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Nadusha1986 [10]
3 years ago
5

Assess the pros and cons of using these two materials, crystalline silicon and amorphous silicon, to manufacture photovoltaic ce

lls. Which material would be used if the main criterion was keeping the cost of materials as low as possible? Which material would be used if the main criterion was getting the best performance from the material? Explain your reasoning.
HELP ME PLZ!

Chemistry
1 answer:
bearhunter [10]3 years ago
6 0

Answer:

Polycrystalline cells are less expensive to make than single-crystalline modules but are also slightly less efficient than the single-crystalline (12% -15%). Also, it has been found that amorphous silicon cells dominate in warm, sunny conditions due to their lower power-loss temperature coefficient [1]

Explanation:

You might be interested in
How much potassium chloride, KCI, is produced during the decomposition of 100.0 grams of potassium chlorate KCIO3? Please show s
Ne4ueva [31]
Answer: the reaction will produce 15.3 g of
KCl.

explanation:
1. write the balanced equation.
2KClO
3
→
2KCl
+
3O
2

2. calculate the moles of
KClO
3
.
Moles of KClO
3
=
25.0
g KClO
3
×
1 mol KClO
3
122.55
g KClO
3
=
0.2046 mol KClO
3

3. calculate the moles of
KCl
.
Moles of KCl
=
0.2046
mol KClO
3
×
2 mol KCl
2
mol KClO
3
=
0.2046 mol KCl

4. calculate the mass of
KCl
.
Mass of KCl
=
0.2046
mol KCl
×
74.55 g KCl
1
mol KCl
=
15.3 g KCl
4 0
3 years ago
Which of these, all of which include earth, most likely contains the most dark matter
mixer [17]

Answer:

Plasma

Explanation:

8 0
3 years ago
When should you<br> use a fume hood?
AfilCa [17]

Answer:

if you are working with hazardous materials.

Explanation:

A properly operating and correctly used fume hood can reduce or eliminate exposure to volatile liquids, dusts, and mists. It is advisable to use a laboratory hood when working with all hazardous substances.

4 0
4 years ago
This answer please :))
r-ruslan [8.4K]

Answer: ya this one

Explanation: this is the one

4 0
3 years ago
The acid dissociation constant Ka of boric acid (H3BO3) is 5.8 times 10^-10. Calculate the pH of a 4.4 M solution of boric acid.
madam [21]

Answer: The pH of a 4.4 M solution of boric acid is 4.3

Explanation:

H_3BO_3\rightarrow H^+H_2BO_3^-

at t=0  cM              0             0

at eqm c-c\alpha        c\alpha          c\alpha  

So dissociation constant will be:

K_a=\frac{(c\alpha)^{2}}{c-c\alpha}

Give c= 4.4 M and \alpha = ?

K_a=5.8\times 10^{-10}

Putting in the values we get:

5.8\times 10^{-10}=\frac{(4.4\times \alpha)^2}{(4.4-4.4\times \alpha)}

(\alpha)=0.000011

[H^+]=c\times \alpha

[H^+]=4.4\times 0.000011=4.8\times 10^{-5}M

Also pH=-log[H^+]

pH=-log[4.8\times 10^{-5}]=4.3

Thus pH of a 4.4 M H_3BO_3 solution is 4.3

3 0
3 years ago
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