At what pressure would 14.4 grams of oxygen gas occupy 7.5 liters at a temperature of 375 k?

1.85 atm.

Explanation:

where, P is the pressure of the gas in atm.

V is the volume of the gas in L (V = 7.5 L).

n is the no. of moles of the gas in mol (n = mass/moar mass = 14.4 g/32.0 g/mol = 0.45 mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K).

T is the temperature of the gas in K (T = 375 K).

∴ P = nRT/V = (0.45 mol)(0.0821 L.atm/mol.K)(375 K)/(7.5 L) = 1.85 atm.

the answer is d.) potassium