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How much heat (in kj) is absorbed when 32.5 g h2o( l) at 100°c and 101.3 kpa is converted to steam at 100°c? (the molar heat of vaporization of water is 40.7 kj/mol.)?

Respuesta :

The molar heat of vaporization of water is 40.7 kj/mol 
The molecular mass of water is 18g/mol
Therefore; the number of moles of water will be; 32.5g/18g/mol = 1.81 mol
Hence; the amount of heat absorbed will be;
1.81 mol × 40.7 kJ/mol = 73.7 kJ
Thus; heat absorbed will be; 73.7kJ 

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