A. Atoms of chlorine (Cl)
Explanation:
The atoms of chlorine are held together by non-polar covalent bonds.
- Covalent bonds are formed between two or more atoms having zero or very small electronegativity difference.
- For homonuclear molecules where the two bonding atom are of the same kind, the electronegativity difference is zero.
- This covalent bond type is called non-polar covalent bond.
- In this bond type, there is equal sharing of the electron pair between the atoms and there is no electrostatic charge on the molecule.
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Covalent bond brainly.com/question/5258547
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It shows the atomic number
Answer:
Explanation:En la tabla periódica los elementos están ordenados de forma que aquellos con propiedades químicas semejantes, se encuentren situados cerca uno de otro. Los elementos se distribuyen en filas horizontales, llamadas períodos.
The given question is incomplete. The complete question is as follows.
Sodium sulfate is slowly added to a solution containing 0.0500 M
and 0.0390 M
. What will be the concentration of
(aq) when
begins to precipitate? What percentage of the
can be separated from the Ag(aq) by selective precipitation?
Explanation:
The given reaction is as follows.

= 0.0390 M
When
precipitates then expression for
will be as follows.
![K_{sp} = [Ag^{+}]^{2}[SO^{2-}_{4}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5BAg%5E%7B%2B%7D%5D%5E%7B2%7D%5BSO%5E%7B2-%7D_%7B4%7D%5D)
![1.20 \times 10^{-5} = (0.0390)^{2} \times [SO^{2-}_{4}]](https://tex.z-dn.net/?f=1.20%20%5Ctimes%2010%5E%7B-5%7D%20%3D%20%280.0390%29%5E%7B2%7D%20%5Ctimes%20%5BSO%5E%7B2-%7D_%7B4%7D%5D)
= 0.00788 M
Now, equation for dissociation of calcium sulfate is as follows.

![K_{sp} = [Ca^{2+}][SO^{2-}_{4}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5BCa%5E%7B2%2B%7D%5D%5BSO%5E%7B2-%7D_%7B4%7D%5D)
![4.93 \times 10^{-5} = [Ca^{2+}] \times 0.00788](https://tex.z-dn.net/?f=4.93%20%5Ctimes%2010%5E%7B-5%7D%20%3D%20%5BCa%5E%7B2%2B%7D%5D%20%5Ctimes%200.00788)
= 0.00625 M
Now, we will calculate the percentage of
remaining in the solution as follows.

= 12.5%
And, the percentage of
that can be separated is as follows.
100 - 12.5
= 87.5%
Thus, we can conclude that 87.5% will be the concentration of
when
begins to precipitate.
Answer:
Second element(Titanium); [Ar] 3d2 4s2
Third element(Vanadium):Ar 3d3 4s2
Explanation:
Given that there are only three d orbitals in universe L instead of five, the electronic configuration of the second and third elements in the first transition series will now look thus;
Second element(Titanium); [Ar] 3d2 4s2
Third transition element(Vanadium):Ar 3d3 4s2
Hence, the electronic configuration of Titanium and Vanadium in universe L is just the same as what it is on earth.