CHEMISTRY

CHEMISTRY

A compound burns in oxygen to produce 8.8 g of CO₂ and 3.6 g of H₂O. What is its empirical formula? (Atomic masses: C =

Given: A compound burns in oxygen to produce 8.8 g of CO₂ and 3.6 g of H₂O. What is its empirical formula? (Atomic masses: C = 12, H = 1, O = 16) These values define the system as per NCERT data. Formula: Mass of C = (12 / 44) × 8.8 = 2.4 g. This is standard NCERT relation. Substitution & Calculation: Mass of H = (2 / 18) × 3.6 = 0.4 g. Moles: C = 2.4 / 12 = 0.2, H = 0.4 / 1 = 0.4. Ratio: 0.2 / 0.2 : 0.4 / 0.2 = 1 : 2. Empirical formula = CH₂. Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.

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CHEMISTRY

Which ion among the following has a 3d⁶ configuration?

Given: Which ion among the following has a 3d⁶ configuration? These values define the system as per NCERT data. Formula: Fe (Z = 26) has 3d⁶ 4s². This is standard NCERT relation. Substitution & Calculation: For Fe^{2+, 2 electrons from 4s are lost, leaving 3d⁶ . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.

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CHEMISTRY

The initial rate of a reaction is 1.2 × 10⁻³mol L^{-1 s^{-1 when [A] = 0.2 M and [B] = 0.3 M. If the order is 1 with res

Given: The initial rate of a reaction is 1.2 × 10⁻³mol L^{-1 s^{-1 when [A] = 0.2 M and [B] = 0.3 M. If the order is 1 with respect to A and 1 with respect to B, what is k in L mol^{-1 s^{-1 ? These values define the system as per NCERT data. Formula: k = fracRate[A][B] = frac1.2 × 10⁻³⁰.2 × 0.3 = 0.02 .. This is standard NCERT relation. Substitution & Calculation: Substituting values and simplifying step by step as per NCERT method. Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.

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CHEMISTRY

For the reaction 1/2H₂(g) + 1/2Cl₂(g) -> HCl(g), if Δ_f H° of HCl(g) is -92.32 kJ/mol, what is Δ_r H?

Given: For the reaction 1/2H₂(g) + 1/2Cl₂(g) -> HCl(g), if Δ_f H° of HCl(g) is -92.32 kJ/mol, what is Δ_r H? These values define the system as per NCERT data. Formula: Δ_r H = [-92.32] - [0 + 0] = -92.32 kJ/mol.. This is standard NCERT relation. Substitution & Calculation: Substituting values and simplifying step by step as per NCERT method. Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.

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CHEMISTRY

The K_{sp of AgI is 8.3 × 10⁻¹⁷. What is its solubility in 0.01 M NaI ?

Given: The K_{sp of AgI is 8.3 × 10⁻¹⁷. What is its solubility in 0.01 M NaI ? These values define the system as per NCERT data. Formula: K_{sp = [Ag+][I-], [I-] = 0.01, 8.3 × 10⁻¹⁷= S · 0.01. This is standard NCERT relation. Substitution & Calculation: S = 8.3 × 10⁻¹⁵M. Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.

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CHEMISTRY

What is the EMF of the ll Zn(s) | Zn²⁺(0.1 M) || Cu²⁺(0.01 M) | Cu(s) at 298 K, given E_{Zn^{2+/Zn⁰ = -0.76 V and E_{Cu^

Given: What is the EMF of the ll Zn(s) | Zn²⁺(0.1 M) || Cu²⁺(0.01 M) | Cu(s) at 298 K, given E_{Zn^{2+/Zn⁰ = -0.76 V and E_{Cu^{2+/Cu⁰ = 0.34 V ? These values define the system as per NCERT data. Formula: E_{ll⁰ = 0.34 - (-0.76) = 1.10 V, n = 2. This is standard NCERT relation. Substitution & Calculation: E_{ll = E_{ll⁰ - 0.059/2 log frac[Zn^{2+][Cu^{2+] = 1.10 - 0.059/2 log 0.1/0.01 . Q = 10, E_{ll = 1.10 - 0.0295 × 1 = 1.0705 V approx 1.07 V . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.

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CHEMISTRY

In neutral or faintly alkaline medium, KMnO₄ oxidizes I^- to:

In neutral or faintly alkaline conditions, 2MnO₄^- + H₂O + I^- -> 2MnO₂ + 2OH^- + IO₃^-, producing iodate ( IO₃^- ). This follows from NCERT principle where relation explains outcome clearly for students.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.

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CHEMISTRY

What is the heat released when 13.5 g of ethanol burns completely? (Δ H° = -1367 kJ/mol, molar mass of C₂H₅OH = 46 g/mol

Given: What is the heat released when 13.5 g of ethanol burns completely? (Δ H° = -1367 kJ/mol, molar mass of C₂H₅OH = 46 g/mol) These values define the system as per NCERT data. Formula: Moles = 13.5 / 46 = 0.2935 mol. This is standard NCERT relation. Substitution & Calculation: Heat = 0.2935 × 1367 = 401.2 kJ. Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.

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CHEMISTRY

The oxidation state of sulfur in SO_3 is:

Given: The oxidation state of sulfur in SO_3 is: These values define the system as per NCERT data. Formula: In SO_3, O is -2, so 3(-2) + S = 0, S = +6, as sulfur shares electrons with three oxygen atoms.. This is standard NCERT relation. Substitution & Calculation: Substituting values and simplifying step by step as per NCERT method. Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.

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CHEMISTRY

A zero-order reaction has a rate constant of 5.0 × 10⁻³mol L^{-1 min^{-1 . How long will it take for 0.25 M to reduce to

Given: A zero-order reaction has a rate constant of 5.0 × 10⁻³mol L^{-1 min^{-1 . How long will it take for 0.25 M to reduce to 0.1 M? These values define the system as per NCERT data. Formula: t = frac[R]_0 - [R]k = frac0.25 - 0.15.0 × 10⁻³= 30 min .. This is standard NCERT relation. Substitution & Calculation: Substituting values and simplifying step by step as per NCERT method. Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.

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