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qaws [65]
3 years 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:
Daniel [21]3 years ago
4 0

Answer:

K = 137.55 atm/M.

Explanation:

  • The relationship between gas pressure and the  concentration of dissolved gas is given by  Henry’s law:

<em>P = (K)(C)</em>

where P is the partial pressure of the gaseous  solute above the solution (P = 1.0 atm).

k is a constant (Henry’s constant).

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

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

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The answer is "0.00172172603".

Explanation:

Given:

Mass (M) = 1.60 \times  10^{-3} \ g\\\\Density (D) = 9.293 \times 10^{-1} \ \frac{g}{cm^3}\\\\Volume (V) =  ?

Formula:

\to \bold{V = \frac{M}{D}}

       = \frac{1.60 \times  10^{-3}}{9.293 \times  10^{-1}} \\\\  = \frac{1.60 \times  10^{1}}{9.293 \times  10^{3}} \\\\ = \frac{1.60 \times  10}{9.293 \times  1000} \\\\ = \frac{1.60 }{9.293 \times  100} \\\\ = \frac{1.60 }{929.3 } \\\\= 0.00172172603

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