Asked by Esra
The solubility of argon in water at 25°C is 0.0150 mol/L. What is the Henry's Law constant for argon if the partial pressure of argon in air is 0.00934 atm?
Answers
Answered by
bobpursley
http://www.tau.ac.il/~tsirel/dump/Static/knowino.org/wiki/Henry%27s_law.html
k=p/c so the number is dependent on the units you choose. If p is i atm, and c in mol/L. In some texts, K is defined differently.
k=.00934/.0150 atm*L/mol
k=p/c so the number is dependent on the units you choose. If p is i atm, and c in mol/L. In some texts, K is defined differently.
k=.00934/.0150 atm*L/mol
Answered by
Angeline
0.61mol/L
Answered by
Jessica Keith
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