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RSB [31]
3 years ago
7

A first-order reaction has a half-life of 29.2 s . how long does it take for the concentration of the reactant in the reaction t

o fall to one-sixteenth of its initial value?
Chemistry
1 answer:
Dmitriy789 [7]3 years ago
5 0

Answer is: it takes 116,8 seconds to fall to one-sixteenth of its initial value

<span> The half-life for the chemical reaction is 29,2 s and is independent of initial concentration.
c</span>₀ - initial concentration the reactant.

c - concentration of the reactant remaining at time.

t = 29,2 s.<span>
First calculate the rate constant k:
k = 0,693 ÷ t = 0,693 ÷ 29,2 s</span> = 0,0237 1/s.<span>
ln(c/c</span>₀) = -k·t₁.<span>
ln(1/16 </span>÷ 1) = -0,0237 1/s · t₁.

t₁ = 116,8 s.

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Explanation:

Given data:

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by putting values,

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What effect does the anion of an ionic compound have on the appearance of the solution?
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Explanation:

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The effect of the anion of an ionic compound on the appearance of the solution is that the anion affects the color of the solution more than the intensity of the color.

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<h3><u>Answer</u>;</h3>

B. 3/2

<h3><u>  Explanation;</u></h3>

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