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: C
Explanation: i did this before
Methane is a hydrocarbon which when burns in air (combustion) produces carbon dioxide and water. The equation for the reaction;
CH4 +2O2 = CO2 +2H2O
When one mole of methane combusts 2 moles of water are formed
Therefore; when 22 moles of methane combusts 44 moles of water are formed (22 ×2)
A compound is when two or more elements are joined together. The compound that is created with two hydrogens and one oxygen is h2o or water.
Answer:
2x+3y=-6
4x+3y=12
3x-y=5
5x+3y=1
Explanation:
standard form: ax + by = C
