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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 10.0 mol of benzene is vaporized at a constant pressure of 1.00 atm and at its normal boiling point of 80.1°C, 339 kJ are absorbed and PΔV for the vaporization process is equal to 29.0 kJ, then
A) ΔE = 310. kJ and ΔH = 339. kJ.
B) ΔE = 368. kJ and ΔH = 339. kJ.
C) ΔE = 339. kJ and ΔH = 310. kJ.
D) ΔE = 339. kJ and ΔH = 368. kJ.
Textbook 
Introductory Chemistry Essentials

Introductory Chemistry Essentials


Edition: 5th
Author:
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hbutler2hbutler2
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