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aalyn [17]
3 years ago
12

for the final step of the copper cycle, zinc, Zn(s), is added to copper sulfate, CuSO4(aq). Elemental copper appears as a solid.

Explain what you think happens to the elemental zinc.
Chemistry
1 answer:
AleksAgata [21]3 years ago
5 0

Elemental zinc displaces the copper

Explanation:

In this reaction the zinc added to the copper sulfate solution has displaced the copper in the compound.

This is a single displacement reaction.

                      Zn + CuSO₄     →   ZnSO₄    +  Cu

This is the reaction of the recycling process.

The reaction is driving by the positions of the elements in the reactivity series of metal.

  • In a single displacement reaction, an atom higher up in the activity series displaces another that is lower.
  • Zinc is higher that copper in the activity series and it reacts with the sulfate.
  • It will then push the copper out as the solid product in the solution.
  • Elements higher up in the series are more reactive

learn more:

Chemical reaction brainly.com/question/6281756

#learnwithbrainly

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The Michaelis constant for pancreatic lipase is 5 mM. At 60 C, lipase is subject to deactivation with a half-life of 8 min. Fat
Dmitriy789 [7]

Explanation:

According to the given data, we will calculate the following.

 Half life of lipase t_{1/2} = 8 min x 60 s/min

                                       = 480 s

Rate constant for first order reaction is as follows.

         k_{d} = \frac{0.6932}{480}

                        = 1.44 \times 10^{-3}s^{-1}


Initial fat concentration S_{o} = 45 mol/m^{3}

                                                = 45 mmol/L

Rate of hydrolysis V_m_{o} = 0.07 mmol/L/s

Conversion X = 0.80

Final concentration (S) = S_{o} (1 - X)

                                      = 45 (1 - 0.80)

                                      = 9 mol/m^{3}

or,                                  = 9 mmol/L

It is given that K_{m} = 5mmol/L

Therefore, time taken will be calculated as follows.

                    t = -\frac{1}{K_{d}}ln[1 - \frac{K_{d}}{V}{K_{M} ln (\frac{S_{o}}{S}) + (S_{o} - S)]

Now, putting the given values into the above formula as follows.

            t = -\frac{1}{K_{d}}ln[1 - \frac{K_{d}}{V}{K_{M} ln (\frac{S_{o}}{S}) + (S_{o} - S)]  

             = -\frac{1}{1.44 \times 10^{-3}s^{-1}}ln[1 - \frac{1.44 \times 10^{-3}s^{-1}}{0.07 mmol/L/s
}{K_{M} ln (\frac{45 mmol/L
}{9 mmol/L
}) + (45 mmol/L - 9 mmol/L
)]

              = 1642.83 s \times \frac{1 min}{60 sec}

              = 27.38 min

Therefore, we can conclude that time taken by the enzyme to hydrolyse 80% of the fat present is 27.38 min.

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Answer:

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Explanation:

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Q = 100 x 4.18 x 15

Q = 6270J

Therefore, the total amount of heat absorbed is 6270J

7 0
3 years ago
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