Ncert Solutions Chemistry Class 11th

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New answer posted•a year ago

Ncert Solutions Chemistry Class 11th Class 11th

Consider the following pairs of electrons

(a) $n = 3, l = 1, m_{l} = 1, m_{s} = + \frac{1}{2}$

(b) $n = 3, l = 2, m_{l} = 1, m_{s} = + \frac{1}{2}$

(a) $n = 3, l = 2, m_{l} = - 2, m_{s} = - \frac{1}{2}$ $_{}_{} \frac{}{}$

(b) $n = 3, l = 2, m_{l} = - 1, m_{s} = - \frac{1}{2}$

(a) $_{}_{} \frac{}{}$ $n = 4, l = 2, m_{l} = 2, m_{s} = + \frac{1}{2}$

(b) $n = 3, l = 2, m_{l} = - 2, m_{s} = + \frac{1}{2}$

The pairs of electrons present in degenerate orbitals is/are :

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Degenerate orbitals must have same value of energy

Orbitals with same n and values are degenerate orbitals.

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New answer posted•a year ago

Ncert Solutions Chemistry Class 11th Class 11th

2.5g protein containing only glycine (C₂H₅NO₂) is dissolved in water to make 500 mL of solution. TH osmotic pressure of this solution at 300 K is found to be 5.03 × 10^-3 bar. The total number of glycine units present in the protein is___________.

(Given : R = 0.083 L bar K^-1 mol^-1)

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$π = \frac{n}{v} R T$

$π = \frac{W}{M \times v} R T$

$5.03 \times 1 0^{- 3} = \frac{2.5}{M \times 0.5} \times 0.083 \times 300$

$M = \frac{2.5 \times 0.083 \times 300}{5.03 \times 0.5} \times 1 0^{3} g / m o l$

Molar mass of protein, M = 24751 g/mol

Molar mass of glycine = 75 g/mol

So; number of glycine units = $\frac{24751}{75} = 330$

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New answer posted•a year ago

Ncert Solutions Chemistry Class 11th Class 11th

For combustion of one mole of magnesium in an open container at 300 K and1 bar pressure, $Δ_{C} H^{Θ} = - 601.70 k J m o l^{- 1},$  the magnitude of change in internal energy for the reaction is__________ kJ. (Nearest integer)

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

For combustion of Mg:

Mg (s) + $\frac{1}{2} O_{2} (g) \to M g O (s)$

Here, $Δ n g = - \frac{1}{2}$

Now using

$Δ H = Δ U + Δ n g R T$

-601.7 = $Δ U + (- \frac{1}{2}) \times 8.3 \times 1 0^{- 3} \times 300$

$Δ U = - 600.45 k J$

So; magnitude of $Δ U$ is 600 kJ (the nearest integer).

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New answer posted•a year ago

Ncert Solutions Chemistry Class 11th Class 11th

In the metallurgical extraction of copper, following reaction is used:

FeO + SiO₂ ® FeSiO₃

FeO and FeSiO₃

FeO and FeSiO₃ respectively are.

7 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

In extraction of copper from copper pyrite (CuFeS₂).

FeS converted to FeO as;

$\underset{G a n g u e}{2 F e S} + 3 O_{2} \overset{Δ}{\to} \underset{G a n g u e}{2 F e O} + 2 S O_{2}$

Then FeO is removed as slag by given reaction

FeO + SiO₂ ® $\underset{S l a g}{F e S i O_{3}}$

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New answer posted•a year ago

Ncert Solutions Chemistry Class 11th Class 11th

Match List – I with List – II:

List – I                                                                               List – II

(a) Negatively charged sol                                            (I) Fe₂O₃. xH₂O

(b) Macromolecular colloid                                          (II) CdS sol

(c) Positively charged sol                                              (III) Starch

(d) Cheese (IV) a gel

Choose the correct answer from the options given below

1 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Metallic suphides i.e CdS are negatively charged colloidal solution.

Starch colloid is macro molecular colloidal solution.

Fe₂O₃. xH₂O sol is positively charged colloidal solution.

Cheese is an example of gel.

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New answer posted•a year ago

Ncert Solutions Chemistry Class 11th Thermodynamics

Elevation in boiling point for 1.5 molal solution of glucose in water is 4 K. The depression in freezing point for 4.5 molal solution of glucose in water is 4K. The ratio of molal elevation constant to molal depression constant (K_b/K_f) is________.

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$? T_{b} = i K_{b} m$

$? T_{f} = i K_{f} m$

$? \frac{4}{4} = \frac{K_{b} * 1.5}{K_{f} * 4.5}$

$\frac{K_{b}}{K_{f}} = 3$

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New answer posted•a year ago

Ncert Solutions Chemistry Class 11th Class 11th

2.2g 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)

5 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

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

$Δ U = q_{p} + w$

$w = - P_{e x t} . Δ V$

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

$Δ U = - 200 + 5 = 195 J$

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New answer posted•a year ago

Ncert Solutions Chemistry Class 11th Class 11th

A white precipitate was formed when BaCl₂ was added to water extract of an inorganic salt. Further, a gas ‘X’ with characteristic odour was released when the formed white precipitate was dissolved in dilute HCl. The anion present in the inorganic salt is

2 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$B a C l_{2} + S O_{3}^{2 -} \overset{}{\to} \underset{W h i t e}{B a S O_{3}} ↓ \overset{d i l . H C l}{\to} \underset{\begin{array}{l} b u r n i n g s u l p h u r \\ l i k e s m e l l \end{array}}{S O_{2} ↑}$