3 years ago

Giving out 15 free pts bc im in a good mood :)

1 answer:

6 0

Answer:

thank you my friend

Explanation:

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sladkih [1.3K]

A correct B incorrect C incorrect E correct

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barxatty [35]

Answer:

[H⁺] = 1.30X10⁻⁴ M

Explanation:

the problem will be solved by using Nernst's equation, which is :

$E_{cell}=E^{0}_{cell}- \frac{0.0592}{n}logQ$

In the given equation

n = 2

Q = $=\frac{[Zn^{+2}][p_{H2}]}{[H^{+}]^{2}}$

Putting values

$0.53= 0.76 - \frac{0.0592}{n}log(\frac{1X1}{[H^{+}]^{2}})$

on calculating

[H⁺] = 1.30X10⁻⁴ M

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lisov135 [29]

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siniylev [52]

Answer: 2,720 cm

Explanation:

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3 years ago