Answer:
The formal charge on nitrogen in
is +1.
Explanation:
The structure of
is as follows.
(In attachment)

From the structure, Nitrogen has no non bonding electrons. Nitrogen has four bonds and each bond corresponds to 2 electrons. Hence, nitrogen have eight bonding electrons and five valence electrons.
![Formal\,charge\,on\,nitrogen = 5-[0+ \frac{8}{2}]= +1](https://tex.z-dn.net/?f=Formal%5C%2Ccharge%5C%2Con%5C%2Cnitrogen%20%3D%205-%5B0%2B%20%5Cfrac%7B8%7D%7B2%7D%5D%3D%20%2B1)
Therefore, The formal charge on nitrogen in
is +1.
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:
The equilibrium constant for the reversible reaction = 0.0164
Explanation:
At equilibrium the rate of forward reaction is equal to the rate of backwards reaction.
The reaction is given as
A ⇌ B
Rate of forward reaction is first order in [A] and the rate of backward reaction is also first order in [B]
The rate of forward reaction = |r₁| = k₁ [A]
The rate of backward reaction = |r₂| = k₂ [B]
(Taking only the magnitudes)
where k₁ and k₂ are the forward and backward rate constants respectively.
k₁ = 0.010 s⁻¹
k₂ = 0.0610 s⁻¹
|r₁| = 0.010 [A]
|r₂| = 0.016 [B]
At equilibrium, the rate of forward and backward reactions are equal
|r₁| = |r₂|
k₁ [A] = k₂ [B] (eqn 1)
Note that equilibrium constant, K, is given as
K = [B]/[A]
So, from eqn 1
k₁ [A] = k₂ [B]
[B]/[A] = (k₁/k₂) = (0.01/0.0610) = 0.0163934426 = 0.0164
K = [B]/[A] = (k₁/k₂) = 0.0164
Hope this Helps!!!
Answer:- The natural abundance of
is 0.478 or 47.8% and
is 0.522 or 52.2% .
Solution:- Average atomic mass of an element is calculated from the atomic masses of it's isotopes and their abundances using the formula:
Average atomic mass = mass of first isotope(abundance) + mass of second isotope(abundance)
We have been given with atomic masses for
and
as 150.919860 and 152.921243 amu, respectively. Average atomic mass of Eu is 151.964 amu.
Sum of natural abundances of isotopes of an element is always 1. If we assume the abundance of
as n then the abundance of
would be 1-n .
Let's plug in the values in the formula:

151.964=150.919860n+152.921243-152.921243n
on keeping similar terms on same side:


negative sign is on both sides so it is canceled:



The abundance of
is 0.478 which is 47.8%.
The abundance of
is = 
= 0.522 which is 52.2%
Hence, the natural abundance of
is 0.478 or 47.8% and
is 0.522 or 52.2% .