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alina1380 [7]
3 years ago
15

Given the information below, which is more favorable energetically, the oxidation of succinate to fumarate by NAD+ or by FAD? Fu

marate + 2H+ + 2e- → Succinate E°´ = 0.031 V NAD+ + 2H+ + 2e- → NADH + H+ E°´ = -0.320 FAD + 2H+ + 2e- → FADH2 E°´ = -0.219
Chemistry
1 answer:
sasho [114]3 years ago
8 0

Answer:

Oxidation by FAD  

Explanation:

1. Oxidation by NAD⁺

Succinate ⇌ Fumarate + <u>2H⁺ + 2e⁻</u>;                  E°´ =  -0.031 V  

<u>NAD⁺ + </u><u>2H⁺ + 2e⁻</u><u> ⇌ NADH + H⁺;                    </u>  E°´ = <u> -0.320 V</u>

Succinate + NAD⁺ ⇌ Fumarate  + NADH + H⁺; E°' =  -0.351 V

2. Oxidation by FAD

Succinate ⇌ Fumarate + 2H⁺ + 2e⁻;        E°´ = -0.031 V  

<u>FAD + 2H⁺ + 2e⁻ ⇌ FADH₂;                   </u>  E°´  = <u>-0.219 V </u>

Succinate + FADH₂ ⇌ Fumarate + FAD; E°' = -0.250 V

Neither reaction is energetically favourable, but FAD has a more positive half-cell potential.

FAD is the stronger oxidizing agent.

The oxidation by FAD has a more positive cell potential, so it is more favourable energetically.  

 

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Mr. Ragusa asks Hassan to make silver crystals from the following reaction.
SVEN [57.7K]

Answer:

Percentage yield = 61.7%

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated:

2AgNO₃ + Cu —> Cu(NO₃)₂ + 2Ag

Next, we shall determine the mass of AgNO₃ that reacted and the mass of Ag produced from the balanced equation. This is illustrated below:

Molar mass of AgNO₃ = 108 + 14 + (16×3)

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Molar mass of Ag = 108 g/mol

Mass of Ag from the balanced equation = 2 × 108 = 216 g

SUMMARY:

From the balanced equation above,

340 g of AgNO₃ reacted to produce 216 g of Ag.

Next, we shall determine the theoretical yield of Ag. This can be obtained as follow:

From the balanced equation above,

340 g of AgNO₃ reacted to produce 216 g of Ag.

Therefore, 51 g of AgNO₃ will react to produce = (51 × 216)/340 = 32.4 g of Ag.

Thus, the theoretical yield of Ag is 32.4 g.

Finally, we shall determine the percentage yield of Ag. This can be obtained as follow:

Actual yield = 20 g

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Percentage yield =?

Percentage yield = Actual yield /Theoretical yield × 100

Percentage yield = 20 / 32.4 × 100

Percentage yield = 61.7%

6 0
2 years ago
A chemist adds of a M zinc nitrate solution to a reaction flask. Calculate the mass in grams of zinc nitrate the chemist has add
weqwewe [10]

corrected question: A chemist adds 135mL of a 0.21M zinc nitrate solution to a reaction flask. Calculate the mass in grams of zinc nitrate the chemist has added to the flask. Round your answer to significant digits.

Answer:

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Explanation:

0.21M means ; 0.21mol/dm³

1dm³=1L , so we can say 0.21mol/L

if 0.21mol of Zinc nitrate is contained in 1L of water

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molar mass of Zn(NO₃)₂=189.36 g/mol

mass= 0.02835 *189.36

mass=5.37g

3 0
2 years ago
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Answer:

Answer: pH = 2.72

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