Answer: Too much base was added
i guessed
Explanation:
Answer:
Look for extra things to do, small details, until you find a big enough one to go off that to continue
Explanation:
n/a
Answer:
C.) HOCl Ka=3.5x10^-8
Explanation:
In order to a construct a buffer of pH= 7.0 we need to find the pKa values of all the acids given below
we Know that
pKa= -log(Ka)
therefore
A) pKa of HClO2 = -log(1.2 x 10^-2)
=1.9208
B) similarly PKa of HF= -log(7.2 x 1 0^-4)= 2.7644
C) pKa of HOCl= -log(3.5 x 1 0^-8)= 7.45
D) pKa of HCN = -log(4 x 1 0^-10)= 9.3979
If we consider the Henderson- Hasselbalch equation for the calculation of the pH of the buffer solution
The weak acid for making the buffer must have a pKa value near to the desired pH of the weak acid.
So, near to value, pH=7.0. , the only option is HOCl whose pKa value is 7.45.
Hence, HOCl will be chosen for buffer construction.
The question is incomplete. Complete question is attached below
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Correct Answer:
Option C i.e. I ~ III < IV < V < II
Reason:
During a nucleophilic subsitution reaction of chloroarenes, Cl- group is replaced by an nucleophile like OH-.
Order of reactivity, during such reactions depends on the electron density on carbon atom that is attached to Cl. Lower the electron density, greater will be the reactivity.Among the provided chloroarenes, electron density on C atom will be minimum in case of compound II, because of presence of electron withdrawing group (-NO2) at ortho and para position. Due to this, there will be large number of resonating structures. This signifies greater electron de-localization, and hence largest reactivity for nucleophilic substitution reaction.
Followed by this, compound V will show greater reactivity, due to presence of -NO2 group at para and one of the ortho position. Compound IV will have less number of resonating structures as compared to compound II and V, hence it will display poor reactivity towards nucleophilic substitution reaction.
Finally, compound 1 and III will minimum reactivity towards nucleophilic substitution reaction, because -NO2 group present at meta position (compound III) will not participate in resonance.