13-32 - fraction of O 2 in air is 0.21 Now we proceed with the data for the mountain lake P O2 =(665 torr(760 torr atm-1(0.21 = 0.18 atm And

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Question 13.32 Sorry I missed the point about partial pressures during our help session on Saturday, but here’s the solution to this problem. The information about sea level data is used to get the mole fraction of oxygen in the air: Since P O2 = 0.21 atm at an assumed normal atmospheric pressure at sea level, the mole
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Unformatted text preview: fraction of O 2 in air is 0.21. Now we proceed with the data for the mountain lake: P O2 = (665 torr)/(760 torr atm-1 ) (0.21) = 0.18 atm And finally, using Henry’s law: S O2 = (1.38 x 10-3 M atm-1 )(0.18 atm) = 2.5 x 10-4 M...
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This note was uploaded on 07/17/2008 for the course CHEM 122 taught by Professor Zellmer during the Winter '07 term at Ohio State.

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