Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

  📘 1 Mole = Kitne Particles? Easy Explanation! "1 mole ka matlab sirf ek bada number nahi hai… ye concept JEE ke Stoichiometry ke sawalon ki chaabi hai! Agar aap mole ka matlab samajh gaye, toh chemical calculations bohot aasaan ho jaayengi." 🧪 What is 1 Mole? 1 mole = 6.022 × 10²³ particles Ye number ko Avogadro Number kehte hain. Particles ka matlab kya hota hai? 👉 Atoms, molecules, ions, electrons – jo bhi aap measure kar rahe ho, ussi ke hote hain particles. ⚖️ 1 Mole & Mass ka Connection: Har substance ka ek molar mass hota hai (grams mein). 👉 1 mole = molar mass in grams 💡 Example: Water (H₂O) ka molar mass = 18g/mol So, 18g H₂O = 1 mole = 6.022 × 10²³ molecules of water 🔁 Conversion Trick: Mole Bridge Ye ek shortcut technique hai: Mass ⇄ Moles ⇄ Number of Particles Jaise: Agar aapke paas mass ho → moles nikaalo → phir number of particles Ya particles ho → moles nikaalo → phir mass 📌 Shortcut Formula: No. of part...

  ⚛️ Bohr’s Atomic Model in 59 Seconds | Doubtify JEE Shorts 📌 Question: What explains the fixed energy levels of electrons in an atom and why do atoms emit light? Bohr’s Atomic Model gives us the first quantum explanation of atomic structure! 📸 Image: 📕 Concept & Formula: Stationary Orbits: Electrons revolve around the nucleus in fixed circular paths called orbits. No radiation is emitted in these orbits. Quantized Angular Momentum: m v r = n h 2 π mvr = \frac{nh}{2\pi} Where: m = mass , v = velocity , r = radius , n = orbit number (1, 2, 3…) , h = Planck's constant Energy Transitions: When an electron jumps between orbits: Δ E = E 2 − E 1 = h ν \Delta E = E_2 - E_1 = h\nu Energy is absorbed or emitted in the form of light/spectral lines Energy of Orbit: E n ∝ − 1 n 2 E_n \propto -\frac{1}{n^2} → Lower orbit → More negative energy → More stability 🔢 Example: An electron jumps from n = 3 to n = 2 , and emits energy. This corre...

  ⚗️ Balance the Equation: Mg + HNO₃ ⟶ Mg(NO₃)₂ + N₂O + H₂O ❓ Question: Balance the following chemical equation: Mg + HNO₃ ⟶ Mg(NO₃)₂ + N₂O + H₂O This question appears simple at first glance, but it actually requires deep understanding of redox reactions and balancing complex equations . 🎯 Key Concepts Tested: Oxidation and reduction processes Balancing redox equations using ion-electron method or oxidation number method Conservation of mass and conservation of charge Identifying oxidizing and reducing agents Understanding product formation in acidic medium (presence of HNO₃) 🧠 Step-by-Step Solution Strategy: Step 1: Identify oxidation states Let’s determine oxidation states of all elements in reactants and products: Mg (0) → Mg²⁺ in Mg(NO₃)₂ → Oxidation (Mg is losing electrons) N in HNO₃: +5 N in N₂O: +1 → Reduction (N is gaining electrons) Step 2: Write unbalanced half-reactions Oxidation (Mg): Mg → Mg²⁺ + 2e⁻ Reduction (N): 2...

🧪 Question: The ratio of mass percent of C and H of an organic compound (CxHyOz) is 6:1. If one molecule of the above compound contains half as much oxygen as required to burn one molecule of compound CxHy completely to CO₂ and H₂O, then what is the empirical formula of the compound? 📌 Why This Question is Important: This question combines two highly testable concepts in JEE Chemistry – combustion analysis and empirical formula derivation . You’re not only working with mass percent ratio , but also applying oxygen balancing logic in a combustion reaction. Such questions are frequently seen in organic chemistry chapters , particularly under Basic Principles of Organic Chemistry and Stoichiometry . Mastering this helps you crack more complex analysis-based questions in JEE Mains and even Advanced . 💡 Concept Used: To find the empirical formula , we first convert the mass percent ratio of C and H into moles , assuming 100 g of substance. Then we apply a stoichiometric appr...

🧮 Question: 100 grams of propane is completely reacted with 1000 grams of oxygen. The mole fraction of carbon dioxide in the resulting mixture is X × 10⁻². The value of x is: 🖼️ Question Image:

  🧪 Balance the Ionic Equation in Alkaline Medium: Cr(OH)₃ + IO₃⁻ → I⁻ + CrO₄²⁻ | Doubtify JEE 📌 Question: Balance the following ionic equation in alkaline medium : Cr(OH)₃ + IO₃⁻ → I⁻ + CrO₄²⁻ 🖼️ Question Image: 🧠 Solution (Image): 🎥 Video Solution: 🔍 Why this Question is Important: This is a frequently asked question from Redox Reactions , commonly seen in JEE Mains and Advanced exams. Knowing how to balance redox reactions in alkaline medium is a core skill every aspirant must master. 📩 Have a Doubt? Drop us a mail at doubtifyqueries@gmail.com or DM us on Instagram . Stay consistent. Watch – Pause – Solve – Repeat. Team Doubtify is with you till the finish line! 💪