Multiple Choice
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.
Correct Answer:
Verified
Correct Answer:
Verified
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