Answer:
RbOH → Rb⁺ + OH⁻
As the hydroxide can gives the OH⁻ in water, it is considered as an Arrhenius's base
Explanation:
Arrhenius theory states that a compound is considered a base, if the compound can generate OH⁻ ions in aqueous solution.
Our compound is the RbOH.
When it is put in water, i can dissociate like this:
RbOH → Rb⁺ + OH⁻
As the hydroxide can gives the OH⁻ in water, it is considered as an Arrhenius's base
Answer:
D
Explanation:
The fact is that the both elements belong to different groups in the periodic table. Mg is in group 12 while Al is in group 13. The outer most electron configuration for Mg2+=[Ne]3s0
Al+= [Ne]3s13p0
It is evident that in Al+, the second electron is to be removed from a filled 3s subshell which is energetically unfavourable. Unlike In Mg2+ where the two electrons are removed. There is always a tendency towards a quick loss of all the valence electrons in the valence shell. It will be more difficult to remove an electron from a filled shell.
Answer:
9.88
Explanation:
As higher is the Ksp, more soluble is the compound. So, Co(OH)₂ is the less soluble hydroxide.
The maximum concentration of it must be 1x10⁻⁶ M, and the reaction is:
Co(OH)₂(s) ⇄ Co⁺²(aq) + 2OH⁻(aq)
So, [Co⁺²] = 1x10⁻⁶M
Ksp = [Co⁺²] *[OH⁻]²
[OH⁻]² = 5.9x10⁻¹⁵/1x10⁻⁶
[OH⁻] = √(5.9x10⁻⁹)
[OH⁻] = 7.6811x10⁻⁵
pOH = -log[OH⁻]
pOH = -log(7.6811x10⁻⁵)
pOH = 4.11
Knowing that pH + pOH = 14
pH = 14 - 4.11
pH = 9.88
Answer:
B. PhCHO
Explanation:
Every organic group shows a characteristic IR absorption at certain wavelength . With the help of these absorption spectra we can identify the group present on organic molecules .
The wave number of 2710 cm⁻¹ is absorbed by aldehyde bond stretching .
The wave number of 1705 cm⁻¹ is shown by conjugated aldehyde . So the most likely compound among given compounds is PhCHO .
Answer:
E° = 0.65 V
Explanation:
Let's consider the following reductions and their respective standard reduction potentials.
Sn⁴⁺(aq) + 2 e⁻ → Sn²⁺(aq) E°red = 0.15 V
Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V
The reaction with the highest reduction potential will occur as a reduction while the other will occur as an oxidation. The corresponding half-reactions are:
Reduction (cathode): Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V
Oxidation (anode): Sn²⁺(aq) → Sn⁴⁺(aq) + 2 e⁻ E°red = 0.15 V
The overall cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an = 0.80 V - 0.15 V = 0.65 V