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6 months ago

Chemistry Class 12th

Match List -I with List -II

List -I (Ores)
(a) Haematite
(b) Magnetite
(c) Calamine
(d) Kaolinite

List-II (Composition)
(i) Fe₃O₄
(ii) ZnCO₃
(iii) Fe₂O₃
(iv) [Al₂(OH)₄Si₂O₅]

Choose the correct answer from the options given below:

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Vishal Baghel

Contributor-Level 10

Hematite: Fe? O?
Magnetite: Fe? O?
Calamine: ZnCO?
Kaolinite: [Al? (OH)? Si? O? ]

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6 months ago

Chemistry Class 12th

The product formed from the following reaction sequence is:

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Vishal Baghel

Contributor-Level 10

kc = [O? ]² / [O? ]³
3 * 10? = [O? ]² / [4 * 10? ²]³
[O? ]² = 192 * 10? ⇒ [O? ] = 4.38 * 10? ³²

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6 months ago

Chemistry Equilibrium

3O₂(g) ⇌ 2O₃(g)
For the above reaction at 298 K, Kc is found to be 3.0 * 10⁻⁵⁹. If the concentration of O₂ at equilibrium is 0.040M the then concentration of O₃ in M is:

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Vishal Baghel

Contributor-Level 10

3O? 2O?

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New answer posted

6 months ago

Chemistry Chemistry Chemical Kinetics

For a first order reaction A → Products, initial concentration of A is 0.1M, which becomes 0.001 after 5 minutes. Rate constant for the reaction min^-1 is:

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Vishal Baghel

Contributor-Level 10

k = (2.303 / 5) log (0.1 / 0.001) = (2.303 * 2) / 5 = 4.606 / 5 = 0.921 min? ¹

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6 months ago

Chemistry Class 12th

Which one of the following is not formed where acetone reacts with 2-pentanone in the presence of dilute NaOH followed by heating?

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Vishal Baghel

Contributor-Level 10

When acetone & 2-pentanone are added in a base for an aldol condensation reaction there are several possibilities that may arise for product formation.

But structure in choice (2) is not possible. It will be possible with 3-pentanone.

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6 months ago

Chemistry Chemistry Electrochemistry

At 298 K, the standard electrode potentials of Cu²⁺/Cu, Zn²⁺ /Zn, Fe²⁺ /Fe and Ag⁺ /Ag are 0.34 V, -0.76 V, -0.44 V and 0.80 V, respectively. On the basis of standard electrode potential, predict which of the following reaction can not occur?

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Vishal Baghel

Contributor-Level 10

Silver will not be able to reduce Cu²? to Cu.

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6 months ago

Chemistry Equilibrium

What mass of 95% pure CaCO₃ will be require neutralise 50 mL of 0.5M HCl solution according to the following reaction?
CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + 2H₂O(l) [Calculate upto second place of decimal point]

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Vishal Baghel

Contributor-Level 10

Moles of HCl = (50 / 1000) * 0.5 = 0.025 moles
So, moles of CaCO? used = 0.025 / 2 = 0.0125 moles = 1.25 g
95% (Total mass of CaCO? ) = 1.25 g
Total mass of CaCO? = 1.25 / 0.95 = 1.32 g

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6 months ago

Chemistry Class 12th

Which statement regarding polymers is not correct?

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