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katen-ka-za [31]
3 years ago
10

Which statement is possible

Chemistry
1 answer:
Mashutka [201]3 years ago
3 0

Answer:

Which statement are you looking for?

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(a) Find the concentration of electrons and holes in a sample of germanium that has a concentration of donor atoms equal to 1015
matrenka [14]

Answer:

a) Germanium = 5.76 x 〖10〗^11 〖cm〗^(-3) , Semiconductor is n-type.

b) Silicon = 2.25 x 〖10〗^5 〖cm〗^(-3) , Semiconductor is n-type.

For clear view of the answers: Please refer to calculation 5 in the attachments section.

Explanation:

So, in order to find out the concentration of holes and electrons in a sample of germanium and silicon which have the concentration of donor atoms equals to 〖10〗^15 〖cm〗^(-3). We first need to find out the intrinsic carrier concentration of silicon and germanium at room temperature (T= 300K).

Here is the formula to calculate intrinsic carrier concentration: For calculation please refer to calculation 1:

So, till now we have calculated the intrinsic carrier concentration for germanium and silicon. Now, in this question we have been given donor concentration (N_d) (N subscript d), but if donor concentration is much greater than the intrinsic concentration then we can write: Please refer to calculation 2.

So, now we have got the concentration of electrons in both germanium and silicon. Now, we have to find out the concentration of holes in germanium and silicon (p_o).  (p subscript o)

Equation to find out hole concentration: Please refer to calculation 3. and Calculation 4. in the attachment section.

Good Luck Everyone! Hope you will understand.  

6 0
3 years ago
A student following the reaction seen here calculated a theoretical yield of 38.3g C₆H₅Cl but when he did the experiment in the
Reil [10]

Answer:

96.1 %

Explanation:

Which teacher do you have lol

7 0
3 years ago
A solution of rubbing alcohol is 75.7 % (v/v) isopropanol in water. How many milliliters of isopropanol are in a 87.7 mL sample
beks73 [17]

Answer:

66.4 mL

Explanation:

A 75.7% (v/v) value, means that f<u>or every 100 mL of rubbing alcohol, there are 75.7 mL of isopropanol.</u>

With the above information in mind, we can s<u>olve the problem by multiplying 87.7 mL by 75.7 %</u>:

87.7 mL * 75.7 / 100 = 66.4 mL

So there are 66.4 mL of isopropanol in 88.7 mL of rubbing alcohol.

5 0
3 years ago
What solution does ?
Ainat [17]
Unsaturated would be the correct answer.
3 0
2 years ago
Read 2 more answers
Help<br> me i am on a quiz
castortr0y [4]
The screen is blank for me can you say the question
6 0
3 years ago
Read 2 more answers
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