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saplinglearning.com California S Sapling -178-CHEM111-01-4360 Dashboard-My Study Life ty, Chico -CHEM 111-Fall7- HILLYARD Activities and Due Dates HW 12 Gra 11/2017 09:55 AM 3.9/1012/7/2017 01:31 PM Calculator -rd Periodic Table Print 17 of 34 Ma Sapling Learning At 298 K, the Henry's law constant for oxygen is 0.00130 M/atm. Air is 21.0% oxygen. At 298 K, what is the solubility of oxygen in water exposed to air at 1.00 atm? Number Tools x 102 At 298 solubility of oxygen in water exposed to air at 0.893 atm? Number If atmospheric pressure suddenly changes from 1.00 atm to 0.893 atm at 298 K, how much oxygen will be released from 3.90 L of water in an unsealed container? Number mol O Previous Give Up & View Solution 0 Next int Check Answer

Explanation / Answer

The partial pressure of oxygen in the atmosphere is 1 atm * 21% O2 = 0.21 atm

1 atm, the O2 in the water is 0.00130 M/atm * 0.21 * 1 atm = 0.000273 mole / L

0.893 atm, the O2 in the water is 0.00130 M/atm * 0.21 * 0.893 atm = 0.000244 mole / L

If you drop the pressure, the change in solubility is 0.000273 M - 0.000244 M = 0.000029 M

Concentration = mole / L, mole = concentration * volume = 0.000029 mole / L * 3.6 L = 0.0001044 mole

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