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AfilCa [17]
2 years ago
14

A certain first-order reaction (A→products) has a rate constant of 9.00×10−3 s−1 at 45 ∘C. How many minutes does it take for the

concentration of the reactant, [A], to drop to 6.25% of the original concentration?
Chemistry
1 answer:
Amiraneli [1.4K]2 years ago
5 0

Answer:

27.8 minutes

Explanation:

The reaction follows a first order

Rate = k[A] = change in concentration/time

k = 9×10^-3s^-1

Let the original concentration of A be y

Concentration of A at time t = 6.25% × y = 0.0625y

Change in concentration = y - 0.0625y = 0.9375y

0.009 × 0.0625y = 0.9375y/t

t = 0.9375y/0.0005625y = 1666.7sec = 1666.7/60 = 27.8 minutes

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The ka of hf is 6.8 x 10-4. what is the ph of a 0.35 m solution of hf?
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When the reaction equation is:

HF ↔ H+   +   F-

and when the Ka expression
= concentration of products/concentration of reactions

so, Ka = [H+][F-]/[HF]

when we assume:

[H+] = [F-] = X

and [HF] = 0.35 - X

So, by substitution:

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∴PH = -㏒0.015

        = 1.8
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2 years ago
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