Answer:
A.
Carbon dioxide
Explanation:
In a tissue that metabolizes glucose via the pentose phosphate pathway, C-1 of glucose would be expected to end up principally in Carbon dioxide
2-Methyl-4-oxo-pentanoic acid is unlikely to produce 2-Methyl-3-butanone upon strong heating.
Upon heating, the β ketoacid becomes unstable and decarboxylates, leading to the formation of the methyl ketone.
A carboxylic acid is an organic acid that contains a carboxyl group (C(=O)OH) attached to an R-group. The general formula of a carboxylic acid is R−COOH or R−CO2H, with R referring to the alkyl, alkenyl, aryl, or other group.
Carboxylic acids occur widely. Important examples include the amino acids and fatty acids. Deprotonation of a carboxylic acid gives a carboxylate anion.
Full question :
Q. Which reactant is unlikely to produce the indicated product upon strong heating?
- A) 2,2-Dimethylpropanedioic acid 2-methylpropanoic acid
- B) 2-Ethylpropanedioic acid Butanoic acid
- C) 2-Methyl-3-oxo-pentanoic acid 3-Pentanone
- D) 2-Methyl-4-oxo-pentanoic acid 2-Methyl-3-butanone
- E) 4-Methyl-3-oxo-heptanoic acid 3-Methyl-2-hexanone
Hence, option (D) is correct.
Learn more about carboxylic acid here : brainly.com/question/26855500
#SPJ4
There are two hydrogen atoms in one molecule of water.
The answer is B) 2
To get the percent yield, we will use this formula:
((Actual Yield)/(Theoretical Yield)) * 100%
Values given: actual yield is 220.0 g
theoretical yield is 275.6 g
Now, let us substitute the values given.
(220.0 grams)/(275.6 grams) = 0.7983
Then, to get the percentage, multiply the quotient by 100.
0.7983 (100) = 79.83%
Among the choices, the most plausible answer is 79.8%
<span>
</span>
