Answer:
Percent Composition of 41K = 6.7302%
Explanation:
The explination is in the image.
NAD serves as the bulk of the oxidative processes in the citric acid cycle's initial electron acceptor.
<h3>What are
electron acceptors in c
itric acid cycle?</h3>
- In the Krebs cycle, which transfers electrons via the electron transport chain with oxygen as the final acceptor, coenzymes like FAD and NAD+ are reduced.
- In a single cycle, three NADH+ and one FADH2 are produced, and when the cycle enters the electron transport chain, 10 ATP is produced.
- The final electron acceptor in the electron transport chain is oxygen. The proton gradient in the intermembrane gap is produced by NADH molecules donating electrons that are then transmitted through a number of different proteins.
<h3>What occurs throughout the citric acid cycle?</h3>
The cycle of citric acid: In the citric acid cycle, a six-carbon citrate molecule is created when an acetyl group from acetyl CoA is joined to a four-carbon oxaloacetate molecule.
Citrate is oxidized over a number of steps, generating two molecules of carbon dioxide for each acetyl group added to the cycle.
learn more about citric acid cycle here
<u>brainly.com/question/14900762</u>
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Answer:
9.82 g of Mg(NO₃)₂
Explanation:
Let's determine the reaction:
2AgNO₃ + MgBr₂ → Mg(NO₃)₂ + 2AgBr
2 moles of nitrate silver reacts with MgBr₂ in order to produce 1 mol of magnesium nitrate and silver bromide.
We determine the moles of AgNO₃
22.5 g . 1mol / 169.87g = 0.132 moles
Ratio is 2:1.
2 moles of silver nitrate can produce 1 mol of magnesium nitrate
Then, our 0.132 moles may produce (0.132 . 1)/ 2 = 0.0662 moles
We convert moles to mass:
0.0662 mol . 148.3 g/ mol = 9.82 g
Answer
is: 0.375 moles are present in 8.4 liters of nitrous oxide at stp.
V(N₂O) = 8.4 L.
V(N₂O) =
n(N₂O) · Vm.
Vm = 22,4 L/mol.<span>
n</span>(N₂O) = V(N₂O) ÷ Vm.
n(N₂O) = 8.4 L ÷ 22.4 L/mol.
n(N₂O) = 0.375 mol.<span>
Vm - molare volume on STP.</span>
