How many kj of heat are needed to completely vaporize 1.30 moles of h2o? the heat of vaporization for water at the boiling point is 40.6 kj/mole?

Explanation:

Heat of vaporization is defined as the amount of heat required to change one mole of a liquid into vapor state without any change in the temperature.

It is known that for 1 mole of water, latent heat of vaporization is 40.6 kJ/mol.

Therefore, heat of vaporization for 1.30 moles will be calculated as follows.

               = 52.78 kJ

Thus, we can conclude that the 52.78 kJ of heat are needed to completely vaporize 1.30 moles of .

a gas, having a molecular weight of 13 kg/kmol and a kinematic viscosity of 0.25 cm2/s, flows through a pipe 0.25 m internal diameter and 5 km long at the rate of 0.4 m3/s and is delivered at atmospheric pressure.

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