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3 years ago

Explain what takes place during the G2 phase of the cell cycle.

Chemistry

1 answer:

mrs_skeptik [129]3 years ago

8 0

Answer:

The cell's DNA is replicated during the S stage as the first step to engaging in mitosis. Once DNA replication is complete, the cell enters the second gap stage, G2. During G2 the correct duplication of the DNA is verified and cell proteins necessary for cell division are produced.

Explanation:

Hope this is helpful

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What do the forms of the highlighted limbs in the illustration tell you about their functions in each animal?

iren2701 [21]

Where are the high lighted limbs and the illustrations

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2 years ago

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The palmetto landfill in greer, sc supplies over 50% of the energy for the bmw plant there. this reduces carbon dioxide emission

Verdich [7]

Hello there!

The answer is 92000 tons per year

The palmetto landfill supplies methane (CH₄) for the BMW plant there. The gas is used to operate the turbines used to power the BMW campus. It reduces carbon dioxide emissions because the combustion of methane produces fewer CO₂ than the combustion of other gases like propane or butane, which are commonly used for powering turbines.

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3 years ago

How many atoms are in 5.2 moles of Na ?

Nastasia [14]

One mole of a substance is defined by Avogadro as consisting of 6.022 x 1023 atoms. This is Avogadro's number. To calculate the number of atoms in two moles of sodium, use dimensional analysis. 2.0 moles Na x 6.022⋅1023g1mol=1.20⋅1024 atoms of Na

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2 years ago

3. 2.15 x 3.1 x 100 =

jeka94

Answer:

666.5

Explanation:

Multiply 2.15 and 3.1 to get 6.665.

6.665×100≈666.5

Multiply 6.665 and 100 to get 666.5.

666.5

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3 years ago

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How many liters of hydrogen gas will be produced at STP from the reaction of 7.179×10^23 atoms of magnesium with 54.219g of phos

Alexeev081 [22]

Answer: The volume of hydrogen gas produced will be, 12.4 L

Explanation : Given,

Mass of $H_3PO_4$ = 54.219 g

Number of atoms of $Mg$ = $7.179\times 10^{23}$

Molar mass of $H_3PO_4$ = 98 g/mol

First we have to calculate the moles of $H_3PO_4$ and $Mg$ .

$\text{Moles of }H_3PO_4=\frac{\text{Given mass }H_3PO_4}{\text{Molar mass }H_3PO_4}$

$\text{Moles of }H_3PO_4=\frac{54.219g}{98g/mol}=0.553mol$

and,

$\text{Moles of }Mg=\frac{7.179\times 10^{23}}{6.022\times 10^{23}}=1.19mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$3Mg+2H_3PO_4\rightarrow Mg(PO_4)_2+3H_2$

From the balanced reaction we conclude that

As, 3 mole of $Mg$ react with 2 mole of $H_3PO_4$

So, 0.553 moles of $Mg$ react with $\frac{2}{3}\times 0.553=0.369$ moles of $H_3PO_4$

From this we conclude that, $H_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $Mg$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2$

From the reaction, we conclude that

As, 3 mole of $Mg$ react to give 3 mole of $H_2$

So, 0.553 mole of $Mg$ react to give 0.553 mole of $H_2$

Now we have to calculate the volume of $H_2$ gas at STP.

As we know that, 1 mole of substance occupies 22.4 L volume of gas.

As, 1 mole of hydrogen gas occupies 22.4 L volume of hydrogen gas

So, 0.553 mole of hydrogen gas occupies $0.553\times 22.4=12.4L$ volume of hydrogen gas

Therefore, the volume of hydrogen gas produced will be, 12.4 L

4 0

3 years ago