How many kj or heat are needed to completely vaporize 23.4 g of h2o? the heat of vaporization for water at boiling point is 40.6 kj/mole. 31.2 52.8 23.4 2.26 none of the above?

Find how many moles of water you are dealing with. There are 18 g per mole.

23.4 g / 18 grams per mole = 1.3 moles of water

40.6 kJ are needed for each mole so 40.6 kJ x 1.3 moles = 52.8 kJ

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CintiaSalazar CintiaSalazar · Answer 2 · 2021-11-23T16:05:51+01:00

Considering the definition of heat of vaporization, the correct answer is second option: 52.8 kJ are needed t ocompletely vaporize 23.4 g of H₂O.

The heat of vaporization for water at boiling point is 40.6 kJ/mole. In other words, the amount of energy required to transform a mole of liquid water into gas at constant temperature is 40.6 kJ.

Molar mass is the amount of mass that a substance contains in one mole. In this case, the molar mass of the water is 18 g/mole. Then the number of moles that 23.4 g of the compound contains can be calculated as:

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

Then you can apply the following rule of three: if 1 mole of water requires a heat of vaporization of 40.6 kJ, 1.3 moles of water, how much heat of vaporization does it require?

[tex]heat=\frac{1.3 molesx40.6 kJ}{1 mole}[/tex]

heat= 52.78 kJ ≅ 52.8 kJ

The correct answer is second option: 52.8 kJ are needed to completely vaporize 23.4 g of H₂O.

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