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drek231 [11]
3 years ago
12

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 n

o change in temperature or pressure.
Chemistry
1 answer:
Annette [7]3 years ago
3 0

Answer:

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

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dsp73

Answer:

12.8 g of O_{2} must be withdrawn from tank

Explanation:

Let's assume O_{2} gas inside tank behaves ideally.

According to ideal gas equation- PV=nRT

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We can also write, \frac{V}{RT}=\frac{n}{P}

Here V, T and R are constants.

So, \frac{n}{P} ratio will also be constant before and after removal of O_{2} from tank

Hence, \frac{n_{before}}{P_{before}}=\frac{n_{after}}{P_{after}}

Here, \frac{n_{before}}{P_{before}}=\frac{0.66mol}{43atm} and P_{after}=17atm

So, n_{after}=\frac{n_{before}}{P_{before}}\times P_{after}=\frac{0.66mol}{43atm}\times 17atm=0.26mol

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7 0
3 years ago
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goblinko [34]
<h3>Answer: 386.67 g/mol </h3>

Explanation:

Molar Mass = Mass ÷ Mole

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3 years ago
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