Answer:
The answer is given below
Explanation:
- β oxidation of stearoyl-CoA------------ 8 FAD reduced
- β oxidation of oleoyl-CoA -------------- 1- 7 FAD reduced and 2- reduced requires δ3,δ2-enoyl-CoA isomerase required for completion of round 4
- in Both ----------- 1- 9 acetyl-CoA produced, 2- 8 NAD+ reduced and 3- β oxidation completed in 8 rounds
Answer:
just answer this and you will have yours
Explanation:Find the area of a circle with a diameter of \color{green}{16}16start color green, 16, end color green.
Either enter an exact answer in terms of \piπpi or use 3.143.143, point, 14 for \piπpi and enter your answer as a decimal.
of octane had been converted to carbon dioxide CO₂.
<h3>Explanation</h3>
Octane has a molar mass of
1.000 gallon of this fuel would have a mass of 2.650 kilograms or 
, which corresponds to 
of octane.
Octane undergoes complete combustion to produce carbon dioxide and water by the following equation:
An incomplete combustion of octane that gives rise to carbon monoxide and water but no carbon dioxide would consume not as much oxygen:
The mass of the product mixture is 
heavier than that of the octane supplied. Thus 
of oxygen were consumed in the combustion. There are 
of oxygen molecules in 
of oxygen.
Let the number of moles of octane that had undergone complete combustion as seen in the first equation be 
(
). The number of moles of octane that had undergone incomplete combustion through the second equation would thus equal 
.
25 moles of oxygen gas is consumed for every two moles of octane that had undergone complete combustion and 17 moles if the combustion is incomplete.
Therefore 
out of the 23.2 moles of octane had undergone complete combustion to produce carbon dioxide.
I think it’s D, silicon Fluoride.
Answer is: a chemical formula of the compound is C₄Cl₄.
CCl is empirical formula of the compound. Empirical formula gives the proportions of the elements present in a compound, so we know that proportion od carbon and chlorine is 1 : 1 (n(C) : n(Cl) = 1 : 1).
M(CCl) = Ar(C) + Ar(Cl) · g/mol.
M(CCl) = 12.0107 + 35.453 · g/mol.
M(CCl) = 47.4637 g/mol; molar mass of empirical formula.
M(compound) = 189.86 g/mol; molar mass of the compound.
M(compound) / M(CCl) = 189.86 g/mol / 47.4637 g/mol.
M(compound) / M(CCl) = 4; chemical formula of the compound have four carbon and four chlorine atoms.
