If atmospheric pressure suddenly changes from 1.00 atm to 0.896 atm at 298 k, how much oxygen will be released from 3.30 l of water in an unsealed container?

At  a temperature of 298 K, the Henry's law constant is 0.00130 M/atm for oxygen. The solubility of oxygen in water 1.00 atm would be calculated as follows:

S = (H) (Pgas) = 0.00130 M / atm x 0.21 atm = 0.000273 M

At 0.890 atm,
S = (H)(Pgas) = 0.00130 M / atm x 0.1869 atm = 0.00024297 M

If atmospheric pressure would suddenly change from 1.00 atm to 0.890 atm at the same temperature, the amount of oxygen that will be released from 3.30 L of water in an unsealed container would be as follows

3.30 L x (0.000273 mol / L) = 0.0012012 mol

3.30 L x (0.00024297 mol / L) = 0.001069068 mol

0.0012012 mol - 0.001069068 mol = 0.000132 mol

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Velocity=total displacement/timefrom house to store is 500 from store to neighbor is 200 therefore total displacement=500+200=700metrestherefore,700metres/910seconds=0.8metres per second