at its boiling temperature, how much heat (in kj) is needed to completely vaporize 23.4 g of h2o? the heat of vaporization for water at the boiling point is 40.6 kj/mole.

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09-12-2022
Chemistry

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at its boiling temperature, how much heat (in kj) is needed to completely vaporize 23.4 g of h2o? the heat of vaporization for water at the boiling point is 40.6 kj/mole.

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rathipushpendra345 rathipushpendra345 · Answer 1 · 2022-12-15T22:34:53-05:00

The right response, according to the definition of the heat of vaporization, is that 52.8 kJ is required to totally evaporate 23.4 g of water.

40.6 kJ/mole is the heat of vaporization for water at its boiling point. In other words, 40.6 kJ of energy is needed to convert a mole of liquid water into a mole of gas at a constant temperature.

The amount of mass that a substance holds in one mole is known as its molar mass. The water in this instance has a molar mass of 18 g/mol. The following formula can be used to determine how many moles 23.4 g of the chemical contains:

[tex]23.4 grams x\frac{1 mole}{18 grams} = 1.3 moles[/tex]

Then you can use the rule of three as follows: How much heat of vaporization is needed to vaporize 1.3 moles of water if 1 mole of water costs 40.6 kJ?

[tex]heat = \frac{1.3 moles * 40.6kJ}{1 mole}[/tex]

52.78 kJ 52.8 kJ = heat

To learn more about vaporization, refer:-

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