The vapor pressure of pure water at 296 k is 2778.5 pa. the vapor forms an ideal gas. 1) in some oil, the equilibrium concentration of water molecules is only 1% as large as in pure water. suppose that the pure water is covered with a layer of that oil. what's the equilibrium vapor pressure of water above the oil layer?

Explanation:

It is given that vapor pressure of pure water at 296 K is 2778.5 Pa.These vapors will result in the formation of an ideal gas.

Now, as water is covered with oil and contains only 1% molecules of water. Hence, the vapor pressure of this mixture will also be equal to the vapor pressure of pure water.

So, vapor pressure of mixture = 1% vapor pressure of pure water

Therefore, =

                                                 = 27.785 Pa

Thus, we can conclude that the equilibrium vapor pressure of water above the oil layer is 27.785 Pa.

question: ray weighs 100 kilograms, his wrestling partner, bobby, weighs 50 kilograms. if ray moves at 4 m s , how fast will bobby have to run at him to apply an equal amount of kinetic energy?

answer: 8 m/s he needs to run twice as fast to keep up so he has some work to do

explanation:   if the mass has units of kilograms and the velocity of meters per second, the kinetic energy has units of kilograms-meters squared per second squared.

question answered by

(jacemorris04)