The equilibrium for the dissolution of the weak base is ;(CH3)2NH(aq) + H2O(l) ⇄ (CH3)2NH3^+(aq) + OH^-(aq)
<h3>What is a weak base?</h3>
A weak base is one that does not ionize completely in solution. As such, a weak base will have a very low base dissociation constant Kb reflecting its minimal dissociation in solution.
The question is incomplete hence we are are unable to work out the equilibrium but in solution it will look like this;
(CH3)2NH(aq) + H2O(l) ⇄ (CH3)2NH3^+(aq) + OH^-(aq)
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Hydrogen has only one electron. It has one proton of nucleus and one outter electron.
Prontosil is a compound produced by the coupled reaction of an aryldiazonium ion and an aromatic compound.
<h3>
What are diazonium compounds?</h3>
These are organic compounds in which there are ionic interactions between the azo group (-N₂⁺) and an anion X⁻.
The general structure is RN₂⁺X⁻.
- R is the lateral chain that might be an aromatic ring, among other options.
The azo group characterizes as being unstable and reactive. This property is because one of the N atoms has a positive charge.
-N⁺≡ N
<h3>What is the coupling reaction of aryldiazonium compounds?</h3>
Aryldiazonium salt reactions can occur in two ways,
- Substitution reactions
- Coupling reactions
Coupling reactions are the aromatic electrophilic substitution, where the aryldiazonium ion acts as an electrophile for an activated aromatic compound to attack it.
The coupling reaction occurs at the azo group level.
In the exposed example,
- the benzene ring with sulfur bonded to oxygen atoms is the coupling component
- the benzene ring with NH₂ and the azo group is the diazonium ion
In the attached files you will find the drawings.
You can learn more about diazonium compounds at
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Answer:
what's the question? is it A
Answer:
See explanation
Explanation:
The noble gas core electron configuration involves writing the inert gas core of an atom followed by the valence electrons. This is shown for the following atoms;
Bismuth;
[Xe]4f14 5d10 6s2 6p3
Chromium;
[Ar]4s1 3d5
Strontium;
[Kr]5s2
Phosphorus;
[Ne]3s2 3p3
2.
Bi
6p- n=6, l= 1, ml= 1, ms= 1/2
Cr
3d- n=3, l=2, ml=2,ms=1/2
Sr
5s- n=5, l=0, ml=0, ms=1/2
P
3p- n=3, l= 1, ml= 1, ms=1/2
3.
a) Tin (Sn) - [Kr] 5s2 4d10 5p2
b) Caesium (Cs)- [Xe] 6s1
c) Copper (Cu)- [Ar] 4s1 3d10