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user100 [1]
3 years ago
15

Please help thankyou

Chemistry
1 answer:
dangina [55]3 years ago
6 0

Hope this serves helpful to u!!

Hope this helps by the way as well

Good luck!!!

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What mass of gold is produced when 23.8 a of current are passed through a gold solution for 36.0 min ?
SVETLANKA909090 [29]

Answer:-

Solution:- The equation for the conversion of gold ions to gold metal is as follows:

Au^3^+(aq)+3e^-\rightarrow Au(s)

From this equation, 1 mol of Gold metal is deposited by 3 mole of electrons.

We know that one mol of electron carries one Faraday that is 96500 Coulombs.

So. 3 moles of electrons would be = 3(96500) = 289500 Coulombs

We could calculate the total Coulombs by using the formula:

q = i*t

where, q is the charge in coulombs, i is the current in ampere and t is time in seconds.

q = 23.8 ampere*36.0 min(\frac{60 sec}{1 min})

q = 51408 ampere per second

Since, 1 ampere per second = 1 coulomb

So, 51408 ampere per second is same as 51408 coulombs.

Let's use this value of q and the info from balanced equation and molar mass(196.967 g epr mol) of gold to calculate the mass of it being deposited while the current is passed as:

51408 coulomb(\frac{1 mol Au}{289500 coulombs})(\frac{196.967 g}{1 mol})

= 35.0 g Au

So, 35.0 g of gold are deposited on passing 23.8 ampere of current for 36.0 min to a gold solution.


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A, i think, I'm not 100 on that 
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Answer:

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