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
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Answer:
P = 13.5 atm
Explanation:
Given that
No. of moles, n = 20 moles
Volume of nitrogen gas = 36.2 L
Temperature = 25°C = 298 K
We need to find the pressure of the gas. Using the ideal gas equation
PV = nRT
Where
R is gas constant, 
So,
so, the pressure of the gas is equal to 13.5 atm.
<span>The half-life of 9 months is 0.75 years.
2.0 years is 2.0/0.75 = 2.67 half-lives.
Each half-life represents a reduction in the amount remaining by a factor of two, so:
A(t)/A(0) = 2^(-t/h)
where A(t) = amount at time t
h = half-life in some unit
t = elapsed time in the same unit
A(t)/A(0) = 2^(-2.67) = 0.157
15.7% of the original amount will remain after 2.0 years.
This is pretty easy one to solve. I was happy doing it.</span>
