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Class 12th Chemistry

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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Class 12th Chemistry

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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Class 12th Physics

Two point charges A and B of magnitude + 8 * 10^-6 and -8 * 10^-6C respectively are placed at a distance d apart. The electric field at the middle point O between the charges is 6.4 * 10⁴ NC^-1. The distance 'd' between the point charges A and B is:

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alok kumar singh

Contributor-Level 10

E₁ (done to q₁) = $\frac{k, q_{1}}{{(d / 2)}^{2}} = \frac{[9 * 1 0^{9} * 8 * 1 0^{? 6}]}{d^{2}} * 4$

E₂ (done to q₂) = $\frac{k ? q_{2}}{{(d / 2)}^{2}} = \frac{[9 * 1 0^{9} * 8 * 1 0^{? 6}]}{d^{2}} * 4$

Enet = E₁+ E₂

$= 2 [\frac{4 * 9 * 1 0^{9} * 8 * 1 0^{? 6}}{d^{2}}]$

Given Enet = 6.4 * 104

$? 2 * [\frac{4 * 9 * 1 0^{9} * 8 * 1 0^{? 6}}{d^{2}}] = 6.4 * 1 0^{4}$

d² = 9 * 10^-6 + 6

d= 3m

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Class 12th Chemistry

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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Class 12th Chemistry

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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Class 12th Chemistry Chemical Kinetics

The spin- only magnetic moment value of the most basic oxide of vanadium among V₂O₃, V₂O₄ and V₂O₅ is _____________B. M (Nearest integer)

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Payal Gupta

Contributor-Level 10

Most basic oxide V₂O3

Here V has +3 O.S. Hence V⁺³ $[A r] 3 d^{2}$

two unpaired e^- in d- subshell

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

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Class 12th Chemistry

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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Class 12th Chemistry

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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Class 12th Chemistry NCERT Exemplar Solutions Class 12th Chapter Two

The moles of methane required to produce 81 g of water after complete combustion is________* 10^-2 mol. (nearest integer)

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Vishal Baghel

Contributor-Level 10

$C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$

2 moles of water produced by 1 mole of methane

Or 36 gm of water produced by 1 mole of CH₄

$∴$ 81 gm of water produced = $\frac{1 * 81}{36}$ = 2.25 mole of CH₄

Mole of CH4 required = 225 * 10^-2

the nearest integer = 225

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Class 12th Chemistry NCERT Exemplar Solutions Class 12th Chapter Two

Reaction of [Co(H₂O)₆]²⁺ with excess ammonia and in the presence of oxygen results into a diamagnetic product. Number of electrons present in t_2g-orbitals of the product is_________.

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Vishal Baghel

Contributor-Level 10

${[C o {(H_{2} O)}_{6}]}^{2 +} + N H_{3 (e x c e s s)} \to \underset{D i a m a g n e t i c n a t u r e}{{[C o {(N H_{3})}_{6}]}^{3 +} + 6 H_{2} O}$

$C o^{3 +} \to 3 d^{6} 4 s^{0}$

$\Rightarrow {t_{2 g}}^{6} e g^{0}$ (form of paired electron)

No. of electron in t_2g = 6

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