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.

Solubility of N₂ in water exposed to air at 25°C is 6.09x10⁻⁴ mol/L

Explanation:

Henry's law → C = K . Partial pressure of gas

where C is solubility

and K is Henry's law constant for nitrogen in water

First of all, let's find out the partial pressure for N₂.

We have the mole fraction so

Mole fraction N₂ = Partial pressure N₂ / Total pressure

0.78 = Partial pressure N₂ / 1.28 atm

0.78 . 1.28 atm = Partial pressure N₂ → 0.9984 atm

C = 6.1x10⁻⁴ M /atm . 0.9984 atm

C = 6.09x10⁻⁴ mol/L

adding or remove a gaseous reactant or product changes the concentrations. if the concentration of reactant or product is increased, the system will shift away from the side in which concentration was increased (i.e. if the concentration of reactants is increased, the system will shift toward the products. if more products are added, the system will shift to form more reactants). conversely, if the concentration of reactant or product is decreased, the system will shift toward the side in which concentration was decreased (i.e. if reactants are removed, the system will shift to form more reactants. if the concentration of products is decreased, the equilibrium will shift toward the products).

explanation:

in 1783 cavendish was working with his hydrogen gas and found that when he burned it and condensed the vapor it was water. this was very significant since it disproved the greek theory of elements completely since water was not a simple substance but the product of two gases. it was lavoisier who hearing of the experiment gave cavendish's gas the name hydrogen ("water-producer") and added that hydrogen burned by combining with oxygen and that water was a hydrogen-oxygen combination.

explanation: