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At 298 K, the Henry's law constant for oxygen is 0.00130 M/atm. Air is 21.0% oxygen.

1) At 298 K, what is the solubility of oxygen in water exposed to air at 1.00 atm?

2) At 298 K, what is the solubility of oxygen in water exposed to air at 0.89 atm?

3) If atmospheric pressure suddenly changes from 1.00 atm to 0.893 atm at 298 K, how much oxygen will be released from 4.70L of water in an unsealed container?

Respuesta :

Answer:

a) 0.000273 mol /L

b) 0.000244 mol/L

c) 0.0001044 mole

Explanation:

The partial pressure of oxygen in the atmosphere  =

1 atm [tex]* 21[/tex]%  of atm O2 [tex]= 0.21[/tex] atm

a) Solubility [tex]0.00130 * 0.21 * 1 = 0.000273[/tex]  mole / L

b) The solubility of oxygen in water exposed to air at 0.89 atm = [tex]0.00130 * 0.21 * 0.89 = 0.000244[/tex] mole / L

c) Oxygen  released from 4.70 L of water in an unsealed container If=  

[tex]0.000273 - 0.000244 = 0.000029[/tex]

Concentration [tex]= 0.000029 * 3.6 = 0.0001044[/tex] mole

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