How much heat energy is required to boil 13.3 g of ammonia, NH3? The molar heat of vaporization of ammonia is 23.4 kJ/mol.?
回答 (1)
2016-02-24 8:24 am
Molar mass of NH₃ = 14 + 1×3 = 17 g/mol
No. of moles of NH₃ = (13.3 g) / (17 g/mol) = 0.782 mol
Heat energy required = (23.4 kJ/mol) × (0.782 mol) = 18.3 kJ
No. of moles of NH₃ = (13.3 g) / (17 g/mol) = 0.782 mol
Heat energy required = (23.4 kJ/mol) × (0.782 mol) = 18.3 kJ
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