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12345 [234]
3 years ago
14

The CO2 produced in one round of the citric acid cycle does not originate in the acetyl carbons that entered that round. If the

acetyl-CoA is labeled with 14C at its carbonyl carbon, how many rounds of the citric acid cycle are required before 14CO2 is released
Chemistry
1 answer:
aleksley [76]3 years ago
7 0

Answer:

It will require<u> second round</u> of the cycle to release 14C0_2

Explanation:

<u>Reason behind the requirement of second round of the cycle to release </u>CO_2 -:

The C4 carbon of succinyl CoA is acetyl from acetyl CoA. Succinyl CoA is converted to succinate, which is then converted to fumarate, fumarate, malate, and eventually oxaloacetate. 14C will be found in oxaloacetate at either C1 or C4. During the second round of the loop, each of these carbons will be converted to carbon dioxide.

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3 years ago
Read 2 more answers
How many grams of carbon are present in 45.0 g of CCl4?
lisabon 2012 [21]
To determine the amount of a certain element in a compound, we use the ratio of the elements from the compound. We calculate is follows:

45.0 g CCl4 ( 1 mol CCl4 / 153.82 g CCl4 ) ( 1 mol C / 1 mol CCl4 ) ( 12.01 g C / 1 mol C ) = 3.5135 g carbon present

Hope this answers the question. Have a nice day.
5 0
3 years ago
Someone Please HELP Me
scoray [572]
Use a calculator to add those thank you ur welcome
7 0
3 years ago
A compound was found to be soluble in water. It was also found that addition of acid to an aqueous solution of this compound res
gogolik [260]

Answer:

c. Cr

Explanation:

Cr^3+(aq) + CO_3^2-(aq) ------------> Cr_2CO_3(s)

The compound is containing CO_3^2- ion.

If acid reacts with it, CO_2 evolves.

H^+ + CO_3^2- ----------> H_2O + CO_2

therefore,  Cr would form a precipitate when added to an aqueous solution of this compound. Cr2CO3 is the precipitate.

3 0
3 years ago
How many grams is 3.35 moles of hcl
lions [1.4K]
Do this

3.35mol HCl | 34.46g HCl
------------------------------------
          1          | 1mol HCl
Multiply all the numbers on top by all the numbers on bottom.

7 0
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