The number of grams of helium in the mixture can be calculated using the ideal gas law.
PV = nRT
n = (PV)/(RT)
n = (1.79 atm * 6.41 L) / (0.08206 L*atm/mol*K * 351.15 K)
n = 0.039 mol
Since the mixture contains 3.66 g of neon, the number of moles of neon is 0.039 mol.
Therefore, the number of moles of helium is 0.039 mol - 0.039 mol = 0 mol.
Since the molar mass of helium is 4.00 g/mol, the number of grams of helium in the mixture is 0 g.
a mixture of neon and helium gases is maintained in a 6.41 l flask at a pressure of 1.79 atm and a temperature of 78 oc. if the gas mixture contains 3.66 grams of neon, the number of grams of helium in the mixture is
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