he solubility of oxygen gas in water at 25 °C and 1.0 atm pressure of oxygen is 0.041 g/L. The solubility of oxygen in water at 2.5 atm and 25 °C is ________ g/L.
1 answer:
Answer:
0.1025 g/L
Explanation:
From the question above,
The solubility of oxygen gas in water at 25°c and 1.0 atm pressure is 0.041g/L
Henry's law states at a constant temperature the solubility of a gas is proportional to the pressure
Since the temperatures are both similar (25°c) then, the solubility of oxygen in water at 2.5 atm can be calculated as follows
= 0.041/1.0 × 2.5
= 0.041×2.5
= 0.1025 g/L
Hence the solubility of oxygen in water at 2.5 atm and 25°c is 0.1025 g/L
You might be interested in
Answer:
Hot and something you do not touch
Explanation:
Answer:
1L of hot water just below the Boling point
Explanation:
asking questions is best to learn please ask more questions
One mole of hydrogen peroxide contains 6.02 x 10^23 molecules of hydrogen peroxide. And each molecule contains 4 atoms, so the answer is 4 x 6.02 x 10^23.
In chemical reactions the reactants are on the left side of the equation and the product are on the right
