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zalisa [80]
3 years ago
6

Calculate the half-life (in s) of a first-order reaction if the concentration of the reactant is 0.0576 M 17.1 s after the react

ion starts and is 0.0249 M 97.8 s after the reaction starts.
Chemistry
1 answer:
garik1379 [7]3 years ago
7 0

Answer:

66.7s

Explanation:

Let's bring out the parameters we were given...

Half life = ?

Initial Concentration = 0.0576 M

Final Concentration = 0.0249 M

Time for the concentration change to occur = 97.8s - 17.1s = 80.7s

Formular for half life (t1/2) is given as;

t1/2 = ln2 / k   ≈   0.693 / k

where k = rate constant

From the formular of first order reactions;

ln[A] = ln[A]o − kt

where [A] = Final Concentration and [A]o = Initial Concentration

Inserting the values, we have;

ln(0.0249) = ln(0.0576) - k(80.7)

Upon solving for k, we have;

-0.8387 = -k(80.7)

k = 0.01039 s−1

t1/2 = 0.693 / k = 0.693 / 0.01039 = 66.7s

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Answer:

pH=12.3\\\\pOH=1.7\\

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Explanation:

Hello there!

In this case, according to the given ionization of magnesium hydroxide, it is possible for us to set up the following reaction:

Mg(OH)_2(s)\rightleftharpoons Mg^{2+}(aq)+2OH^-(aq)

Thus, since the ionization occurs at an extent of 1/3, we can set  up the following relationship:

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Best regards!

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3 years ago
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