If 9.00 g helium gas is added to a 1.00 L ballon containing 1.00 g of helium gas, what is the new volume of the ballon? Assume no change in temperature or pressure.

10 L

Explanation:

We'll begin by calculating the number of mole in 9 g and 1 g of He. This can be obtained as follow:

Mass of He = 9 g

Molar mass of he= 4 g/mol

Mole of He =?

Mole = mass /Molar mass

Mole of He = 9/4

Mole of He = 2.25 moles

Mass of He = 1 g

Molar mass of he= 4 g/mol

Mole of He =?

Mole = mass /Molar mass

Mole of He = 1/4

Mole of He = 0.25 moles

Finally, we shall determine the new volume of the balloon as follow:

Initial mole (n₁) = 0.25 mole

Initial volume (V₁) = 1 L

Final mole (n₂) = 0.25 + 2.25 = 2.5 moles

Final volume (V₂) =?

V₁/n₁ = V₂/n₂

1 / 0.25 = V₂ / 2.5

4 = V₂ / 2.5

Cross multiply

V₂ = 4 × 2.5

V₂ = 10 L

Therefore, the new volume of the balloon is 10 L

its mg hope this

1. they vary with a higher number because boiling is way hotter than melting