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o-na [289]
3 years ago
15

The reduced coenzymes generated by the citric acid cycle donate electrons in a series of reactions called the electron transport

chain. The energy from the electron transport chain is used for oxidative phosphorylation. Which compounds donate electrons to the electron transport chain? H 2 O H2O FADH 2 FADH2 ADP ADP NADH NADH NAD + NAD+ O 2 O2 FAD FAD ATP ATP Which compound is the final electron acceptor? NADH NADH ATP ATP H 2 O H2O NAD + NAD+ FADH 2 FADH2 FAD FAD O 2 O2 ADP ADP Which compounds are the final products of the electron transport chain and oxidative phosphorylation? ADP ADP NADH NADH O 2 O2 NAD + NAD+ FADH 2 FADH2 ATP ATP H 2 O H2O FAD
Chemistry
1 answer:
PtichkaEL [24]3 years ago
8 0

Answer:

The reduced coenzymes generated by the citric acid cycle donate electrons in a series of reactions called the electron transport chain. The energy from the electron transport chain is used for oxidative phosphorylation.

a)The compounds that donate electrons to the electron transport chain are NADH and . FADH2

b) O2 is the final electron acceptor.

c) The final products of the electron transport chain and oxidative phosphorylation are NAD+, H2O, ATP and FAD

Explanation:

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We make a basic solution by mixing 50. mL of 0.10 M NaOH and 50. mL of 0.10 M Ca(OH)2. It requires 250 mL of an HCl solution to
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<u>Answer:</u> The correct answer is Option 5.

<u>Explanation:</u>

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M=\frac{n_1M_1V_1+n_2M_2V_2}{V_1+V_2}

where,

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n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of the Ca(OH)_2

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n_1=1\\M_1=0.10M\\V_1=50mL\\n_2=2\\M_2=0.1\\V_2=50mL  

Putting all the values in above equation, we get:

M=\frac{(1\times 0.1\times 50)+(2\times 0.1\times 50)}{50+50}\\\\M=0.15M

  • To calculate the molarity of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is HCl

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base.

We are given:

n_1=1\\M_1=?M\\V_1=250mL\\n_2=1\\M_2=0.15M\\V_2=100mL

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1\times M_1\times 250=1\times 0.15\times 100\\\\M_1=0.06M

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