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denis-greek [22]
3 years ago
11

What is the final electron acceptor in cellular respiration?

Chemistry
1 answer:
Artemon [7]3 years ago
5 0
<span>The potential of NADH and FADH2 is converted to more ATPthrough an electron transport chain with oxygen as the "terminal electron acceptor". Most of the ATP produced by aerobic cellular respiration is made by oxidative phosphorylation.

Hope this helps. :)

</span>
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What has more kinetic energy? 120 180, 105
bija089 [108]

Answer:

If those values are in terms of mass, weight or velocity

Kinetic energy will be in the order 180 » 120 » 105

5 0
3 years ago
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Organisms that are most similar belong in a ,
solmaris [256]

Answer:

The answer is D, organisms that are mostly related belong in a species.

Explanation:

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3 years ago
Suppose the formation of iodine proceeds by the following mechanism:
NARA [144]

Answer:

Overall reaction

H2(g) + 2ICI(g) -----> I2(g) +2HCl(g)

Overall Rate = k1[H2] [ICl]

Explanation:

Overall reaction

H2(g) + 2ICI(g) -----> I2(g) +2HCl(g)

The overall reaction is the sum of the two two reactions shown in the question. After the two reactions are summed up properly, this overall reaction equation his obtained.

Since K1<<K2 it means that step 1 is slower than step 2. Recall that the rate if reaction depends on the slowest step of the reaction. Hence

Overall Rate = k1[H2] [ICl]

3 0
3 years ago
Hydrocarbons that contoins least double bond are
Helen [10]

Answer:

Chemistry has taught that ;

Unsaturated hydrocarbon molecules that contain one or more double bonds are called alkenes.

Alkynes are  unsaturated hydrocarbons due to the fact that some of the carbons are connected to fewer than four neighbouring atoms.

Note : Alkenes contain at least one double bond, while alkynes contain at least one triple bond.

Explanation:

3 0
3 years ago
Assuming a car (with a 70-L) gas tank can hold approximately 50,000 (5.00 * 10^4) g of octane(C8H18) or 50,000 (5.00 * 10^4) g o
konstantin123 [22]

Answer:

- From octane: m_{CO_2}=1.54x10^5gCO_2

- From ethanol: m_{CO_2}=9.57x10^4gCO_2

Explanation:

Hello,

At first, for the combustion of octane, the following chemical reaction is carried out:

C_8H_{18}+\frac{25}{2} O_2\rightarrow 8CO_2+9H_2O

Thus, the produced mass of carbon dioxide is:

m_{CO_2}=5.00x10^4gC_8H_{18}*\frac{1molC_8H_{18}}{114gC_8H_{18}}*\frac{8molCO_2}{1molC_8H_{18}}*\frac{44gCO_2}{1molCO_2} \\\\m_{CO_2}=1.54x10^5gCO_2

Now, for ethanol:

C_2H_6O+3O_2\rightarrow 2CO_2+3H_2O

m_{CO_2}=5.00x10^4gC_2H_6O*\frac{1molC_2H_6O}{46gC_2H_6O}*\frac{2molCO_2}{1molC_2H_6O}*\frac{44gCO_2}{1molCO_2} \\\\m_{CO_2}=9.57x10^4gCO_2

Best regards.

3 0
3 years ago
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