Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning
ISBN 10: 1133949649
ISBN 13: 978-1-13394-964-0

Chapter 11 Intermolecular Forces and Liquids - Study Questions - Page 435f: 65


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Work Step by Step

From the previous question, we know: $ln\ P=-2804.8\ K/(T_C+273)+11.70$ At 20$°C$, this equation gives: $P=8.39\ atm$
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