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A city's water supply is contaminated with a toxin at a concentration of 0.63 mg/L. Fortunately, this toxin decomposes to a safe mixture of products by first-order kinetics with a rate constant of 0.27 day–1. How long will it take for half of the toxin to decompose?

Answer :

jufsanabriasa

Answer:

t = 2,57 days

Explanation:

A reaction follows the first-order kinetics when attend the formula:

[tex]ln\frac{[A]_t}{[A]_0} = -kt[/tex]

Where [A]₀ is the initial concentration of the toxin (0,63mg/L) and [A]t is the half of the initial concentration (0,315mg/L). k is 0,27 day⁻¹ and t is the time the descomposition takes. Replacing:

[tex]ln\frac{0,315}{0,63} = -0,27day^{-1}t[/tex]

t = 2,57 days

I hope it helps!

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