O iodine (I) because scientists use it for a lot of experiments
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
Dmitri Mendeleev Explanation:The modern periodic tables is credited primarily to the russian chemist Dmitri Mendeleev. Mendeleev's table is based on the periodic law, which states that when elements are arranged in order of increasing mass, their properties recur periodically.
Answer:
1. Balanced chemical for the given chemical reaction is :
number of mole of sulphuric acid produced = 12.5 mol
number of mole of oxygen required=6.25 mol
Explanation:
1. Balanced chemical for the given chemical reaction is :
from above balanced equation it is clearly that ,
2 mole of sulpher dioxide gives 2 mole of sulphuric acid
1 mole of sulpher dioxide gives 1 mole of sulphuric acid
therefore,
12 .5 mole of sulpher dioxide will givs 12.5 mole of sulphuric acid
number of mole of sulphuric acid produced = 12.5 mol
2 mole of sulphur dioxide needs 1 mole of oxygen gas
so,
1 mole of sulphur dioxide needs 0.5 mole of oxygen gas
therefore 12.5 of sulphur dioxide needs mole of oxygen
number of mole of oxygen required=6.25 mol