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

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.