Air is a mixture of gases that is about78.0percent n2 by volume. when air is at standard pressure and 25.0 degreec, the n2 component will dissolve in water with a solubility of 4.88 x 10^-4m. what is the value of henry's law constant for n2 under these conditions? express the constant numerically inmoles per liter per atmosphere. =6.26? 10^-4 mol/(l. atm) correct part b as a scuba diver descends under water, thepressure increases. at a total air pressure of 2.73 atm and a temperature of 25.0 degreec, what is the solubility of n2 in a diver's blood? [use the value of the henry's lawconstant calculated in part a, 6.26 x 10^-4mol/(l. atm). assume that the composition of the air in the tank is the sameas on land and that all of the dissolved nitrogen remains in theblood. express your answer numerically inmoles per liter. solubility =mol/l

Explanation:

(a) According to the Henry's law, the solubility is equal to the product between the Henry's law constant and the partial pressure of a gas:

Air is at standard pressure:

Therefore, its partial pressure is:

Solving for the Henry's constant:

(b) Using the constant we've found in the previous part, we know that:

In this case, the percentage is kept the same, however, the total air pressure is:

Substituting the variables:

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

there is none because earth is stationary and flat.