In the situation where an ion has a charge of -2, the Lewis structure is affected because it must add two additional electrons.
The Lewis structure is a term to refer to the graphical representation that shows the pairs of electrons in dashes or points of bonds between:
- The atoms of a molecule
- The lone electron pairs that may exist
Additionally, the charge of an ionic compound is calculated using the difference in electrical charge between the electrons. On the other hand, an ion is formed by adding or removing electrons, so the magnitude of the resulting charge depends on the number of electrons added or removed.
According to the above, if an ionic compound has a -2 charge, the Lewis structure must add two electrons.
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Answer: Potassium Iodide, KI
Explanation:
Flame test colors:
Li+ = Crimson Red
Na+ = Bright Orange-Yellow
K+ = Lilac
Addition of nitric acid and silver nitrate (HNO3 and AgNO3),
Cl- = White precipitate
Br- = Creamy precipitate
I- = Yellow Precipitate
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Explanation:
will dissociate into ions as follows.
Hence, 
for this reaction will be as follows.
![K_{sp} = [Pb^{2+}][Br^{-}]^{2}](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5BPb%5E%7B2%2B%7D%5D%5BBr%5E%7B-%7D%5D%5E%7B2%7D)
We take x as the molar solubility of when we dissolve x moles of solution per liter.
Hence, ionic molarities in the saturated solution will be as follows.
![[Pb^{2+}]](https://tex.z-dn.net/?f=%5BPb%5E%7B2%2B%7D%5D)
= ![[Pb^{2+}]_{o}](https://tex.z-dn.net/?f=%5BPb%5E%7B2%2B%7D%5D_%7Bo%7D)
+ x
![[Br^{-}]^{2}](https://tex.z-dn.net/?f=%5BBr%5E%7B-%7D%5D%5E%7B2%7D)
= ![[Br^{-}]_{o}](https://tex.z-dn.net/?f=%5BBr%5E%7B-%7D%5D_%7Bo%7D)
+ 2x
So, equilibrium solubility expression will be as follows.
= ![([Pb^{2+}]_{o} + x)([Br^{-}]_{o} + 2x)^{2}](https://tex.z-dn.net/?f=%28%5BPb%5E%7B2%2B%7D%5D_%7Bo%7D%20%2B%20x%29%28%5BBr%5E%7B-%7D%5D_%7Bo%7D%20%2B%202x%29%5E%7B2%7D)
Each sodium bromide molecule is giving one bromide ion to the solution. Therefore, one solution contains = 0.10 and there will be no lead ions. So, = 0
So, ![[Br^{-}]_{o} + 2x](https://tex.z-dn.net/?f=%5BBr%5E%7B-%7D%5D_%7Bo%7D%20%2B%202x)
will approximately equals to .
Hence, ![K_{sp} = x[Br^{-}]^{2}_{o}](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20x%5BBr%5E%7B-%7D%5D%5E%7B2%7D_%7Bo%7D)
x = 
M
Thus, we can conclude that molar solubility of is M.
You have to use Avogadro's number (6.02x10^23 molecules/mole) to find the number of moles each reactant starts off with.
moles of Fe and O₂:
12 atoms/(6.02x10^23 atoms/mole)=1.99x10^-23 mol Fe
6 molecules/(6.02x10^23 molecules/mole)=9.967x10^-24 mol <span>O₂
</span>Then you find the limiting reagent by finding how much product each given amount of reactant can make. Which ever one produces the least amount of product is the limiting reagent.
amount of Fe₂O₃ produced:
<span>(1.99x10^-23 mol Fe)x(2mol/4mol)= 9.967x10^-24mol Fe</span>₂O₃<span>
</span>(9.967x10^-24 mol O₂)x(2mol/3mol)= 6.645x10^-24 mol Fe₂O₃<span>
</span>since oxygen produces the leas amount of product, oxygen is the limiting reagent. since we know that oxygen is the limiting reagent we can use the amount of product formed with oxygen to find the amount of iron used.
6.645x10^-24 mol Fe₂O₃x(4mol/2mol)=1.329x10^-23 mol Fe consumed
<span> find the amount left over by subtracting the original amount of Fe by the amount consumed in the reaction.
</span>1.993x10^-23-1.329x10^-23= 6.645x10^-23mol Fe left
find the number of atoms by multiplying that by Avogadro's number.
<span>(6.645x10^-23mol)x(6.02x10^23 atoms/mol)=4 atoms
</span>therefore 4 atoms of Fe will be left over after the reaction happens.
I hope this helps.
