Class 11th

The hybridization of P exhibited in PF₅ is sp^xd^y. The value of y is_________

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Class 11th Structure of Atom

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Hybridization of P in PF₅ is sp³d, so value of y = 1

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If the work function of a metal is 6.63 * 10^-¹⁹J, the maximum wavelength of the photon required to remove a photoelectron from the metal is _______nm. (Nearest integer)

[Given h = 6.63 * 10^-³⁴ Js, and c = 3 * 10⁸ ms^-¹]

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Contributor-Level 9

Work function, $?_{0} = 6.63 * 1 0^{- 19} J$

Threshold wavelength $λ_{0} = ?$

Using ; $?_{0} = \frac{h c}{λ_{0}}$

$λ_{0} = \frac{h c}{?_{0}}$

=300 * 10^-9 m = 300 nm

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Class 11th Redox Reactions

A 2.0g sample containing MnO₂ is treated with HCl liberating Cl₂. The Cl₂ gas is passed into a solution of Kl and 60.0mL of 0.1M Na₂S₂O₃ is required to titrate the liberated iodine. The percentage of MnO₂ in the sample is_______. (Nearest integer)

[Atomic masses (in u) Mn = 55; Cl = 35.5; O = 16, I = 127, Na = 23, K = 39, S = 32]

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$M n O_{2} + 4 H C l \to M n C l_{2} + C l_{2} + 2 H_{2} O$

$\begin{array}{l} C l_{2} + 2 K l \to 2 K C l + I_{2} \\ I_{2} + 2 N a_{2} S_{2} O_{3} \to 2 N a l + N a_{2} S_{4} O_{6} \end{array}$

Here, meq of MnO₂ = meq of Na₂S₄O₆

$\frac{w}{E} * 1000 = M * n - f a c t o r * V$

$\frac{w}{87 / 2} * 1000 = 0.1 * 1 * 60$

Mass of MnO₂ in sample = 0.261 g

Percentage of MnO₂ in sample = $\frac{0.261}{2} * 100$

= 13.05%

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Class 11th Maths

Let 3, 6, 9, 12,…. upto 78 terms and 5, 9, 13, 17,…. upto 59 terms be two series. Then, the sum of the terms common to both the series is equal to…………

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Class 11th Chemistry Organic Chemistry

Contributor-Level 10

First common term to both AP's is 9

t₇₈ of $(3, 6, 9, . . . . . .) = 78 * 3 = 234$

t₅₉ of $(5, 9, 13, . . . . . . . .) = 5 + (5 - 1) 4 = 237$

n^th common term $\leq 234$

9 + (n – 1) 12 234

n < $\frac{237}{12} \Rightarrow n = 19$

Now sum of 19 terms with a = 9, d = 12

$= \frac{19}{2} (2.9 + (19 - 1) \cdot 12) = 2223$

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In the given reaction

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Class 11th Chemistry Hydrocarbon

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Which one of the following structures are aromatic in nature?

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Kindly consider the following Image

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.2g of nitrous oxide (N₂O) gas is cooled at a constant pressure of 1 atm from 310 K to 270 K causing the compression of the gas from 217.1mL to 167.75mL. The change in thermal energy of the process, $Δ U$ is '-x' J. The value of 'x' is________. [Nearest Integer]

(Given : atomic mass of N = 14 g mol^-1 and of O = 16 g mol^-1.

Molar heat capacity of N₂O is 100 J K^-1 mol^-1)

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Contributor-Level 10

Moles of N₂O = $\frac{2.2}{44} = \frac{1}{20}$

$Δ H = n C_{p} Δ T = \frac{1}{20} * 100 (- 40) = - 200 J$

$Δ U = q_{p} + w$

$w = - P_{e x t} . Δ V$

$w = - 1 \frac{(167.75 - 217.1)}{1000} * 101.3 J = + 5 J$

$Δ U = - 200 + 5 = 195 J$

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Given below are two statements:

Statement I : The pentavalent oxide of group-15 element, E₂O₅, is less acidic than trivalent oxide, E₂O₃, of the same element.

Statement II : The acidic character of trivalent oxide of group 15 elements, E₂O₃ decreases down the group.

In light of the above statements, choose most appropriate answer form the options given below:

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Class 11th Structure of Atom

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Pentavalent oxides of group – 15 elements, E₂O₅ is more acidic than trivalent oxides, E₂O₃ of the same element.

Acidic strength of trivalent oxides decreases down the group as metallic strength increases.

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A box contains 0.90g of liquid water in equilibrium with water vapour at 27°C. The equilibrium vapour pressure of water at 27°C is 32.0 Torr. When the volume of the box is increased, some of the liquid water evaporates to maintain the equilibrium pressure. If all the liquid water evaporates, then the volume of the box must be________ litre. [Nearest Integer]

(Given : R = 0.082 L atm K^-1 mole^-1)

(Ignore the volume of the liquid water and assume water vapours behave as an ideal gas.)

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Contributor-Level 10

V = $\frac{n R T}{P} = \frac{0.90 * 0.0821 * 300}{18 * 32} * 760 = 29.21 ≅ 29$

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Nitrogen gas is obtained by thermal decomposition of

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