Use Henry's law and the solubilities given below to calculate the total volume of nitrogen and oxygen gas that should bubble out of 1.1L of water upon warming from 25 ∘C to 50 ∘C. Assume that the water is initially saturated with nitrogen and oxygen gas at 25 ∘C and a total pressure of 1.0 atm. Assume that the gas bubbles out at a temperature of 50 ∘C. The solubility of oxygen gas at 50 ∘C is 27.8 mg/L at an oxygen pressure of 1.00 atm. The solubility of nitrogen gas at 50 ∘C is 14.6 mg/L at a nitrogen pressure of 1.00 atm. Assume that the air above the water contains an oxygen partial pressure of 0.21 atm and a nitrogen partial pressure of 0.78 atm.

For "C" where did you get 3E-4 M from?

To calculate the total volume of nitrogen and oxygen gas that bubbles out of the water upon warming, we will use Henry's law, which states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.

Step 1: Calculate the number of moles of nitrogen and oxygen gas initially dissolved in 1.1 L of water at 25 °C and a total pressure of 1.0 atm.
Given that the mole fraction of nitrogen gas in air is 0.78 and the mole fraction of oxygen gas is 0.21, we can calculate the number of moles of nitrogen gas (n_N2) and oxygen gas (n_O2) using the ideal gas law:
n_N2 = (0.78)(1.0 atm)(1.1 L)/(0.0821 atm L/mol K)(298 K)
n_O2 = (0.21)(1.0 atm)(1.1 L)/(0.0821 atm L/mol K)(298 K)

Step 2: Calculate the mass of nitrogen and oxygen gas initially dissolved.
Using the solubilities given at 25 °C (0.782 mg/L for nitrogen and 0.21 mg/L for oxygen), we can calculate the mass of nitrogen gas (m_N2) and oxygen gas (m_O2) initially dissolved in 1.1 L of water:
m_N2 = (0.782 mg/L)(1.1 L)
m_O2 = (0.21 mg/L)(1.1 L)

Step 3: Calculate the number of moles of nitrogen and oxygen gas dissolved at equilibrium.
Since the water is saturated with nitrogen and oxygen gas at 25 °C and a total pressure of 1.0 atm, the number of moles of nitrogen (n_N2') and oxygen (n_O2') at equilibrium is the same as in step 1.

Step 4: Calculate the number of moles of nitrogen and oxygen gas bubbling out upon warming.
The number of moles of nitrogen (n_N2_out) and oxygen (n_O2_out) gas bubbling out upon warming can be calculated by subtracting the number of moles at equilibrium (step 3) from the number of moles initially dissolved (step 1):
n_N2_out = n_N2 - n_N2'
n_O2_out = n_O2 - n_O2'

Step 5: Convert the number of moles of nitrogen and oxygen gas to volume at 50 °C.
Using the ideal gas law at 50 °C, we can calculate the volume of nitrogen gas (V_N2_out) and oxygen gas (V_O2_out) bubbling out:
V_N2_out = n_N2_outRT/P
V_O2_out = n_O2_outRT/P

where R is the ideal gas constant (0.0821 L·atm/mol·K), T is the temperature in Kelvin (50 °C + 273 = 323 K), and P is the pressure of the gas (1.0 atm).

Step 6: Solve for the total volume of nitrogen and oxygen gas that bubbles out.
The total volume of nitrogen and oxygen gas that bubbles out upon warming is the sum of the volumes of nitrogen and oxygen gas:
Total volume = V_N2_out + V_O2_out

Performing the calculations with the given values will give you the final result.

To solve this problem, we can use Henry's Law, which states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.

First, let's calculate the partial pressures of oxygen and nitrogen in the water at 25 °C using the given solubilities and partial pressures:

Solubility of oxygen at 25 °C = 27.8 mg/L at 1.00 atm
Solubility of nitrogen at 25 °C = 14.6 mg/L at 1.00 atm

Partial pressure of oxygen at 25 °C = 27.8 mg/L / 1.0 mg/L/atm = 27.8 atm
Partial pressure of nitrogen at 25 °C = 14.6 mg/L / 1.0 mg/L/atm = 14.6 atm

Next, let's use Henry's Law to calculate the solubilities of oxygen and nitrogen in water at 50 °C using the given solubilities at 1.00 atm and the partial pressures in the air:

Partial pressure of oxygen in air = 0.21 atm
Partial pressure of nitrogen in air = 0.78 atm

Solubility of oxygen at 50 °C = (Partial pressure of oxygen in air) * (Solubility of oxygen at 25 °C / Partial pressure of oxygen at 25 °C)
Solubility of nitrogen at 50 °C = (Partial pressure of nitrogen in air) * (Solubility of nitrogen at 25 °C / Partial pressure of nitrogen at 25 °C)

Solubility of oxygen at 50 °C = 0.21 atm * (27.8 mg/L / 27.8 atm) = 0.21 mg/L
Solubility of nitrogen at 50 °C = 0.78 atm * (14.6 mg/L / 14.6 atm) = 0.78 mg/L

Now, let's calculate the amount of oxygen and nitrogen that bubbles out of the water when it is warmed from 25 °C to 50 °C. We can subtract the solubilities at 50 °C from the solubilities at 25 °C:

Amount of oxygen bubbled out = (Solubility of oxygen at 25 °C) - (Solubility of oxygen at 50 °C)
Amount of nitrogen bubbled out = (Solubility of nitrogen at 25 °C) - (Solubility of nitrogen at 50 °C)

Amount of oxygen bubbled out = 27.8 mg/L - 0.21 mg/L = 27.59 mg/L
Amount of nitrogen bubbled out = 14.6 mg/L - 0.78 mg/L = 13.82 mg/L

Finally, let's calculate the total volume of oxygen and nitrogen gas that bubbles out of 1.1 L of water:

Total volume of oxygen bubbled out = (Amount of oxygen bubbled out) / (Solubility of oxygen at 50 °C)
Total volume of nitrogen bubbled out = (Amount of nitrogen bubbled out) / (Solubility of nitrogen at 50 °C)

Total volume of oxygen bubbled out = (27.59 mg/L) / (0.21 mg/L) = 131.38 L
Total volume of nitrogen bubbled out = (13.82 mg/L) / (0.78 mg/L) = 17.71 L

Therefore, the total volume of nitrogen and oxygen gas that should bubble out of 1.1 L of water upon warming from 25 °C to 50 °C is approximately 131.38 L of oxygen gas and 17.71 L of nitrogen gas.

p = KcC

p = 0.21 when saturated and Kc = about 769 atm*L/mol but you should look this up in your tables and use that number here. Also, I have estimated all of my other answers; you definitely should go through the calculations yourself and refine each of the steps.
C for the sat'd solution at 25C is
about 3E-4M. Convert that to mol.
3E-4 mols/L x 1.1 L = about 3E-4 mols and that x 32 = about 9.5E-3 g which is about 9.5 mg in the 1.1 L @ 25C for O2.

At 50C, oxygen is 27.8 mg/L @ 1atm or
27.8mg/L x 1.1 L = about 30 mg @ 1 atm. If the pressure is to be 0.21 above the solution at equilibrium the solubility will be decreased to about 30 x 0.21 atm/1 atm = about 6.5 mg. So 9.5mg initially - 6.5 mg at 50C is about 3 mg O2 that should bubble out.

N2 can be done the same way, then add the partial pressure of O2 to that of N2 to find total P. Check my thinking.