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Delvig [45]
3 years ago
11

Pyruvate is the end product of glycolysis. Its further metabolism depends on the organism and on the presence or absence of oxyg

en. Draw the structure of the product from each reaction as it would exist at pH 7. Include the appropriate hydrogen atoms. Reaction A: aerobic conditions in humans or yeast

Chemistry
1 answer:
murzikaleks [220]3 years ago
5 0

The given question is incomplete. The image present in the question for Reaction A is attached below along with the answer.

Explanation:

Pyruvate molecule reacts with Coenzyme A in the presence of oxygen and it results in the formation of acetyl Coenzyme A and carbon dioxide.

The enzyme pyruvae dehydrogenase helps in catalyzing this reaction. As in this biochemical reaction NAD^{+} gets converted into NADH.

This reaction is shown in the image attached below.  

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1. If a person's mass is 56 kg, what is this weight in pounds?<br> 56kg
Natali [406]

Answer:

123lb

Explanation:

lb= 2.2046 kg

56×2.2046= 123lb

3 0
3 years ago
The activation energy for a reaction is changed from 184 kJ/mol to 59.0 kJ/mol at 600. K by the introduction of a catalyst. If t
11111nata11111 [884]

Answer:

The catalyzed reaction will take 2.85 seconds to occur.

Explanation:

The activation energy of a reaction is given by:                                                        

k = Ae^{-\frac{E_{a}}{RT}}

For the reaction without catalyst we have:

k_{1} = Ae^{-\frac{E_{a_{1}}}{RT}}   (1)

And for the reaction with the catalyst:

k_{2} = Ae^{-\frac{E_{a_{2}}}{RT}}   (2)

Assuming that frequency factor (A) and the temperature (T) are constant, by dividing equation (1) with equation (2) we have:                      

\frac{k_{1}}{k_{2}} = \frac{Ae^{-\frac{E_{a_{1}}}{RT}}}{Ae^{-\frac{E_{a_{2}}}{RT}}}

\frac{k_{1}}{k_{2}} = e^{\frac{E_{a_{2}} - E_{a_{1}}}{RT}    

\frac{k_{1}}{k_{2}} = e^{\frac{59.0 \cdot 10^{3}J/mol - 184 \cdot 10^{3} J/mol}{8.314 J/Kmol*600 K} = 1.31 \cdot 10^{-11}    

Since the reaction rate is related to the time as follow:

k = \frac{\Delta [R]}{t}

And assuming that the initial concentrations ([R]) are the same, we have:

\frac{k_{1}}{k_{2}} = \frac{\Delta [R]/t_{1}}{\Delta [R]/t_{2}}

\frac{k_{1}}{k_{2}} = \frac{t_{2}}{t_{1}}

t_{2} = t_{1}\frac{k_{1}}{k_{2}} = 6900 y*1.31 \cdot 10^{-11} = 9.04 \cdot 10^{-8} y*\frac{365 d}{1 y}*\frac{24 h}{1 d}*\frac{3600 s}{1 h} = 2.85 s

Therefore, the catalyzed reaction will take 2.85 seconds to occur.

I hope it helps you!                            

4 0
3 years ago
Balance the following Equation Show your work.
7nadin3 [17]

Answer:

1) alr balanced

2) 2 2 1

3) 2 1 1 2

4) 2 3 2

5) 2 1 2 1

6) 1 6 2 3

7) 2 2 1

8) 1 2 1 2

9) 2 5 4 6

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3 0
3 years ago
What have all living things evolved from?
EastWind [94]

Answer:

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3 0
3 years ago
Read 2 more answers
Use the solubility graph to answer the following questions.
Akimi4 [234]

The required amount of KClO₃ to prepare a saturated solution is 48 grams.

<h3>What is solubility?</h3>

The ability of a material, the solute, to create a solution with another substance, the solvent, is known as solubility.

Given graph is plotted between the temperature and solubility of the substance at 100g of water, so for the given points calculation will be:

  • According to the graph for the KClO₃ we require almost 48g of KClO₃ to prepare a saturated solution.
  • From the graph it is clear that at 40°C 40g of KCl will dissolve in 100g of water then amount of KCl which will dissolve in 68g of water will be calculated as:

x = (40)(68) / (100) = 27.2g

  • On the basis of the graph, if 75 grams of calcium chloride is added to 100 g of water at 25°C then the obtained solution is the unsaturated as they have les solute then the limit.
  • Based on the graph cerium sulfate behaves like a precipitate which will not dissolve at any temperature and present in the physical state.
  • We can add more 30g to make the KCl solution saturated.

Hence, we require almost 48g of KClO₃ to prepare a saturated solution of KClO₃.

To know more about solubility, visit the below link:

brainly.com/question/16903071

#SPJ1

4 0
2 years ago
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