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ICE Princess25 [194]
3 years ago
13

Which of the following is the major regulator of oxygen consumption during oxidative phosphorylation?

Chemistry
2 answers:
Klio2033 [76]3 years ago
8 0

Answer:

f.ADP

Explanation:

  • <u>Oxidative phosphorylation </u>occurs during the electron transport chain which is the final phase of cellular respiration. It depends on the hydrogen ion concentration gradient generated and is maintained by the electron transport chain.
  • <u>The oxidative phosphorylation acquires energy of high-energy electrons to synthesize ATP. During this process ADP is the major regulator of oxygen consumption.</u>
  • The regulation by ADP is referred to as the acceptor or respiratory control.
aalyn [17]3 years ago
5 0

Answer: F. ADP

Explanation:

The level of ADP is the most important factor in the determination of the rate of Oxidative Phosphorylation.

The rate of oxygen utilization by the Mitochondria is increased when ADP is added and usually returns to its initial level when the added ADP has being converted to ATP.

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Answer:

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What is the density of chlorine gas (MM = 71.0glmol) at 1.50 atm and 25.0C
IrinaVladis [17]

Answer: D=4.35g/L

Explanation:

The formula for density is D=\frac{M}{V}. M is mass in grams and V is volume in liters.

Since we are give pressure and temperature, we can use the ideal gas law to find moles/volume. FInding moles/volume would give us the base for density. All we would have to do is convert moles to grams.

Ideal Gas Law: PV=nRT

\frac{n}{V} =\frac{P}{RT}

\frac{n}{V}=\frac{1.50 atm}{(0.08206Latm/Kmol)(25+274.15K)}

\frac{n}{V} =\frac{0.061309mol}{L}

Now that we have moles, we can use molar mass of chlorine gas to find grams.

0.061309mol*\frac{71.0g}{mol} =4.3529g

With our grams, we can find our density.

D=\frac{4.3529g}{L}

We need correct significant figures so our density is:

D=\frac{4.35g}{L}

5 0
3 years ago
Can an element with eight valence electrons still be reactive?
raketka [301]
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3 years ago
Some plants can grow from both blank and blank
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A gas mixture at 535.0°C and 109 kPa absolute enters a heat exchanger at a rate of 67.0 m3/hr. The gas leaves the heat exchanger
SVEN [57.7K]

Answer:

the heat rate required to cool down the gas from 535°C until 215°C is -2.5 kW.

Explanation:

assuming ideal gas behaviour:

PV=nRT

therefore

P= 109 Kpa= 1.07575 atm

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T= 215 °C = 488 K

R = 0.082 atm L /mol K

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since the changes in kinetic and potencial energy are negligible, the heat required is equal to the enthalpy change of the gas:

Q= n* Δh = 0.5 mol/s * (- 5 kJ/mol) =2.5 kW

7 0
3 years ago
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