20 public questions tagged with this topic.
Cyclopentane (C₅H₁₀) requires 15/2 O₂ (C₅H₁₀ + 15/2 O₂ -> 5CO₂ + 5H₂O), less than pentane’s 8 O₂.
Ref: NCERT Class 11 Chemistry > Chapter 9: Hydrocarbons > Topic: Alkanes - Properties - Conformations and Reactions
Complete combustion of hydrocarbons produces carbon dioxide and water. For C₄H₁₀ (butane): 2C₄H₁₀ + 13O₂ -> 8CO₂ + 10H₂O.
Ref: NCERT Class 11 Chemistry > Chapter 9: Hydrocarbons > Topic: Alkanes - Properties - Conformations and Reactions
For C₄H₁₀ (butane): C₄H₁₀ + 13/2 O₂ -> 4CO₂ + 5H₂O, producing 4 CO₂ and 5 H₂O.
Ref: NCERT Class 11 Chemistry > Chapter 9: Hydrocarbons > Topic: Alkanes - Properties - Conformations and Reactions
5 carbons (from 5 CO₂). CH₃CH₂CH₂CH₂CH₃ (pentane) has 9 σ (8 C-H, 1 C-C from single bonds), no π, and fits C₅H₁₂.
Ref: NCERT Class 11 Chemistry > Chapter 8: Organic Chemistry - Some Basic Principles and Techniques > Topic: Fundamental Concepts - Reaction Mechanism - Fission and Reaction Intermediates
Δ ng = 1 - 1 = 0, RT = 2.48 kJ, Δ H = Δ U + Δ ng RT, Δ U = -393.5 - 0 = -393.5 kJ/mol.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Third Law of Thermodynamics and Applications
Moles = 60 / 46 = 1.304 mol, heat released = 1.304 × 1366.7 = 1782.2 kJ.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: First Law of Thermodynamics and Enthalpy and Internal Energy
Moles = 90 / 180 = 0.5 mol, heat released = 0.5 × 2808 = 1404 kJ.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: First Law of Thermodynamics and Enthalpy and Internal Energy
Reaction: 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O. Moles of C₄H₁₀ ≈ 0.1724 mol. 2 mol produce 8 mol CO₂; 0.1724 mol produce ≈ 0.6896 mol. Volume = 0.6896 × 22.4 ≈ 15.45 L.
Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Stoichiometry and Stoichiometric Calculations and Limiting Reagent
Reaction: 2C₃H₆ + 9O₂ → 6CO₂ + 6H₂O. Moles of C₃H₆ = 15/42 ≈ 0.3571 mol. 2 mol C₃H₆ produce 6 mol CO₂; 0.3571 mol produce ≈ 1.0714 mol. Volume = 1.0714 × 22.4 ≈ 24 L.
Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Stoichiometry and Stoichiometric Calculations and Limiting Reagent
Reaction: 2C₃H₆ + 9O₂ → 6CO₂ + 6H₂O. Moles of C₃H₆ ≈ 0.1857 mol. 2 mol need 9 mol O₂; 0.1857 mol need ≈ 0.8357 mol. Volume = 0.8357 × 22.4 ≈ 18.72 L.
Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Reactions in Solutions and Mass Percentage and Numericals
Mass of C ≈ 0.84 g; mass of H = 0.14 g. Total = 0.98 g. Moles: C = 0.07, H = 0.14; ratio = 1 : 2; empirical formula = CH₂, mass = 14 g/mol. n = 98/14 = 7; molecular formula = C₇H₁₄.
Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Mole Concept and Molar Masses and Percentage Composition
Moles of gas ≈ 0.1339 mol; molar mass ≈ 28 g/mol. Mass of C ≈ 1.44 g; C per molecule ≈ 1. H atoms ≈ 4 (adjusted). Molecular formula = C₂H₄.
Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Mole Concept and Molar Masses and Percentage Composition
