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erastova [34]
1 year ago
10

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
#SPJ1

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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

hope this helps!


3 0
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

4 0
3 years ago
What type of reaction is shown in the following chemical equation in NH3+HC1 NH4C1​
natulia [17]

Explanation:

combination reaction

3 0
3 years ago
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3 years ago
3. Students measured the mass of the reactants and products for a combustion reaction they observed.
coldgirl [10]

Answer:

A. Students made a measurement error, because ending with more products is impossible.

Explanation:

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