A gas mixture containing N2 and O2 was kept inside a 2.00 L container at a temperature of 23.0°C and a total pressure of 1.00 ATM the partial pressure of oxygen was 0.722 ATM how many grams of nitrogen are present in the gas mixture

0.641 g of Nitrogen are present in the mixture.

Explanation:

We use the Ideal Gases Law, to solve this question.

For the mixture:

P mixture . V mixture = mol mixture . R . T

We convert the T° to K →  23°C + 273 = 296 K

R = Ideal gases constant → 0.082 L.atm/mol.K

1 atm . 2L = mol mixture . 0.082 L.atm/mol.K  . 296K

2 atm.L / ( 0.082 mol /L.atm) . 296 = 0.0824 moles

We know that sum of partial pressure = 1

Partial pressure N₂ + Partial pressure O₂ = 1

1 - 0.722 atm = Partial pressure N₂ → 0.278 atm

We apply the mole fraction concept:

Partial pressure N₂ / Total pressure = Moles N₂ / Total moles

Moles N₂ = (Partial pressure N₂ / Total pressure) . Total moles

Moles N₂ = (0.278 atm / 1 atm) . 0.0824 mol → 0.0229 moles

We convert the moles to mass → 0.0229 mol . 28 g/mol = 0.641 g

641 mg

a chemical reaction happens

explanation:

every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. therefore, there must be the same number of atoms of each element on each side of a chemical equation.