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xz_007 [3.2K]
2 years ago
5

Please, can i get some help with these question? Can anyone please answer what exactly i should write?

Chemistry
2 answers:
pav-90 [236]2 years ago
8 0
Use photomath. I would really recommend it! Hope this helps!
FinnZ [79.3K]2 years ago
5 0
"Photomath" was the answer for your question, Buddy
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Calculate the Zn conc. of Zn/Zn++ // Cl/Cl- 0.1M Emf=2.21v
Leto [7]

Answer:

Option a. = 0.01 M

Explanation:

To do this, we need to gather the data:

E = 2.21 V

[Cl⁻] = 0.1 M

And the Redox reaction taking place is the following:

Zn(s) + Cl₂(g) <-------> Zn²⁺(aq) + 2Cl⁻(aq)       Q = [Zn] [Cl]²

E° Cl⁻/Cl₂ = 1.36 V

E° Zn/Zn²⁺ = -0.76 V

According to this, the expression to use will be the Nernst equation, and we can assume we are working at 25 °C, therefore, the Nernst equation will be:

E = E° - (0.059/n) logQ

E = E° - (0.059/n) ln([Cl⁻]² * [Zn²⁺])   (1)

From there, we can solve for Zn later.

First, we need to write the semi equation of oxidation and reduction, and get the standard potential of the cell:

Zn(s) --------> Zn²⁺(aq) + 2e⁻       E₁° = 0.76 V

Cl₂(g) + 2e⁻ -----------> 2Cl⁻(aq)   E₂° = 1.36 V

---------------------------------------------------------------

Zn(s) + Cl₂(g) -------> Zn²⁺(aq) + 2Cl⁻(aq)    E° = 0.76 + 1.36 = 2.12 V

Now, let's replace in (1) and then, solve for [Zn]:

2.21 = 2.12 - (0.059/2) log ([0.1]² * [Zn])

2.21 - 2.12 = -0.0295 log (0.01[Zn])

- 0.09 / 0.0295 = log (0.01[Zn])

-3.0508 = log (0.01[Zn])

10^(-3.0508) = 0.01[Zn]

8.8961x10⁻⁴ = 0.01[Zn]

[Zn²⁺] = 0.08896 M

This value can be rounded to 0.1 M. so the correct option will be option A.

5 0
3 years ago
Which of the following choices represents elements with the most similar
matrenka [14]

Answer:

Explanation:

F Cl Br belongs to the Same group

5 0
3 years ago
What is neutralization reaction?
Verizon [17]
It's basically when a base and an acid of some kind, come together to make H2O or Water. 
7 0
3 years ago
Read 2 more answers
Click the "draw structure" button to launch the drawing utility. under certain reaction conditions, 2,3−dibromobutane reacts wit
REY [17]

Answer:

Explanation:

According to this. Let's analize the possible products a, b and c.

First, the problem states that we have 2 eq. of base, For this case, let's assume it's KOH. Now, As we are doing a reaction with base, means that this reaction can only take places under conditions of SN2 and E2, a fast reaction that is taking place in only 1 step.

With this in mind, let's analyze product a. This states that it has two sp hybridized carbon, in other words, a triple bond between two carbons. So the product is with no doubt, an alkyne.

Product b has only one sp hybridized carbon, which means that this carbon should cannot be an alkyne because we need two carbon atoms. The only way to have one atom of C sp hybridized, is with two double bonds, so product b would have to a alkene with two double bonds.

Product c do not have sp hybridized carbon, therefore, it only has two double bonds in two different Carbon atoms, so it's another alkene with two double bonds, but in two different atoms of carbon.

Picture attached show the product a, b and c. Hope this can help

4 0
2 years ago
Balance in basic solution: O2(g) + Cr³+ (aq) → H₂O2 (1) + Cr₂O7²- (aq)
likoan [24]

cr2o72+H2o2-(r3+o2)

Explanation:

r62o72+H2o2-(r3+o2)

3 0
3 years ago
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