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2 years ago

How many moles of magnesium are in 3. 01 x 10^22 atoms of magnesium.

Chemistry

1 answer:

marshall27 [118]2 years ago

5 0

Answer:

Explanation:

1 mol of anything is 6.02 * 10^23 atoms (in this case)

x mol of 3.01 * 10^22

Set up the proportion

1/x = 6.02*10^23 / 3.01 * 10^22 Cross multiply

x*6.02 * 10^23 = 1 * 3.01 * 10^22 Divide by 6.02*10^23

x = 3.01 * 10^22 / 6.02*10^23

x = 1/(2 * 10)

x = 1/20 mol

x = 0.05 mol

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We are given:

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$CH_4(g)+CCl_4(g)\rightleftharpoons 2CH_2Cl_2(g)$

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At eqllm: 0.0410-x 0.0651-x 2x

We are given:

Equilibrium concentration of carbon tetrachloride = $6.02\times 10^{-2}M=0.0602M$

Evaluating the value of 'x', we get:

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Now, equilibrium concentration of methane = $0.0410-x=[0.0410-0.0049]=0.0361M$

Equilibrium concentration of $CH_2Cl_2=2x=[2\times 0.0049]=0.0098M$

The expression of $K_{eq}$ for the above reaction follows:

$K_{eq}=\frac{[CH_2Cl_2]^2}{[CH_4]\times [CCl_4]}$

Putting values in above expression, we get:

$K_{eq}=\frac{(0.0098)^2}{0.0361\times 0.0603}\\\\K_{eq}=0.044$

Hence, the value of $K_{eq}$ is 0.044

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