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
Protons is the number of the element on the periodic table. Electrons are the same number of protons and i'm sorry I do not know neutrons, I hope I helped!
It would typically be around 5000 seconds(83.33) minutes for the water to freeze.
Explanation:
2H2O2 => 2H2O + O2
Moles of hydrogen peroxide = 0.150dm³ * (0.02mol/dm³) = 0.003mol .
Moles of oxygen = 0.0015mol.
Volume of oxygen = 0.0015mol * (22.4dm³/mol) = 0.0336dm³.
