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Consider the following equations.
N2H4(l) + O2(g) N2(g) + 2 H2O(l) ÄH = -622.2 kJ
H2(g) + 1/2 O2(g) H2O(l) ÄH = -258.5 kJ
H2(g) + O2(g) H2O2(l) ÄH = -187.8 kJ
Use this information to calculate the enthalpy change for the reaction shown below.
N2H4(l) + 2 H2O2(l) N2(g) + 4 H2O(l) ÄH = ?

I got -764.kj. ia that correct?

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