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Tems11 [23]
3 years ago
8

Identify the source of the electrons that travel down the electron transport chain. explain why oxygen is the final electron acc

eptor in aerobic cellular respiration.
Chemistry
1 answer:
satela [25.4K]3 years ago
6 0
1) The electrons that travel down the electron transport chain come from the NADH and FADH2 molecules produced in the three previous stages of cellular respiration : glycolysis, pyruvate oxidation, and the <span>citric acid cycle.

</span>2) At the end of the electron transport chain is the Oxygen  that will accept electrons and picks up protons to form water.
If the oxygen molecule is not there the electron transport chain will stop running, and ATP will no longer be produced.



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If aluminum has a density of 2.7 g/cm3, what is the volume, in cubic
Pachacha [2.7K]

Answer:

20cm3

Explanation:

density = mass/volume, so volume = mass/density

volume = (54g)/(2.7g/cm3)= 20cm3

6 0
3 years ago
Which statement about mixtures and solutions is incorrect?
harkovskaia [24]
B)all  solutions are mixtures

5 0
4 years ago
A piece of copper (12.0 g) is heated to 100.0 °C. A piece of chromium (also 12.0 g) is chilled in an ice bath to 0 °C. The speci
Gennadij [26K]

Answer:

(a) Slightly greater than 20.0 °C

(b) T_F=293.46K=20.3^oC

Explanation:

Hello,

In this case, since we are talking about the equilibrium temperature that will be reached when the copper, chromium and water samples get in contact, the following equation is useful to describe such situation:

\Delta H_{water}+\Delta H_{copper}+\Delta H_{chromium}=0\\

Thus, in terms of masses, heat capacities and temperatures we consider the final temperature as the unknown:

m_{water}Cp_{water}(T_F-T_{water})+m_{copper}Cp_{copper}(T_F-T_{copper})+m_{chromium}Cp_{chromium}(T_F-T_{chromium})=0In such a way, by knowing that the heat capacities of copper and chromium are 0.386 and 0.45 J/(g°C) respectively, by solving for the equilibrium temperature one has:

T_F=\frac{m_{water}Cp_{water}T_{water}+m_{Cu}Cp_{Cu}T_{Cu}+m_{Cr}Cp_{Cr}T_{Cr}}{m_{water}Cp_{water}+m_{Cu}Cp_{Cu}+m_{Cr}Cp_{Cr}}

T_F=\frac{200.0g*4.184\frac{J}{g*K}* 293.15K+12.0g*0.386\frac{J}{g*K} *373.15K+12.0g*0.45\frac{J}{g*K}*273.15K}{200.0g*4.184\frac{J}{g*K}+12.0g*0.386\frac{J}{g*K} +12.0g*0.45\frac{J}{g*K}}\\\\T_F=293.46K=20.3^oC

Hence, the resulting temperature of water turns out slightly greater than 20.0 °C.

Best regards.

7 0
3 years ago
A student adds 5400.0mL of 0.15M NaOH to 201.2mL of 4.0M NaOH. What is the final [NaOH]?
finlep [7]

Answer:

0.287 M

Explanation:

Multiply the concentration of each solution by the volume of each (in liters) to get the moles of NaOH in that solution.

0.15 M • 5.4000 L = 0.81 mol NaOH

4.0 M • 0.2012 L = 0.80 mol NaOH

Add the mol of NaOH together to get the total --> 0.81 + 0.80 = 1.61 mol NaOH

Divide by the total volume of solution (5400.0 mL + 201.2 mL = 5,601.2 mL = 5.6012 L)

1.61 mol / 5.6012 L = 0.287 M NaOH

6 0
3 years ago
Select the best answer.
madreJ [45]
Answer: All of these I’m pretty sure

Explanation: It just makes sense
8 0
3 years ago
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