When Ksp = [A2+] [S2-]
when A is the metal: Fe, Ni, Pb, and Cu
When we have [S2-] = 0.1 m and we have Ksp for each metal So by substitution in Ksp formula we can get [A2+] for each metal and compare its value with solution concentration 0.01 M, when we have a concentration more than 0.01 M So there are no sulfides precipitates
- [Fe2+] = Ksp/[S2-]
by substitution with Fe2+ Ksp value:
= 6x10^2 / 0.1
= 6x10^3 M
when [Fe2+] > 0.01 M
∴ no precipitate- [Ni2+] = Ksp /[S2-]
by sustitution with Ni Ksp value :
= 8x10^-1 / 0.1
= 8 M
When [Ni2+] > 0.01 M
∴ no precipitate-[Pb2+] = Ksp / [S2-]
by substitution with Pb Ksp value:
= 6x10^-7 / 0.1
= 6 x 10^-6 M
when [Pb2+] < 0.01 M
∴PbS will be precipited-[Cu2+] = Ksp / [S2-]
by substitution with Cu2+ Ksp value:
= 6x10^-16 / 0.1
= 6x10^-15 M
when [Cu2+] < 0.01 M
∴ CuS will be precipited∴The sulfides precipitates are CuS & PbS
Answer:
d. a brittle solid.
Explanation:
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In this case, since cesium fluoride is widely known as an ionic compound due to the large electronegativity difference, it is possible to discard a. due to its crystalline structure, b. because conducting solids are metals, c. because ionic compounds are not likely to be gases and e. because even when it is soluble in water, the problem is not referring to an aqueous solution; therefore, the answer is d. a brittle solid.
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Hey! The best way I learned is if when I write it out put a space where a number should be
Kd = (concentration in one solvent) / (concentration in other solvent)
Because the solute is equally soluble in both, concentration in water = concentration in hexane. Therefore:
Kd = 1