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Butoxors [25]
3 years ago
7

20.352 mL of chlorine under a pressure of 680. mm Hg are

Chemistry
2 answers:
Lunna [17]3 years ago
7 0

Answer:

0.01144L or 1.144x10^-2L

Explanation:

Data obtained from the question include:

V1 (initial volume) = 20.352 mL

P1 (initial pressure) = 680mmHg

P2 (final pressure) = 1210mmHg

V2 (final volume) =.?

Using the Boyle's law equation P1V1 = P2V2, the volume of the container can be obtained as follow:

P1V1 = P2V2

680 x 20.352 = 1210 x V2

Divide both side by 1210

V2 = (680 x 20.352)/1210

V2 = 11.44mL

Now we need to convert 11.44mL to L in order to obtain the desired result. This is illustrated below:

1000mL = 1 L

11.44mL = 11.44/1000 = 0.01144L

Therefore the volume of the container is 0.01144L or 1.144x10^-2L

Semmy [17]3 years ago
4 0

Answer:

0.011437L

Explanation:

In the question, we are told that under a pressure of 680mmHg, chlorine gas occupies a volume of 20.352mL and then the pressure is changed to 1210mmHg at constant temperature.

Boyle's law states that the volume of a fixed mass of gas is directly proportional to the pressure of the gas at constant temperature.

Mathematically;

P1V1=P2V2

P1= initial pressure

V1= initial volume

P2= final pressure

V2= final volume

We will apply Boyle's law to get the new volume.

From, the relationship P1V1=P2V2

We make V2 subject of formula

V2= (P1V1)/P2

Given;

P1=680mmHg

V1=20.352mL= 20.352/1000L= 0.020352L

P2=1210mmHg

V2=(680×0.020352)/1210

V2=0.011437L

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Answer:

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Explanation:

Given that,

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Put all the values,

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Answer:

equal to M

Explanation:

The mass of the fully melted mass and the initial solid will be the same. So, the mass of the melt is equal to M.

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We are correct to say that in the heating process no mass was destroyed or added in melting the solid.

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What happens to the cell membrane during exocytosis?
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Answer:

Endocytosis and Exocytosis: Differences and Similarities

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by Nicole Gleichmann

Endocytosis and Exocytosis: Differences and Similarities

Endocytosis and exocytosis are the processes by which cells move materials into or out of the cell that are too large to directly pass through the lipid bilayer of the cell membrane. Large molecules, microorganisms and waste products are some of the substances moved through the cell membrane via exocytosis and endocytosis.

Why is bulk transport important for cells?

Cell membranes are semi-permeable, meaning they allow certain small molecules and ions to passively diffuse through them. Other small molecules are able to make their way into or out of the cell through carrier proteins or channels.

But there are materials that are too large to pass through the cell membrane using these methods. There are times when a cell will need to engulf a bacterium or release a hormone. It is during these instances that bulk transport mechanisms are needed.

Endocytosis and exocytosis are the bulk transport mechanisms used in eukaryotes. As these transport processes require energy, they are known as active transport processes.

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Cell membranes are comprised of a lipid bilayer. The walls of vesicles are also made up of a lipid bilayer, which is why they are capable of fusing with the cell membrane. This fusion between vesicles and the plasma membrane facilitates bulk transport both into and out of the cell.

What is endocytosis? Endocytosis definition and purposes

Endocytosis is the process by which cells take in substances from outside of the cell by engulfing them in a vesicle. These can include things like nutrients to support the cell or pathogens that immune cells engulf and destroy.

Endocytosis occurs when a portion of the cell membrane folds in on itself, encircling extracellular fluid and various molecules or microorganisms. The resulting vesicle breaks off and is transported within the cell.

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Explanation:

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