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2 years ago

During the breakdown of fatty acids, not all acetyl coa enters the citric acid cycle, and your body converts some acetyl coa to?

1 answer:

7 0

During the breakdown of fatty acids, not all acetyl CoA enters the citric acid cycle, and your body converts some acetyl CoA to ketone bodies.

Acids are molecules or ions that can donate protons, known as Bronsted-Lowry acids, or form covalent bonds with electron pairs, known as Lewis acids. The first category of acids is proton donors or Brønsted-Lowry acids.

Acids are used as catalysts in technical and organic chemistry. For example, sulfuric acid is used in very large amounts in the alkylation process for producing gasoline. Some acids, such as sulfuric acid, phosphoric acid, and hydrochloric acid, also cause dehydration and condensation reactions. Although about 500 amino acids have been identified in nature, only 20 amino acids make up the proteins found in the human body.

Learn more about acid here

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5. What concentration of acid must be added to change the pH of 1 mM phosphate buffer from 7.4 to 7.3 (pKas of the phosphate buf

mr_godi [17]

Explanation:

According to the Henderson-Hasselbalch equation, the relation between pH and $pK_{a}$ is as follows.

pH = $pK_{a} + log \frac{base}{acid}$

where, pH = 7.4 and $pK_{a}$ = 7.21

As here, we can use the $pK_{a}$ nearest to the desired pH.

So, 7.4 = 7.21 + $log \frac{base}{acid}$

0.19 = $log \frac{base}{acid}$

$\frac{base}{acid}$ = 1.55

1 mM phosphate buffer means $[HPO_{4}]$ + $[H_{2}PO_{4}]$ = 1 mM

Therefore, the two equations will be as follows.

$\frac{HPO_{4}}{H_{2}PO_{4}}$ = 1.55 ............. (1)

$[HPO_{4}]$ + $[H_{2}PO_{4}]$ = 1 mM ........... (2)

Now, putting the value of $[HPO_{4}]$ from equation (1) into equation (2) as follows.

1.55 $[H_{2}PO_{4}] + [tex][H_{2}PO_{4}]$ = 1 mM

2.55 $[H_{2}PO_{4}]$ = 1 mM

$[H_{2}PO_{4}]$ = 0.392 mM

Putting the value of $[H_{2}PO_{4}]$ in equation (1) we get the following.

0.392 mM + $[HPO_{4}]$ = 1 mM

$[HPO_{4}]$ = (1 - 0.392) mM

$[HPO_{4}]$ = 0.608 mM

Thus, we can conclude that concentration of the acid must be 0.608 mM.

7 0

3 years ago

How many grams of material is lost in the aqueous phase if two extractions are carried out on 100 mL of a 5% (m/v) aqueous solut

san4es73 [151]

Answer:

4.94g of material

Explanation:

Partition coefficient (Kp) of a substance is defined as the ratio between concentration of organic solution and aqueous solution, that is:

Kp = 8 = Concentration in Ethyl acetate / Concentration in water

100mL of a 5% solution contains 5g of material in 100mL of water. Thus:

8 = X / 100mL / (5g-X) / 100mL

Where X is the amount of material in grams that comes to the organic phase.

8 = X / 100mL / (5g-X) / 100mL

8 = 100X / (500-100X)

4000 - 800X = 100X

4000 = 900X

4.44g = X

Thus, in the first extraction you will lost 4.44g of material from the aqueous phase.

And will remain 5g-4.44g = 0.56g.

In the second extraction:

8 = X / 100mL / (0.56g-X) / 100mL

8 = 100X / (56-100X)

448 - 800X = 100X

448 = 900X

0.50g = X

In the second extraction, you will extract 0.50g of material

Thus, after the two extraction you will lost:

4.44g + 0.50g = 4.94g of material

6 0

3 years ago

How can I calculate the valence electrons of ions?

DIA [1.3K]

First of all, you should know the valence electron of atom.

Then add charge, In case of -ve ion & subtract, in case of +ve ion.

8 0

3 years ago

An ideal gas has a volume of 14.5 L in a flexible balloon at 660.0 mmHg at

Alex787 [66]

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4 0

3 years ago

What is formed when atoms of different elements combined.

ki77a [65]

Answer:A molecule is formed when two or more atoms join together chemically.

Explanation:

6 0

3 years ago