The symbol for xenon (xe) would be a part of the noble gas notation for the element cesium.
For writing the electronic configuration of any element by using the preceding noble gas configuration, we simply use the symbols of noble gas belongs to the previous period of that particular elements. We can't use the symbol of noble gas of same period from which the element belong.
A is the wrong option because the noble gas in the preceding period to the period from which antimony belongs is krypton.
The actual electronic configuration of antimony is as follow:
[Kr] 4d10 5s2 5p3
B is correct option because the noble gas in the preceding period to the period from which Cesium belongs is Xenon.
The actual electronic configuration of Cesium is as follow:
[Xe] 6s1
Thus, we concluded that the symbol for xenon (xe) would be a part of the noble gas notation for the element cesium.
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Answer:
Cl⁻ was oxidized.
Explanation:
- 4HCl + MnO₂ → Cl₂ + 2H₂O + MnCl₂
Oxidation can be defined as the process in which the oxidation number of a substance increases.
On the left side of the equation, Cl has a charge of -1 (in HCl); while on the right side of the equation Cl has a charge of 0 in Cl₂.
Thus, Cl⁻ was oxidized.
The unbalanced equations are the equations with different atomic numbers on the sides of the reaction. The unbalanced reaction is Na + Cl₂ → NaCl
<h3>What are balanced equations?</h3>
Balanced equations are the chemical reaction representation that has an equal number of the atomic number of the same species on the left and the right side of the reaction.
An unbalanced equation between sodium metal and chloride can be shown as:
Na + Cl₂ → NaCl
The equation is unbalanced as the number of chloride ions is more on the reactant side than the product side.
The balanced reaction will be:
2 Na + Cl₂ 2NaCl
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"John Dalton" is the one scientist among the following choices given in the question that <span>came up with the first widely recognized atomic theory. The correct option among all the options that are given in the question is the first option or option "A". I hope that this is the answer that has come to your help.</span>
Answer: Concentration of 
in the equilibrium mixture is 0.31 M
Explanation:
Equilibrium concentration of 
= 0.729 M
The given balanced equilibrium reaction is,
Initial conc. x 0 0
At eqm. conc. (x-2y) M (y) M (3y) M
The expression for equilibrium constant for this reaction will be:
3y = 0.729 M
y = 0.243 M
![K_c=\frac{[y]\times [3y]^3}{[x-2y]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5By%5D%5Ctimes%20%5B3y%5D%5E3%7D%7B%5Bx-2y%5D%5E2%7D)
Now put all the given values in this expression, we get :
concentration of in the equilibrium mixture = 
Thus concentration of in the equilibrium mixture is 0.31 M
