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According to the reaction equation:
and by using ICE table:
CN- + H2O ↔ HCN + OH-
initial 0.08 0 0
change -X +X +X
Equ (0.08-X) X X
so from the equilibrium equation, we can get Ka expression
when Ka = [HCN] [OH-]/[CN-]
when Ka = Kw/Kb
= (1 x 10^-14) / (4.9 x 10^-10)
= 2 x 10^-5
So, by substitution:
2 x 10^-5 = X^2 / (0.08 - X)
X= 0.0013
∴ [OH] = X = 0.0013
∴ POH = -㏒[OH]
= -㏒0.0013
= 2.886
∴ PH = 14 - POH
= 14 - 2.886 = 11.11
Answer:
60 moles of NaF
Explanation:
The balanced equation for the reaction is given below:
Al(NO3)3 + 3NaF —> 3NaNO3 + AlF3
From the balanced equation above,
3 moles of NaF reacted to produce 1 mole of AlF3.
Therefore, Xmol of NaF will react to produce 20 moles of AlF3 i.e
Xmol of NaF = 3 x 20
Xmol of NaF = 60 moles
Therefore, 60 moles of NaF are required to produce 20 moles of AlF3.
Answer:
11.31 g.
Explanation:
Molarity is defined as the no. of moles of a solute per 1.0 L of the solution.
M = (no. of moles of solute)/(V of the solution (L)).
<em>∴ M = (mass/molar mass)of NaCl/(V of the solution (L)).</em>
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<em>∴ mass of NaCl remained after evaporation of water = (M)(V of the solution (L))(molar mass)</em> = (0.45 M)(0.43 L)(58.44 g/mol) = <em>11.31 g.</em>