On a molar basis, air is about 78% N2 . What is the Henry's law constant for , N2 , in water if the solubility of, N2 , in water is .000535 mole/ L, on a day when the atmospheric pressure is 745 torr?
First convert pressure into atm
P=745/760
=.98
S=k*p
Solubility=Henry constant*pressure
kh=S/p
=.000535/.98
=5.45*10-4
This is the value of Henry constant
On a molar basis, air is about 78% N2 . What is the Henry's law constant...
Air is a mixture of gases that is about 78.0% N2 by volume. When air is at standard pressure and 25.0 ∘C, the N2 component will dissolve in water with a solubility of 4.88×10−4 M. What is the value of Henry's law constant for N2 under these conditions?
Determine the solubility of N2 in water exposed to air at 25°C if the atmospheric pressure is 1.2 atm. Assume that the mole fraction of nitrogen is 0.78 in air and the Henry's law constant for nitrogen in water at this temperature is 6.1 × 10-4 M/atm.
10) Air is a mixture of gases that is about 78.0% N2 by volume. When air is at standard pressure and 25.0 °C, the N2 component will dissolve in water with a solubility of 4.88x10-4 M. What is the value of Henry's law constant for N2 under these conditions?
At 298 K, the Henry's law constant for oxygen is 0.00130 M/atm. Air is 21.0% oxygen. At 298 K, what is the solubility of oxygen in water exposed to air at 0.892 atm? If atmospheric pressure suddenly changes from 1.00 atm to 0.892 atm at 298 K, how much oxygen will be released from 4.60 L of water in an unsealed container?
A solution reaches a concentration of 0.014 molar N2 in water. What is the applied pressure in torr of the gas? (Use 6.25*10-4 M/atm as the Henry's law constant kH for N2)
how could you detect the presence of
At 298 K, the Henry's law constant for oxygen is 0.00130 M/atm. Air is 21.0% oxygen. At 298 K, what is the solubility of oxygen in water exposed to air at 1,00 atm? Number At 298 K, what is the solubility of oxygen in water exposed to air at 0.893 atm? Number If atmospheric pressure suddenly changes from 1.00 atm to 0.893 atm at 298 K, how much oxygen will be released from...
The Henry's Law constant for oxygen dissolving in water is 1.7 x 10 mol/L mmHg at 25 °C. Assume that atmospheric pressure is 760 mmHg and that the mole fraction of oxygen in air is 0.18. What is the concentration (mol/L) of oxygen in water that is exposed to the air at 25 °C? a) 2.2 x 10 mol/L b) 1.3 x 103 mol/L c) 2.3 x 104 mol/L d) 1.2 x 10 mol/L b c d
17. What is the Henry's law constant for O2 if the solubility of O2 in water is 0.570 g/L at a pressure of 18.5 atm?
Use Henry's law to determine the molar solubility of helium at a pressure of 1.1 atm and 25 degree C. Henry's law constant for helium gas in water at 25 degree C is 3.70-10^-4 M/atm. You can treat Henry's law constant as an equilibrium constant. The goal of this problem is to find the concentration of helium dissolved in water that would be in equilibrium with the helium pressure over the water of 1.1 atm according to K = [He/P_He].
need help with part b please
Henry's Law Henry's Law states that the solubility of a gas in a liquid (at a constant temperature) is entirely dependent on the partial pressure of that gas above the liquid. Increasing the pressure of increases the presence of that gas in the solution. We have talked about soda, carbonated with CO). These sodas are carbonated by placing the soda under a high COpressure Considering Henry's Law for gas solubility, consider the top of...