Answer:
<u>STEP I</u>
This is the balanced equation for the given reaction:-

<u>STEP II</u>
The compounds marked with (aq) are soluble ionic compounds. They must be
broken into their respective ions.
see, in the equation KOH, H2SO4, and K2SO4 are marked with (aq).
On breaking them into their respective ions :-
- 2KOH -> 2K+ + 2OH-
- H2SO4 -> 2H+ + (SO4)2-
- K2SO4 -> 2K+ + (SO4)2-
<u>STEP III</u>
Rewriting these in the form of equation

<u>STEP </u><u>IV</u>
Canceling spectator ions, the ions that appear the same on either side of the equation
<em>(note: in the above step the ions in bold have gotten canceled.)</em>

This is the net ionic equation.
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- KOH has been taken as aqueous because the question informs us that we have a solution of KOH. by solution it means that KOH has been dissolved in water before use.
[Alkali metal hydroxides are the only halides soluble in water ]
Answer:
The four resonance structures of the phenoxide ion are shown in the image attached
The conjugate base of cyclohexanol has only one resonance contributor, while
the conjugate base of phenol has four resonance contributors.
Explanation:
In organic chemistry, it is known that structures are more stable if they possess more resonance contributors. The greater the number of contributing canonical structures, the more stable the organic specie. Since the phenoxide ion has four contributing canonical structures, it is quite much more stable than cyclohexanol having only one contributing structure to its conjugate base. Hence the PKa(acid dissociation constant) of phenol is lesser than that of cyclohexanol. The conjugate base of phenol is stabilized by resonance.
Answer:... I'm sorry.. I think you're missing something
Explanation:
Answer:
Here's what I get.
Explanation:
The MO diagrams of KrBr, XeCl, and XeBr are shown below.
They are similar, except for the numbering of the valence shell orbitals.
Also, I have drawn the s and p orbitals at the same energy levels for both atoms in the compounds. That is obviously not the case.
However, the MO diagrams are approximately correct.
The ground state electron configuration of KrF is

KrF⁺ will have one less electron than KrF.
You remove the antibonding electron from the highest energy orbital, so the bond order increases.
The KrF bond will be stronger.
Answer:
The reducing agent is Zn.
Explanation:
Let's consider the reaction between zinc and hydrochloric acid.
Zn(s) + 2 HCl(aq) ⇄ ZnCl₂(aq) + H₂(g)
This is a redox reaction, which can be divided in 2 half-reactions: reduction and oxidation.
In the reduction, H⁺ gains electrons and it is considered the oxidizing agent.
2H⁺ + 2 e⁻ ⇒ H₂
In the oxidation, Zn loses electrons and it is considered the reducing agent.
Zn ⇒ Zn²⁺ + 2 e⁻