The solubility of oxygen in lakes high in the rocky mountains is affected by the altitude. if the solubility of o 2 from the air is 2.67 × 10 –4 m at sea level and 25°c, what is the solubility of o 2 at an elevation of 12,000 ft where the atmospheric pressure is 0.657 atm?

assume the temperature is 25°c, and that the mole fraction of o 2 in air is 0.209 at both 12,000 ft and at sea level. 1.75 × 10–4 m 2.67 × 10–4 m 3.66 × 10–5 m 4.06 × 10–4 m

S = 1,75x10⁻⁴M

Explanation:

It is possible to answer this question using Henry's law that states that the amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid. The formula is:

S = k×Pi

Where S is solubility, K is Henry's constant and Pi is partial pressure.

At sea level, Pi of O₂ is:

1atm×0,209 = 0,209 atm. Replacing:

2,67x10⁻⁴M = k×0,209atm

k = 1,28x10⁻³M/atm

At 12,000ft Pi of O₂ is:

0,657atm×0,209 = 0,137 atm. Replacing:

S = 1,28x10⁻³M/atm×0,137atm

S = 1,75x10⁻⁴M

I hope it helps!

dynamic equilibrium is the answer.

warm front forms when a moist, warm air mass slides up and over a cold air mass. as the warm air mass rises, it condenses into a broad area of clouds. a warm front brings gentle rain or light snow, followed by warmer, milder weather.

hope i !