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3 years ago

Find the velocity of a swimmer who swims exactly 220 m toward the shore in 72 s.

Chemistry

1 answer:

Aleksandr-060686 [28]3 years ago

5 0

Answer: 3.06m/s

Explanation:

Distance covered by swimmer = 220 m

Time taken by swimmer = 72 s.

Velocity = ?

Recall that velocity is the rate of change of distance per unit time. it is measured in metres per second.

Velocity = Distance / time taken

Velocity = 220m / 72seconds

Velocity = 3.055 m/s (approximate to 3.06m/s)

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statuscvo [17]

<u>Answer:</u> The value of $K_{eq}$ is 0.044

<u>Explanation:</u>

We are given:

Initial moles of methane = $4.10\times 10^{-2}mol=0.0410moles$

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

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

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

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

Concentration of carbon tetrachloride = $\frac{0.0651}{1.00}=0.0651M$

The given chemical equation follows:

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

<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:

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

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

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<h2>Question:</h2>

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<h2>Answer:</h2>

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<h2>For example:</h2>

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