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3 years ago

PLEASE HELP ASAP

Chemistry

1 answer:

Kruka [31]3 years ago

4 0

The mass stays constant as a substance changes from a liquid to a gas.

The Law of Conservation of Mass states that, in ordinary chemical reactions, mass is neither destroyed nor created.

That is, the mass of the reactants must equal the mass of the products.

2H₂O(ℓ) ⟶ 2H₂O(g)

1 g 1 g

If the mass of liquid water is 1 g, the mass of the water vapour must be 1 g.

Even though the water vapour is a gas and you can’t see it, it still has a mass

of 1 g.

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Answer:

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Explanation:

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Suppose that 0.1000 mole each of H2and I2are placed in a 1.000-L flask, stoppered, and the mixture is heated to 425oC. At equili

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Answer: The value of equilibrium constant for the given reaction is 56.61

Explanation:

We are given:

Initial moles of iodine gas = 0.100 moles

Initial moles of hydrogen gas = 0.100 moles

Volume of container = 1.00 L

Molarity of the solution is calculated by the equation:

$\text{Molarity of solution}=\frac{\text{Number of moles}}{\text{Volume}}$

$\text{Molarity of iodine gas}=\frac{0.1mol}{1L}=0.1M$

$\text{Molarity of hydrogen gas}=\frac{0.1mol}{1L}=0.1M$

Equilibrium concentration of iodine gas = 0.0210 M

The chemical equation for the reaction of iodine gas and hydrogen gas follows:

$H_2+I_2\rightleftharpoons 2HI$

Initial: 0.1 0.1

At eqllm: 0.1-x 0.1-x 2x

Evaluating the value of 'x'

$\Rightarrow (0.1-x)=0.0210\\\\\Rightarrow x=0.079M$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[HI]^2}{[H_2][I_2]}$

$[HI]_{eq}=2x=(2\times 0.079)=0.158M$

$[H_2]_{eq}=(0.1-x)=(0.1-0.079)=0.0210M$

$[I_2]_{eq}=0.0210M$

Putting values in above expression, we get:

$K_c=\frac{(0.158)^2}{0.0210\times 0.0210}\\\\K_c=56.61$

Hence, the value of equilibrium constant for the given reaction is 56.61

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