What is the mass of oxygen gas in a 12.2 L container at 16.0°C and 8.15 atm?
Answer in units of g.

grams O₂ = 134 grams

Explanation:

PV = nRT => n = PV/RT

P = 8.15atm

V = 12.2 Liters

R = 0.08206L·atm/mol·K

T = 16.0°C + 273 = 289K

n = (8.15atm)(12.2L)/(0.08206L·atm/mol·K)(289K) = 4.2 moles O₂

grams O₂ = 4.2 moles O₂ x 32g/mol = 134 grams O₂

the answer would be b i believe

your answer is a. the enthalpy of formation for a pure element in its standard state is always positive.