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Solutions
C . ΔT = (1.23 kJ/K)(6.12 K) = 7.53 kJ
Molar mass of NH4NO3 = 80 g/mol
When 1 g NH4NO3 decomposed, 7.53 kJ of heat is released
When 80 g which is 1 mol NH4NO3 decompose, the heat will be;
(80 g/mol x 7.53 kJ) / 1 g = 602.4 kJ/mol
is the molar heat of decomposition for ammonium nitrate.