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CHEM HELP!!!!!! How many grams of Kr are in a 8.27 L cylinder at 27.0 °C and 8.67 atm?
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- Roger the MoleLv 73 years agoFavorite Answer
n = PV / RT = (8.67 atm) x (8.27 L) / ((0.08205746 L atm/K mol) x (27.0 + 273.15) K) = 2.9112 mol
(2.9112 mol) x (83.798 g Kr/mol) = 244 g
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