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Class 12th Chemistry

150g of acetic acid was contaminated with 10.2g ascorbic acid $(C_{6} H_{8} O_{6})$ to lower down its freezing point by (x * 10^-¹)°C. The value of x is _________. (Nearest integer)

(Given K_f = 3.9 K kg mol^-¹; molar mass of ascorbic acid = 176 g mol^-¹]

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Raj Pandey

Contributor-Level 9

$m = \frac{w * 1000}{m o l e c u l a r w t * w_{s o l v e n t}} = \frac{10.2 * 1000}{176 * 150}$

$= \frac{10200}{176 * 150} = 0.386$

$Δ T_{f} = K_{f} * m$

3.9 * 0.386

$∴ x * 1 0^{- 1} = 15.05 * 1 0^{- 1}$

Ans = 15

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Class 12th Chemistry

An element M crystallises in a body centred cubic unit cell with a cell edge of 300 pm. The density of the element is 6.0g cm^-³. The number of atoms in 180g of the element is_________* 10²³ (Nearest integer)

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Raj Pandey

Contributor-Level 9

B.C.C structure

a = 300 pm = 300 * 10^-¹² m

d = 6g/cm³

z = 2

$d = \frac{Z * M}{a^{3}}$

$6 = \frac{2 * A}{{(300 * 1 0^{- 10})}^{3}} = \frac{2 * A}{27 * 1 0^{- 24}}$

$∴ A t o m s o f M$ = 3.69 * 6.022 * 10²³

...more

B.C.C structure

a = 300 pm = 300 * 10^-¹² m

d = 6g/cm³

z = 2

$d = \frac{Z * M}{a^{3}}$

$6 = \frac{2 * A}{{(300 * 1 0^{- 10})}^{3}} = \frac{2 * A}{27 * 1 0^{- 24}}$

$∴ A t o m s o f M$ = 3.69 * 6.022 * 10²³

= 22.22 * 10²³

the nearest integer = 22

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Class 12th Ncert Solutions Chemistry Class 12th

The compound(s) that is (are) removed as slag during the extraction of copper is:

(A) CaO

B) FeO

C) Al₂O₃

(D) ZnO

(E) NiO

Choose the correct answer from the options given below.

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Raj Pandey

Contributor-Level 9

$F e O + \underset{(A c i d i c f l u x)}{S i O_{2}} \to \underset{(S l a g)}{F e S i O_{3}}$

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Class 12th Chemistry

The IUPAC nomenclature of an element with electronic configuration [Rn] 5f¹⁴6d¹7s² is:

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Raj Pandey

Contributor-Level 9

IUPAC nomenclature of element with atomic no. 103 is uniltrium

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Class 12th Ncert Solutions Chemistry Class 12th

Match the List-I with List-II

List-I	List-II
(A) N₂(g) + 3H₂(g) ® 2NH₃(g)	(I) Cu
(B) CO(g) + 3H₂(g) ® CH₄(g) + H₂O(g)	(II) Cu/ZnO – Cr₂O₃
(C) CO(g) + H₂(g) ® HCOH(g)	(III) Fe_xO_y + K₂O + Al₂O₃
(D) CO(g) + 2H₂(g) ® CH₃OH(g)	(IV) Ni

Choose the correct answer from the options given below:

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Raj Pandey

Contributor-Level 9

Uses of catalyst is very specific for a particular reaction.

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Class 12th Chemistry

The depression in freezing point observed for a formic acid solution of concentration 0.5 mL L^-1 is $0.0405 ° C .$ Density of formic acid is 1.05 g mL^-1. The Van't Hoff factor of the formic acid solution is nearly: (Given for water $K_{f} = 1.86 k k g m o l^{- 1}$ )

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Raj Pandey

Contributor-Level 9

Let we take $1 l$ of solution

Mass of solute = Volume * Density

= 0.5 $m l * 1.05 g m / m l$

= 0.525 gram

Mass of solution = 1 kg. [considering very dilute solution]

Mass of solvent = 1000 – 0.525 = 999.475 gram

$\Rightarrow i = 1.9$

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2 months ago

Class 12th Nuclei

A sample contains 10^-2kg each of two substances A and B with half lives 4s and 8s respectively. The ration of their atomic weights is 1 : 2. The ratio of the amounts of A and B after16s is $\frac{x}{100}$ . The value of x is ______.

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alok kumar singh

Contributor-Level 10

$f o r A, t_{\frac{1}{2}} = 4 s e c$

$= m A_{0} e^{-} \frac{0.693}{4} * 16$

$m_{A} = m_{A 0} e^{- 4 * 0.693}$ -(1)

for B,

for B, $t_{\frac{1}{2}} = 8 s e c$

$m_{B} = m_{B_{0}} e^{- 2 * 0.693}$ -(2)

$\frac{m_{A}}{m_{B}} = \frac{m_{A O}}{m_{B O}} \frac{e^{- 4} * 0.693}{e^{- 2} * 0.693}$

$\frac{m_{A}}{m_{B}} = \frac{25}{100} = \frac{x}{100} x = 25$

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2 months ago

Class 12th Physics Wave Optics

Sodium light of wavelength 650nm and 655 nm is used to study diffraction at a single slit of aperture 0.5 mm. The distance between the slit and the screen is 2.0 m. The separation between the positions of the first maximum of diffraction pattern obtained in the two cases is _________* 10^-⁵ m.

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alok kumar singh

Contributor-Level 10

For maxima = y = (2n + 1) $\frac{λ}{2} \frac{D}{a}$

For 1^st maxima for l₁ wavelength (n = 1)

$y_{1} = \frac{3 λ_{1}}{2} \frac{D}{a}$ - (1)

First maxima for l₂ wavelength

$y_{2} = \frac{3}{2} λ_{2} \frac{D}{a}$ - (2)

$\Rightarrow y_{2} - y_{1} = \frac{3}{2} \frac{D}{a} [λ_{2} - λ_{1}]$

$= \frac{3}{2} * \frac{2 * 5 * 1 0^{- 9}}{0.5 * 1 0^{- 3}}$

$= \frac{3}{1 0^{- 3}} * 1 0^{- 8}$

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2 months ago

Class 12th Physics Wave Optics

A sample contains 10^-2kg each of two substances A and B with half lives 4s and 8s respectively. The ration of their atomic weights is 1 : 2. The ratio of the amounts of A and B after16s is $\frac{x}{100}$ . The value of x is ______.

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New answer posted

2 months ago

Class 12th Physics

As shown in the figure an inductor of inductance 200 mH is connected to an AC source of emf 220 V and frequency 50 Hz. The instantaneous voltage of the source is 0 V when the peak value of current is $\frac{\sqrt[]{a}}{π} A .$ The value of a is __________.

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alok kumar singh

Contributor-Level 10

Let l = l₀ cos wt

Then v = v₀ sinwt

at t = 0, v = 0

but l = l₀

$l_{r m s} = \frac{l_{0}}{\sqrt[]{2}}$

$V_{r m s} = l_{r m s} z$

$\Rightarrow 220 = \frac{l_{0}}{\sqrt[]{2}} (X_{L})$

$\Rightarrow 220 = \frac{l_{0}}{\sqrt[]{2}} (2 π * 50 * 200 * 1 0^{- 3})$

$\Rightarrow 220 = \frac{l_{0}}{\sqrt[]{2}} (20 π)$

$l_{0} = \frac{220 \sqrt[]{2}}{20 π}$

$l_{0} = \frac{11 \sqrt[]{2}}{π} = \frac{\sqrt[]{a}}{π}$

a = 121 * 2

a = 242

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