When the enthalpy change of the reaction is positive, the reaction is exothermic.
Answer:
Li2S>Na2S>K2S>Rb2S>Cs2S
Explanation:
Lattice Energy is a type of potential energy which may be defined in as, energy that is required to break apart an ionic solid and convert its component atoms into gaseous ions. More technically, it is a measure of the energy contained in the crystal lattice of a compound, equal to the energy that would be released if the component ions were brought together from infinity.
Lattice energy of ionic compounds decreases with increasing ionic sizes. Since the anion is the same, we consider increase in cation sizes. We discover that for gtoi 1 cations; Li+<Na+<K+<Rb+<Cs+. The ionic lattice energy decreases in the reverse order. Li2S will show the largest lattice energy since it contains the smallest cation in the list.
Answer:
number of Carbons = 2 times number of Hydrogens
Therefore, for alkene with 10 carbons, it will have 2 x 10 = 20 hydrogens.
Its chemical formula will be
C10H20
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
Solution :
We know, pH of this solution :
Hence, this is the required solution.