How many seconds are required to deposit 0.110 grams of magnesium metal from a solution that contains Mg2 ions, if a current of
1 answer:
Answer: The time required to deposit such amount of Mg is 886secs.
Explanation: According to Faraday Law of Electrolysis, the mass of a substance deposited is directly proportional to quantity of electricity passed.
He also stated that;
96500C(1Farday) of electricity is required to deposit 1 mole of any metal.
For +2e- ==>Mg
193000C(2Faraday) of electricity is required to deposit 1mole of Magnesium metal (1 mole of Mg=24g)
Which implies;
19300C will liberate 24g
xCwill liberate 0.110g
Where x is the amount of electricity to deposit 0.110g
x = (19300 × 0.110)/24
x = 884.6C
Recall that Q =It
Where Q is the quantity of electricity, I is the current and t is the time taken
884.6= 0.998 × t
t= 884.6/0.998
t= 886.3secs.
Therefore the time require is 886.3s.
You might be interested in
Answer:
Fr = 26.83 [N]
Explanation:
To solve this problem we must use the Pythagorean theorem, since the forces are vector quantities, that is, they have magnitude and density. Therefore the Pythagorean theorem is suitable for the solution of this problem.