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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:

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Solid copper is burned in oxygen gas to make solid copper (ii) oxide
Olin [163]

Answer:

The copper oxide can then react with the hydrogen gas to form the copper metal and water. When the funnel is removed from the hydrogen stream, the copper was still be warm enough to be oxidized by the air again.

Explanation:

8 0
4 years ago
5. Write a net ionic equation that occurs in a Na2HPO4/NaH2PO4 buffer solution when: A) a small amount of HCl is added (2 points
labwork [276]

Answer: (A) H_{3}O^{+}(aq) + HPO^{2-}_{4}(aq) \rightarrow H_{2}PO^{2-}_{4}(aq) + H_{2}O(l)

(B) H_{2}PO^{-}_{4}(aq) + OH^{-}(aq) \rightarrow HPO^{2-}_{4}(aq) + H_{2}O(l)

Explanation:

(A) As we know that HCl is a strong acid and when it is added to an aqueous solution then it leads to increase in the concentration of hydrogen ions. And, when an acid or base is added to a solution then any resistance by the solution in changing the pH of the solution is known as a buffer.

This means that addition of buffer into the given solution will not cause much change in the concentration of H_{3}O^{+} in large amount.

As both the buffer components are salt then they will remain dissociated as follows.

       Na_{2}HPO_{4}(aq) \rightarrow 2Na^{+}(aq) + HPO^{2-}_{4}(aq)

 NaH_{2}PO_{4}(aq) \rightarrow Na^{+}(aq) + H_{2}PO^{-}_{4}(aq)

Hence, net ionic equation will be as follows.

       H_{3}O^{+}(aq) + HPO^{2-}_{4}(aq) \rightarrow H_{2}PO^{2-}_{4}(aq) + H_{2}O(l)

(B)  When we add small amount of sodium hydroxide into the solution then there will occur an increase in concentration of hydroxide ions into the solution. But then due to the presence of buffer there will occur not much change in concentration and the acid will get converted into salt.

     NaOH(aq) \rightarrow Na^{+}(aq) + OH^{-}(aq)

The net ionic equation is as follows.

        H_{2}PO^{-}_{4}(aq) + OH^{-}(aq) \rightarrow HPO^{2-}_{4}(aq) + H_{2}O(l)

7 0
3 years ago
Given the atomic mass of select elements, calculate the molar mass of each salt. Element Molar mass (g/mol) Beryllium (Be) 9.012
jeka94

Answer:

1. 266.22 g/mol

2. 168.81 g/mol

3. 223.35 g/mol

4. 199.88 g/mol

Explanation:

For you to calculate the molar mass of the salt you need to sum the molar masses of every element in the salt.

In the first salt, PdBr2_{123}, the subscript 2 means that there are 2 atoms of Br. So for you to calculate the molar mass of the salt you need to sum the molar mass of Pd and 2 times the molar mass of Br, as follows:

106 g/mol + 2(79.90 g/mol) = 266.22 g/mol

In the second salt BeBr2_{123} there are 2 atoms of Br and 1 of Be, so the molar mass is:

9.012 g/mol +2(79.90 g/mol) = 186.22 g/mol

In the third salt CuBr2_{123} there are 2 atoms of Br and 1 of Cu, so the molar mass is:

63.55 g/mol + 2(79.90 g/mol) = 223.35 g/mol

And in the fourth salt CaBr2_{123} there are 2 atoms of Br and 1 of Ca, so the molar mass is:

40.08 g/mol + 2(79.90 g/mol) = 199.88 g/mol

6 0
3 years ago
A sample of nitrous Oxide has a volume of 1 £.20 at 1500 mm HG and -10°C. Calculators pressure when it’s value is compressed 250
ExtremeBDS [4]

Answer:

nitric acid will be if you get you will pixel purpose

3 0
3 years ago
An object weighing 9.6 g is placed in a graduated cylinder displacing the volume from 10.0 ML to 13.2 ML What is its density in
tigry1 [53]
Density = mass/ volume
mass= 9.6 g
Volume of the object = 13.2-10.0 =3.2 ml=3.2 cm³
1ml =1 cm³

D=9.6g/3.2 cm³ = 3.0 g/cm³

3 0
3 years ago
Read 2 more answers
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