Chemistry: Molecular Approach (4th Edition)

Published by Pearson
ISBN 10: 0134112830
ISBN 13: 978-0-13411-283-1

Chapter 5 - Exercises - Page 240: 27

Answer

a. $$ 809.0 \space mmHg$$ b. $$ 1.064 \space atm$$ c.$$ 809.0 \space torr$$ d. $$ 107.9 \space kPa$$

Work Step by Step

a. $$ 31.85 \space in Hg \times \frac{1 \space atm}{29.92 \space in Hg} \times \frac{760 \space mmHg}{1 \space atm} = 809.0 \space mmHg$$ b. $$ 31.85 \space in Hg \times \frac{1 \space atm}{29.92 \space in Hg}= 1.064 \space atm$$ c.$$ 31.85 \space in Hg \times \frac{1 \space atm}{29.92 \space in Hg}\times \frac{760 \space torr}{1 \space atm} = 809.0 \space torr$$ d. $$ 31.85 \space in Hg \times \frac{1 \space atm}{29.92 \space in Hg}\times \frac{101.325 \space kPa}{1 \space atm} = 107.9 \space kPa$$
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