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Ksenya-84 [330]
3 years ago
12

PLEASE ANSWER!! I NEED HELP :)

Chemistry
1 answer:
elena-14-01-66 [18.8K]3 years ago
7 0

Explanation:

To create a p-type semiconductor, group 14 semiconductors should be d o ped by atoms from group 13.

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PLEASE HELP
nata0808 [166]

Answer:

Is this math? Cause as a fourth grader, I can do Algebra, but not this.

Explanation:

7 0
1 year ago
PLEASE HELP IM SO CONFUSED How does potassium need to be modified either on site or in a factory to make it useful. This should
3241004551 [841]


Potassium is not found  free in nature but is found in the form of potash. Potash is the ore of potassium and this ore is mined from deep down the earth or can sometimes  be found on the surface. Potash was mostly formed as sea water receded and left deposits.

Potash is usually in the form of potassium salts such potassium chloride and  potassium sulphate.  The potash  is mined then taken to the factory where it is crushed and purified  by removing such impurities as clay.

The now purified potassium salts are subjected to a process called electrolysis where  potassium metal is obtained from its salt. 

5 0
3 years ago
Write a balanced equation for the reaction of ca with hcl
Aleksandr-060686 [28]
Ca(s)+2Hcl(aq) ------>CaCl2(s)+H2(g)
3 0
3 years ago
What types of properties change during a chemical reaction?
ryzh [129]
The chemical composition 

4 0
3 years ago
Read 2 more answers
How many molecules of CBr4 are in 250 grams of CBr4
Kazeer [188]

Answer:- 4.54*10^2^3 molecules.

Solution:- The grams of tetrabromomethane are given and it asks to calculate the number of molecules.

It is a two step unit conversion problem. In the first step, grams are converted to moles on dividing the grams by molar mass.

In second step, the moles are converted to molecules on multiplying by Avogadro number.

Molar mass of CBr_4  = 12+4(79.9)  = 331.6 g per mol

let's make the set up using dimensional analysis:

250g(\frac{1mol}{331.6g})(\frac{6.022*10^2^3molecules}{1mol})

= 4.54*10^2^3 molecules

So, there will be 4.54*10^2^3 molecules in 250 grams of CBr_4 .


8 0
3 years ago
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