Ask a New Question
Menu

N2H4(g) + H2(g) 2 NH3(g) H1 = –1876 kJ

3 H2(g) + N2(g) 2 NH3(g) H2 = –922 kJ

The H f for the formation of hydrazine: 2 H2(g) + N2(g) N2H4(g) will be _____ kJ/mol.

3 answers

You can't have an equation without arrows. No arrows means we don't know the reactants from the products.
95.4 kJ/mol
can you elaborate on how you got that browmf0x?
Similar Questions
  1. N2H4(g) + H2(g) ---> 2 NH3(g) H1 = –1876 kJ3 H2(g) + N2(g) ---> 2 NH3(g) H2 = –922 kJ The H f for the formation of
    1. answers icon 1 answer
  2. N2H4(g) + H2(g) 2 NH3(g) H1 = –1876 kJ3 H2(g) + N2(g) 2 NH3(g) H2 = –922 kJ The H f for the formation of hydrazine: 2 H2(g)
    1. answers icon 1 answer
  3. Please :)N2H4(g) + H2(g) 2 NH3(g) H1 = –1876 kJ 3 H2(g) + N2(g) 2 NH3(g) H2 = –922 kJ The H f for the formation of
    1. answers icon 0 answers
  4. Please :)N2H4(g) + H2(g) 2 NH3(g) H1 = –1876 kJ 3 H2(g) + N2(g) 2 NH3(g) H2 = –922 kJ The H f for the formation of
    1. answers icon 1 answer
more similar questions