20 Liters of a mixture of N2 gas and CH4 gas was made up with a pressure of 700. Mm Hg and a temperature of 300. C. If the partial pressure of N2 gas was 250. Mm Hg, how many grams of CH4 were added to the mixture?

4.04g of CH₄ are in the mixture

Explanation:

In a mixture of gases, the total pressure is equal to partial pressure of each gas in the mixture.

In the problem, total pressure is 700mmHg and there are just 2 gases (N₂ ans CH₄) where the partial pressure of N₂ is 250mmHg. Thus, pressure of CH₄ is:

700mm Hg - 250mm Hg = 450mmHg. In atm:

450 mmHg ₓ (1 atm / 760mm Hg) = 0.5921 atm

Using PV = nRT

Where P is pressure (0.5921atm), V is volume (20L), R is gas constant (0.082atmL/molK), T is absolute temperature (300°C + 273.15 = 573.15K), it is possible to obtain moles -n- of the gas, thus:

PV / RT = n

0.5921atm×20L / 0.082atmL/molK×573.15K = n

0.252 moles CH₄ = n

As molar mass of CH₄ is 16.04 g/mol:

0.252 moles CH₄ × (16.04g / 1mol) = 4.04g of CH₄ are in the mixture

answer: cellular respiration takes in food and uses it to create atp, a chemical which the cell uses for energy. usually, this process uses oxygen, and is called aerobic respiration. it has four stages known as glycolysis, link reaction, the krebs cycle, and the electron transport chain.