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You have a 450.0 mL balloon filled with 0.0184 moles of Helium gas. The volume of the balloon will change, but the pressure of the gas inside will always match the 1.00 atm pressure outside. If the balloon is kept at a constant 25oC, how many moles of gas must you remove to shrink the balloon down to 200.0 mL?
- hcbiochemLv 71 month ago
V1/n1 = V2/n2
450.0 mL / 0.0184 mol = 200.0 mL / n2
n2 = 8.17X10^-3 mol
moles removed = 0.0184-8.17X10^-3 = 0.0102 mols He removed