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Novosadov [1.4K]3 years ago

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Answer:

01.

Explanation:

Half the acceleration. Its heavier and moves slower. If it moved the same acceleration, the forces would also have to be doubled since the mass was.

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A gas mixture contains 320mg methane, 175 mg argon, 225 mg nitrogen (N2). The partial pressure of argon at 300K is 12.52 kPa. Wh

svet-max [94.6K]

Answer: The volume and total pressure of the mixture is 0.876 L and 89.36 kPa respectively.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For methane:

Given mass of methane = 320 mg = 0.3 g (Conversion factor: 1 g = 1000 mg)

Molar mass of methane = 16 g/mol

Putting values in equation 1, we get:

$\text{Moles of methane}=\frac{0.3g}{16g/mol}=0.019mol$

For argon:

Given mass of argon = 175 mg = 0.175 g

Molar mass of argon = 40 g/mol

Putting values in equation 1, we get:

$\text{Moles of argon}=\frac{0.175g}{40g/mol}=0.0044mol$

For nitrogen:

Given mass of nitrogen = 225 mg = 0.225 g

Molar mass of nitrogen = 28 g/mol

Putting values in equation 1, we get:

$\text{Moles of nitrogen}=\frac{0.225g}{28g/mol}=0.0080mol$

To calculate the volume of the mixture, we use the equation:

PV = nRT ......(2)

We are given:

Partial pressure of argon = 12.52 kPa

Temperature = 300 K

R = Gas constant = $8.31\text{L kPa }mol^{-1}K^{-1}$

n = number of moles of argon = 0.0044 moles

Putting values in equation 2, we get:

$12.52kPa\times V=0.0044mol\times 8.31\text{L kPa }mol^{-1}K^{-1}\times 300K\\\\V=\frac{0.0044\times 8.31\times 300}{12.52}=0.876L$

Now, calculating the total pressure of the mixture by using equation 2:

Total number of moles = [0.019 + 0.0044 + 0.0080] mol = 0.0314 mol

V= volume of the mixture = 0.876 L

Putting values in equation 2, we get:

$P\times 0.876L=0.0314mol\times 8.31\text{L kPa }mol^{-1}K^{-1}\times 300K\\\\P=\frac{0.0341\times 8.31\times 300}{0.876}=89.36kPa$

Hence, the volume and total pressure of the mixture is 0.876 L and 89.36 kPa respectively.

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