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Zigmanuir [339]
3 years ago
10

Which of the following is considered a drawback to using wind energy as a source of power?

Chemistry
1 answer:
Olin [163]3 years ago
5 0

Answer:

Option (3)

Explanation:

Wind energy is directly derived from the wind. In the places where wind blowing is quite frequent, there wind mills are being set up, and the turbines in it rotates due to the prevailing wind. Due to this continuous motion of turbines, it collects the wind energy and it is being transferred into electrical energy.

It is cost-effective and does not produce any kind of pollution and is completely a renewable energy, that it can generated again and again.

It does have certain drawbacks also, because <u>the area may sometime do not experience constant wind, due to which it cannot store energy. So frequent wind blowing areas are the best place to set up windmills</u>.

Thus, the correct answer is option (3)

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A chemist designs a galvanic cell that uses these two half-reactions: half-reaction standard reduction potential (s)(aq)(aq)(l)
miv72 [106K]

Answer:

Reduction (cathode): Cu²⁺(⁺aq) + 2 e⁻ → Cu(s)  

Oxidation (anode): Zn(s) → Zn²⁺(⁺aq) + 2 e⁻        

Cu²⁺(⁺aq) + Zn(s) → Cu(s) + Zn²⁺(⁺aq)

E°cell = 1.10 V

Explanation:

<em>The half-reactions are missing, but I will propose some to show you the general procedure and then you can apply it to your equations.</em>

<em>Suppose we have the following half-reactions.</em>

<em>Cu²⁺(⁺aq) + 2 e⁻ → Cu(s)   E°red = 0.34 V</em>

<em>Zn²⁺(⁺aq) + 2 e⁻ → Zn(s)    E°red = -0.76 V</em>

<em />

To identify how to make a spontaneous cell, we need to consider the standard reduction potentials (E°red). The half-reaction with the higher E°red will occur as a reduction (in the cathode), whereas the one with the lower E°red will occur as an oxidation (in the anode).

Reduction (cathode): Cu²⁺(⁺aq) + 2 e⁻ → Cu(s)   E°red = 0.34 V

Oxidation (anode): Zn(s) → Zn²⁺(⁺aq) + 2 e⁻        E°red = -0.76 V

To get the overall equation we add both half-reactions.

Cu²⁺(⁺aq) + Zn(s) → Cu(s) + Zn²⁺(⁺aq)

The standard cell potential (E°cell) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E°cell = E°red, cat - E°red, an

E°cell = 0.34 V - (-0.76 V) = 1.10 V

Since E°cell > 0, the reaction is spontaneous.

5 0
2 years ago
1. How many moles of LiBr are present in 100 mL of 1.25M LIBr solution?
Svet_ta [14]

Answer:

https://www.webassign.net/question_assets/wertzcams3/appendix.pdf

7 0
3 years ago
Two moles of gas A spontaneously convert to 3 moles of gas b in a container where the temperature and pressure are held constant
erica [24]

At constant temperature and pressure, If the amount of gas increases to the given value, its volume also increases to 20.85L.

<h3>What is Avogadro's law?</h3>

Avogadro's law states that "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules."

It is expressed as;

V₁/n₁ = V₂/n₂

Given the data in the question;

  • Initial amount of gas n₁ = 2moles
  • Initial volume v₁ = 13.9L
  • Final amount of gas n₁ = 3moles
  • Final volume v₂ = ?

V₁/n₁ = V₂/n₂

V₁n₂ = V₂n₁

V₂ = V₁n₂ / n₁

V₂ = (13.9L × 3moles) / 2moles

V₂ = 41.7molL / 2mol

V₂ = 20.85L

At constant temperature and pressure, If the amount of gas increases to the given value, its volume also increases to 20.85L.

Learn more about Avogadro's law here: brainly.com/question/15613065

#SPJ1

4 0
1 year ago
What is the molarity of a solution in which 0.732 moles of hcl are dissolved in 0.975 liters of solution?
posledela
Molarity is moles divided by liters so do .732 divided by .975 liters.
6 0
3 years ago
Which of the following would dissolve in water?<br> CCl4<br> LiCl<br> CH4<br> PCl6
babymother [125]
Like dissolves like
so water is polar

CCl4 is nonpolar
LiCl is polar
CH4 is nonpolar
PCl6 is nonpolar

so LiCl would dissolve
7 0
3 years ago
Read 2 more answers
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