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
D. reproductive cells uwu
Answer:
the name for NO is nitrogen monoxide
Explanation:
Molar mass of
= 39.1 + 35.5 + 3(16.0) = 122.6 g
Molar mass of KCl = 39.1 + 35.5 = 74.6 g
Molar mass of
= 32.0 g
According to the equation, 2 moles of
reacts to give 3 moles of oxygen.
Therefore, 2 (122.6) = 245.2 g of
will give 3 (32.0) = 96.0 g of oxygen. Thus, 245.2 g of
gives 96.0 g of oxygen.
(a) Calculate the amount of oxygen given by 2.72 g of
as follows.
of
(b) Calculate the amount of oxygen given by 0.361 g of
as follows.
of
c) Calculate the amount of oxygen given by 83.6 kg
as follows.
of 
Convert kg into grams as follows.
= 32731 g of 
(d) Calculate the amount of oxygen given by 22.5 mg of
as follows.

Convert mg into grams as follows.
of 