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Question
When 1.00 mol of benzene is vaporized at a constant pressure of 1.00 atm and at its normal boiling point of 80.1C, 33.9 kJ are absorbed and P
V for the vaporization process is equal to 2.90 kJ, then
A) E = 31.0 kJ and H = 33.9 kJ.
B) E = 36.8 kJ and H = 33.9 kJ.
C) E = 33.9 kJ and H = 31.0 kJ.
D) E = 33.9 kJ and H = 36.8 kJ.
Answer
This answer is hidden. It contains 1 characters.
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