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Strike441 [17]
3 years ago
15

Look at the diagram below , which shows gas particles in a container. If the piston was lowered, so that the size of the contain

er decreases, what would happen to the pressure? Explain your reasoning. In your explanation, include which gas law (Boyle's law, Charles' law, or Gay-Lussac's law) this involves.

Chemistry
1 answer:
raketka [301]3 years ago
3 0

The pressure of the gas is expected to increase in accordance to Boyle's law.

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

Boyle's law states that, the volume of a given mass of gas is inversely proportional to its pressure at constant temperature.

By implication, when the piston is lowered and the volume of the gas is decreased, the pressure of the gas is expected to increase in accordance to Boyle's law.

Learn more about Boyle's law: brainly.com/question/1437490

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m₀=60 g

w=100m₁/m₀

w=100*50/60=83.3%
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write the spontaneous net cell reaction. include physical states. net cell reaction: 2al(s) 2br {2}(l)
ElenaW [278]

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What is a net cell reaction?

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What is a chemical reaction?

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8 0
1 year ago
Given the value of the equilibrium constant (Kc) for the equation (a), calculate the equilibrium constant for equation (b)
matrenka [14]

Answer: The value of equilibrium constant for new reaction is 1.92\times 10^{-25}

Explanation:

The given chemical equation follows:

O_2(g)\rightarrow \frac{2}{3}O_3(g)+\frac{1}{2}O_2(g)  

The equilibrium constant for the above equation is 5.77\times 10^{-9}

We need to calculate the equilibrium constant for the equation of 3 times of the above chemical equation, which is:

3O_2(g)\rightarrow 2O_3(g)

The equilibrium constant for this reaction will be the cube of the initial reaction.

If the equation is multiplied by a factor of '3', the equilibrium constant of the new reaction will be the cube of the equilibrium constant of initial reaction.

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K_{eq}'=(5.77\times 10^{-9})^3=1.92\times 10^{-25}

Hence, the value of equilibrium constant for new reaction is 1.92\times 10^{-25}

5 0
3 years ago
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