Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

  ❓ Concept Limiting Reagent & Water Volume Trick in 60 Seconds! How do you quickly find how much water forms in a combustion reaction without writing long stoichiometry tables ? This 60-second hack covers: ✔ Limiting reagent shortcut ✔ Quick mole-ratio conversion ✔ Mass → volume trick for liquid water ✍️ Short Explanation Let’s break the entire idea into three ultra-fast steps , using the popular combustion example: C 4 H 10 + 13 2 O 2 → 4 C O 2 + 5 H 2 O \mathrm{C_4H_{10}} + \frac{13}{2} \mathrm{O_2} \rightarrow 4\mathrm{CO_2} + 5\mathrm{H_2O} This trick works for any reaction that forms liquid water . 🔹 Step 1 — Identify the Limiting Reagent FAST Instead of computing everything, compare “ moles available : moles required ”. Example: 174 kg butane → 174000 58 ≈ 3000 \frac{174000}{58} \approx 3000 320 kg O₂ → 320000 32 = 10000 \frac{320000}{32} = 10000 Reaction needs 6.5 mol O₂ per mol C₄H₁₀ . Demand for O₂ to consume all butane: 3000 × 6.5 = 19500 ...

  ❓ Question Evaluate the integral: ∫ 0 π ( x + 3 ) sin ⁡ x 1 + 3 cos ⁡ 2 x   d x 🖼️ Question Image ✍️ Short Solution We split the integral into two parts: I = ∫ 0 π x sin ⁡ x 1 + 3 cos ⁡ 2 x   d x + 3 ∫ 0 π sin ⁡ x 1 + 3 cos ⁡ 2 x   d x . I = \int_{0}^{\pi} \frac{x\sin x}{1+3\cos^{2}x}\,dx + 3\int_{0}^{\pi} \frac{\sin x}{1+3\cos^{2}x}\,dx. Let: I 1 = ∫ 0 π x sin ⁡ x 1 + 3 cos ⁡ 2 x   d x , I 2 = ∫ 0 π sin ⁡ x 1 + 3 cos ⁡ 2 x   d x . I_1=\int_0^\pi \frac{x\sin x}{1+3\cos^2x}\, dx,\qquad I_2=\int_0^\pi \frac{\sin x}{1+3\cos^2x}\, dx. 🔹 Step 1 — Simplify I 1 I_1 ​ using substitution symmetry Let J = ∫ 0 π x sin ⁡ x 1 + 3 cos ⁡ 2 x   d x . J=\int_0^\pi \frac{x\sin x}{1+3\cos^2x}\, dx. Substitute: x = π − t , d x = − d t , sin ⁡ ( π − t ) = sin ⁡ t , cos ⁡ ( π − t ) = − cos ⁡ t . x=\pi - t,\qquad dx=-dt,\qquad \sin(\pi - t)=\sin t,\qquad \cos(\pi - t)=-\cos t. Thus: J = ∫ 0 π ( π − t ) sin ⁡ t 1 + 3 cos ⁡ 2 t   d t . J=\int_0^\pi \frac{(\pi - t)\sin t}{1+3\cos^2 t}\,...

  ❓ Question Why is Al³⁺ stable , Tl⁺ highly stable , and Tl³⁺ a strong oxidising agent? Let’s break the entire concept in one crisp explanation. ✍️ Short Explanation The inert pair effect refers to the reluctance of the ns² electrons (the "inert pair") to participate in bonding for heavy p-block elements . This means as we go down a group, especially in groups 13–16 , the lower oxidation state becomes more stable , and the higher oxidation state becomes less stable . Let's apply this to Group 13: Al → Ga → In → Tl . 🔹 1. Why Al³⁺ is Stable Aluminium is in the 3rd period . Relativistic effects are very small. The 3s² electrons are not inert — they easily participate in bonding. So Al prefers oxidation state +3 (Al³⁺). 👉 Al³⁺ is very stable, does not get reduced easily. 🔹 2. Why Tl⁺ is More Stable than Tl³⁺ Thallium (Tl) is in the 6th period . The 6s² electrons are held tightly due to relativistic contraction. They behave like ...

  ❓ Question The number of paramagnetic metal complex species among the following — that have the same number of unpaired electrons — is asked for: [ Co(NH 3 ) 6 ] 3 + ,    [ Co(C 2 O 4 ) 3 ] 3 − ,    [ MnCl 6 ] 3 − ,    [ Mn(CN) 6 ] 3 − ,    [ CoF 6 ] 3 − ,    [ Fe(CN) 6 ] 3 − ,    [ FeF 6 ] 3 − We must find which of these are paramagnetic and then see how many of them share the same number of unpaired electrons. 🖼️ Question Image ✍️ Short Solution (stepwise) We analyse each complex: General steps: Determine metal oxidation state from complex charge. Write the d-electron count. Decide whether ligand is strong-field (low-spin) or weak-field (high-spin). Count unpaired electrons in octahedral splitting. 1. [ Co(NH 3 ) 6 ] 3 + [ \text{Co(NH}_3)_6 ]^{3+} Co oxidation: + 3 +3  → d 6 d^6 . NH₃ is a borderline/relatively strong ligand for Co³⁺ → typically low-spin . Low-spin d 6 d^6 d 6 → t 2 g 6 e g 0 t_{2g}^6 e_g^0  → 0 unpaired → di...

  ❓ Question One litre buffer solution was prepared by adding 0.10 mol each of NH₃ and NH₄Cl in deionised water. What is the change in pH on addition of 0.05 mol of HCl to the above solution? (Treat the solution volume change on addition as negligible — 1 L final volume.) 🖼️ Question Image ✍️ Short Solution (idea) This is a classical buffer problem . Use the Henderson–Hasselbalch equation for the NH₄⁺/NH₃ buffer: pH = p K a + log ⁡ [ base ] [ acid ]​ For the NH₄⁺/NH₃ system, p K a = 14 − p K b ( NH 3 ) \text{p}K_a = 14 - \text{p}K_b(\text{NH}_3) . Use p K b ( NH 3 ) = 4.75 \text{p}K_b(\text{NH}_3)=4.75  ⇒ p K a ( NH 4 + ) = 9.25 \text{p}K_a(\text{NH}_4^+)=9.25 . Initial moles (in 1 L): n NH 3 = 0.10 →  [ b a s e ] = 0.10   M [{\rm base}]=0.10\ \mathrm{M} n NH 4 + = 0.10 n_{\text{NH}_4^+}=0.10  → [ a c i d ] = 0.10   M [{\rm acid}]=0.10\ \mathrm{M} So initial pH: pH initial = 9.25 + log ⁡ 0.10 0.10 = 9.25 + 0 = 9.25. Add 0.05 mo...

  ❓ Question In Dumas' method , 292 mg of an organic compound released 50.0 mL of nitrogen gas (N₂) at 300 K and 715 mm Hg total pressure. Aqueous tension at 300 K = 15 mm Hg . Find the percentage composition of nitrogen (N) in the organic compound. (Nearest integer) 🖼️ Question Image ✍️ Short Solution Step 1 — Correct the gas pressure for water vapor Total pressure above the collected gas includes water vapor. The partial pressure of dry N₂ is: P N 2 = P total − P H 2 O = 715 − 15 = 700  mm Hg . Step 2 — Use ideal gas law to find moles of N₂ Use P V = n R T. Convert volume to litres:  V = 50.0  mL = 0.0500  L V = 50.0\ \text{mL} = 0.0500\ \text{L} . Use gas constant in mmHg·L units: R = 62.3637   L \ mmHg \ mol − 1 \ K − 1 . So n N 2 = P N 2   V R   T = 700 × 0.0500 62.3637 × 300 . Calculate numerator: 700 × 0.0500 = 35.0 700\times0.0500 = 35.0 . Denominator: 62.3637 × 300 = 18709.11 62.3637\times300 = 18709.11 . Thus n N 2 = 35...

❓ Question 1 M aqueous solutions of each of C u ( N O 3 ) 2 ,   A g N O 3 ,   H g 2 ( N O 3 ) 2 ,   M g ( N O 3 ) 2 \mathrm{Cu(NO_3)_2},\ \mathrm{AgNO_3},\ \mathrm{Hg_2(NO_3)_2},\ \mathrm{Mg(NO_3)_2}  are electrolysed using inert electrodes . Statement (I): With increasing (more negative) cathode potential (i.e. increasing applied voltage), the sequence of deposition of metals on the cathode will be Ag, Hg and Cu . Statement (II): Magnesium will not be deposited at cathode; instead oxygen gas will be evolved at the cathode. Decide which statement(s) are correct and explain. 🖼️ Question Image ✍️ Short Solution Key idea: Which species is reduced at the cathode depends on standard (or actual) reduction potentials. The species with the most positive reduction potential is reduced first (at the least negative cathode potential). As the applied cathode potential is driven more negative, species with lower reduction potentials begin to reduce. 🔹 Standard reduction pot...