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NISA [10]
3 years ago
7

Hydrocyanic acid has a Ka of 4.0

Chemistry
1 answer:
Akimi4 [234]3 years ago
7 0

Answer:

0.002 %

Explanation:

Given that:

K_{a}=4.0\times 10^{-10}

Concentration = 1.0 M

Consider the ICE take for the dissociation of Hydrocyanic acid as:

                                      HCN    ⇄     H⁺ +        CN⁻

At t=0                            1.0                -              -

At t =equilibrium        (1.0-x)                x           x            

The expression for dissociation constant of Hydrocyanic acid is:

K_{a}=\frac {\left [ H^{+} \right ]\left [ {CN}^- \right ]}{[HCN]}

4.0\times 10^{-10}=\frac {x^2}{1.0-x}

x is very small, so (1.0 - x) ≅ 1.0

Solving for x, we get:

x = 2×10⁻⁵  M

Percentage ionization = \frac {2\times 10^{-5}}{1.0}\times 100=0.002 \%

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Answer:

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6 0
3 years ago
The activity of a certain isotope dropped from 3200 ci to 800 ci in 24.0 years. what is the half-life of this isotope (in years)
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Ln(800/3200) = - kt
t = 24 years.
ln(0.25) = -k*24
(- 1.3863) = -k*24
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ln(0.5) = -k*t
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I don't know if you can just look at the question and know the answer. If 24 years is a quarter life then is it obvious that the 1/2 life is 12 years? It might be, but the method I've used works for sure. 


3 0
3 years ago
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