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Ncert Solutions Chemistry Class 12th

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Ncert Solutions Chemistry Class 12th Class 12th

Match List – I with List – II:

List – I (Ore) List – II (Composition)

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Vishal Baghel

Contributor-Level 10

Ore Formula

Siderite FeCO₃

Malachite &n

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Ncert Solutions Chemistry Class 12th Class 12th

Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R.

Assertion (A) : The first ionization enthalpy for oxygen is lower than that of nitrogen

Reason (R) : The four electrons in 2p orbitals of oxygen experience more electron electorn repulsion.

In the light of the above statements, choose the correct answer from the options given below:

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Vishal Baghel

Contributor-Level 10

1st ionization energy of N>O in oxygen atom, 2 electrons out of 4 electron of 2p orbital resulting in an increased electron repulsion.

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Ncert Solutions Chemistry Class 12th Class 12th

The number of oxygen's present in a nucleotide formed from a base, that is present only in RNA is_______________.

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Payal Gupta

Contributor-Level 10

Consider the following image

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Ncert Solutions Chemistry Class 12th Class 12th

Compound 'P' on nitration with dil. HNO₃ yields two isomers (A) and (B). These isomers can be separated by steam distillation. Isomers (A) and (B) show the intramolecular and intermolecular hydrogen bonding respectively. Compound (P) on reaction with conc. HNO₃ yields a yellow compound 'C'___________

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Payal Gupta

Contributor-Level 10

Intermolecular H- bonding and intra-molecular H- bonding producing compound may be the phenol derivatives.

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Ncert Solutions Chemistry Class 12th Class 12th

On complete combustion 0.30g of an organic compound gave 0.20g of carbon dioxide and 0.10 g of water. The percentage of carbon in the given organic compound is __________. (Nearest integer)

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Payal Gupta

Contributor-Level 10

Complete combustion of compound produces 0.2 gm CO₂

Hence wt of carbon in 0.2 gm CO₂

$= (\frac{12}{44} * 0.2) g m$

Therefore % of carbon in compound

$= \frac{w t o f c a r b o n * 100}{w t o f c o m p o u n d} = \frac{12 * 0.2 * 100}{44 * 0.3} = \frac{2400}{44 * 3} = 18.18 \approx 18 %$

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Ncert Solutions Chemistry Class 12th Class 12th

The spin only magnetic moment value of an octhahedral complex among CoCl₃.4NH₃, NiCl₂.6H₂O and PtCl₄.2HCl, which upon reaction with excess of AgNO₃ gives 2 moles of AgCl is____________B. M. (Nearest integer)

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Payal Gupta

Contributor-Level 10

For precipitation of two moles of AgCl

Two Cl will produce as a free anion

CoCl₃.4NH₃ complex will $[C O {(N H_{3})}_{4} C l_{2}]$ Cl (will not give 2Cl)

$P t C l_{4} . 2 H C l \to$ complex will be H2 [PtCl6] will not any Cl

$N i C l_{2} . 6 H_{2} O \to [N i {(H_{2} O)}_{6}] C l_{2}$ will produce two Cl ion.

${[N i {(H_{2} O)}_{6}]}^{+ +} + 2 C l^{-} \to_{A g N O_{3}}^{} 2 A g C l (s)$ precipitate formation

$N i^{2 +} \to [A r] 3 d^{8} 4 s^{0}$

$μ = \sqrt[]{2 (2 + 2)} = \sqrt[]{8} = 2.84 B M = 3$

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Ncert Solutions Chemistry Class 12th Class 12th

2.0g of H₂ gas is adsorbed on 2.5g platinum powder at 300K and 1 bar pressure. The volume of the gas adsorbed per gram of the adsorbent is ______________mL.

(Given : R = 0.083 L bar K^-¹ mol^-¹)

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Payal Gupta

Contributor-Level 10

Volume of H₂adsorbed = $\frac{n R T}{P} = \frac{2 * 0.083 * 300}{2 * 1} = 24.9 l i t = 24900 m l$

Therefore volume of gas adsorbed per gram of the adsorbent = $\frac{24900}{2.5} = 9960$

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Ncert Solutions Chemistry Class 12th Class 12th

A flask is filled with equal moles of A and B. The half lives of A and B are 100s and 50s respectively and are independent of the initial concentration. The time required for the concentration of A to be four times that of B is ______________s.

(Given : In 2 = 0.693)

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Payal Gupta

Contributor-Level 10

Process is based upon simultaneous disintegration hence,

$\frac{0.693}{100} * t = 2.303 l o g_{10} \frac{A_{0}}{A_{t}}$ ………….(i)

and $\frac{0.693}{50} * t = 2.303 l o g_{10} \frac{B_{0}}{B_{t}}$ ………….(ii)

from equation (i) and (ii)

$0.693 t [\frac{1}{50} - \frac{1}{100}] = [l o g \frac{B_{0}}{B_{t}} - l o g \frac{A_{0}}{A_{t}}] * 2.303$

Here; A0 = B0 and $A_{t} = 4 * B_{t}$

Therefore $0.693 t [\frac{1}{100}] = 2.303 [l o g (\frac{B_{0}}{B_{t}} * \frac{A_{t}}{A_{0}})]$

$∴ t = \frac{2.303 * 0.3010 * 2 * 100}{693} = 200 s$

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Ncert Solutions Chemistry Class 12th Class 12th

13.3 Account for the following:

(i) pKb of aniline is more than that of methylamine.

(ii) Ethylamine is soluble in water whereas aniline is not.

(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.

(iv) Although amino group is o– and p– directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.

(v) Aniline does not undergo Friedel-Crafts reaction.

(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines.

(vii) Gabriel phthalimide synthesis is pre

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Ncert Solutions Chemistry Class 12th Class 12th

13.3 Account for the following:

(i) pKb of aniline is more than that of methylamine.

(ii) Ethylamine is soluble in water whereas aniline is not.

(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.

(iv) Although amino group is o– and p– directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.

(v) Aniline does not undergo Friedel-Crafts reaction.

(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines.

(vii) Gabriel phthalimide synthesis is pre

...more

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