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Chemistry Class 12th

The reaction of H₂O₂ with potassium permanganate in acidic medium leads to the formation of mainly.

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Vishal Baghel

Contributor-Level 10

$2 M n O_{4}^{-} + 6 H^{+} + 5 H_{2} O_{2} \to 2 M n^{2 +} + 8 H_{2} O + 5 O_{2}$

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Chemistry Class 12th

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

(A) CaO

(B) FeO

(C) Al₂O_{3 &n}

...more

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Vishal Baghel

Contributor-Level 10

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

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Chemistry Class 12th

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

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Vishal Baghel

Contributor-Level 10

IUPAC nomenclature of element with atomic no. 103 is uniltrium.

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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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Vishal Baghel

Contributor-Level 10

Uses of catalyst is very specific for a particular reaction.

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Chemistry Equilibrium

20 mL of 0.1 M NH₄Oh is mixed with 40 mL of 0.05 M HCl. The pH of the mixture is nearest to: (Given K_b (NH₄OH) = 1 * 10^-5, log2 = 0.30, log3 = 0.48, log5 = 0.69, log7 = 0.84, log11 = 1.04

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Vishal Baghel

Contributor-Level 10

$[N H_{4} C l] = \frac{2}{60} = \frac{1}{30} M$

$p H = 7 - \frac{1}{2} P K_{b} - \frac{1}{2} l o g C$

$= 7 - \frac{5}{2} - \frac{1}{2} l o g (\frac{1}{30}) = 5.24$

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Chemistry Solutions

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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Vishal Baghel

Contributor-Level 10

Let we take 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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Chemistry Structure of Atom

Which of the following sets of quantum numbers is not allowed?

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Vishal Baghel

Contributor-Level 10

Value of n & $l$ can't be same

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a year ago

Chemistry Class 11th

Given below are two statements. One is labeled as Assertion A and, the other is labeled as Reason R.

Assertion A: [6] Annulene, [8] Annulene and cis-[10] Annulene, as respectively aromatic, not-aromatic and aromatics.

Reason R: Planarity is one of the requirements of aromatic systems.

In the light of the above statements, choose the correct answer from the options given below:

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alok kumar singh

Contributor-Level 10

Please find the solution below:

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Chemistry Class 12th

Match List I with List II:

List I (molecule)		List II (hybridization; shape)
A.	XeO₃	I.	sp³d; linear
B.	XeF₂	II.	sp³; pyramidal
C.	XeOF₄	III.	sp³d³; distorted octahedral
D.	XeF₆	IV.	sp³d³; square pyramidal

Choose the correct answer from the options given below

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Payal Gupta

Contributor-Level 10

$X e O_{3} \Rightarrow s p^{3}, P y r a m i d a l$

XeF₂ sp³d₂, square pyramidal

XeF₆sp³d³, Distorted octahedral

XeOF₄ sp³d², square pyramidal

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a year ago

Chemistry Class 12th

SO₂Cl₂ on reaction with excess of water results into acidic mixture

SO₂Cl₂ + H₂O -> H₂SO₄ + 2HCl

16 moles of NaOH is required for the complete neutralization of the resultant acidic mixture. The number of moles of SO₂Cl₂ used is:

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Vishal Baghel

Contributor-Level 10

$S O_{2} C l_{2} + 2 H_{2} O \to H_{2} S O_{4} + 2 H C l$

Let a moles of SO2Cl2 is taken

Then no. of moles of H2SO4 = a moles

No. of moles of HCl = 2a moles

No. of moles of NaOH required = 2a + 2a = 4a = 16

$\Rightarrow a = 4 m o l e s$

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