List I | List II | ||
(A) | Weak intermolecular forces of attraction | I | Hexamethylenedia mine + adipic acid |
(B) | Hydrogen bonding | II | AlEt3 + TiCl4 |
(C) | Heavily branched polymer | III | 2-chloro-1, 3-butadiene |
(D) | High density polymer | III | phenol+ formaldehyde |
choose the correct answer from the options given below:
1. A. Weak intermolecular forces of attraction: 2–chloro–1,3–butadiene (chloroprene) is the monomer of neoprene, which has weak intermolecular forces of attraction, being an elastomer. Correct match: III. 2–chloro–1,3–butadiene.
2. B. Hydrogen bonding: Hexamethylenediamine reacts with adipic acid to form Nylon-6,6. The presence of the amide group allows hydrogen bonding between polymer chains. Correct match: I. Hexamethylenediamine + adipic acid.
3. C. Heavily branched polymer: - Phenol reacts with formaldehyde to form Bakelite, which is a heavily branched (cross-linked) polymer. Correct match: IV. Phenol + formaldehyde.
4. D. High density polymer: - High-density polyethylene is prepared using the Ziegler-Natta catalyst (\( \text{AlEt}_3 + \text{TiCl}_4 \)). Correct match: II. \( \text{AlEt}_3 + \text{TiCl}_4 \). ### Final Matching: \[ \text{A–III, B–I, C–IV, D–II.} \]
Calculate the potential for half-cell containing 0.01 M K\(_2\)Cr\(_2\)O\(_7\)(aq), 0.01 M Cr\(^{3+}\)(aq), and 1.0 x 10\(^{-4}\) M H\(^+\)(aq).
Let \( T_r \) be the \( r^{\text{th}} \) term of an A.P. If for some \( m \), \( T_m = \dfrac{1}{25} \), \( T_{25} = \dfrac{1}{20} \), and \( \displaystyle\sum_{r=1}^{25} T_r = 13 \), then \( 5m \displaystyle\sum_{r=m}^{2m} T_r \) is equal to: