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chem1211 hw 10

# chem1211 hw 10 - 1 Convert 574.5 mmHg to atm torr Pa and...

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1. Convert 574.5 mmHg to atm, torr, Pa, and kPa. (1 atm = 1.01325 x 10 5 Pa) 0.7559 atm 574.5 torr 76592.0 Pa 76.59 kPa 574.5 mmHg x (1atm/760 mmHg) = 0.7559 atm 574.5 mmHg = 574.5 torr 0.7559 atm x (101325Pa/1atm) = 76592 Pa 76592Pa x (1kPa/1000Pa) = 76.59 kPa 2. A balloon that containes 1.40 liters of air at 725 torr is taken under water to a depth at which the pressure is 2.68 atm. Calculate the new volume of the balloon. Assume the temperature remains constant. 0.498 L First, change 725 torr to atm 725 torr x (1 atm / 760 torr) = 0.954 atm V 2 = P 1 V 1 /P 2 = (0.954 atm) x (1.40 L) / (2.68 atm) = 0.498 L 3. A 51.0 L sample of gas collected in the upper atmosphere at a pressure of 52.9 torr is compressed into a 165.0 mL container at the same temperature. What is the new pressure in atm? 21.5 atm To what volume would the original sample have had to be compressed in order to exert a pressure of 10.0 atm? 355.0 mL Use Boyle's Law: P = [52.9torr x (1atm/760torr) x 51.0 L]/[165.0 mL x (1L/1000mL)] = 21.5 atm V ={ [52.9torr x (1atm/760torr) x 51.0 L]/10.0 atm } x (1000mL/1L) = 355 mL 4. A scuba diver exhales 3.19 L of air while swimming at a depth of 20.0 m where the sum of atmospheric and water pressure is 3.06 atm. By the time the bubbles of air rise to the

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chem1211 hw 10 - 1 Convert 574.5 mmHg to atm torr Pa and...

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