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  ⚗️ 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...

🔥 Combustion Analysis – Percentage of Oxygen in an Organic Compound | JEE Chemistry | Doubtify JEE

🧪 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:

     🧲 Solutions  | Chemistry  |  Doubtify JEE 🧮 Question: The density of the NaOH solution is 1.2 g/cm³. What is the molality of the solution? 🖼️ Question Image: ⚗️ Concept Overview: This is a fundamental problem from the Solutions chapter of Physical Chemistry. It tests your understanding of molality , density , and the method of converting between different concentration units. In many JEE Main and Advanced problems, students get confused between molality (mol/kg) and molarity (mol/L) . This problem helps to eliminate that confusion and improve conceptual clarity. 🧠 Step-by-Step Explanation: Let’s assume we are working with 1000 cm³ = 1 L of NaOH solution. Density = 1.2 g/cm³ So, total mass of 1 L solution = 1.2   g/cm³ × 1000   cm³ = 1200   g 1.2 \, \text{g/cm³} \times 1000 \, \text{cm³} = 1200 \, \text{g} Let’s say the solution is x% NaOH by mass (you may be given the percentage or molarity in the full version of the question – for no...

🧪 Redox Reactions | Chemistry |  JEE Mains 📌 Question: Balance the equation by ion-electron method in basic medium: Zn + NO 3 − → Zn 2 + + NH 4 +​ This is a classic redox equation where zinc is oxidized and nitrate is reduced. The balancing must be done using the ion-electron method under basic medium conditions . 🖼️ Question Image: 🧠 Solution (Image): 🎥 Video Solution: 🔍 Why this Question is Important: This question is frequently asked in JEE Mains and Advanced , especially in the chapter Redox Reactions . It tests key skills like: Identifying oxidation and reduction half-reactions Using the ion-electron method properly in basic medium Applying concepts of oxidation states, electrons, and balancing H₂O/OH⁻ ions Understanding this concept is crucial for solving more advanced redox problems in competitive exams. 📩 Have a Doubt? Drop us a mail at  doubtifyqueries@gmail.com  or DM us on  Instagram . Stay consistent. Watch – Pause – Solve – Repeat. ...