A 3.0-L gas mixture contains 30.% oxygen and 70.% nitrogen. Use the ideal gas law to determine the number of moles of oxygen at 2.0 atm and 400-K.

The number of moles of oxygen gas comes out to be 0.0548 mole

Explanation:

Given volume of gas = V = 3.0 L

The mixture contains 30 % oxygen gas by mole.

Pressure of mixture of gas =  P = 2.0 atm

Temperature = T = 400 K

Assuming n be the total number of moles of the mixture of gas.

The ideal gas equation is shown below

The mixture contains 30% oxygen gas by mole

Number of moles of oxygen gas is 0.0548 mole

the electron configuration that you gave me has 16 electrons ergo it has 16 protons ergo it is the element sulfur.