1
JEE Main 2021 (Online) 24th February Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
Match List - I and List - II.

List-I List-II
(a) JEE Main 2021 (Online) 24th February Evening Shift Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 159 English 1 (i) $$\mathrm{Br}_{2} / \mathrm{NaOH}$$
(b) JEE Main 2021 (Online) 24th February Evening Shift Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 159 English 2 (ii) $$\mathrm{H}_{2} / \mathrm{Pd}-\mathrm{BaSO}_{4}$$
(c) JEE Main 2021 (Online) 24th February Evening Shift Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 159 English 3 (iii) $$\mathrm{Zn}(\mathrm{Hg}) / \mathrm{H}$$ Conc. $$\mathrm{HCl}$$
(d) JEE Main 2021 (Online) 24th February Evening Shift Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 159 English 4 (iv) $$\mathrm{Cl}_{2}$$ Red $$\mathrm{P}, \mathrm{H}_{2} \mathrm{O}$$


Choose the correct answer from the options given below :
A
(a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
B
(a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
C
(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
D
(a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
2
JEE Main 2021 (Online) 24th February Evening Shift
Numerical
+4
-1
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The formula of a gaseous hydrocarbon which requires 6 times of its own volume of O2 for complete oxidation and produces 4 times its own volume of CO2 is CxHy. The value of y is _____________.
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3
JEE Main 2021 (Online) 24th February Evening Shift
Numerical
+4
-1
Change Language
The solubility product of PbI2 is 8.0 $$\times$$ 10$$-$$9. The solubility of lead iodide in 0.1 molar solution of lead nitrate is x $$\times$$ 10$$-$$6. mol/L. The value of x is __________. (Rounded off to the nearest integer) [Given $$\sqrt 2 $$ = 1.41]
Your input ____
4
JEE Main 2021 (Online) 24th February Evening Shift
Numerical
+4
-1
Change Language
1.86 g of aniline completely reacts to form acetanilide. 10% of the product is lost during purification. Amount of acetanilide obtained after purification (in g) is __________ $$\times$$ 10$$-$$2.
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