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CHEMISTRY

Latest questions in this category.

45 questions

CHEMISTRY

The number of unpaired electrons in Ti^{3+ (Z = 22) is:

Given: The number of unpaired electrons in Ti^{3+ (Z = 22) is: These values define the system as per NCERT data. Formula: Ti: 3d² 4s². This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Ti^{3+ : 3d¹ (loses 2 from 4s, 1 from 3d). 1 unpaired electron. 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: Some Basic Concepts, Structure of Atom and Periodicity, Topic: Mole concept, atomic models and periodic trends.

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CHEMISTRY

Which of the following molecules has a bond order of 1 according to molecular orbital theory?

Given: Which of the following molecules has a bond order of 1 according to molecular orbital theory? These values define the system as per NCERT data. Formula: Bond order = 1/2 (10 - 8) = 1 .. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: For F₂ : (sigma 1s)² (sigma^* 1s)² (sigma 2s)² (sigma^* 2s)² (sigma 2p_z)² (π 2p_x)² (π 2p_y)² (π^* 2p_x)² (π^* 2p_y)² . 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: Some Basic Concepts, Structure of Atom and Periodicity, Topic: Mole concept, atomic models and periodic trends.

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CHEMISTRY

How many pi bonds are present in the Câ‚‚Hâ‚‚ molecule?

Câ‚‚Hâ‚‚ has a triple bond between carbons (1 sigma, 2 pi), totaling 2 pi bonds. This follows from NCERT principle where the relation explains the outcome clearly for students in simple steps.

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

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CHEMISTRY

What is the ll potential of a hydrogen electrode in a solution with pH = 4 at 298 K, given E⁰ = 0 V ?

Given: What is the ll potential of a hydrogen electrode in a solution with pH = 4 at 298 K, given E⁰ = 0 V ? These values define the system as per NCERT data. Formula: For H+ + e- -> 1/2 H₂, E = E⁰ - 0.059/1 log frac1[H^+]. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: [H^+] = 10⁻⁴, E = 0 - 0.059 × 4 = -0.236 V . 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: Solutions, Topic: Concentration terms and colligative properties.

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CHEMISTRY

Which molecule has a double covalent bond formed by sharing four electrons between different atoms?

In Oâ‚‚, two oxygen atoms share two pairs of electrons (4 electrons), forming a double covalent bond between identical atoms, but CO has a triple bond, Hâ‚‚ a single bond, and HI a single bond. The question specifies "different atoms, " but among options, Oâ‚‚ is the closest fit for a double bond.

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

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CHEMISTRY

A gaseous reaction follows Rate = k p_A² with k = 0.2 atm^{-1 s^{-1 and p_A = 0.5 atm . What is the rate in atm s^{-1 ?

Given: A gaseous reaction follows Rate = k p_A² with k = 0.2 atm^{-1 s^{-1 and p_A = 0.5 atm . What is the rate in atm s^{-1 ? These values define the system as per NCERT data. Formula: Rate = k p_A² = 0.2 × (0.5)² = 0.2 × 0.25 = 0.05 .. 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

Calculate the wavenumber of light with wavelength 5800 Å . (1 Å = 10 ⁻¹⁰ m)

Given: Calculate the wavenumber of light with wavelength 5800 Å . (1 Å = 10 ⁻¹⁰ m) These values define the system as per NCERT data. Formula: Wavenumber barnu = 1/lambda. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: lambda = 5800 Å = 5800 × 10⁻¹⁰ m = 5.8 × 10⁻⁷ m . barnu = frac15.8 × 10⁻⁷= 1.724 × __10POW₆__m^{-1 . 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

What is the entropy change when two gases mix isothermally?

Mixing of gases increases disorder due to increased randomness, so Δ S > 0. This follows from NCERT principle where the relation explains the outcome clearly for students in simple steps.

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

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CHEMISTRY

A reaction has k = 1.5 × 10⁻⁴ s^{-1 at 300 K and E_a = 54 kJ mol^{-1 . What is the pre-exponential factor A in s^{-

Given: A reaction has k = 1.5 × 10⁻⁴ s^{-1 at 300 K and E_a = 54 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: 1.5 × 10⁻⁴= A e^{-54000/(8.314 × 300), A = 1.5 × 10⁻⁴/ e^{-21.63 = 4.2 × __10POW₅__. 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 Henry's law constant for a gas in water is 8 × __10POW₃__bar at 298 K. What is the mole fraction of the gas under

Given: The Henry's law constant for a gas in water is 8 × __10POW₃__bar at 298 K. What is the mole fraction of the gas under a partial pressure of 0.4 bar? These values define the system as per NCERT data. Formula: p = K_H · x. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: 0.4 = 8 × __10POW₃__ · x . x = 0.4/8 × __10POW₃__= 5 × 10⁻⁵. 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: Some Basic Concepts, Structure of Atom and Periodicity, Topic: Mole concept, atomic models and periodic trends.

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