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The heat of vaporization for water is 40.7 kJ/mol. How much heat energy must 150.0 g of water absorb to boil away completely? Use the formula

339.2 kJ

381.6 kJ

610.5 kJ

6,105 kJ

Sagot :

breathingforward

Answer:

B. 381.6 works

Explanation:

The answer can be any of these but I and the answer above have approved of option B.

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View image breathingforward

The heat of vaporization for water is 40.7 kJ/mol, the heat energy will be  6,105 kJ.

What is heat of vaporization?

The heat of vaporization was described as the quantity of heat required to convert 1 gram of liquid into such a vapor without raising the liquid's temperature.

By using the formula:

q = m ΔH

where, q is heat m is mass and ΔH is heat of vaporization.

Given data:

ΔH = 40.7 kJ/mol

m = 150.0 g

Now, put the value of given data in heat of vaporization formula.

q = 150.0 g × 40.7 g J/mol

q =  6,105 kJ

Therefore, the heat will be 6,105 kJ.

To know more about heat of vaporization.

https://brainly.com/question/13372553

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