Answer:
8.59 g
2.25 g
Explanation:
According to the given situation the calculation of grams of PbO and grams of NaCL is shown below:-
Moles of Pb(OH)CL is
= 0.0385 mol
Mass of PbO needed is
After solving the above equation we will get
= 8.59 g
Mass of NaCL needed is
After solving the above equation we will get
= 2.25 g
Therefore we have applied the above formula.
i would say Psychoanalysis so you need to call Frasier Crane lol
hope it helps
The electrophilic bromination or chlorination of benzene requires Lewis acid along with the halogen.
<h3>
What is bromination of benzene?</h3>
The bromination or chlorination of benzene is an example of an electrophilic aromatic substitution reaction.
During the reaction, the bromine forms a sigma bond to the benzene ring, yielding an intermediate. Subsequently a a proton is removed from the intermediate to form a substituted benzene ring.
This reaction is achieved with the help of Lewis acid as catalysts.
Thus, the electrophilic bromination or chlorination of benzene requires Lewis acid along with the halogen.
Learn more about bromination of benzene here: brainly.com/question/26428023
Missing in your question Ka2 =6.3x10^-8
From this reaction:
H2SO3 + H2O ↔ H3O+ + HSO3-
by using the ICE table :
H2SO3 ↔ H3O + HSO3-
intial 0.6 0 0
change -X +X +X
Equ (0.6-X) X X
when Ka1 = [H3O+][HSO3-]/[H2SO3]
So by substitution:
1.5X10^-2 = (X*X) / (0.6-X) by solving this equation for X
∴ X = 0.088
∴[H2SO3] = 0.6 - 0.088 = 0.512
[HSO3-] = [H3O+] = 0.088
by using the ICE table 2:
HSO3- ↔ H3O + SO3-
initial 0.088 0.088 0
change -X +X +X
Equ (0.088-X) (0.088+X) X
Ka2= [H3O+] [SO3-] / [HSO3-]
we can assume [HSO3-] = 0.088 as the value of Ka2 is very small
6.3x10^-8 = (0.088+X)*X / 0.088
X^2 +0.088 X - 5.5x10^-9= 0 by solving this equation for X
∴X= 6.3x10^-8
∴[H3O+] = 0.088 + 6.3x10^-8
= 0.088 m ( because X is so small)
∴PH= -㏒[H3O+]
= -㏒ 0.088 = 1.06
Answer:
<em>1 mole is equal to 1 moles NaOH, or 39.99711 grams.</em>
Explanation:
<em>Hope this helps have a nice day :)</em>
