12 public questions tagged with this topic.
Cr 2 O 7 2– oxidizes H 2 S to S (yellow precipitate) and reduces to Cr 3+ (green).
Ref: NCERT Class 12 Chemistry > Chapter 4: d and f Block Elements > Topic: Oxidation States and Colour and Magnetic Properties of d-Block
Three molecules collide → molecularity = 3.
Ref: NCERT Class 12 Chemistry > Chapter 3: Chemical Kinetics > Topic: Temperature Dependence - Arrhenius Equation and Activation Energy
Δ[B]/Δt = (1/2) × (−Δ[A]/Δt) = 0.02 mol L⁻¹ min⁻¹.
Ref: NCERT Class 12 Chemistry > Chapter 3: Chemical Kinetics > Topic: Pseudo First Order and Complex Reactions
−Δ[A]/Δt = 2 × (0.16/4) = 0.08 mol L⁻¹ min⁻¹.
Ref: NCERT Class 12 Chemistry > Chapter 3: Chemical Kinetics > Topic: Collision Theory - Effective Collisions and Catalyst Effect
Δ[C]/Δt = 3 × (−Δ[B]/Δt) = 0.09 mol L⁻¹ min⁻¹.
Ref: NCERT Class 12 Chemistry > Chapter 3: Chemical Kinetics > Topic: Collision Theory - Effective Collisions and Catalyst Effect
S (-2 in H₂S) becomes 0 in S, losing electrons.
Ref: NCERT Class 11 Chemistry > Chapter 7: Redox Reactions > Topic: Redox Titrations and Equivalent Weight and Normality
S (+6 in FeSO₄) becomes +4 in SO₂ and +6 in SO₃, but Fe (+2 to +3) is oxidized.
Ref: NCERT Class 11 Chemistry > Chapter 7: Redox Reactions > Topic: Applications of Redox Reactions and Disproportionation
Let PC = 2x , PA = 1 - x , PB = 2 - 3x . Kp = ((PC)²/PA (PB)³) = ((2x)²/(1 - x)(2 - 3x)³) = 0.125 , 4x² = 0.125 (1 - x)(2 - 3x)³ , x ≈ 0.25 , PC = 0.5 atm .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier
Let PC = 2x , PA = 2 - 2x , PB = 1 - x , total pressure = 3 - x . Kp = ((PC)²/PA² PB) = ((2x)²/(2 - 2x)² (1 - x)) = 16 , (4x²/4(1 - x)² (1 - x)) = 16 , (x²/(1 - x)³) = 4 , (x/1 - x) = 2 , x = 2 - 2x , 3x = 2 , x = (2/3) , PC = 2 × (2/3) = 1.33 atm .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Physical Equilibrium - Solid-Liquid Gas-Liquid and Henry's Law
Initial: [A] = 1 M , [B] = 3 M , [C] = 0 . Let 3x be moles of C formed, so A decreases by x , B by 2x . At equilibrium: [A] = 1 - x , [B] = 3 - 2x , [C] = 3x . Kc = ([C]³/[A][B]²) = ((3x)³/(1 - x)(3 - 2x)²) = 27 , (27x³/(1 - x)(3 - 2x)²) = 27 , (x³/(1 - x)(3 - 2x)²) = 1 . Solving, x³ = (1 - x)(3 - 2x)² , test x = 0.5 : (0.5)³ = 0.125 , (1 - 0.5)(3 - 1)² = 0.5 × 4 = 2 (not equal). Solving numerically, x ≈ 0.75 , [C] = 3 × 0.75 = 2.25 M .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Physical Equilibrium - Solid-Liquid Gas-Liquid and Henry's Law
Initial: [H₂] = 0.2 M , [I₂] = 0.3 M , [HI] = 0 . Let 2x = [HI] , [H₂] = 0.2 - x , [I₂] = 0.3 - x . Kc = ([HI]²/[H₂][I₂]) = ((2x)²/(0.2 - x)(0.3 - x)) = 50 , 4x² = 50 (0.06 - 0.5x + x²) , 46x² - 25x + 3 = 0 , x ≈ 0.15 , [HI] = 0.3 M .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Ionic Equilibrium - Acids Bases and pH and Ionization of Weak Acids Bases
Condensation reactions involve the removal of water to form bonds, such as peptide and glycosidic bonds.
Ref: NCERT Class 11 Biology Chapter 9: Biomolecules Chemical Analysis and Biomolecules - Amino Acids Lipids
