Answer:
a minimum of <em>1</em><em>0</em><em>,</em><em>0</em><em>0</em><em>0</em><em> </em>years
The first dissociation for H2X:
H2X +H2O ↔ HX + H3O
initial 0.15 0 0
change -X +X +X
at equlibrium 0.15-X X X
because Ka1 is small we can assume neglect x in H2X concentration
Ka1 = [HX][H3O]/[H2X]
4.5x10^-6 =( X )(X) / (0.15)
X = √(4.5x10^-6*0.15)
∴X = 8.2 x 10-4 m
∴[HX] & [H3O] = 8.2x10^-4
the second dissociation of H2X
HX + H2O↔ X^2 + H3O
8.2x10^-4 Y 8.2x10^-4
Ka2 for Hx = 1.2x10^-11
Ka2 = [X2][H3O]/[HX]
1.2x10^-11= y (8.2x10^-4)*(8.2x10^-4)
∴y = 1.78x10^-5
∴[X^2] = 1.78x10^-5 m
Answer:
Explanation:
first write the equilibrium equaion ,
⇄
assuming degree of dissociation =1/10;
and initial concentraion of =c;
At equlibrium ;
concentration of
is very small so can be neglected
and equation is;
=
composiion ;
1,3-pentadiene has two double bonds which are conjugated, which undergo electrophilic addition reaction on reacting with .
The structure of 1,3-pentadiene is shown in the image.
When strong acid such as reacts with 1,3-pentadiene, the electrophilic addition reaction can occur either on double bond at 1,2-position or at 3,4-position. The reaction that occurs is shown in the image.