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
In order from most to least similar:
1. Germanium
2. Lead
3. Phosphorus
4. Chlorine
The elements in the same column as the element you have are the most similar. The rows are not. For example, though chlorine and magnesium are in the same row, they have very different properties, whereas chlorine and fluorine more similar
Leftover: approximately 11.73 g of sulfuric acid.
<h3>Explanation</h3>
Which reactant is <em>in excess</em>?
The theoretical yield of water from Al(OH)₃ is lower than that from H₂SO₄. As a result,
- Al(OH)₃ is the limiting reactant.
- H₂SO₄ is in excess.
How many <em>moles</em> of H₂SO₄ is consumed?
Balanced equation:
2 Al(OH)₃ + 3 H₂SO₄ → Al₂(SO₄)₃ + 6 H₂O
Each mole of Al(OH)₃ corresponds to 3/2 moles of H₂SO4. The formula mass of Al(OH)₃ is 78.003 g/mol. There are 15 / 78.003 = 0.19230 moles of Al(OH)₃ in the five grams of Al(OH)₃ available. Al(OH)₃ is in excess, meaning that all 0.19230 moles will be consumed. Accordingly, 0.19230 × 3/2 = 0.28845 moles of H₂SO₄ will be consumed.
How many <em>grams</em> of H₂SO₄ is consumed?
The molar mass of H₂SO₄ is 98.076 g.mol. The mass of 0.28845 moles of H₂SO₄ is 0.28845 × 98.076 = 28.289 g.
How many <em>grams</em> of H₂SO₄ is in excess?
40 grams of sulfuric acid H₂SO₄ is available. 28.289 grams is consumed. The remaining 40 - 28.289 = 11.711 g is in excess. That's closest to the first option: 11.73 g of sulfuric acid.
<u>One s orbital</u> and <u>one p orbital</u> are the exact types of atomic orbitals of the central atom mix to form hybrid orbitals in CS₂
<h3>
What is atomic orbital?</h3>
An atomic orbital is a function used in atomic theory and quantum mechanics to explain the position and wave-like behaviour of an electron in an atom. This function can be used to determine the likelihood of discovering any atom's electron in any particular area surrounding the nucleus.
The physical area or space where the electron may be calculated to be present, as predicted by the specific mathematical shape of the orbital, is referred to as an atomic orbital.
The three quantum numbers n, l, and 
which correspond to the electron's energy, angular momentum, and an angular momentum vector component, are used to describe all orbitals in an atom (magnetic quantum number).
Learn more about Atomic Orbital
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Answer: Concentration of 
at equilibrium = 0.08 M
Concentration of 
at equilibrium =0.08 M
Concentration of 
= 0.64M
Explanation:
Moles of 
= 4.00 mole
Volume of solution = 5.00 L
Initial concentration of =
Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as 
For the given chemical reaction:
Initial conc. 0.8 M 0 M 0 M
At eqm. conc. (0.8-2x) M (x) M (x) M
The expression for is written as:
![K_c=\frac{[H_2]^1[I_2]^1}{[HI]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BH_2%5D%5E1%5BI_2%5D%5E1%7D%7B%5BHI%5D%5E2%7D)
Concentration of at equilibrium = x M = 0.08 M
Concentration of at equilibrium = x M = 0.08 M
Concentration of = (0.8-2x) = 
