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seropon [69]
3 years ago
15

How cholestrol transport fatty acids in body ?​

Chemistry
1 answer:
sleet_krkn [62]3 years ago
4 0

In order to use the energy stored in fat, the body breaks dow triglycerides into fatty acids, which individual cells burn for energy

You might be interested in
A compound is found to have a composition of 35.880% Cr, 21.076% P, and 43,545% O
stepan [7]

Explanation:

Cr=35.880/51=0.73≈0.7

P=21.076/31=0.67≈0.7

O=43.543/16=2.71

Divide each by 0.7 u get

Cr=1 ,P =1, O=4

Empirical formula is CrPO4

3 0
2 years ago
What kinds of things could a magnet make another object or magnet do without touching it?​
Sidana [21]

Answer:

Magnets can be used to make other magnets and things made of iron move without being touched. Something that has been electrically charged can make other things move without touching them.

Explanation:

(can someone help me with my question?)

8 0
3 years ago
Consider the reaction Mg(s) + I2 (s) → MgI2 (s) Identify the limiting reagent in each of the reaction mixtures below:
Lapatulllka [165]

Answer:

a) Nor Mg, neither I2 is the limiting reactant.

b) I2 is the limiting reactant

c) <u>Mg is the limiting reactant</u>

<u>d) Mg is the limiting reactant</u>

<u>e) Nor Mg, neither I2 is the limiting reactant.</u>

<u>f) I2 is the limiting reactant</u>

<u>g) Nor Mg, neither I2 is the limiting reactant.</u>

<u>h) I2 is the limiting reactant</u>

<u>i) Mg is the limiting reactant</u>

Explanation:

Step 1: The balanced equation:

Mg(s) + I2(s) → MgI2(s)

For 1 mol of Mg we need 1 mol of I2 to produce 1 mol of MgI2

a. 100 atoms of Mg and 100 molecules of I2

We'll have the following equation:

100 Mg(s) + 100 I2(s) → 100MgI2(s)

This is a stoichiometric mixture. <u>Nor Mg, neither I2 is the limiting reactant.</u>

b. 150 atoms of Mg and 100 molecules of I2

We'll have the following equation:

150 Mg(s) + 100 I2(s) → 100 MgI2(s)

<u>I2 is the limiting reactant</u>, and will be completely consumed. There will be consumed 100 Mg atoms. There will remain 50 Mg atoms.

There will be produced 100 MgI2 molecules.

c. 200 atoms of Mg and 300 molecules of I2

We'll have the following equation:

200 Mg(s) + 300 I2(s) →200 MgI2(s)

<u>Mg is the limiting reactant</u>, and will be completely consumed. There will be consumed 200 I2 molecules. There will remain 100 I2 molecules.

There will be produced 200 MgI2 molecules.

d. 0.16 mol Mg and 0.25 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

<u>Mg is the limiting reactant</u>, and will be completely consumed. There will be consumed 0.16 mol of I2. There will remain 0.09 mol of I2.

There will be produced 0.16 mol of MgI2.

e. 0.14 mol Mg and 0.14 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

This is a stoichiometric mixture. <u>Nor Mg, neither I2 is the limiting reactant.</u>

There will be consumed 0.14 mol of Mg and 0.14 mol of I2. there will be produced 0.14 mol of MgI2

f. 0.12 mol Mg and 0.08 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

<u>I2 is the limiting reactant</u>, and will be completely consumed. There will be consumed 0.08 moles of Mg. There will remain 0.04 moles of Mg.

There will be produced 0.08 moles of MgI2.

g. 6.078 g Mg and 63.455 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 6.078 grams / 24.31 g/mol = 0.250 moles

Number of moles I2 = 63.455 grams/ 253.8 g/mol = 0.250 moles

This is a stoichiometric mixture. <u>Nor Mg, neither I2 is the limiting reactant.</u>

There will be consumed 0.250 mol of Mg and 0.250 mol of I2. there will be produced 0.250 mol of MgI2

h. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 2.00 grams/ 253.8 g/mol = 0.00788 moles

<u>I2 is the limiting reactant</u>, and will be completely consumed. There will be consumed 0.00788 moles of Mg. There will remain 0.03322 moles of Mg.

There will be produced 0.00788 moles of MgI2.

i. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 20.00 grams/ 253.8 g/mol = 0.0788 moles

<u>Mg is the limiting reactant</u>, and will be completely consumed. There will be consumed 0.0411 moles of Mg. There will remain 0.0377 moles of I2.

There will be produced 0.0411 moles of MgI2.

4 0
3 years ago
Enter your answer in the provided box. Consider the reaction: N2(g) + 3H2(g) → 2NH3(g) Suppose that a particular moment during t
Mashcka [7]

Answer:

0.0253 M/s

Explanation:

From the reaction

N₂ + 3H₂ → 2NH₃

The rate of reaction can be written as

Rate = - \frac{d[N_2]}{dt} = - \frac{1}{3} \frac{d[H_2]}{dt} = + \frac{1}{2} \frac{d[NH_3]}{dt}

From the above rate equation we can conclude that the rate of reaction of N₂ is equal to one third of the rate of reaction of H₂,

So,

Rate of reaction of molecular nitrogen = \frac{1}{3} \times 0.0759

Upon calculation, we get rate of reaction of molecular nitrogen = 0.0253 M/s

4 0
3 years ago
The volume of a gas is the same as its _____.
AleksandrR [38]

The volume of a gas is the same as its CONTAINER.

Gases generally has no shape and no definite volume. When a gas is placed in a container, the gas usually takes the shape and the volume of the container, that is, the gas fills up all the available spaces in the container. Thus, the volume of a gas will always be the same as its container. This is in contrast with solids, which have definite shape and volume and liquids, which have definite volume but no fixed shape.

8 0
3 years ago
Read 2 more answers
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