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
Halogens (atoms with 7 valence electrons) and Hydrogen
or generally, atoms with their shells almost full
Answer:
41.44 g
Explanation:
First of all, we must put down the equation of the reaction;
Number of moles of CaO = 33g/56 g/mol = 0.59 moles
Number of moles of H20 = 10g/18 g/mol = 0.56 moles
Since the reaction is in 1:1 mole ration, H2O is the limiting reactant
Hence;
mass of Ca(OH)2 produced = 0.56 moles * 74 g/mol = 41.44 g
The molarity of (HNO₃) that was used if 2.00 L must be used to prepare 4.5 L of a 0.25M HNO₃ solution is 0.563 M
<u><em>calculation</em></u>
This is calculated usind M₁V₁=M₂V₂ formula
where,
M₁( molarity ₁) = ?
V₁( volume ₁) = 2.00 L
M₁ (molarity ₂) = 0.25M
V₂( volume₂) = 4.5 L
make M₁ the subject of the formula by diving both side of the formula by V₁
M₁ is therefore = M₂V₂/V₁
M₁ =[ (0.25 M x 4.5 L) / 2.00 L ] =0.563 M
Answer:
D. Number of electrons
