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Question
When 5.00 mol of benzene is vaporized at a constant pressure of 1.00 atm and at its normal boiling point of 80.1C, 169.5 kJ are absorbed and P
V for the vaporization process is equal to 14.5 kJ then
A) E = 155.0 kJ and H = 169.5 kJ.
B) E = 184.0 kJ and H = 169.5 kJ.
C) E = 169.5 kJ and H = 184.0 kJ.
D) E = 169.5 kJ and H = 155.0 kJ.
Answer
This answer is hidden. It contains 1 characters.
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