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12 Dec 2019

Fish need at least 4 O2 dissolved for survival.What is this concentration in mol/L?What partial pressure of O2 above the water is needed to obtain this concentration at 10 degrees celcius ? (The Henry's law constant for O2 at this temperature is 1.71x 10^-3 mol/L*atm .)

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Patrina Schowalter
Patrina SchowalterLv2
13 Dec 2019

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Related questions

The solubility of oxygen, O2, in water is 7.20 ✕ 10−4 mol/L at 0°C when the nitrogen pressure above water is 0.554 atm. Calculate the solubility of oxygen in water when the partial pressure of oxygen above water is 1.083 atm at 0°C? The Henry's Law constant for oxygen is 1.30 ✕ 10−3 mol/ L atm.

Part #1:

Which of the following would DECREASE the amount of gas dissolved in your tank?

A) Increase the air pressure above the tank

B) Decrease the Temperature

C) Increase the Temperature

Part #2:

At a particular temperature, you determine that in 50.4 L of water, 0.703 grams of N2 gas are dissolved. You measure the pressure above the tank to be 0.771 atm. Assuming the maximum amount of gas has been dissolved, calculate the Henry's Law constant in the units of mol/(L*atm). Assume that there is ONLY nitrogen in the air above the tank.

Part #3:

You determine that at room temperature (25 oC) the Henry’s Law constant of O2 gas is 1.3 * 10-3 mol/(L*atm). If you have a 50.4 L of water in your fish tank, and the air pressure is 0.771 atm in the room, how many grams of O2 gas could be dissolved? (enter the number only, without units)

Given that the solubility of O2 in water at 25oC is 8.70 ppm, calculate (a) the concentration of dissolved O2 in mol L-1 and (b) the value of the Henry’s law constant, in mol L-1 kPa-1, for O2 at this temperature? The atmospheric pressure is 1.00 atm. (Hint: the mole fraction of oxygen in air is 0.210).

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