# General Chemistry Chapter 6 (Chang)

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Question description

1) Calculate ∆H(rxn) for N2H4(l) + O2(g) = N2(g) 2H2O(l)

​2NH3(g) + 3N2O(g) = 4N2(g) + 3H2O(l) ∆H(rxn) = -1010 kJ/mole

N2O(g) + 3H2(g) = N2H4(l) + H2O(l) ∆H(rxn) = -317 kJ/mole

2NH3 + (1/2) O2(g) = N2H4(l) + H2O(l) ∆H(rxn) = -143 kJ/mole

H2(g) + (1/2) O2(g) = H2O (l) ∆H(rxn) = -286 kJ/mole

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Anonymous
Outstanding Job!!!!

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