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Chemistry Chemistry Some Basic Concepts of Chemistry

Compound A contains 8.7% Hydrogen, 74% Carbon and 17.3% Nitrogen. The molecular formula of the compound is,

Given : Atomic masses of C,H and N are 12, 1 and 14 amu respectively. The molar mass of the compound A is 162 g mol^-1.

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alok kumar singh

Contributor-Level 10

Empirical formula is C₅H₇N

Empirical mass = 81

Molecular mass = 162

So, molecular formula is C₁₀H₁₄N₂

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a year ago

Chemistry 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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a year ago

Chemistry 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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a year ago

Chemistry 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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a year ago

Chemistry 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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a year ago

Chemistry 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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a year ago

Chemistry 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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a year ago

Chemistry 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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a year ago

Chemistry Class 11th

50 mL of 0.1 M CH₃COOH is being titrated against 0.1 M NaOH. When 25 mL of NaOH has been added, the pH of the solution will be_________* 10^-². (Nearest integer)

(Given : pKa (CH₃-COOH) = 4.76)

log 2 = 0.30

log 3 = 0.48

log 5 = 0.69

log 7 = 0.84

log 11 = 1.04

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Payal Gupta

Contributor-Level 10

Here, total meq of acetic acid = 50 * 0.1 = 5

And total meq of NaOH = 25 * 0.1 = 2.5

After neutralization process

Meq of left acetic acid = 2.5

And meq of formed CH₃COONa = 2.5

$p H = p K a + l o g_{10} \frac{[S]}{[A]}$

$p H = 4.76 + l o g_{10} \frac{2.5}{2.5} = 4.76 = 476 * 1 0^{- 2}$

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New answer posted

a year ago

Chemistry Structure of Atom

A 0.5 percent solution of potassium choride was found to freeze at -0.24°C. The percentage dissociation of potassium chloride is__________. (Nearest integer)

(Molal depression constant for water is 1.80 K kg mol^-¹ and molar mass of KCl is 74.6 g mol^-¹)

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Payal Gupta

Contributor-Level 10

0.5 % KCl solution has molality (m) = $\frac{0.5 * 1000}{74.5 * 99.5}$

$K C l_{(a q)} ? K_{(a g)}^{+} + C l_{(a q)}^{-}$

1 - α α α

And I = $(1 - α + α + α) = 1 + α$

$i = \frac{Δ T f}{k f * m} = \frac{0.24 * 74.5 * 99.5}{1.8 * 0.5 * 1000} = 1 + α$

1.976 = 1 + α

$∴ α = 0.976$

% = 97.6%

the nearest 98.

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