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.
Answer:
Option D. 1×10¯¹⁰ M
Explanation:
From the question given above, the following data were obtained:
pH = 10
Hydrogen ion concentration, [H+] =?
The hydrogen ion concentration, [H+] of the Ba(OH)2 solution can be obtained as follow:
pH = – Log [H+]
pH = 10
10 = – Log [H+]
Divide both side by – 1
– 10 = Log [H+]
Take the antilog of – 10
[H+] = antilog (– 10)
[H+] = 1×10¯¹⁰ M
Therefore, the hydrogen ion concentration, [H+] of the Ba(OH)2 solution is 1×10¯¹⁰ M
The naturally occurring isotopes of Li are Li-6 of mass 6.015121 amu and Li-7 of mass 7.016003 amu. The atomic mass of Li is 6.9409 amu, the percent abundance can be calculated using the following relation.
Atomic mass=m(Li-6 )×%(Li-6 )+m(Li-7 )×%(Li-7 )
Let the percent abundance of Li-6 be X thus, that of Li-7 will be 1-X, putting the values,

Or,

Or,
X=0.075
Thus, 
Thus, percent abundance of Li-6 is 0.075 or 7.5 % and that of Li-7 is 0.925 or 92.5%.
Answer:
0.0185 min⁻¹
Explanation:
Using integrated rate law for first order kinetics as:
Where,
is the concentration at time t
is the initial concentration
Given:
20.0 % of the initial values is left which means that 0.20 of
is left. So,
= 0.20
t = 87.0 min
<u>k = 0.0185 min⁻¹</u>