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Separation Processes MCQ’s – Rate Based Model for Leaching
If the density of CuO in ore ρ = 0.05
It is leached with H2SO4 = 0.5M
Ore particle have radius r=0.5
Effective diffusivity De= 6*10-6
Molecular weight Mb=79.6
B=0.5
Cab=0.01
Calculate the time required for total leaching.
a) 3.8h
b) 4.6h
c) 7.8h
d) 6.4h
View Answer
Answer: a
Explanation: T= ρr2/6DeMbbCab, hence t=3.79h.
If the density of CuO in ore ρ = 0.05
It is leached with H2SO4 = 0.6M
Ore particle have radius r=0.5
Effective diffusivity De= 6*10-6
Molecular weight Mb=79.6
B=0.5
Cab=0.01
Calculate the time required for total leaching.
a) 3.8h
b) 4.56h
c) 7.87h
d) 9.4h
View Answer
Answer: b
Explanation: T= ρr2/6DeMbbCab, hence t=4.56h.
If the density of CuO in ore ρ = 0.05
It is leached with H2SO4 = 0.5M
Ore particle have radius r=0.56
Effective diffusivity De= 6*10-6
Molecular weight Mb=79.6
B=0.5
Cab=0.01
Calculate the time required for total leaching.
a) 4.2h
b) 4.3h
c) 4.76h
d) 6.4h
View Answer
Answer: c
Explanation: T= ρr2/6DeMbbCab, hence t=4.76h.
If the density of CuO in ore ρ = 0.08
It is leached with H2SO4 = 0.5M
Ore particle have radius r=0.5
Effective diffusivity De= 6*10-6
Molecular weight Mb=79.6
B=0.5
Cab=0.01
Calculate the time required for total leaching.
a) 6.08h
b) 6.32h
c) 6.35h
d) 6.4h
View Answer
Answer: a
Explanation: T= ρr2/6DeMbbCab, hence t=6.08h.
If the density of CuO in ore ρ = 0.06
It is leached with H2SO4 = 0.5M
Ore particle have radius r=0.7
Effective diffusivity De= 6*10-6
Molecular weight Mb=79.6
B=0.5
Cab=0.01
Calculate the time required for total leaching.
a) 10.2h
b) 10.42h
c) 10.54h
d) 6.4h
View Answer
Answer: b
Explanation: T= ρr2/6DeMbbCab, hence t=10.42h.
If the density of CuO in ore ρ = 0.03
It is leached with H2SO4 = 0.5M
Ore particle have radius r=0.5
Effective diffusivity De= 6*10-6
Molecular weight Mb=79.6
B=0.5
Cab=0.01
Calculate the time required for total leaching.
a) 2.1h
b) 2.28h
c) 7.8h
d) 6.4h
View Answer
Answer: b
Explanation: T= ρr2/6DeMbbCab, hence t=2.28h.
If the density of CuO in ore ρ = 0.04
It is leached with H2SO4 = 0.5M
Ore particle have radius r=0.5
Effective diffusivity De= 6*10-6
Molecular weight Mb=79.6
B=0.5
Cab=0.01
Calculate the time required for total leaching.
a) 3.8h
b) 4.75h
c) 7.8h
d) 6.4h
View Answer
Answer: b
Explanation: T= ρr2/6DeMbbCab, hence t=4.75h.
If the density of CuO in ore ρ = 0.05
It is leached with H2SO4 = 0.5M
Ore particle have radius r=0.66
Effective diffusivity De= 6*10-6
Molecular weight Mb=79.6
B=0.5
Cab=0.01
Calculate the time required for total leaching.
a) 6.12h
b) 6.22h
c) 6.62h
d) 6.4h
View Answer
Answer: c
Explanation: T= ρr2/6DeMbbCab, hence t=6.62h.
If the density of CuO in ore ρ = 0.05
It is leached with H2SO4 = 0.5M
Ore particle have radius r=0.5
Effective diffusivity De= 4.44*10-6
Molecular weight Mb=79.6
B=0.5
Cab=0.01
Calculate the time required for total leaching.
a) 5.13h
b) 5.3h
c) 5.65h
d) 6.4h
View Answer
Answer: a
Explanation: T= ρr2/6DeMbbCab, hence t=5.13h.
If the density of CuO in ore ρ = 0.05
It is leached with H2SO4 = 0.5M
Ore particle have radius r=0.5
Effective diffusivity De= 9*10-6
Molecular weight Mb=79.6
B=0.5
Cab=0.01
Calculate the time required for total leaching.
a) 2.53h
b) 3.4h
c) 5.56h
d) 6.78h
View Answer
Answer: a
Explanation: T= ρr2/6DeMbbCab, hence t=2.53h.