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
It is known as the medium
Answer: The answer is B: equals mass divided by volume.
just did the quiz
hope this helps :D
Answer: alkaline earth metals (group-IIA)
Explanation:
The element which donates the electron is known as electropositive element and forms a positively charged ion called as cation. The element which accepts the electrons is known as electronegative element and forms a negatively charged ion called as anion.
Alkaline earth metals donate 2 valence electrons to acquire noble gas configuration.
For example: Berrylium is the first alkaline earth metal with atomic number of 4 and thus has 4 electrons
Electronic configuration of berrylium:
Berrylium atom will loose two electrons to gain noble gas configuration and form berrylium cation with +2 charge.
Thus Elements donate 2 electron to produce a cation with a 2+ charge are alkaline earth metals.