Chemistry (4th Edition)

Published by McGraw-Hill Publishing Company
ISBN 10: 0078021529
ISBN 13: 978-0-07802-152-7

Chapter 10 - Section 10.1 - Properties of Gases - Checkpoint - Page 428: 10.1.2

Answer

(b), (c) and (d) are correct.

Work Step by Step

(a) $$0.80 atm \times \frac{101,325 \ Pa}{1 \ atm} \times \frac{1 \ mmHg}{133.322 \ Pa} = 608 \ mmHg$$ (b) $$4180 mmHg \times \frac{133.322 \ Pa}{1 \ mmHg} = 5.573 \times 10^{5} \ Pa$$ (c) $$433 torr \times \frac{133.322 \ Pa}{1 \ torr} \times \frac{1 \ mmHg}{133.322 \ Pa} = 433 mmHg $$ (d) $$2.300 atm \times \frac{101,325 \ Pa}{1 \ atm} \times \frac{1 \ torr}{133.322 \ Pa} = 1748 torr$$ (e) $$5.5 atm \times \frac{101,325 \ Pa}{1 \ atm} = 5.6 \times 10^5 Pa$$
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