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64 public questions tagged with this topic.

CHEMISTRY

The initial rate of a reaction is 2.0 × 10⁻³ mol L^{-1 s^{-1 when [A] = 0.5 M and [B] = 0.2 M. If the rate law is Ra

Given: The initial rate of a reaction is 2.0 × 10⁻³ mol L^{-1 s^{-1 when [A] = 0.5 M and [B] = 0.2 M. If the rate law is Rate = k[A][B], what is k in L mol^{-1 s^{-1 ? These values define the system as per NCERT data. Formula: k = fracRate[A][B] = frac2.0 × 10⁻³⁰.5 × 0.2 = 0.02 .. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Substituting values like 1.2 × 10⁻⁵, 236 J kg⁻¹ K⁻¹, CH₃CH₂NH₂ etc. into the formula and simplifying step by step. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Relevant Chemistry topic covering principles and examples as per NCERT.

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CHEMISTRY

For the reaction 2A + 3B -> C, the concentration of A decreases from 0.14 M to 0.10 M in 20 seconds. What is the average

Given: For the reaction 2A + 3B -> C, the concentration of A decreases from 0.14 M to 0.10 M in 20 seconds. What is the average rate of formation of C in mol L^{-1 s^{-1 ? These values define the system as per NCERT data. Formula: Rate = -1/2 fracΔ[A]Δ t = -1/3 fracΔ[B]Δ t = fracΔ[C]Δ t. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Δ[A] = 0.10 - 0.14 = -0.04 mol L^{-1, Δ t = 20 s . -fracΔ[A]Δ t = 2.0 × 10⁻³, so fracΔ[C]Δ t = frac2.0 × 10⁻³²= 1.0 × 10⁻³. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Relevant Chemistry topic covering principles and examples as per NCERT.

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CHEMISTRY

For the reaction 3A -> 2B + C, the concentration of A decreases from 0.27 M to 0.24 M in 15 seconds. What is the average

Given: For the reaction 3A -> 2B + C, the concentration of A decreases from 0.27 M to 0.24 M in 15 seconds. What is the average rate of formation of B in mol L^{-1 s^{-1 ? Formula: Rate = -1/3 fracΔ[A]Δ t = 1/2 fracΔ[B]Δ t = fracΔ[C]Δ t. Substitution & Calculation: Δ[A] = 0.24 - 0.27 = -0.03 mol L^{-1, Δ t = 15 s . -fracΔ[A]Δ t = 2.0 × 10⁻³, so fracΔ[B]Δ t = 2/3 × 2.0 × 10⁻³= 1.33 × 10⁻³. Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Chemistry Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII)Topic: Mole concept, atomic structure, chemical formulas like H₂O, CO₂, CH₃CH₂NH₂ and periodic trends.

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CHEMISTRY

A reaction has k = 6.0 × 10⁻⁶ s^{-1 at 300 K and E_a = 50 kJ mol^{-1 . What is the pre-exponential factor A in s^{-

Given: A reaction has k = 6.0 × 10⁻⁶ s^{-1 at 300 K and E_a = 50 kJ mol^{-1 . What is the pre-exponential factor A in s^{-1 ? (R = 8.314 J/mol · K) These values define the system as per NCERT data. Formula: k = A e^{-E_a/RT. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: 6.0 × 10⁻⁶= A e^{-50000/(8.314 × 300), A = 6.0 × 10⁻⁶/ e^{-20.06 = 3.2 × __10POW₃__. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Relevant Chemistry topic covering principles and examples as per NCERT.

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CHEMISTRY

The rate constant of a reaction is 3.0 × 10⁻³ s^{-1 at 25°C and E_a = 62 kJ mol^{-1 . What is the rate constant at

Given: The rate constant of a reaction is 3.0 × 10⁻³ s^{-1 at 25°C and E_a = 62 kJ mol^{-1 . What is the rate constant at 35°C? (R = 8.314 J/mol · K) These values define the system as per NCERT data. Formula: log frack_23.0 × 10⁻³= 62000/2.303 × 8.314 ( 10/298 × 308 ) = 0.352. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: k_2 = 3.0 × 10⁻³ × __10POW₀__.352 = 6.75 × 10⁻³ approx 6.8 × 10⁻³. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Relevant Chemistry topic covering principles and examples as per NCERT.

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PHYSICS

A solid sphere of mass 1 kg and radius 0.5 m has a kinetic energy of 5 J . What is its angular speed?

Given: A solid sphere of mass 1 kg and radius 0.5 m has a kinetic energy of 5 J . What is its angular speed? These values define the system as per NCERT data. Formula: I = 2/5 M R² = 2/5 × 1 × (0.5)² = 0.1 kg m². This is standard NCERT relation. Substitution & Calculation: K = 1/2 I omega² Rightarrow 5 = 1/2 × 0.1 × omega² Rightarrow 10 = 0.1 omega² Rightarrow omega² = 100 Rightarrow omega = 10 rad/s . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

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CHEMISTRY

A reaction has the rate law Rate = k[A][B]⁰ . If [A] is increased by a factor of 4, what happens to the rate?

Given: A reaction has the rate law Rate = k[A][B]⁰ . If [A] is increased by a factor of 4, what happens to the rate? Formula: Rate = k[A], new rate = k(4[A]) = 4k[A]. Substitution & Calculation: Rate increases by a factor of 4. Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Chemistry Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII)Topic: Mole concept, atomic structure, chemical formulas like H₂O, CO₂, CH₃CH₂NH₂ and periodic trends.

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