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Advocard [28]
3 years ago
7

How many grams of argon would it take to fill a large light bulb with a volume of 0.745 L at STP?

Chemistry
1 answer:
EastWind [94]3 years ago
8 0

Answer:

Mass = 1.33 g

Explanation:

Given data:

Mass of argon required = ?

Volume of bulb = 0.745 L

Temperature and pressure = standard

Solution:

We will calculate the number of moles of argon first.

Formula:

PV = nRT

R = general gas constant = 0.0821 atm.L/mol.K

By putting values,

1 atm ×0.745 L = n × 0.0821 atm.L/mol.K× 273.15 K

0.745 atm. L = n × 22.43 atm.L/mol

n = 0.745 atm. L / 22.43 atm.L/mol

n = 0.0332 mol

Mass of argon:

Mass = number of moles × molar mass

Mass = 0.0332 mol × 39.95 g/mol

Mass = 1.33 g

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Answer:

P_{He}=9.06torr

Explanation:

Hello there!

In this case, we can identify the solution to this problem via the Dalton's rule because the partial pressure of helium is given by:

P_{He}=x_{He}P_T

Whereas the mole fraction of helium is calculated by firstly obtaining the moles and then the mole fraction:

n_{He}=8.00g\frac{1mol}{4.00g}=2.00mol\\\\ x_{He}=\frac{n_{He}}{n_{He}+n_{Ar}} \\\\ x_{He}=\frac{2.00mol}{2.00mol+8.60mol}\\\\x_{He}=0.189

Then, we calculate the partial pressure as shown below:

P_{He}=0.189 *48.0torr\\\\P_{He}=9.06torr

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