If 43.1 g of O2 and 6.8 g of CO2 are placed in a 13.7 L container at 34 degrees C, what is the mixture of gasses? answer must be in atm

The total pressure of the gas mixture = 2.76 atm

Note: The question is not complete. The complete question is as follow:

If 43.1 g of O2 and 6.8 g of CO2 are placed in a 13.7 L container at 34 degree Celsius , what is the pressure of the mixture of gases?

Explanation:

Mass of O₂ gas = 43.1 g, molar mass of O₂ gas = 32.0 g/mol

Number of moles of O₂ gas = 43.1/32.0 = 1.347 moles

Mass of CO₂ gas = 6.8 g, molar mass of CO₂ gas = 44.0 g

Number of moles of CO₂ gas = 6.8/44 = 0.155 moles

Total number of moles of gas mixture, n = (1.347 + 0.155) = 1.502 moles

Volume of gas mixture, V = 13.7 L

Temperature of gas mixture, T = 34 °C = (273.15 + 34) K = 307.15 K

Pressure of gas mixture = ?

Molar gas constant, R = 0.0821 liter·atm/mol·K.

Using the ideal gas equation: PV =nRT

P = nRT/V

P = (1.502 × 0.0821 × 307.15) / 13.7

P = 2.76 atm

Therefore, the total pressure of the gas mixture = 2.76 atm

idk

explanation:

dasdad

answer: the rate of diffusion increases with increase in temperature.

increase in temperature, increases the molecular/particles speed (kinetic energy. diffusion means movement of particles through the material.

generally, the heavier the particle the more it interacts with the environment and more the temperature matters. so higher temperature results in higher diffusion.