Answers by visitors named: Aris

(2.00 L) x (273 / (20 + 273)) / (22.414 L/mol) = 0.083139 mol air in the "empty" container (7.20 g H2O) / (18.01532 g H2O/mol) = 0.39966 mol H2O added Supposing all the water vaporized: PV = nRT P = nRT / V = (0.083139 mol + 0.39966 mol) x (0.08205746 L atm/K mol) x (530 + 273 K) / (2.00 L) = 15.9 atm 15.9atm x 101.3kPa = 1611.31

1/3 equals to 4/12, 1/4 equals to 3/12. 3/12 plus 4/12 equals 7/12. Therefore 5/12 left

Sulfur