What quantity of copper will be deposited by the same quantity of electricity that deposited 9.0g of aluminum? (A = 27, Cu = 64)

Rhontommy

3 months ago

Read the question very well and it's very easy to solve
Atomic mass of Aluminium -----> Oxidation number of Aluminium × Number of faraday
:• 27--->3×96500

Given mass of Al--->Quantity of electricity (unknown)
9---->X
so let write it out
27––---------›289500
9---------------›X
cross multiply
27X =289500×9
X= 2605500/27
X= 96500( Which is the quantity of electricity of aluminium)

Using the same formula for copper
since we are using the same quantity of electricity of aluminium for copper it will go like this :
64———›193000
X————›96500
cross multiply
193000X=64×96500
X= 6176000/193000
X=32
Our answer is D=32

Rhontommy

3 months ago

I mean B- 32

112shalom

3 years ago

Can I get the solution to this please thanks

CHRIST IS KING

2 months ago

Correct

Cpedro

4 years ago

It's C

Pabloxto

10 years ago

d ansa is a bcos

27 - 64*3

9 - x

so wen u cross multiply

9*64*3/27=64