The vaporization of ethanol at its normal boiling point, 351 K, has S = 110. J/K
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The vaporization of ethanol
at its normal boiling point, 351 K, has ΔS = 110. J/K · mol. Calculate ΔE for the vaporization process at 1 atm and 351 K.
Transcribed Image Text:
C₂H5OH(1)→ C₂H5OH(g)
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E 1 atm 110 JK mol 351 K 2932 K ...View the full answer
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Related Book For
Chemistry
ISBN: 9781305957404
10th Edition
Authors: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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