Ncert Solutions Chemistry Class 12th

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7 months ago

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R
Raj Pandey

Contributor-Level 9

When we add cryolite (Na? AlF? ) in the extraction of aluminium, the melting point of alumina decreases

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7 months ago

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R
Raj Pandey

Contributor-Level 9

Functional group isomers

 

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7 months ago

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A
alok kumar singh

Contributor-Level 10

Haematite: Fe? O?

Bauxite: Al? O? ·xH? O

Magnetite: Fe? O?

Malachite: CuCO? ·Cu (OH)?

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7 months ago

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R
Raj Pandey

Contributor-Level 9

Reaction Sequence:

o   NO? + CH? COOH → HNO? + CH? COO?

o   Sulphanilic acid (+ NH? CH? COO? ) + HNO? → N-acetylated diazonium intermediate + 2H? O

The intermediate + another molecule → Red Azodye + NH? + CH? COOH

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A
alok kumar singh

Contributor-Level 10

100 mol of KBr is doped with 10? mol of SrBr?

Therefore, 1 mol of KBr contains 10? mol of SrBr?

Molar mass of KBr = 119 g/mol .

119 g of KBr contains 10? mol of SrBr?

1 g of KBr contains (10? / 119) mol of SrBr?

Each Sr²? ion introduced creates one cation vacancy to maintain electrical neutrality.

Number of cation vacancies = (moles of SrBr? ) * (Avogadro's number)

Number of vacancies = (10? / 119) * (6.023 * 10²³) = 5.06 * 10¹?

The answer provided in the document is "5 (Rounded off)", which likely refers to the coefficient 5.06.

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7 months ago

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A
alok kumar singh

Contributor-Level 10

For the dissociation of K? [Fe (CN)? ]? 4K? + [Fe (CN)? ]? , the number of ions produced (n) is 5.

The degree of dissociation (α) is related to the van't Hoff factor (i) by α = (i-1)/ (n-1).

Given α = 0.4: 0.4 = (i - 1) / (5 - 1) => 1.6 = I - 1 => I = 2.6

Using the depression in freezing point formula (ΔTf = i·Kf·m), and equating the ΔTf for two different solutions:
ΔTf (K? [Fe (CN)? ]) = ΔTf (A)
i? ·Kf·m? = i? ·Kf·m?
(2.6) * [ (18.1 / M) / (100-18.1)/1000] = (1) * [ (w? /M? ) / (W? /1000)]
The problem simplifies to finding the molar mass M of solute A:
2.6 * [18.1 / (M * 81.9)] * 1000 = [w? /M? ] * [1000/W? ]
Assuming the secon

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A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

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V
Vishal Baghel

Contributor-Level 10

3CaO + 2Al → 3Ca + Al? O?
ΔH? = ΣΔH? (Products) - ΣΔH? (Reactants)
ΔH? = [ (0) + (-1675) ] - [ (3 * -635) + (0) ]
ΔH? = -1675 - (-1905) = -1675 + 1905 = 230 kJ
Ans = 230

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7 months ago

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V
Vishal Baghel

Contributor-Level 10

Cr? O? ²? + Fe²? - (H? )-> Cr³? + Fe³?
(n=6) (n=1)
Meq Cr? O? ²? = Meq Fe²?
20 * 0.03 * 6 = 15 * M * 1
M = (20 * 0.03 * 6) / 15 = 0.24 = 24 * 10? ²
Ans = 24

New answer posted

7 months ago

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V
Vishal Baghel

Contributor-Level 10

P? + 3NaOH + 3H? O → PH? + 3NaH? PO? (A)
Here, NaH? PO? is a reducing agent.
NaH? PO? + AgNO? + H? O → Ag + HNO? + NaH? PO?
Equivalents of NaH? PO? = equivalents of Ag (n-factor of NaH? PO? = 4)
1 mole * 4 = n moles of Ag * 1
So, moles of Ag = 4.

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