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8 months ago

Chemistry Chemistry Chemical Bonding and Molecular Structure

Arrange the following in the decreasing order of their covalent character:

(A) LiCl
(B) NaCl

(C) KCl

(D) CsCl

Choose the most appropriate answer from the options given below:

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alok kumar singh

Contributor-Level 10

Covalent character is $L i C l > N a C l > K C l > C s C l .$ As the cationic size increases polarization decreases.

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8 months ago

Chemistry Structure of Atom

Which of the following statements are correct?

(A) The electronic configuration of Cr is [Ar] 3d⁵4s¹.

(B) The magnetic quantum number may a negative value.

(D) The total number of nodes are given by (n-2)

Choose the most appropriate answer

...more

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alok kumar singh

Contributor-Level 10

(A) Cr = [Ar]3d⁵4s¹

(B) m = $- l t o + l$

(D) Total nodes = n – 1

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8 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter Five

$C_{6} H_{12} O_{6} \overset{Z y m a s e}{\to}$ A $\to_{Δ}^{N a O l} B + C H l_{3}$

The number of carbon atoms present in the product B is______________.

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Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

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8 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter Five

Observe structure of the following compounds

The total number of structures / compounds which possess asymmetric atoms is________.

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Vishal Baghel

Contributor-Level 10

No. of compounds containing asymmetric carbon are

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Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter Five

Kjeldahl's method was used for the estimation of nitrogen in an organic compound. The ammonia evolved from 0.55g of the compound neutralized 12.5 mL of 1M H₂SO₄ solution. The percentage of nitrogen in the compound is_________ (Nearest Integer)

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Vishal Baghel

Contributor-Level 10

Millie q. of H₂SO₄ used by NH₃ = 12.5 * 1 * 2 = 25

So millimoles of N = 25

Moles of N = 25 * 10^-3

Wt of N = 14 * 25 * 10^-3

% of N = $\frac{14 * 25 * 1 0^{- 3}}{0.55} * 100 = 63.66 ≅ 64$

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8 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter Five

An acidified magnate solution undergoes disproportionation reaction. The spin- only magnetic moment value of the product having manganese in higher oxidation state is________ B.M. (Nearest Integer)

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Vishal Baghel

Contributor-Level 10

$3 M n O_{4}^{2 -} + 4 H^{+} \to_{}^{} 2 M n O_{4}^{-} + M n O_{2} + 2 H_{2} O$

No. of unpaired electron in Mn⁷⁺ = 0

$μ = 0 B . M$

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8 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter Five

The number of terminal oxygen atoms present in the product B obtained from the following reaction is_____________.

$F e C r_{2} O_{4} + N a_{2} C O_{3} + O_{2} \to A + F e_{2} O_{3} + C O_{2}$

$A + H^{+} \to B + H_{2} O + N a^{+}$

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Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

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8 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter Five

The activation energy of one of the reactions is a biochemical process is 532611 J mol^-1 when the temperature falls from 310 K to 300 K, the change in rate constant observed is k₃₀₀ = x * 10^-3 k₃₁₀. The value of x is___________.

[Given: In 10 = 2.3 R = 8.3 JK^-1 mol^-1]

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Vishal Baghel

Contributor-Level 10

$l n (\frac{K_{2}}{K_{1}}) = \frac{E a}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})$

$l n (\frac{K_{2}}{K_{1}}) = \frac{532611}{8.3} * (\frac{10}{310 * 300})$

Where, K₂ is at 310 K and K₁ at 300K

$l n (\frac{K_{2}}{K_{1}}) = 6.9 = 3 * l n 10$

$l n (\frac{K_{2}}{K_{1}}) = l n 1 0^{3}$

K₂ = K₁ * 10³

K₁ = K₂ * 10^-3

$∴$ x = 1

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8 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter Five

A dilute solution of sulphuric acid is electrolyzed using a current of 0.10 A for 2 hours to produce hydrogen and oxygen gas. The total volume of gases produced at STP is __________ cm³. (Nearest Integer).

[Given: Faraday constant F = 96500 C mol^-1 at STP, molar volume of an ideal gas is 22.7 L mol^-1]

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Vishal Baghel

Contributor-Level 10

At anode (oxidation)

2H₂O $\to_{}^{}$ O₂ (g) + 4H⁺ + 4e^-

At cathode (Reduction)

2H⁺ + 2e^- $\to_{}^{}$ H₂ (g)

No. of gm- equivalents = $\frac{i * t}{96500} = \frac{0.1 * 2 * 60 * 60}{96500} = 0.00746$

$V_{O_{2}} = \frac{0.00746}{4} * 22.7 = 0.0423 L$

$V_{H_{2}} = \frac{0.00746}{2} * 22.7 = 0.0846 L$ $V_{t o t a l} ≅ 127 m L$

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8 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter Five

1.2 mL of acetic acid is dissolved in water to make 2.0L of solution. The depression in freezing point observed for this strength of acid is 0.0198°C. The percentage of dissociation of the acid is_____________. (Nearest Integer).

[Given Density of acetic acid is 1.02g mL^-1

Molar mass of acetic acid is 60g mol^-1]

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Vishal Baghel

Contributor-Level 10

Mass = d * v = 1.02 * 1.2 = 1.224gm

Moles of acetic acid = 0.0204 moles in 2L

So molality = 0.0102 mol/kg

$Δ T_{f} = i * K_{f} * m$

i = 1 + a for acetic acid

0.0198 = (1 + a) * 1.85 * 0.0102

α = 0.04928 = 5%

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