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This section includes 10 Mcqs, each offering curated multiple-choice questions to sharpen your Separation Processes knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
What is the total area of all cell pairs if:Volumetric flow rate(Q) = 5Difference between feed and diluate ion concentration is Δc=15Current density is i=10Current efficiency is e=0.75 |
| A. | 12.5F |
| B. | 10F |
| C. | 2F |
| D. | 23F |
| Answer» C. 2F | |
| 2. |
What is the total area of all cell pairs if:Volumetric flow rate(Q) = 1Difference between feed and diluate ion concentration is Δc=18Current density is i=10Current efficiency is e=0.9 |
| A. | 1.125F |
| B. | 1.5F |
| C. | 2F |
| D. | 3F |
| Answer» D. 3F | |
| 3. |
What is the total area of all cell pairs if:Volumetric flow rate(Q) = 1Difference between feed and diluate ion concentration is Δc=20Current density is i=10Current efficiency is e=0.6 |
| A. | 1.125F |
| B. | 1.33F |
| C. | 0.33F |
| D. | 0.44F |
| Answer» C. 0.33F | |
| 4. |
What is the total area of all cell pairs if:Volumetric flow rate(Q) = 2Difference between feed and diluate ion concentration is Δc=10Current density is i=10Current efficiency is e=0.5 |
| A. | 1.125F |
| B. | 1.5F |
| C. | 2F |
| D. | 4F |
| Answer» E. | |
| 5. |
What is the total area of all cell pairs if:Volumetric flow rate(Q) = 1Difference between feed and diluate ion concentration is Δc=10Current density is i=10Current efficiency is e=0.8 |
| A. | 1.125F |
| B. | 1.5F |
| C. | 2F |
| D. | 3F |
| Answer» B. 1.5F | |
| 6. |
Why is the electrode rinse solution acidic? |
| A. | To avoid the corrosion of electrodes |
| B. | To avoid the formation of salts |
| C. | To neutralize the OH- ions |
| D. | To make the electrodes acidic |
| Answer» D. To make the electrodes acidic | |
| 7. |
What is true about electrodialysis? |
| A. | Most easily oxidizable species is oxidized at anode |
| B. | Most easily reducible species is oxidized at cathode |
| C. | Most easily oxidizable species is reduced at anode |
| D. | The species don’t get oxidized or reduced since the electrodes are inert. |
| Answer» B. Most easily reducible species is oxidized at cathode | |
| 8. |
Why are the elctrodes in electrodialysis neither oxidized nor reduced? |
| A. | Because they act as catalyst |
| B. | Because they are chemically inert |
| C. | Because no acid is present |
| D. | There is no acid-base reaction involved in the process |
| Answer» C. Because no acid is present | |
| 9. |
How is the arrangement of electrodialysis? |
| A. | 2 membranes arranged in alternating series pattern |
| B. | 4 membranes arranged in alternating series pattern |
| C. | 2 membranes arranged in simultaneous manner |
| D. | 4 membranes arranged back to back |
| Answer» C. 2 membranes arranged in simultaneous manner | |
| 10. |
What principle is used in electrodialysis? |
| A. | Magnetic field and permeable membranes |
| B. | Electric field and cation selective membranes |
| C. | Electric fields and anion selective membranes |
| D. | Electric fields and ion selective membranes |
| Answer» E. | |