Silver can be electroplated at the cathode of an electrolysis cell by the half-reaction. Ag+(aq)+e−→Ag(s) Part A Silver can be e
1 answer:
Answer: Mass of silver deposited at the cathode is 37.1g
Explanation: According to Faraday Law of Electrolysis, the mass of substance deposited at the electrode (cathode or anode) is directly proportional to quantity of electricity passed through the electrolyte
Faraday has found that to liberate one gm eq. of substance from an electrolyte, 96500C of electricity is required.
+e− ==> Ag(s)
Given that
Current (I) = 8.5A
Time (t) = 65 *60 = 3900s
Quantity of electricity passed = 8.5*3900 =33150C
Molar mass of Ag= 108g
96500C will liberate 108g
33150C will liberate Xg
Xg= (108*33150)/96500
=37.1g
Therefore the mass of Ag deposited at the cathode is 37.1g.
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Answer:
m = 1.5 gram
Explanation:
Given that,
Density of protein gelatin, d = 3 g/L
The volume of protein gelatin, V = 0.5 L
We need to find the mass of the protein gelatin. The density of an object is given by :
d = m/V
Where
m is mass
So, the required mass is 1.5 gram.
Answer:
Magnesium
0.003mole
Explanation:
The problem here entails we find the metal in the carbonate.
For group 2 member, let the metal = X;
The carbonate is XCO₃;
If we sum the atomic mass of the elements in the metal carbonate, we should arrive at 84g/mol
Atomic mass of C = 12g/mol
O = 16g/mol
Atomic mass of X + 12 + 3(16) = 84
Atomic mass of X = 84 - 60 = 24g/mol
The element with atomic mass of 24g is Magnesium
B.
Number of moles in 0.3g of CaCO₃:
Molar mass of CaCO₃ = 40 + 12 + 3(16) = 100g/mol
Number of moles =
Number of moles = = 0.003mole