285.8 kJ mol^(-1)`. The `Delta_(f)H^(Theta)` for hydrazine isA. `-336.4 kJ mol^(-1)`B. `+50.6 kJ mol^(-1)`C. `-622.2kJ mol^(-1)`D. `+336.4 kJ mol^(-1)`
Answer» `DeltaH = DeltaH_(P) - DeltaH_(R)`

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