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Class 11th Chemistry

In the structure of SF₄, the lone pair of electrons on S is in.

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alok kumar singh

Contributor-Level 10

Geometry of SF₄ is trigonal bipyramidal, in which there is one lonepair which occupy equatorial position as,

There are two lone pair – bond pair repulsions at 90°

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Class 11th Structure of Atom

Consider the following statements:

(A) The principal quantum number 'n' is a positive integer with values of 'n' = 1, 2, 3,….

(B) The azimuthal quantum number $' l'$ for a given 'n' (principal quantum number) can have values as $' l' = 0, 1, 2, . . . n$

(C) Magnetic orbital quantum number $' m_{l}'$ for a particular $' l'$ (azimuthal quantum number) has $(2 l + 1)$ values.

(D) $\pm 1 / 2$ are two possible orientations of electron spin.

(E) For $l = 5,$ there will be a total of 9 orbitals

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alok kumar singh

Contributor-Level 10

$•$ Values of principal quantum number, n = 1, 2, 3, .

$•$ Values of azimuthal quantum number, $l = 0, 1, 2 . . . . . . . . . . ., n - 1$

$•$ Number of values of magnetic quantum number are $2 l + 1 .$

$•$ Values of spin quantum number are $\pm \frac{1}{2} .$

$•$ Number of orbitals for particular value of $l a r e 2 l + 1, s o f o r l = 5,$ no. of orbitals = 11.

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Class 11th 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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Class 11th Work, Energy and Power

A car cover AB distance with first one-third at velocity $υ_{1} m s^{- 1},$ second one-third at $υ_{2} m s^{- 1}$ and last one-third at $υ_{3} m s^{- 1} .$ If then the average velocity of the car is__________ ms^-1.

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alok kumar singh

Contributor-Level 10

$t_{1} = \frac{L}{3 V_{1}}$

$t_{2} = \frac{L}{3 V_{2}}$

$t_{3} = \frac{L}{3 V_{3}}$

$V_{a m g} = \frac{L}{t_{1} + t_{2} + t_{3}} = \frac{L}{\frac{L}{3 V_{1}} + \frac{L}{3 V_{2}} + \frac{L}{3 V_{3}}}$

$V_{a m g} = \frac{3}{\frac{1}{v_{1}} + \frac{1}{v_{2}} + \frac{1}{v_{3}}}$

$= \frac{3}{\frac{1}{11} + \frac{1}{2 * 11} + \frac{1}{3 * 11}}$

= $\frac{3 * 11 * 6}{6 + 3 + 2}$

= 3 * b

= 18 m/sec

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4 months ago

Class 11th Units and Measurement

The speed of light in media 'A' and 'B' are 2.0 * 10¹⁰ cm/s and 1.5 * 10¹⁰ cm/s respectively. A ray of light enters from the medium B to an incident angle 'q'. If the ray suffers total internal reflection, then

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alok kumar singh

Contributor-Level 10

$\frac{s i n θ_{C}}{V_{B}} = \frac{s i n 90 °}{V_{A}} \Rightarrow s i n θ_{C} = \frac{V_{B}}{V_{A}} = \frac{1.5 * 1 0^{10}}{2.0 * 1 0^{10}} = \frac{3}{4}$

According to question, we can write

$θ > θ = s i n^{- 1} (\frac{3}{4})$

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4 months ago

Class 11th Physics

What will be the effect on the root mean square velocity of oxygen molecules if the temperagture is doubled and oxygen molecule dissociates into atomic oxygen?

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alok kumar singh

Contributor-Level 10

V_rms = $\sqrt[]{\frac{3 R T}{M}}$

V_rms' = $\sqrt[]{\frac{3 R T_{f}}{m_{f}}} = \sqrt[]{\frac{3 R * 2 T * 2}{M}}$ $[\begin{matrix} G i v e n, T_{f} = 2 T & M_{f} = \frac{M}{2} \end{matrix}]$

$= 2 \sqrt[]{\frac{3 R T}{M}}$

V_rms' = 2V_rms

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Class 11th Chemistry

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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4 months ago

Class 11th Physics

A sample of an ideal gas is taken through the cyclic process ABCA as shown in figure. It absorbs, 40 J of heat during the part AB, no heat during BC and rejects 60 J of heat during CA. A work of 50 J is done on the during the part BC. The internal energy of the gas at A is 1560 J. The work done by the gas during the part CA is:

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alok kumar singh

Contributor-Level 10

Given

Q_AB = +40J

Q_BC = 0J

Q_CA = -60J

W_CA =?

W_BC = -50J (on the gas)

U_A = 1560J

1^st Law on path A to B

W_AB + $Δ U = Q$

Þ 0 + $[U_{B} - U_{A}] = 40$

Þ U_B = U_A + 40

= 1560 + 40

U_B = 1600 J

1^st Low on path B

...more

Given

Q_AB = +40J

Q_BC = 0J

Q_CA = -60J

W_CA =?

W_BC = -50J (on the gas)

U_A = 1560J

1^st Law on path A to B

W_AB + $Δ U = Q$

Þ 0 + $[U_{B} - U_{A}] = 40$

Þ U_B = U_A + 40

= 1560 + 40

U_B = 1600 J

1^st Low on path B to C

Q_BC = $Δ u + w_{B C}$

Þ 0 = U_c – U_D – 50

U_C = U_B + 50

= 1600 + 50

= 1650 J

U_A – U_C = $Δ U$ (Path C to A) = 1560-1650

= -90 J

1st Low for path C to A

Q_CA = $Δ U_{C A} + w_{C A} + w_{C A}$

Þ -60 = -90 + w_CA

W_CA = +30J

less

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Class 11th 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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4 months ago

Class 11th Chemistry Some Basic Concepts of Chemistry

For complete combustion of methanol

$C H_{3} O H (1) + \frac{3}{2} O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (I)$

The amount of heat produced as measured by bomb calorimeter is 726 kJmol^-¹ at 27°C. The enthalpy of combustion for the reaction is -x kJ mol^-¹, where x is_________

(Nearest integer)

(Given : R = 8.3 J K^-¹ mol^-¹).

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Payal Gupta

Contributor-Level 10

$C H_{3} O H_{(l)} + \frac{3}{2} O_{2 (g)} \to C O_{2 (g)} + 2 H_{2} O (l)$

$Δ n = - 0.5$

$Δ H = Δ E + Δ n R T = - 726 - 0.5 * \frac{8.3}{1000} * 300 = - 726 - 1.24 = - 727.24 \approx - 727 k J / m o l$

Hence, x = 727 (the nearest integer)

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