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Calculate the standard reaction enthalpy for the following reaction. N2H4(l) + H2(g) -----> 2NH3(g) Given: N2H4(l) + O2(g) -------> N2 (g) + 2H2O(g) DH = -543 kJ x mol-1 2H2(g) + O2(g) ---------> 2H2O(g) DH = -484 kJ x mol-1 N2(g) + 3H2(g) ---------> 2NH3(g) DH = -92.2 kJ x mol-1

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