Separation Processes Questions and Answers – Rate Based Model for Leaching

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This set of Separation Processes Multiple Choice Questions & Answers (MCQs) focuses on “Rate Based Model for Leaching”.

1. 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.
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2. 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.

3. 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.

4. 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.

5. 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.
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6. 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.

7. 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.

8. 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.

9. 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.
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10. 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.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn