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1 year ago

PLEASE ANSWER WILL MARK BRAINLIST

Chemistry

1 answer:

Harman [31]1 year ago

8 0

To find the experimental value I believe we add up all the temperatures found by each trial and divide by the number of trials

When added up and divided up this is 50.7 C

Then we continue with accepted value minus experimental value over accepted

53.0 - 50.7 / 53.0 x 100

4.3% percent error

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Answer: Option (B) is the correct answer.

Explanation:

Expression for the given decomposition reaction is as follows.

$N_{2}O_{4} \rightarrow 2NO_{2}$

Let us assume that x concentration of $N_{2}O_{4}$ is present at the initial stage. Therefore, according to the ICE table,

$N_{2}O_{4} \rightarrow 2NO_{2}$

Initial : x 0

Change : - 0.1 $2 \times 0.1$

Equilibrium : (x - 0.1) 0.2

Now, expression for $K_{p}$ of this reaction is as follows.

$K_{p} = \frac{P^{2}_{NO_{2}}}{P_{N_{2}O_{4}}}$

Putting the given values into the above formula as follows.

$K_{p} = \frac{P^{2}_{NO_{2}}}{P_{N_{2}O_{4}}}$

$2 = \frac{(0.2)^{2}}{(x - 0.1)}$

$2 \times (x - 0.1) = (0.2)^{2}$

x = 0.12

This means that $P_{N_{2}O_{4}}$ = x = 0.12 atm.

Thus, we can conclude that the initial pressure in the container prior to decomposition is 0.12 atm.

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3 years ago

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2 years ago

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3 years ago

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Answer:

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2 years ago

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Explanation:

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