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2 years ago

How many moles of neon gas are there 25.0 L of gas or 5.0°C with a pressure of 89.9 KPa?

Chemistry

1 answer:

vladimir2022 [97]2 years ago

8 0

Volume=25L
Temperature=5°C
Pressure=89.9kPa

Now

$\\ \rm\rightarrowtail PV=nRT$

$\\ \rm\rightarrowtail n=\dfrac{PV}{RT}$

$\\ \rm\rightarrowtail n=\dfrac{89.9(25)}{5R}$

$\\ \rm\rightarrowtail n=89.9(5)/R$

$\\ \rm\rightarrowtail n=449.5/R mol$

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It is winter where jenna lives. Its been snowing all day, but now the snow has changed to sleet and then to freezing rain. What

Vladimir79 [104]

Hi!

Answer:

The change in temperature.

Explanation:

Snow forms at 0 °C, when water vapor converts directly into solid ice crystals. Sleet forms when raindrops fall through a layer of air colder than 0 °C. This means that from the time it was snowing to the time it started sleeting the air has gotten warmer but one layer of air stayed cold, hence the formation of sleet. Freezing rain is rain that freezes when it hits a cold surface. This means that from the time it was sleeting to the time there was freezing rain the air had completely warmed and is now above 0 °C but the ground and all other surfaces are still cold.

I hope this helps, as this happens all the time where I live! :)

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3 years ago

Which of the following most influenced the changing nature of classification systems

Lyrx [107]

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3 years ago

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IrinaK [193]

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2 years ago

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What happens to the cell membrane during exocytosis?

Cloud [144]

Answer:

Endocytosis and Exocytosis: Differences and Similarities

ARTICLE Apr 28, 2020

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.

Vesicle function in endocytosis and exocytosis

During bulk transport, larger substances or large packages of small molecules are transported through the cell membrane, also known as the plasma membrane, by way of vesicles – think of vesicles as little membrane sacs that can fuse with the cell membrane.

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.

8 0

2 years ago

1. Which of the following nuclear equations is correctly balanced? (choices attached)

lawyer [7]

1. Balancing nuclear equations

Answer:

$_{18}^{37}\text{Ar} + _{-1}^{0}\text{e} \rightarrow _{17}^{37}\text{Cl}$

Explanation:

The main point to remember in balancing nuclear equations is that the sums of the superscripts (the mass numbers) and the subscripts (the nuclear charges) must balance.

Mass numbers: 37 + 0 = 37; balanced.

Charges: 18 + 1 = 17; balanced

B is wrong. Mass numbers not balanced. 6 +2(1) ≠ 4 + 3.

C is wrong. Mass numbers not balanced. 254 + 4 ≠ 258 + 2(1).

D is wrong. Mass numbers not balanced. 14 + 4 ≠ 17 + 2.

===============

2. Amount remaining

Answer:

D. 5.25 g

Explanation:

The half-life of Th-234 (24 da) is the time it takes for half the Th to decay.

After one half-life, half (50 %) of the original amount will remain.

After a second half-life, half of that amount (25 %) will remain, and so on.

We can construct a table as follows:

No. of Fraction Amount

half-lives t/(da) remaining remaining/g

1 24 ½ 21.0

2 48 ¼ 10.5

3 72 ⅛ 5.25

4 96 ¹/₁₆ 2.62

We see that 72 da is three half-lives, and the amount of Th-234 remaining is 5.25 g.

===============

3. Calculating the half-life

Answer:

a. 2.6 min

Explanation:

The fraction of the original mass remaining is 1.0 g/4.0 g ≈ ¼.

We saw from the previous table that it takes two half-lives to decay to ¼ of the original amount.

2 half-lives = 5.2 min Divide both sides by 2

1 half-life = 5.2 min/2 = 2.6 min

7 0

3 years ago