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zhenek [66]
2 years ago
8

The molar heat of vaporization for methane, CH4, is 8. 53 kJ/mol. How much energy is absorbed when 54. 8 g of methane vaporizes

at its boiling point? Use q equals n delta H. 6. 42 kJ 29. 1 kJ 137 kJ 467 kJ.
Chemistry
1 answer:
alexandr402 [8]2 years ago
7 0

Molar heat of vaporization is defined as the heat absorbed by one mole of substance to convert from liquid to gas.

<h3>How do you calculate the heat of vaporization?</h3>

The formula used to calculate the heat of vaporization is:

\rm Q &= n \times \Delta H

Where,

Q = Amount of Heat

n = number of moles of a substance

\Delta H = molar enthalpy of fusion

Now, to calculate the moles of methane:

\rm Moles &= \dfrac{Mass \;of\; methane }{Molar\; Mass\; of \; Methane} &= \dfrac{54.8\;g}{16\;g/mol}

Moles = 3.425 mol

Now, 1 mol of methane absorbs = 8.53 KJ

3.425 mol of methane absorbs = 3.424 \times 8.53 &= 29.1 KJ

Thus, the energy is absorbed till the methane vaporizes at its boiling point is 29.1 KJ.

Learn more about <u>vaporization </u>here:

brainly.com/question/2491083

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