Answer:
Without doing any calculations it is possible to determine that silver chromate is more soluble than <u>C,D ,</u> and silver chromate is less soluble than <u>none</u> .
It is not possible to determine whether silver chromate is more or less soluble than <u>A,B</u> by simply comparing Kspvalues
Explanation:
For this kind of question you need to obtain the balanced equation for the solubility, first, you will write the dissociation equation for every molecule, obtaining the ions in which it dissociates. Then you substitute in the Kps equation
![A_{n}B_{m} -> nA^{m+}+mB^{n-} \\K_{s}=[A^{m}]^{n}[B^{n}]^{m}](https://tex.z-dn.net/?f=A_%7Bn%7DB_%7Bm%7D%20-%3E%20nA%5E%7Bm%2B%7D%2BmB%5E%7Bn-%7D%20%5C%5CK_%7Bs%7D%3D%5BA%5E%7Bm%7D%5D%5E%7Bn%7D%5BB%5E%7Bn%7D%5D%5E%7Bm%7D)
Considering the corresponding stoichiometry you'll substitute in your dissociation equations as follow.
![Ag_{2}CrO_{4}(s)->2Ag^{+}(aq)+CrO_{4} ^{2-}\\ s ->2s+s\\s->[2s]^{2}[s]=4s^{3} \\\\A) PbF_{2}(s)->Pb^{2+} (aq)+2F^{-}\\s->s+2s\\s->[s][2s]^{2}=4s^{3}\\\\\\B)Ag_{2}SO_{3}(s)->2Ag^{+}(aq)+SO_{3} ^{2-}\\ s ->2s+s\\s->[2s]^{2}[s]=4s^{3} \\\\\\C)NiCO_{3}(s)->Ni^{2+}(aq)+CO_{3} ^{2-}\\ s ->s+s\\s->[s][s]=s^{2} \\\\\\D)AgCl(s)->Ag^{+}(aq)+Cl^{-}(aq)\\ s ->s+s\\s->[s][s]=s^{2} \\](https://tex.z-dn.net/?f=Ag_%7B2%7DCrO_%7B4%7D%28s%29-%3E2Ag%5E%7B%2B%7D%28aq%29%2BCrO_%7B4%7D%20%5E%7B2-%7D%5C%5C%20%20%20%20%20%20%20%20s%20%20-%3E2s%2Bs%5C%5Cs-%3E%5B2s%5D%5E%7B2%7D%5Bs%5D%3D4s%5E%7B3%7D%20%20%5C%5C%5C%5CA%29%20%20PbF_%7B2%7D%28s%29-%3EPb%5E%7B2%2B%7D%20%28aq%29%2B2F%5E%7B-%7D%5C%5Cs-%3Es%2B2s%5C%5Cs-%3E%5Bs%5D%5B2s%5D%5E%7B2%7D%3D4s%5E%7B3%7D%5C%5C%5C%5C%5C%5CB%29Ag_%7B2%7DSO_%7B3%7D%28s%29-%3E2Ag%5E%7B%2B%7D%28aq%29%2BSO_%7B3%7D%20%5E%7B2-%7D%5C%5C%20%20%20%20%20%20%20%20s%20%20-%3E2s%2Bs%5C%5Cs-%3E%5B2s%5D%5E%7B2%7D%5Bs%5D%3D4s%5E%7B3%7D%20%20%5C%5C%5C%5C%5C%5CC%29NiCO_%7B3%7D%28s%29-%3ENi%5E%7B2%2B%7D%28aq%29%2BCO_%7B3%7D%20%5E%7B2-%7D%5C%5C%20%20%20%20%20%20%20%20s%20%20-%3Es%2Bs%5C%5Cs-%3E%5Bs%5D%5Bs%5D%3Ds%5E%7B2%7D%20%20%5C%5C%5C%5C%5C%5CD%29AgCl%28s%29-%3EAg%5E%7B%2B%7D%28aq%29%2BCl%5E%7B-%7D%28aq%29%5C%5C%20%20%20%20%20%20%20%20s%20%20-%3Es%2Bs%5C%5Cs-%3E%5Bs%5D%5Bs%5D%3Ds%5E%7B2%7D%20%20%5C%5C)
IIn this case, the silver chromate has a Kps of 
, same as the compound in options A and B, comparing these numbers you can't determine which one is bigger. Finally, options C and D have a kps of 
, this value is smaller than silver chromate's kps.
I hope you find this information useful! good luck!
Answer:You can get the valence electrons in an atom's electronic arrangement by consulting the periodic table:
Explanation:
The Group 1 atoms have 1 valence electron.
The Group 2 atoms have 2 valence electrons.
The Group 3 atoms have 3 valence electrons.
The Group 4 atoms have 4 valence electrons.
Answer:
In the 5th cycle rinse, the residual concentration of the solution is < 0.00001M
Explanation:
In each rinse cycle, the dilution that you are doing of the solution is from 1.00mL to 10.00mL, that is a dilution of 10
In the first rinse the concentration must be of 0.9M 10 = 0.09M
2nd = 0.009M
3rd = 0.0009M
4th = 0.00009M
5th = 0.000009M →
<h3>In the 5th cycle rinse, the residual concentration of the solution is < 0.00001M</h3>
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