All categories

3 years ago

How much ATP is produced from a single glucose molecule in each chemical pathway?

Chemistry

2 answers:

vagabundo [1.1K]3 years ago

6 0

Glycolysis yields 2 ATP molecules, Kreb's cycle yields 2 ATP molecules, ETS yields 34 ATP, molecules

AURORKA [14]3 years ago

3 0

<h2>Answer : 38 ATP molecules</h2><h3>Explanation :</h3>

After each cellular respiration cycle there are ATP molecules obtained after oxidizing each glucose molecule

Out of which 2 ATP molecules are obtained from glycolysis cycle, 2 ATP molecules from the Krebs cycle, and about 34 ATP molecules from the electron transport system.

So, in all 2 + 2 + 34 = 38 ATP molecules.

You might be interested in

How many electrons does the oxygen atom need to become stable?

NemiM [27]

Answer:

2 electrons

Explanation:

Oxygen has 6 valence electrons and to be stable it needs 8. That means it needs 2 more electrons to have a full octet.

4 0

2 years ago

Read 2 more answers

Calculate the partial pressure of each gas and the total pressure if the temperature of the gas is 15 ∘C.

umka2103 [35]

Answer:

P(oxygen gas) = 3.75atm

P(helium) = 19.2atm

Ptotal = 22.95atm

Explanation:

8 0

3 years ago

Will mark as Brainliest.

SVETLANKA909090 [29]

<span>If I done the math correctly it is 3729J because you multiply 16.5 g by the 2260 J/g and get 3729 J</span>

4 0

3 years ago

The vapor pressure of pure water at 296 K is 2778.5 Pa. The vapor forms an ideal gas. 1) In some oil, the equilibrium concentrat

Murljashka [212]

Explanation:

It is given that vapor pressure of pure water at 296 K is 2778.5 Pa.These vapors will result in the formation of an ideal gas.

Now, as water is covered with oil and contains only 1% molecules of water. Hence, the vapor pressure of this mixture will also be equal to the vapor pressure of pure water.

So, vapor pressure of mixture = 1% vapor pressure of pure water

Therefore, $\text{(Vapor pressure)}_{mixture}$ = $\frac{1}{100} \times 2778.5 Pa$

= 27.785 Pa

Thus, we can conclude that the equilibrium vapor pressure of water above the oil layer is 27.785 Pa.

3 0

3 years ago

Be sure to answer all parts. one of the most important industrial sources of ethanol is the reaction of steam with ethene derive

lions [1.4K]

Answer: 2.17x10⁻³ atm

Explanation:

First, we must write the balanced chemical equation for the process:

C₂H₄(g) + H₂O(g) ⇌ C₂H₅OH(g)

The chemical reactions that occur in a closed container can reach a state of <u>chemical equilibrium</u> that is characterized because the concentrations of the reactants and products remain constant over time. The <u>equilibrium constant</u> of a chemical reaction is the value of its reaction quotient in chemical equilibrium.

The equilibrium constant (K) is expressed as <u>the ratio between the molar concentrations (mol/L) of reactants and products.</u> Its value in a chemical reaction depends on the temperature, so it must always be specified.

<u>We will use the the equilibrium constant Kc of the reaction to calculate partial pressure of ethene.</u> The constant Kc for the above reaction is,

Kc = $\frac{[C_{2} H_{5}OH]}{[H_{2}O][C_{2} H_{4}]}$

According to the law of ideal gases,

PV = nRT

where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the gas constant (0.082057 atm L / mol K) .

We can use the ideal gas law to determine the molar concentrations ([x] = n / V) from the gas pressures of ethanol and water, assuming that all gases involved behave as ideal gases. In this way,

PV = nRT → P = (n/V) RT → P = [x] RT → [x] = P / RT

So,

$[C_{2} H_{5}OH] = \frac{200 atm}{0.082057 \frac{atm L}{mol K} x 600 K } = 4.06 \frac{mol}{L}$

$[H_{2}O] = \frac{400 atm}{0.082057 \frac{atm L}{mol K} x 600 K } = 8.12 \frac{mol}{L}$

So, the molar concentration of ethene (C₂H₄) will be,

$[C_{2} H_{4}] = \frac{[C_{2} H_{5}OH]}{[H_{2}O] x Kc} = \frac{4.06 \frac{mol}{L} }{8.12 \frac{mol}{L}x9.00 x 10^{3} \frac{L}{mol} } = 5.56 x 10^{-5}\frac{mol}{L}$

Then, according to the law of ideal gases,

$P_{C_{2} H_{4}} = [C_{2} H_{4}]RT = 5.56 x 10^{-5} \frac{mol}{L} x 0.082057 \frac{atm L}{mol K} x 600 K = 2.17x10^{-3} atm$

So, when the partial pressure of ethanol is 200 atm and the partial pressure of water is 400 atm, the partial pressure of ethene at 600 K is 2.17x10⁻³ atm.

7 0

3 years ago