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Fofino [41]
3 years ago
10

Type in the correct values to correctly represent the valence electron configuration of magnesium: AsB A = B =

Chemistry
2 answers:
erik [133]3 years ago
7 0
<h2>Answer : The correct answer will be A =3 and B =2 as the electron configuration of Mg will be as [Ne] 3s^{2} where A is 3 and B is 2.</h2><h3>Explanation : </h3>

Magnesium has atomic number as 12. The complete electron configuration will be as 1s^{2}2s^{2}2p^{6}3s^{2}.

which can be abbreviated as[Ne] 3s^{2} as the electron configuration resembles to that of neon element.

In the outermost shell the electron in 3 s orbital is only 2. So, therefore 3 is the orbital of S and there are 2 electrons in it.

Alexxandr [17]3 years ago
5 0
The correct answer would be,
A= 3 
B=2
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Myth: A dead organism is the same as a nonliving thing in science.<br><br> Fact:<br><br> Evidence:
Ilya [14]

I recently did this assignment!

Instructions: Read each myth (untruth). Reword it to make a factual statement. Then, give two to three reasons why the myth is untrue. Use complete sentences and support your answer with evidence, using your own words.  

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Answer:

Myth: A dead organism is the same as a nonliving thing in science.

o Fact: In science, dead is the same as nonliving.

o Evidence: Things that are nonliving never had the characteristics of life, and never will. Things that are dead once did have the characteristics of life, but when they die, they lose some of the characteristics. That is why dead and non-living are NOT the same thing.

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4 0
3 years ago
Read 2 more answers
How many mL of a stock 50% (w/v) KNO3 solution are needed to prepare 250 mL of a 20% (w/v) KNO3 solution?
Andre45 [30]

Answer:

100ml of a stock 50% KNO3 solutions are needed to prepare 250ml of a 20% KNO3 solution.

Explanation:

In the given question it is mentioned that

     S1=50%

      V2=250ml

      S2= 20%

We all know that

                     V1S1=V2S2

                     ∴V1=  V2×S2÷S1

                     ∴V1=  V2S2×1/S1

                      ∴V1= 250×20÷50

                       ∴V1= 100ml

 

6 0
4 years ago
Give the half equation to show what happens to oxygen when it is rusting
ycow [4]

Answer:

½O 2 + 2e - + H 2O → 2OH.

Explanation:

Redox reactions - Higher

In terms of electrons:

oxidation is loss of electrons

reduction is gain of electrons

Rusting is a complex process. The example below show why both water and oxygen are needed for rusting to occur. They are interesting examples of oxidation, reduction and the use of half equations:

iron loses electrons and is oxidised to iron(II) ions: Fe → Fe2+ + 2e-

oxygen gains electrons in the presence of water and is reduced: ½O2 + 2e- + H2O → 2OH-

iron(II) ions lose electrons and are oxidised to iron(III) ions by oxygen: 2Fe2+ + ½O2 → 2Fe3+ + O2-

3 0
3 years ago
A solution is prepared by dissolving 23.7 g of cacl2 in 375 g of water. the density of the resulting solution is 1.05 g/ml. the
alekssr [168]
<span>The density of the solution =1.05 g/ml.
</span><span>The total mass of the resulting solution is = 398.7 g (CaCl2 + water)
</span>
Find moles of CaCl2 and water.
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Moles of CaCl2 = 23.7 / 110  = 0.22
so, moles of Cl- ion = 2 x 0.22 = 0.44  (because each molecule of CaCl2 will give two Cl- ions)
Moles of water = 375 / 18 = 20.83

Now, Mole fraction of CaCl2 =  (moles of CaCl2) / (total moles)

total moles = moles of Cl- ions + moles of Ca2+ ions + moles of water

                  = 0.44 + 0.22 + 20.83

=21.49


So, mole fraction = 0.44 / (21.49) = 0.02

Guess what !!! density is not used. No need

4 0
3 years ago
Read 2 more answers
What is the difference between pure and applied chemistry?
bekas [8.4K]

Answer:

Explanation:

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Pure chemistry deals with the study of matter, matter transformations, and interactions between the different materials of the world, for only the sake of gaining empirical knowledge about the various substances that exist in the world. It does not really seek to apply this knowledge to do anything industrial.

Applied chemistry is the study of chemistry with the aim of utilizing this knowledge to solve the various problems that man faces. This approach of study is not for knowledge sake alone, rather it is for industrial application

3 0
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