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1 year ago

Hydrogenation of cholesterol

Chemistry

1 answer:

sp2606 [1]1 year ago

8 0

The molecular structure of the fat is changed when lipids like cholesterol are exposed to hydrogenation.

What is cholesterol?
Any member of the lipids class of specific chemical compounds is a kind of cholesterol. It is a sterol, a class of lipid (or modified steroid). All animal cells produce cholesterol, which is a crucial part of the membranes that make up those cells. It is a yellowish crystalline solid when chemically separated. It is possible to increase the saturation of fatty acids by adding hydrogen to unsaturated fatty acid chains, which connects the hydrogen atoms to the saturation sites.

Hence, the molecular structure of the fat is changed when lipids like cholesterol are exposed to hydrogenation.

To know more about cholesterol from the given link
brainly.com/question/841110
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How many mL of a stock solution of 2.00 M KNO3 are needed to prepare 100.0 mL of 0.15M KNO3? with work plz

Tatiana [17]

Using the law of dilution :

Mi x Vi = Mf x Vf

2.00 x Vi = 0.15 x 100.0

2.00 x Vi = 15

Vi = 15 / 2.00

Vi = 7.5 mL

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3 years ago

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

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3 years ago

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Explanation:

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3 years ago

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3 years ago

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3 years ago