Answer:-
The reaction of 2-bromopropane reacts with sodium iodide in acetone is an example of Sn2 reaction.
The I - attacks from backside to give the transition state for both.
If we compare the transition state for cyclobromopropane 2-bromopropane then we see in case of cyclobromopropane transition state, one of the H is very close to the incoming I -.
This results in steric strain and less stability of the transition state. Hence 2-bromopropane reacts with sodium iodide in acetone over 104 times faster than bromocyclopropane.
% m/v of the solution = 0.2%m/v
<h3>Further explanation</h3>
or
% m/v = g solute / 100 ml solution
Concentration : 2 g sodium borate / L water
For 10 L⇒20 g Sodium borate
I think the answers to this is d
Answer : The concentrations of hydroxide and hydronium ions in a solution with a pH of 10.2 are, and respectively.
Explanation : Given,
pH = 10.2
pH : It is defined as the negative logarithm of the hydrogen ion concentration.
First we have to calculate the hydrogen ion concentration
Now put the value of pH in this formula, we get the hydrogen ion concentration.
Now we have to calculate the pOH of the solution.
Now put the value of pH, we get the value of pOH.
Now we have to calculate the hydroxide ion concentration
Now put the value of pOH in this formula, we get the hydroxide ion concentration.
Therefore, the concentrations of hydroxide and hydronium ions in a solution with a pH of 10.2 are, and respectively.