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Kay [80]
3 years ago
6

What is the solubility of methylacetylene (in units of grams per liter) in water at 25 °C, when the C3H4 gas over the solution h

as a partial pressure of 0.301 atm? kH for C3H4 at 25 °C is 9.23×10-2 mol/L·atm.
Chemistry
1 answer:
Luda [366]3 years ago
7 0

Answer:

The solubility of methylacetylene is 0,11 g L⁻¹

Explanation:

Henry's law is a gas law that states that the amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid.

The formula is:

C = kH P

Where C is solubility of the gas (In mol/L)

kH is Henry constant (9,23x10⁻² mol L⁻¹ atm⁻¹)

An P is partial pressure (0,301 atm)

Solving, C = 2,78x10⁻³ mol L⁻¹. In grams per liter:

2,78x10⁻³ mol L⁻¹ₓ \frac{40 g}{mol} = <em>0,11 g L⁻¹</em>

<em></em>

I hope it helps!

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_H2 SiCl2 + _H2O −−→ _H8 Si4O4 + _HCl
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pls mark brainliest right

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