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.
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Answer:
Potential energy = 73.575 kJ
Kinetic energy = 135kJ
Total mechanical energy = 208.575 kJ
Explanation:
The potential energy of a body is given by the expression, PE = mgh, where m is the mass of the body, g is the acceleration due to gravity value and h is the height of the body.
The kinetic energy of a body is given by , where v is the velocity and m is the mass of body.
Total mechanical energy = Kinetic energy + Potential energy
PE = mgh = 75*9.81*100 = 73575 J = 73.575 kJ
Total mechanical energy = Kinetic energy + Potential energy = 135+73.575
= 208.575 kJ