only reactants are present
If equllibrium constant is greater than 10^3 then product
predominates
If equllibrium constant is greater than 10^ -3 then reactant predominates
in between 10^3 and 10^-3 the reaction is in equllibrium
and we see that the above value is 10^-31, it is very small so the answer is reactants are presents
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:
24000
Explanation:
Given that:
The numbers of carbon atoms = 8000
According to the Einstein model of solid;
Einstein’s model assumes that each atom of the solid vibrates around its position of equilibrium independently of the other atoms (which amounts to neglecting the interactions between particles.
Number of oscillators = 3(8000)
Number of oscillators = 24000
Electrons are orbiting the nucleus in the fxed way paths located in solid sphere