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2 years ago

8

What is the new volume of a sample of neon if 5.23 L of the gas, which was originally at 214 mmHg, is compressed until the press

ure increases to 796 mmHg?

1 answer:

Yuliya22 [10]2 years ago

8 0

This problem can be solved using Boyle's law:p1v1=p2v2

so 5.23*214=796*v2 v2=(5.23*214)/796=1.4L

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<u>Answer:</u> The value of $K_c$ is coming out to be 0.412

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

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<u>For hydrogen gas:</u>

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Given mass of hydrogen sulfide = 72.6 g

Molar mass of hydrogen sulfide = 34 g/mol

Putting values in equation 1, we get:

$\text{Moles of hydrogen sulfide}=\frac{72.6g}{34g/mol}=2.135mol$

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$Sb_2S_3(s)+3H_2(g)\rightarrow 2Sb(s)+3H_2S(g)$

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At eqllm: 2.944-x 5-3x 2x 3x

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Equilibrium moles of hydrogen sulfide = 2.135 moles

Calculating for 'x', we get:

$\Rightarrow 3x=2.135\\\\\Rightarrow x=\frac{2.135}{3}=0.712$

Equilibrium moles of hydrogen gas = (5 - 3x) = (5 - 3(0.712)) = 2.868 moles

Volume of the container = 25.0 L

Molarity of a solution is calculated by using the formula:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume}}$

The expression of $K_c$ for above equation, we get:

$K_c=\frac{[H_2S]^3}{[H_2]^3}$

The concentration of solids and liquids are not taken in the expression of equilibrium constant.

$K_c=\frac{(\frac{2.135}{25})^3}{(\frac{2.868}{25})^3}\\\\K_c=0.412$

Hence, the value of $K_c$ is coming out to be 0.412

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