1s2 2s2 2p2 i hope this helps
<span>6s²4f¹⁴5d¹⁰6p²
6 shows that the element is in the 6 period,
6p² shows that the element is in the 14th group. (1 and 2 groups have s -electrons as last ones, 13 group has s²p¹, and 14 group has s²p²)
The element is Pb.
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Answer:
![[Ag^{+}]=4.2\times 10^{-2}M](https://tex.z-dn.net/?f=%5BAg%5E%7B%2B%7D%5D%3D4.2%5Ctimes%2010%5E%7B-2%7DM)
Explanation:
Given:
[AgNO3] = 0.20 M
Ba(NO3)2 = 0.20 M
[K2CrO4] = 0.10 M
Ksp of Ag2CrO4 = 1.1 x 10^-12
Ksp of BaCrO4 = 1.1 x 10^-10

![Ksp=[Ba^{2+}][CrO_{4}^{2-}]](https://tex.z-dn.net/?f=Ksp%3D%5BBa%5E%7B2%2B%7D%5D%5BCrO_%7B4%7D%5E%7B2-%7D%5D)
![1.2\times 10^{-10}=(0.20)[CrO_{4}^{2-}]](https://tex.z-dn.net/?f=1.2%5Ctimes%2010%5E%7B-10%7D%3D%280.20%29%5BCrO_%7B4%7D%5E%7B2-%7D%5D)
![[CrO_{4}^{2-}]=\frac{1.2\times 10^{-10}}{(0.20)}= 6.0\times 10^{-10}](https://tex.z-dn.net/?f=%5BCrO_%7B4%7D%5E%7B2-%7D%5D%3D%5Cfrac%7B1.2%5Ctimes%2010%5E%7B-10%7D%7D%7B%280.20%29%7D%3D%206.0%5Ctimes%2010%5E%7B-10%7D)
Now,

![Ksp=[Ag^{+}]^{2}[CrO_{4}^{2-}]](https://tex.z-dn.net/?f=Ksp%3D%5BAg%5E%7B%2B%7D%5D%5E%7B2%7D%5BCrO_%7B4%7D%5E%7B2-%7D%5D)
![1.1\times 10^{-12}=[Ag^{+}]^{2}](6.0\times 10^{-10})](https://tex.z-dn.net/?f=1.1%5Ctimes%2010%5E%7B-12%7D%3D%5BAg%5E%7B%2B%7D%5D%5E%7B2%7D%5D%286.0%5Ctimes%2010%5E%7B-10%7D%29)
![[Ag^{+}]^{2}]=\frac{1.1\times 10^{-12}}{(6.0\times 10^{-10})}= 1.8\times 10^{-3}](https://tex.z-dn.net/?f=%5BAg%5E%7B%2B%7D%5D%5E%7B2%7D%5D%3D%5Cfrac%7B1.1%5Ctimes%2010%5E%7B-12%7D%7D%7B%286.0%5Ctimes%2010%5E%7B-10%7D%29%7D%3D%201.8%5Ctimes%2010%5E%7B-3%7D)
![[Ag^{+}]=\sqrt{1.8\times 10^{-3}}=4.2\times 10^{-2}M](https://tex.z-dn.net/?f=%5BAg%5E%7B%2B%7D%5D%3D%5Csqrt%7B1.8%5Ctimes%2010%5E%7B-3%7D%7D%3D4.2%5Ctimes%2010%5E%7B-2%7DM)
So, BaCrO4 will start precipitating when [Ag+] is 4.2 x 1.2^-2 M
Answer:
The precipitate is CuS.
Sulfide will precipitate at [S2-]= 3.61*10^-35 M
Explanation:
<u>Step 1: </u>Data given
The solution contains 0.036 M Cu2+ and 0.044 M Fe2+
Ksp (CuS) = 1.3 × 10-36
Ksp (FeS) = 6.3 × 10-18
Step 2: Calculate precipitate
CuS → Cu^2+ + S^2- Ksp= 1.3*10^-36
FeS → Fe^2+ + S^2- Ksp= 6.3*10^-18
Calculate the minimum of amount needed to form precipitates:
Q=Ksp
<u>For copper</u> we have: Ksp=[Cu2+]*[S2-]
Ksp (CuS) = 1.3*10^-36 = 0.036M *[S2-]
[S2-]= 3.61*10^-35 M
<u>For Iron</u> we have: Ksp=[Fe2+]*[S2-]
Ksp(FeS) = 6.3*10^-18 = 0.044M*[S2-]
[S2-]= 1.43*10^-16 M
CuS will form precipitates before FeS., because only 3.61*10^-35 M Sulfur Ions are needed for CuS. For FeS we need 1.43*10^-16 M Sulfur Ions which is much larger.
The precipitate is CuS.
Sulfide will precipitate at [S2-]= 3.61*10^-35 M
Answer:
Explanation:
C) What is the multiplicity of Proton-alpha's signal in this scenario when there are 2 identical protons "next door"?
Based on n+1 rule. Here n=2 (identical beta protons).
2+1=3
So the multiplicity of alpha proton is triplet, .
D) For molecules containing only single bonds (we'll discuss the influence of double bonds in a future lecture), what is the adjective that describes the position of protons that split a "next door neighbor's" signal?
The meaning of the adjective is this: the multiplicity of beta protons is singlet only (no spliting) in absence of alpha proton . But beta protons splits as doublet (n=1) in the presence of alpha proton,
E) How many bonds connect these "splitting next door neighbors"?
There are 3 bonds in between alpha and beta protons in a molecule.
F) What is the multiplicity of the Proton-betas' signal?
Following the n+1 rule, here n=1 (1 alpha proton) so 1+1=2. Hence it is a doublet.