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8 months ago

Chemistry Class 12th

The spin only magnetic moment value of an octhahedral complex among CoCl₃.4NH₃, NiCl₂.6H₂O and PtCl₄.2HCl, which upon reaction with excess of AgNO₃ gives 2 moles of AgCl is____________B. M. (Nearest integer)

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Payal Gupta

Contributor-Level 10

For precipitation of two moles of AgCl

Two Cl will produce as a free anion

CoCl₃.4NH₃ complex will $[C O {(N H_{3})}_{4} C l_{2}]$ Cl (will not give 2Cl)

$P t C l_{4} . 2 H C l \to$ complex will be H2 [PtCl6] will not any Cl

$N i C l_{2} . 6 H_{2} O \to [N i {(H_{2} O)}_{6}] C l_{2}$ will produce two Cl ion.

${[N i {(H_{2} O)}_{6}]}^{+ +} + 2 C l^{-} \to_{A g N O_{3}}^{} 2 A g C l (s)$ precipitate formation

$N i^{2 +} \to [A r] 3 d^{8} 4 s^{0}$

$μ = \sqrt[]{2 (2 + 2)} = \sqrt[]{8} = 2.84 B M = 3$

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8 months ago

Chemistry Chemistry Chemical Kinetics

The spin- only magnetic moment value of the most basic oxide of vanadium among V₂O₃, V₂O₄ and V₂O₅ is _____________B. M (Nearest integer)

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Payal Gupta

Contributor-Level 10

Most basic oxide V₂O3

Here V has +3 O.S. Hence V⁺³ $[A r] 3 d^{2}$

two unpaired e^- in d- subshell

$μ = \sqrt[]{2 (2 + 2)} = \sqrt[]{8} = 2.84 B M \approx 3$

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8 months ago

Chemistry Class 12th

2.0g of H₂ gas is adsorbed on 2.5g platinum powder at 300K and 1 bar pressure. The volume of the gas adsorbed per gram of the adsorbent is ______________mL.

(Given : R = 0.083 L bar K^-¹ mol^-¹)

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Payal Gupta

Contributor-Level 10

Volume of H₂adsorbed = $\frac{n R T}{P} = \frac{2 * 0.083 * 300}{2 * 1} = 24.9 l i t = 24900 m l$

Therefore volume of gas adsorbed per gram of the adsorbent = $\frac{24900}{2.5} = 9960$

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8 months ago

Chemistry Class 12th

A flask is filled with equal moles of A and B. The half lives of A and B are 100s and 50s respectively and are independent of the initial concentration. The time required for the concentration of A to be four times that of B is ______________s.

(Given : In 2 = 0.693)

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Payal Gupta

Contributor-Level 10

Process is based upon simultaneous disintegration hence,

$\frac{0.693}{100} * t = 2.303 l o g_{10} \frac{A_{0}}{A_{t}}$ ………….(i)

and $\frac{0.693}{50} * t = 2.303 l o g_{10} \frac{B_{0}}{B_{t}}$ ………….(ii)

from equation (i) and (ii)

$0.693 t [\frac{1}{50} - \frac{1}{100}] = [l o g \frac{B_{0}}{B_{t}} - l o g \frac{A_{0}}{A_{t}}] * 2.303$

Here; A0 = B0 and $A_{t} = 4 * B_{t}$

Therefore $0.693 t [\frac{1}{100}] = 2.303 [l o g (\frac{B_{0}}{B_{t}} * \frac{A_{t}}{A_{0}})]$

$∴ t = \frac{2.303 * 0.3010 * 2 * 100}{693} = 200 s$

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8 months ago

Chemistry Class 11th

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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8 months ago

Chemistry 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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8 months ago

Chemistry 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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8 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter Two

The moles of methane required to produce 81 g of water after complete combustion is________* 10^-2 mol. (nearest integer)

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Vishal Baghel

Contributor-Level 10

$C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$

2 moles of water produced by 1 mole of methane

Or 36 gm of water produced by 1 mole of CH₄

$∴$ 81 gm of water produced = $\frac{1 * 81}{36}$ = 2.25 mole of CH₄

Mole of CH4 required = 225 * 10^-2

the nearest integer = 225

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8 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter Two

Reaction of [Co(H₂O)₆]²⁺ with excess ammonia and in the presence of oxygen results into a diamagnetic product. Number of electrons present in t_2g-orbitals of the product is_________.

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Vishal Baghel

Contributor-Level 10

${[C o {(H_{2} O)}_{6}]}^{2 +} + N H_{3 (e x c e s s)} \to \underset{D i a m a g n e t i c n a t u r e}{{[C o {(N H_{3})}_{6}]}^{3 +} + 6 H_{2} O}$

$C o^{3 +} \to 3 d^{6} 4 s^{0}$

$\Rightarrow {t_{2 g}}^{6} e g^{0}$ (form of paired electron)

No. of electron in t_2g = 6

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8 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter Two

Catalyst A reduces the activation energy for a reaction by 10 kJ mol^-1 at 300K. The ratio of rate constants, $\frac{k_{T}, C a t a l y s e d}{k_{T}, U n c a t a l y s e d} i s e^{x} .$ The value of x is_________. (nearest integer)

[Assume that the pre-exponential factor is same in both the cases. Given R = 8.31 JK^-1 mol^-1]

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Vishal Baghel

Contributor-Level 10

$K_{c a t} = A e^{-} \frac{{(E a)}_{c a t}}{R T} a n d K_{u n c a t} = A e^{-} \frac{{(E a)}_{u n c a t}}{R T}$

$\frac{K_{c a t}}{K_{u n c a t}} = e^{\frac{{(E a)}_{u n c a t} - {(E a)}_{c a t}}{R T}} = e^{\frac{10 * 1000}{8.314 * 300}} = e^{4.009} = e^{x}$

$∴ x = 4$

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