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Allisa [31]
2 years ago
14

A container with a volume 2. 0 L is filled with a gas at a pressure of 1. 5 atm. By decreasing the volume of the container to 1.

0 L, what is the resulting pressure? Type in your answer using the correct number of significant figures. Remember to use the formula for Boyle's law: P1V1 = P2V2 atm.
Chemistry
1 answer:
worty [1.4K]2 years ago
5 0

The resulting pressure of the gas after decreasing the initial volume from 2 L to 1 L is 3 atm.

<h3>What is Boyle's Law?</h3>

According to the Boyle's Law at constant temperature, pressure of the gas is inversely proportional to the volume of that gas.

For the given question we use the below equation is:

P₁V₁ = P₂V₂, where

P₁ = initial pressure of gas = 1.5 atm

V₁ = initial volume of gas = 2 L

P₂ = final pressure of gas = ?

V₂ = final volume of gas = 1 L

On putting all these values on the above equation, we get

P₂ = (1.5atm)(2L) / (1L) = 3 atm

Hence required pressure of the gas is 3 atm.

To know more about Boyle's Law, visit the below link:
brainly.com/question/469270

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At 25°C, the equilibrium constant Kc for the reaction in thesolvent CCl4 2BrCl &lt;----&gt; Br2 + Cl2 is 0.141. If the initial c
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<u>Answer:</u> The equilibrium concentration of bromine gas is 0.00135 M

<u>Explanation:</u>

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Initial concentration of chlorine gas = 0.0300 M

Initial concentration of bromine monochlorine = 0.0200 M

For the given chemical equation:

                   2BrCl\rightleftharpoons Br_2+Cl_2

<u>Initial:</u>          0.02               0.03

<u>At eqllm:</u>    0.02-2x     x     0.03+x

The expression of K_c for above equation follows:

K_c=\frac{[Br_2]\times [Cl_2]}{[BrCl]^2}

We are given:

K_c=0.141

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Neglecting the value of x = -0.96 because, concentration cannot be negative

So, equilibrium concentration of bromine gas = x = 0.00135 M

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