Chemistry, 24.07.2019 01:30 bettybales1986

What mass of copper is deposited when a current of 10.0a is passed through a solution of copper(ii) nitrate for 30.6 seconds?

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Data Given:

Time = t = 30.6 s

Current = I = 10 A

Faradays Constant = F = 96500

Chemical equivalent = e = 63.54/2 = 31.77 g

Amount Deposited = W = ?

Solution:
According to Faraday's Law,

W = I t e / F

Putting Values,

W = (10 A × 30.6 s × 31.77 g) ÷ 96500

W = 0.100 g

Result:
0.100 g of Cu²⁺ is deposited.

Answer from: Quest

Most likely a is the answer - because you generate 38 atp's but also consume an atp. each atp gives approx. less than 4 kcal so probably a is the closest answer.

Answer from: Quest

explanation:

two years ago, when picking flowers in the mountains back of yosemite valley, i found a book. it was blotted and storm-beaten; all of its outer pages were mealy and crumbly. the paper seemed to dissolve like the snow beneath which it had been buried, however; many of the inner pages were well preserved. and although all were more or less stained and torn, whole chapters were easily readable. this is the condition of the great open book of yosemite glaciers today; its granite pages have been torn and blurred by the same storms that wasted the castaway book.

two years ago, when picking flowers in the mountains back of yosemite valley; i found a book. it was blotted and storm-beaten, all of its outer pages were mealy and crumbly. the paper seemed to dissolve like the snow beneath which it had been buried, however, many of the inner pages were well preserved. and although all were more or less stained and torn; whole chapters were easily readable. this is the condition of the great open book of yosemite glaciers today, its granite pages have been torn and blurred by the same storms that wasted the castaway book.

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What mass of copper is deposited when a current of 10.0a is passed through a solution of copper(ii)...

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