Asked by guy
Argon makes up 0.93% by volume of air. Calculate its solubility in water a 20 C and 1.0 atm. The henry's law constant for argon under these conditions is 0.0015 mol/L*atm
Answers
Answered by
DrBob222
C = K*p
C = (0.0015 mol/L*atm)*0.0093 = ? M.
C = (0.0015 mol/L*atm)*0.0093 = ? M.
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