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Class 12th Surface Chemistry

Consider the ions /molecule $O_{2}^{+}, O_{2}, O_{2}^{-}, O_{2}^{2 -}$

For increasing bond order the correct option is

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Payal Gupta

Contributor-Level 10

$B o n d o r d e r = \frac{e^{-} o f B o n d i n g M O - e^{-} o f A B M O}{2}$

Hence Bond order for $O_{2}^{+} = \frac{10 - 5}{2} = 2.5$

$O_{2} = \frac{10 - 6}{2} = 2.0$

$O_{2}^{-} = \frac{10 - 7}{2} = 1.5$

$O_{2}^{- -} = \frac{10 - 8}{2} = 1.0$

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Class 12th Chemical Bonding and Molecular Structure

If the radius of the 3^rd Bohr's orbit of hydrogen atom r₃ and the radius of 4^th Bohr's orbit is r₄. Then

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Payal Gupta

Contributor-Level 10

Radius of Bohr's orbit $R_{n} = 0.529 \overset{o}{A} * \frac{n^{2}}{z}$

Radius of Bohr's orbit for hydrogen, $R_{n} = 0.529 \overset{o}{A} * n^{2}$

For third orbit (R₃) = $0.529 \overset{o}{A} * 9$ = r₃ and

Fourth orbit (R₄) = $0.59 \overset{o}{A} * 16 = r_{4}$

$∴ r_{4} = \frac{16 r_{3}}{9}$

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Class 12th Classification of Elements and Periodicity in Prop

An evacuated glass vessel weights 40.0g when empty, 135.0g when filled with a liquid of density 0.95 g mL^-¹ and 40.5 g when filled with an ideal gas at 0.82 atm at 250 K. The molar mass of the gas in g mo^-¹ is:

(Given: R = 0.082 L atm K^-¹ mol^-¹)

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Payal Gupta

Contributor-Level 10

Wt of liquid = 135 – 40 = 95 gm

Volume of liquid = $\frac{w t .}{d e n s i t y} = \frac{95}{0.95} = 100 m l$

Hence volume of vessel = 100 ml = 0.1 lit from ideal gas equation,

$M = \frac{d R T}{P} = \frac{0.5}{0.1} * \frac{0.0821 * 250}{0.82}$

$∴ M = 125 g / m o l$

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Class 12th Some Basic Concepts of Chemistry

A commercially sold conc. HCl is 35% HCl by mass. If the density of this commercial acid is 1.46 g/mL, the molarity of this solution is

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Payal Gupta

Contributor-Level 10

$M o l a l i t y = \frac{m o l e s o f s o l u t e * 1000}{w t o f s o l v e n t (g m)}$

$M o l a l i t y = \frac{35 * 1000 * 1.46}{36.5 * 100}$ = 14.0 M

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Class 12th Dual Nature of Radiation and Matter

An ideal fluid of density 800kgm^-3, flows smoothly through a bent pipe (as shown in figure) that tapers in cross-sectional area from a to $\frac{a}{2} .$ The pressure difference between the wide and narrow sections of pipe is 4100 Pa. At wider section, the velocity of fluid is $\frac{\sqrt[]{x}}{6} m s^{- 1}$ for x =________. (Given g = 10 ms^-²)

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Payal Gupta

Contributor-Level 10

$ρ = 800 k g / m^{3}$

P₁ – P₂ = 4100 Pa

Bernoulli's question b/w (1) & (2)

$\frac{P_{1}}{ρ g} + \frac{v_{1}^{2}}{2 g} + z_{1} = \frac{P_{2}}{ρ g} + \frac{v_{2}^{2}}{2 g} + z_{2}$

$\frac{P_{1} - P_{2}}{ρ g} + \frac{v_{1}^{2}}{2 g} + 1 = \frac{v_{2}^{2}}{2 g} + 0$ -(1)

Equation of continuity

A₁v₁ = A₂v₂

v₂ = 2v₁-(2)

from (1) & (2)

$\Rightarrow \frac{P_{1} - P_{2}}{ρ g} + \frac{v_{1}^{2}}{2 g} + 1 = \frac{{(2 v_{1})}^{2}}{2 g}$

$\frac{4100}{800 * 10} + \frac{v_{1}^{2}}{2 g} + 1 = \frac{4 v_{1}^{2}}{2 g}$

$\frac{12100}{8000} = \frac{3 v_{1}^{2}}{2 g}$

$\Rightarrow v_{1}^{2} = \frac{2 * 10}{3} * \frac{12100}{8000}$

$= \frac{2}{3} * \frac{121}{8}$

x = 363

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

Given below are two statements:

Statement I : Phenols are weakly acidic

Statement II : Therefore they are freely soluble in NaOH solution and are weaker acids than alcohols and water

Choose the most appropriate options:

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alok kumar singh

Contributor-Level 10

Phenol is weakly acidic compound but it is more acidic than alcohol and water because its conjugate base is more stable. Hence statement -I is correct while statement -II is wrong.

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Class 12th Thermodynamics

A 110 V, 50Hz, AC source is connected in the circuit (as shown in figure). The current through the resistance $55 Ω,$ at resonance in the circuit, will be__________A.