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o-na [289]
4 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]4 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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A student dissolves 15.0 g of ammonium chloride(NH4Cl) in 250. 0 g of water in a well-insulated open cup. She then observes the
iren2701 [21]

Answer:  

1) Endothermic.  

2) Q_{rxn}=4435.04J  

3) \Delta _rH=15.8kJ/mol

Explanation:  

Hello there!  

1) In this case, for these calorimetry problems, we can realize that since the temperature decreases the reaction is endothermic because it is absorbing heat from the solution, that is why the temperature goes from 22.00 °C to 16.0°C.  

2) Now, for the total heat released by the reaction, we first need to assume that all of it is released by the solution since it is possible to assume that the calorimeter is perfectly isolated. In such a way, it is also valid to assume that the specific heat of the solution is 4.184 J/(g°C) as it is mostly water, therefore, the heat released by the reaction is:

Q_{rxn}=-(15.0g+250.0g)*4.184\frac{J}{g\°C}(16.0-20.0)\°C\\\\ Q_{rxn}=4435.04J    

3) Finally, since the enthalpy of reaction is calculated by dividing the heat released by the reaction over the moles of the solute, in this case NH4Cl, we proceed as follows:

\Delta _rH=\frac{ Q_{rxn}}{n}\\\\\Delta _rH= \frac{ 4435.04J}{15.0g*\frac{1mol}{53.49g} } *\frac{1kJ}{1000J} \\\\\Delta _rH=15.8kJ/mol

Best regards!  

Best regards!

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3 years ago
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