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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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Answer:

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A 7.00-mL portion of 8.00 M stock solution is to be diluted to 0.800 M. What will be the final volume after dilution? Enter your
amm1812

Explanation:

The number of moles of solute present in liter of solution is defined as molarity.

Mathematically,         Molarity = \frac{\text{no. of moles}}{\text{Volume in liter}}

Also, when number of moles are equal in a solution then the formula will be as follows.

                     M_{1} \times V_{1} = M_{2} \times V_{2}

It is given that M_{1} is 8.00 M, V_{1} is 7.00 mL, and M_{2} is 0.80 M.

Hence, calculate the value of V_{2} using above formula as follows.

                    M_{1} \times V_{1} = M_{2} \times V_{2}

                 8.00 M \times 7.00 mL = 0.80 M \times V_{2}

                      V_{2} = \frac{56 M. mL}{0.80 M}

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Answer:

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