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Svetach [21]
3 years ago
12

A gas occupies a volume of 67.54 liters at 158°C and 4.87 atm pressure. Calculate the number of moles of this gas.

Chemistry
1 answer:
Katarina [22]3 years ago
6 0

Answer:

The number of moles of the gas is 9.295 moles or 9.30 moles

Explanation:

We use PV = nRT

Where P = 4.87 atm;

V = 67.54 L

R= 0.0821Latm/molK

T = 158 C = 158 +273 K = 431 K

the number of moles can be obtained by substituting the values in the respective columns and solve for n

n = PV / RT

n = 4.87 * 67.54 / 0.0821 * 431

n = 328.9198 / 35.3851

n = 9.295moles

The number of moles is approximately 9.30moles.

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A student ran the following reaction in the laboratory at 311 K:CH4(g) + CCl4(g) 2CH2Cl2(g)When she introduced 4.10×10-2 moles o
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<u>Answer:</u> The value of K_{eq} is 0.044

<u>Explanation:</u>

We are given:

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Initial moles of carbon tetrachloride = 6.51\times 10^{-2}mol=0.0651moles

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Concentration of a substance is calculated by:

\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}

So, concentration of methane = \frac{0.0410}{1.00}=0.0410M

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The given chemical equation follows:

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<u>Initial:</u>          0.0410    0.0651

<u>At eqllm:</u>     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:

\Rightarrow (0.0651-x)=0.0602\\\\\Rightarrow x=0.0049M

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

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Hello,

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If you need anymore help feel free to ask me!

Hope this helps!

~Nonportrit 
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3 years ago
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