Chemistry: Molecular Approach (4th Edition)

Published by Pearson

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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