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kondaur [170]
2 years ago
7

Nguyên tố A có cấu hình electron phân lớp cuối cùng là 4p3. A phải

Chemistry
1 answer:
Studentka2010 [4]2 years ago
4 0

Answer: ?

Explanation: I translated this and It makes no sense.

Tôi đã dịch điều này và nó không có ý nghĩa.

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Explain ionization suppressants as applied to atomic spectrometry
levacccp [35]

An ionization suppressor is an alkali metal capable of preventing ionization, which can be used in atomic spectroscopy to determine matter composition.

<h3>What is ionization?</h3>

Ionization refers to the phenomena capable of converting neutral atoms/molecules to electrically charged atoms/ions.

Ionization is a process by which radiation (e.g., alpha, beta, gamma rays) can pass energy to inert matter.

Some examples of ionization suppressors include salts of alkali metals (for example, potassium), which can be used in atomic spectroscopy to determine matter composition.

Learn more about ionization here:

brainly.com/question/1445179

8 0
1 year ago
*URGENT*
DaniilM [7]

Answer:

d hope this helps u hhwjs GD jehehj

4 0
3 years ago
Read 2 more answers
Science 6th grade please help me :)
Gnom [1K]
I’m pretty sure the answer is D :)
6 0
3 years ago
Read 2 more answers
!!!!!PLEASE HELP!!!!!
ollegr [7]

Answer:

Difference in the potential energy of the reactants and products

Explanation:

 

The products have a lower potential energy than the reactants, and the sign of ΔH is negative. In an endothermic reaction, energy is absorbed. The products have a higher potential energy than the reactants, and the sign of ΔH is positive.

3 0
2 years ago
Under the right conditions aluminum will react with chlorine to produce aluminum chloride.
salantis [7]

Answer:

m_{Al}=9.42gAl

Explanation:

Hello there!

In this case, according to the given chemical reaction:

2 Al + 3 Cl2 --> 2 AlCl3

Whereas there is a 2:3 mole ratio of aluminum to chlorine; it will be possible for us to calculate the required grams of aluminum by using the equality 22.4 L = 1 mol, the aforementioned mole ratio and the atomic mass of aluminum (27.0 g/mol) to obtain:

m_{Al}=11.727LCl_2*\frac{1molCl_2}{22.4LCl_2}*\frac{2molAl}{3molCl_2}  *\frac{27.0gAl}{1molAl} \\\\m_{Al}=9.42gAl

Regards!

8 0
2 years ago
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