Answer:
C
Explanation:
the enthalpy of reaction is independent of the reaction path
Answer:
Total number of ATP molecules generated from a 32-carbon fatty acid = 206 ATP molecules
Explanation:
A 32 carbon fatty acid which undergoes complete beta-oxidation assuming that the fatty acid is fully saturated will pass through the beta-oxidation cycle 14 times to produce the following:
15 molecules of acetylCoA, 14 molecules of FADH₂, and 14 molecules of NADH.
Each of the 15 acetylCoA molecules can be further oxidized in the citric acid cycle to yield the following: 15 × 3 NADH; 15 × 1 FADH₂, and 15 ATP molecules from the substrate level phosphorylation occuring at the succinylCoA synthetase catalyzed-reaction.
Total FADH₂ produced = 15 + 14 = 29 molecules of FADH₂
Total NADH produced = 45 + 14 = 59 molecules of NADH
The FADH₂ and NADH will each donate a pair of electrons to the electron transfer flavoprotein and mitochondrial NADH dehydrogenase respectively of the electron transport chain, and about 1.5 and 2.5 molecules of ATP are generated respectively when these electrons are transfered to molecular oxygen.
Thus, number of molecules of ATP generated by 29 molecules of FADH₂ = 1.5 × 29 = 43.5 molecules of ATP.
Number of molecules of ATP generated by 59 molecules of NADH = 2.5 × 59 = 147.5
Sum of ATP generated from FADH₂ and NADH = 43.5 + 147.5 = 191 ATP molecules
Total number of ATP molecules generated = 191 + 15 = 206 ATP molecules
Total number of ATP molecules generated from a 32-carbon fatty acid = 206 ATP molecules
Explanation:
<h3>yes, Radioactive decay involves the emission of a particle and/or energy as one atom changes into another. In most instances, the atom changes its identity to become a new element.</h3>
Answer:
- 13.56 g of sodium chloride are theoretically yielded.
- Limiting reactant is copper (II) chloride and excess reactant is sodium nitrate.
- 0.50 g of sodium nitrate remain when the reaction stops.
- 92.9 % is the percent yield.
Explanation:
Hello!
In this case, according to the question, it is possible to set up the following chemical reaction:
Thus, we can first identify the limiting reactant by computing the yielded mass of sodium chloride, NaCl, by each reactant via stoichiometry:
Thus, we infer that copper (II) chloride is the limiting reactant as it yields the fewest grams of sodium chloride product. Moreover the formed grams of this product are 13.56 g. Then, we take 13.56 g of sodium chloride to compute the consumed mass sodium nitrate as it is in excess:
Therefore, the leftover of sodium nitrate is:
Finally, the percent yield is computed via:
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