a) Aqua regia is prepared by mixing conc. HCl and conc. HNO3 in 3: 1 molar ratio.
b) Reaction of gold with aqua regia produces an anion having Au in +3 oxidation state.
c) Reaction of gold with aqua regia produces NO2 in the absence of air
d) The yellow colour of aqua regia is due to the presence of NOCl & Cl2.
a) Aqua regia is HCl & HNO3 (conc.) in a 3 : 1
b) Oxidation state of Au in [AuCl4]- is +3.
c) NOCl/NO is formed
d) NOCl is yellow in colour
Answer: (a, b, d)
Answer: a, b
Answer: c, d
a) Teflon is formed by polymerization of tetrafluoroethene.
b) Natural rubber is the trans from of polyisoprene.
c) Cellulose contains only α-D-glucose linkage
d) Nylon-6 contains amide linkage.
a)Fact
b) Natural rubber is Cis form of polyisoprene
c) Cellulose contains B – 1, 4 – glycosidic linkage
d) Nylon 6 contains amide linkage.
Solution: (a, d)
Consider Q, R and S as major products
Answer: a, b
Zn + Hot conc.
Zn + conc.
(precipitate) X + Y
Choose the correct option(s)
a) R is a V-shaped molecule
b) Z is dirty white in colour
c) Bond order of Q is 1 in its ground state
d) The oxidation state of Zn in T is +1
a) SO2 is v shaped.
b) ZnS is dirty white in colour.
c) Bond order of H2 is 1.
d) Oxidation state of Zn in Na2ZnO2 is +2.
Answer: a, b, c
a) R is [Au(CN4)](-)
b) Z is [Zn(CN)4]2-
c) Q is O2
d) Y is Zn
Answer: b, c, d
a) The subshell is 4d.
b) The number of angular nodes in it is 2.
c) The numbers of radial nodes in it is 3.
d) The nuclear charge experienced in n = 4 is 2e less than that in n = 1, where e is electric charge.
3.4 = 13.6 x 4/n2
n = 2
l = 2
a) Subshell is 4 d
b) Number of angular nodes is 2
c) Number of radial nodes is 1.
d) Nuclear charge would be the same.
Answer: a, b
Total 10
Balancing
∴ Mass of H2O = 288
Number of moles of H2O = 900/18 = 50
n1/(n1+50) = 0.05
(n1 is No. of moles of urea)
=> n1 = 2.63
Weight of urea = 2.63 × 60 = 157.8 g
Total weight = 157.8 + 900 = 1057.8g
Volume = 1057.8/1.2 = 881.5 cm3
Molarity = 2.63/0.8 = 2.99
Total 6 – OH groups.
Initial pressure was 1 atm, while the final pressure was 1.45 atm at time y × 103 sec calculate ‘y’.
|
t = 0 |
1 |
0 |
0 |
|
t = t |
1 – P |
P |
P/2 |
|
t = ∞ |
0 |
1 |
0.5 |
P0 = 1 atm, Pt = 1.45 atm, P∞ = 1 atm
= 2.3 x 103
This implies, y = 2.3
This is the Cis form of [Mn(en)2Cl2]
Therefore, No. of N–Mn–Cl bonds = 6
Which of the following is correct
a) P I
b) P II
c) P V
d) P III
Answer: (c)
a) S IV
b) R I
c) R II
d) S III
K.E. directly proportional to Z2/n2
Answer: (c)
Answer the question no. 17 & 18 on the basis of information given in Column – I & Column – II. Match the reactant in column – I with the possible intermediates and products of Column – II.
17. Which of the following is correct?
a) P – II, III; S – II, III
b) P – II, IV; S – II, III
c) P – III, VI; S – II, III
d) P – I, III; S – IV, V
Answer: (a)
a) Q – I, IV, VI; R – II, III, V
b) Q – I, III, VI; R – II, IV, V
c) Q – I, II, VI; R – II, III, VI
d) Q – I, IV, V; R – III, I, V
Answer: (a)
Video Lessons – Paper 2 Chemistry
JEE Advanced 2019 Chemistry Paper 2 Solutions
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