(may be broke/outdated!)
Commercial electrolytic cells for producing aluminum operate at 5 volts and 100,000 Amperes. How long does it take to produce 1000 kilograms of aluminum?
a. 1 x 10^15 seconds
b. 9 x 10^6 seconds
c. 7 x 10^3 seconds
d. 3.3 x 10^4 seconds
e. 1 x 10^5 seconds
Unspillable.com is your gateway into to the intriguing world of energy mysteries, revolutionary fuels, and the science driving alternative energy solutions. Delve into the depths of covert agendas and unexplained phenomena surrounding energy sources while uncovering the latest advancements in renewable technologies. From debunking conspiracies to exploring the cutting edge of sustainable energy, we’re your go-to resource for unraveling the secrets shaping our energy landscape.
© 2024 All Rights Reserved
One Response
This can be calculated from the following equatiom , derived by michael faraday
m = QM/(Fn)
where
m is the mass of the substance produced at the electrode (in grams), (1,000, 000)
Q is the total electric charge that passed through the solution (in coulombs),
n is the valence number of the substance as an ion in solution (electrons per ion), (+3)
F is Faraday’s constant, 96485 coulombs per mole
M is the molar mass of the substance -assumed to be aluminium oxide-(in grams per mole), 101.96 g/mol
Q=mFn/M= 1000000 *96485*3/101.96= 2838907415 coulombs
bur Q=I*t and t= Q/I = 2838907415/100000= 2.8389 x 10 ^4 , which is close to d