Answer:
a) galvanic cell
b)electrolytic cell
c) i) K=6.27x10'34
ΔG°=198790 J
ii) K=3.58x10'-34
ΔG°= 191070 J
d) E°=0.278 v
ΔG°= -26827 J
Explanation:
a) There are two kinds of an electrochemical cell, the first is called "galvanic cells", and the second "electrolytic cell".
The fuel cells are capable of produce electric energy through chemical reactions. These reactions are often spontaneous. So, the galvanic cell has a negative value for Gibbs free energy.
b) The electrolytic cell increases the value of Gibbs energy, to positive values, due to the reactions are not spontaneous.
c) i) look image attached
ii) k = look image attached
ΔG° = -nFE° = - 6 X 95500 J/vmole x (-0.33 v)
ΔG° =-191070
d) E°= 0.0592 v/n x lg K
E°= 0.0592V / 1 X log 5.0X10'4
E°= 0.278 v
ΔG° = -nFE° = -1 x 96500 J/ vmole x 0.278v
ΔG° = -26827 J
1.question, 2.observe, 3.hypothesize, 4.experiment 5.conclusion, 6. record.
For this problem we can use half-life formula and radioactive decay formula.
Half-life formula,
t1/2 = ln 2 / λ
where, t1/2 is half-life and λ is radioactive decay constant.
t1/2 = 8.04 days
Hence,
8.04 days = ln 2 / λ
λ = ln 2 / 8.04 days
Radioactive decay law,
Nt = No e∧(-λt)
where, Nt is amount of compound at t time, No is amount of compound at t = 0 time, t is time taken to decay and λ is radioactive decay constant.
Nt = ?
No = 1.53 mg
λ = ln 2 / 8.04 days = 0.693 / 8.04 days
t = 13.0 days
By substituting,
Nt = 1.53 mg e∧((-0.693/8.04 days) x 13.0 days))
Nt = 0.4989 mg = 0.0.499 mg
Hence, mass of remaining sample after 13.0 days = 0.499 mg
The answer is "e"
Keep trying you will get if you keep trying ok