Solution

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The given polymer is melamine polymer and the monomers are Melamine and Formaldehyde.

(iii) Natural rubber becomes soft at high temperatures. Vulcanization causes sulphur to form cross links at the reactive sites of double bonds of natural rubber thus the rubber gets stiffened.

Solution

(i) Tranquilizers are used in sleeping pills.

(ii) Anionic detergents are used in toothpastes.

(iii) The control of sweetness of food is difficult while using alitame, hence its use as artificial sweetener is not recommended.
or

(i) Broad-spectrum antibiotics – Antibiotics which kill or inhibit a wide range of gram-positive and gram-negative bacteria are called broad-spectrum antibiotics.
e.g. – Chloramphenicol / Vancomycin / Ofloxacin
(ii) Disinfectants – The chemicals which either kill or prevent the growth of microorganisms are called disinfectants.
e.g. Chlorine or Sulfur dioxide
(iii) Cationic detergents – They are quaternary ammonium salts of amines with acetates, chlorides or bromides as anions.
e.g. Cetyltrimethylammonium bromide

Solution

(CH₃)₃C–I
Reason: Reactivity towards SN1 depends upon the ease of formation of carbocation. Since C–I bond is weaker than C–Br bond, hence in case of (CH₃)₃C–I, carbocation will be formed easily.

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(iii) Dextro and laevo rotatory isomers of Butan-2-ol are enantiomers of each other and both have same boiling point and hence they cannot be separated by fractional distillation.

Solution

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Solution

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Solution

(i) Amylose :
● Amylose constitutes about 20% of the starch.
Amylopectin constitutes about 80% of the starch.
● Amylose is straight chain polymer of D-glucose.
Amylopectin is a branched chain polymer of D-glucose units.
● Amylose stains blue with iodine.
Amylopectin stains reddish brown with iodine.
● α and β amyloses can hydrolyse amylose
α and β amyloses can hydrolyse α−1−>4 glycosidic linkage but cannot at branch points of amylopectin.
● Amylose is more soluble in water.
Amylopectin is water insoluble.

● Glycosidic linkage is joint monosaccharide units.
Peptide linkage is joint amino acid units.
● Glycosidic linkage is formed by the reaction between aldehyde or ketone group and alcoholic groups.
Peptide linkage is formed by the reaction between amino acids and carboxyl groups.
● Glycosidic linkage is a –O– linkage.
Peptide linkage is a –CONH– linkage.

● In fibrous protein polypeptide chains run parallel to give fibre-like structure.
In globular protein polypeptide chains coil around to give spherical shape.
● Fibrous proteins are generally composed of long and narrow strands and have a structural role.
Globular proteins generally have a more compact and rounded shape and have functional roles.
● Fibrous proteins are generally insoluble in water such as keratin.
Globular proteins are usually soluble in water such as insulin.
or
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Solution

Solution

The cell can be represented as:
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(i) When the external opposite potential is less than 2.71 V then electron flows from Mg rod to Cu rod hence current flows from Cu to Mg i.e., the direction of flow of current is from cathode to anode.
(ii) When the external opposite potential is greater than 2.71 V then electron flows from Cu rod to Mg rod and current flows from Mg to Cu i.e., the direction of flow of current is from anode to cathode.
or
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(b) (i) The electrolyte A is a strong electrolyte, and the electrolyte B is a weak electrolyte.
(ii) On extrapolation, for electrolyte, A limiting value of conductance is obtained that is conductance at zero concentration.
The curve obtained for a strong electrolyte shows that there is a small decrease in molar conductivity with increase in concentration. In other words, the molar conductivity is increased only slightly on dilution (for observing dilution effects, go towards zero on X-axis). A strong electrolyte is completely dissociated in solution and thus, furnishes all ions for conductance. However, at higher concentrations, the dissociated ions are close to each other and thus, the inter-ionic attractions are greater. These forces retard the motion of the ions and thus, conductivity is low. With a decrease in concentration (dilution), the ions move away from each other thereby feeling less attractive forces from the counterions. This results in an increase in molar conductivity with dilution. The molar conductivity approaches a maximum limiting value at infinite dilution designated as ∧ₘ⁰.
For electrolyte B:
The curve obtained for B shows that there is a large increase in the value of molar conductivity with dilution, especially near-infinite dilution. This is because as the solution of a weak electrolyte is diluted, its ionization is increased. This results in more number of ions in solution and thus, there is an increase in molar conductivity. However, the conductance of a weak electrolyte never approaches a limiting value. Or in other words, it is not possible to find conductance at zero concentration.

Solution

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(c) –OH group of phenol is activating group which increases the electron density at ortho/para position within the benzene ring so that electrophile can easily attack at ortho/para position in phenol than in benzene.
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a. (i) This discrepancy is caused by differences in the amount to which molecules are associated with one another. In o-nitrophenol, the nitro and hydroxyl groups in the same molecule form a hydrogen bond (intra-molecular), resulting in the least amount of association with nearby molecules, whereas in p-nitrophenol, the nitro and hydroxyl groups of adjacent molecules form hydrogen bonds (intermolecular), resulting in long-range molecule association. As a result, breaking the intermolecular hydrogen bonds requires a lot of energy, resulting in a higher boiling point than 2-nitrophenol molecules.
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(ii) As a strong base, sodium methoxide removes a proton from one of the methyl groups of t-butyl chloride, forming a primary carbanion that soon loses Cl – to form a double bond, yielding 2-methyl propene.
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Solution

(i) In the vapour phase, sulphur partly exists as S₂ molecule and S₂ molecule like O₂ molecule has two unpaired electrons in antibonding n orbital, hence it shows paramagnetic behaviour.

(ii) Due to small size of N, there is strong interelectronic repulsion of the non-bonding electrons and as a result the N–N single bond is weaker than P–P single bond.

(iii) Ozone (O₃) is thermodynamically less stable than dioxygen (O₂) because decomposition of ozone into dioxygen results in the liberation of heat (ΔH is negative) and increase in entropy (ΔS is positive). These two effects reinforce each other.
(b) (i) When Cu is added to dil.HNO₃ Nitrogen monoxide (NO) is released-
3Cu + 8HNO₃ (dil.) → 3Cu(NO₃)₂ + 2NO + 4H₂O

(ii) When Cu is added to conc. HNO₃ Nitrogen dioxide (NO₂) is released:
Cu + 4HNO₃ (conc.) → Cu(NO₃)₂ + 2NO₂ + 2H₂O
or
(a) (i) 4H₃PO₃ ——3H₃PO₄ + PH₃
(ii) Structure of XeF4
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(b) (i) Fluorine has a lower negative electron gain enthalpy due to its small size. Hydration energy, bond dissociation energy, and electron gain enthalpy all affect oxidising power. Fluorine has a high hydration energy due to its small size, hence F₂ is a powerful oxidising agent.

(ii) As the electronegativity of an element decreases, the acidic strength of the oxide decreases, resulting in a decrease in acidic character from N₂O₃ to Bi₂O₃.
(c) 5SO₂ + 2MnO₄⁻ + 2H₂O —>5SO₄²⁻+ 4H⁺ + 2Mn²⁺