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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 40% carbon, 6.7% hydrogen, and 53.3% oxygen. What is the empirical formula?
Ahat [919]

Answer:

                           O        H        C

Moles in 100g  3.33   6.65    3.33

Ratio                  1.00   2.00    1.00

Possible empirical formula = OH_{2}C

6 0
3 years ago
Read 2 more answers
Isotopes of carbon differ with the respect to the number of
hodyreva [135]

Answer:

Neutrons.

Explanation:

Isotopes can be defined as the atom of an element that has the same number of protons but different number of neutrons. This ultimately implies that, the isotopes of an element have the same atomic number (number of protons) but different atomic mass (number of nucleons).

The isotope of an element is denoted by X^{A}_{Z}

Where; X is the symbol of the element.

A is the atomic mass or number of nucleons.

Z is the atomic number or number of protons.

<em>Therefore, the number of neutrons = A - Z</em>

<em>Isotopes of carbon differ with respect to the number of neutrons.</em>

<em>Basically, there are three (3) Isotopes of Carbon and these are;</em>

<em>1. Carbon-12: it has an atomic mass of 12 with 6 numbers of proton and neutron respectively. </em>

<em>2. Carbon-13: it has an atomic mass of 13 with 6 numbers of proton and 7 numbers of neutron. </em>

<em>3. Carbon-14: it has an atomic mass of 14 with 6 numbers of proton and 8 numbers of neutron.  </em>

8 0
3 years ago
Consider the following intermediate reactions.
Alja [10]

2.1648 kg of CH4 will generate 119341 KJ of energy.

Explanation:

Write down the values given in the question

CH4(g) +2 O2 → CO2(g) +2 H20 (g)

ΔH1 = - 802 kJ

2 H2O(g)→2 H2O(I)

ΔH2= -88 kJ

The overall chemical reaction is

CH4 (g)+2 O2(g)→CO2(g)+2 H2O (I) ΔH2= -890 kJ

CH4 +2 O2 → CO2 +2 H20

(1mol)+(2mol)→(1mol+2mol)

Methane (CH4) = 16 gm/mol

oxygen (O2) =32 gm/mol

Here 1 mol CH4 ang 2mol of O2 gives 1mol of CO2 and 2 mol of 2 H2O

which generate 882 KJ /mol

Therefore to produce 119341 KJ of energy

119341/882 = 135.3 mol

to produce 119341 KJ of energy, 135.3 mol of CH4 and 270.6 mol of O2 will require

=135.3 *16

=2164.8 gm

=2.1648 kg of CH4

2.1648 kg of CH4 will generate 119341 KJ of energy

4 0
3 years ago
Read 2 more answers
Assuming that an acetic acid solution is 12% by mass and that the density of the solution is 1.00 g/mL, what volume of 1 M NaOH
Doss [256]

Explanation:

Let us assume that total mass of the solution is 100 g. And, as it is given that acetic acid solution is 12% by mass which means that mass of acetic acid is 12 g and 88 g is the water.

Now, calculate the number of moles of acetic acid as its molar mass is 60 g/mol.

    No. of moles = \frac{mass}{\text{molar mass}}

                           = \frac{12 g}{60 g/mol}

                           = 0.2 mol

Molarity of acetic acid is calculated as follows.

              Density = \frac{mass}{volume}

                 1 g/ml = \frac{100 g}{volume}

                    volume = 100 ml

Hence, molarity = \frac{\text{no. of moles}}{volume}

                           = \frac{0.2 mol}{0.1 L}

                           = 2 mol/l

As reaction equation for the given reaction is as follows.

     NaOH + CH_{3}COOH \rightarrow CH_{3}COONa + H_{2}O

So,          moles of NaOH = moles of acetic acid

Let us suppose that moles of NaOH are "x".

          x \times 1 M = 10 mL \times 2 M     (as 1 L = 1000 ml)

                        x = 20 L

Thus, we can conclude that volume of NaOH required is 20 ml.                    

6 0
3 years ago
A 0.595 g sample of a metal, M, reacts completely with sulfuric acid according to the reaction M ( s ) + H 2 SO 4 ( aq ) ⟶ MSO 4
zalisa [80]

Answer:

molar mass M(s) = 65.326 g/mol

Explanation:

  • M(s) + H2SO4(aq) → MSO4(aq) + H2(g)

∴ VH2(g) = 231 mL = 0.231 L

∴ P atm = 1.0079 bar

∴ PvH2O(25°C) = 0.03167 bar

Graham´s law:

⇒ PH2(g) = P atm - PvH2O(25°C)

⇒ PH2(g) = 1.0079 bar - 0.03167 bar = 0.97623 bar = 0.9635 atm

∴ nH2(g) = PV/RT

⇒ nH2(g) = ((0.9635 atm)(0.231 L))/((0.082 atmL/Kmol)(298 K))

⇒ nH2(g) = 9.1082 E-3 mol

⇒ n M(s) = ( 9.1082 E-3 mol H2(g) )(mol M(s)/mol H2(g))

⇒ n M(s) = 9.1082 E-3 mol

∴ molar mass M(s) [=] g/mol

⇒ molar mass M(s) =   (0.595 g) / (9.1082 E-3 mol)

⇒ molar mass M(s) = 65.326 g/mol

7 0
3 years ago
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