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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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+<br> c)<br> FeCl3 +<br> NH4OH<br> Fe(OH)3<br> NHACI
bixtya [17]

The question is incomplete, the complete question is:

Write the net ionic equation for the below chemical reaction:

(c): FeCl_3+3NH_4OH\rightarrow Fe(OH)_3+3NH_4CI

<u>Answer:</u> The net ionic equation is Fe^{3+}(aq)+3OH^{-}(aq)\rightarrow Fe(OH)_3(s)

<u>Explanation:</u>

Net ionic equation is defined as the equations in which spectator ions are not included.

Spectator ions are the ones that are present equally on the reactant and product sides. They do not participate in the reaction.

(c):

The balanced molecular equation is:

FeCl_3(aq)+3NH_4OH(aq)\rightarrow Fe(OH)_3(s)+3NH_4Cl(aq)

The complete ionic equation follows:

Fe^{3+}(aq)+3Cl^-(aq)+3NH_4^+(aq)+3OH^{-}(aq)\rightarrow Fe(OH)_3(s)+3NH_4^+(aq)+3Cl^-(aq)

As ammonium and chloride ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

Fe^{3+}(aq)+3OH^{-}(aq)\rightarrow Fe(OH)_3(s)

5 0
3 years ago
List the number of significant figures for the following numbers:
Dennis_Churaev [7]

Answer:

a) 2

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c( 1

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3 0
3 years ago
Which of the following represents positron emission? Select one: a. 3015P ---&gt; 3014Si + ___ b. 23892U ---&gt; 23490Th + ___ c
xxTIMURxx [149]
The simple trick which one can consider in such problem where it is asked for positron emission is :
<span>When the atomic number goes DOWN by one and mass number remains unchanged, then a positron is emitted.
</span>
<span>a.  </span>P^{30}_{15}  ----\ \textgreater \   Si^{30}_{14} + e^{0}_{1}^+
<span>
Here the atomic number decreases by one.

Similarly, options b and d are eliminated.

Option c is also not the answer.
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</span>
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