1Part 1 — Where chemistry startsFREE
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2Part 2 — Bonding and equations🔒 Locked
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Selected exam questions
Questions chosen for this lecture, exactly as they appear in the model.
5Question 5 — Model 3
Question 5
Study the following table:
What process was used to dress the ore?
| Properties Before Dressing | Properties After Dressing | |
|---|---|---|
| Mass of ore | 4 Kg | 3.85 Kg |
| % of iron | 45% | 69% |
✓ Correct Answer: (b) Roasting
📚 Detailed Explanation
Detailed Step-by-Step Analysis
1. Understanding the Data
- Before the process: The percentage of iron in the ore is 45%.
- After the process: The mass of the ore decreases to 3.85 Kg, and the percentage of iron increases significantly to 69%.
2. Identifying the Process (Roasting)
-
Roasting (b) is a chemical process that involves heating the ore strongly in the air.
- The main goals of roasting are to dry the ore, remove moisture, expel volatile impurities as gases, and raise the percentage of iron in the ore to 69% by converting it into Hematite (Fe2O3).
- The exact target value of 69% matches the textbook standard for iron ore after the roasting step:
2FeCO3 (Siderite: 48.5%) + ½O2 →Δ Fe2O3 (Hematite: 69%) + 2CO2(g)
2Fe2O3·3H2O (Limonite: 40%) →Δ 2Fe2O3 (Hematite: 69%) + 3H2O(g)
3. Why the Other Options Are Incorrect
- (a) Crushing & (c) Sintering: These are purely physical processes aimed at changing the particle size of the ore. They do not alter the chemical percentage of iron or cause volatile mass loss to this extent.
- (d) Reduction: Reduction occurs inside the Blast Furnace or Midrex Furnace to produce pure iron metal, not to dress or prepare the ore.
12Question 12 — Model 3
Question 12
Study the tubes (A) and (B) in the illustrated figure:
Which of the following describes the result of the reaction in each tube?
✓ Correct Answer: (a)
📚 Detailed Explanation
- Both tubes are at the same temperature (50°C) with equal masses of zinc.
- Tube A has HCl at 1M (higher concentration) while Tube B has HCl at 0.5M (lower concentration).
- Higher concentration → higher reaction rate → zinc disappears faster in tube A. ✓
- Upon complete dissolution of equal masses of zinc, the same amount of H₂ is produced in both tubes (since it depends on the mass of zinc, not the acid concentration, assuming excess acid).
Zn + 2HCl → ZnCl₂ + H₂↑
30Question 30 — Model 3
Question 30
An alcohol that cannot be prepared by catalytic hydration. All of the following apply to this alcohol except:
✓ Correct Answer: (b)
📚 Detailed Explanation
- An alcohol that cannot be prepared by catalytic hydration = Methanol (CH₃OH), because catalytic hydration requires an alkene, and there's no C₁ alkene.
- (a) Can be prepared by the general method (halide + KOH) — technically yes for methyl halide ✓
- (b) Methanol (CH₃OH) has NO structural isomers — it's the simplest alcohol ✗
- (c) Methanol is toxic and causes blindness/death ✓
- (d) Methanol reacts with acids to form esters (e.g., methyl salicylate) ✓
Homework
Quiz You need 60% to unlock the next lecture.
1.The element which has the electronic configuration [Ar] 3d¹⁰ 4s² is …..
Explanation: Total electrons = 18 (Ar) + 10 + 2 = 30 → zinc (Zn). Copper is 29 and iron is 26.
2.The oxidation number of chromium in Cr₂O₃ is …..
Explanation: Oxygen is −2 and there are 3 atoms = −6. The compound is neutral, so 2 × Cr = +6 ⇒ Cr = +3.
3.The transition element which has the electronic configuration [Ar] 4s¹ 3d¹⁰ is …….
Explanation: 18 + 1 + 10 = 29 → copper (Cu). This is the famous anomaly: one 4s electron moves to 3d so that the d sublevel is completely filled and more stable.
4.The transition element which has only one oxidation state (+3) is …..
Explanation: Scandium is [Ar] 4s² 3d¹. Losing these 3 electrons gives it the stable argon configuration, so +3 is its only oxidation state. Zinc has only +2 but it is not a true transition element.
5.The element which can give the oxidation state (+7) is ……..
Explanation: Manganese is [Ar] 4s² 3d⁵, so its valence electrons = 2 + 5 = 7 and it can lose them all (as in KMnO₄). Chromium reaches only +6.
6.The coinage metals are …..
Explanation: The coinage metals are gold, silver and copper (group IB). They were used for minting coins because they are inert and resist corrosion.
7.The elements of the first transition series have several oxidation states, because they lose electrons from ………. sublevels.
Explanation: The energy difference between 4s and 3d is very small, so the element loses the 4s electrons first and then the 3d ones — this is the reason for the multiple oxidation states.
8.The highest oxidation state of chromium, according to its electronic configuration, is …..
Explanation: Chromium is [Ar] 4s¹ 3d⁵, so its valence electrons = 1 + 5 = 6 (as in K₂Cr₂O₇).
9.The number of single (unpaired) electrons in Fe³⁺ is …..
Explanation: Iron (26) is [Ar] 4s² 3d⁶. On ionisation it loses the two 4s electrons first, then one 3d electron ⇒ Fe³⁺ = [Ar] 3d⁵. The five d orbitals are half filled, so there are 5 unpaired electrons.
10.Atomic number of the element in which the number of filled orbitals equals the number of half-filled orbitals plus one, in the 3d sublevel.
Explanation: We need: filled = half-filled + 1. For 24 (3d⁵): 0 filled, 5 half — no. For 27 (3d⁷): 2 filled, 3 half — no. For 28 (3d⁸): 3 filled, 2 half → 3 = 2 + 1 ✓. For 29 (3d¹⁰): 5 filled, 0 half — no.
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