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Home Q & A Board Science-Related Homework Help Chemistry (Moderator: Laser_3)
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Whelan Whelan
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8 years ago
When 2.00 mol of benzene is vaporized at a constant pressure of 1.00 atm and at its normal boiling point of 80.1°C, 67.8 kJ are absorbed and PΔV for the vaporization process is equal to 5.80 kJ then
A) ΔE = 62.0 kJ and ΔH = 67.8 kJ.
B) ΔE = 73.6 kJ and ΔH = 67.8 kJ.
C) ΔE = 67.8 kJ and ΔH = 73.6 kJ.
D) ΔE = 67.8 kJ and ΔH = 62.0 kJ.
Textbook 
Introductory Chemistry Essentials

Introductory Chemistry Essentials


Edition: 5th
Author:
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hbutler2hbutler2
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8 years ago Edited: 2 years ago, bio_man
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3 years ago
Thank you.
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