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Vinil7
14 days ago
14

Assuming that the solubility of radon in water with 1 atm pressure of the gas over the water at 30 ∘C is 7.27×10−3M, what is the

Henry's law constant for radon in water at this temperature?
Chemistry
1 answer:
alisha [2.8K]14 days ago
4 0

Response:

K = 137.55 atm/M.

Justification:

  • The correlation of gas pressure to the concentration of solubilized gas is expressed by Henry’s law:

P = (K)(C)

where P refers to the partial pressure of the solute gas above the solvent (P = 1.0 atm).

k is a fixed value (Henry’s constant).

C is the concentration of the solubilized gas (C = 7.27 x 10⁻³ M).

Thus, K = P/C = (1.0 atm)/(7.27 x 10⁻³ M) = 137.55 atm/M.

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