1
JEE Main 2020 (Online) 9th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
5 g of zinc is treated separately with an excess of

(a) dilute hydrochloric acid and
(b) aqueous sodium hydroxide.

The ratio of the volumes of H2 evolved in these two reactions is :
A
1 : 2
B
1 : 1
C
1 : 4
D
2 : 1
2
JEE Main 2020 (Online) 9th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Consider the following reactions,

JEE Main 2020 (Online) 9th January Evening Slot Chemistry - Compounds Containing Nitrogen Question 135 English
The compound [P] is :
A
JEE Main 2020 (Online) 9th January Evening Slot Chemistry - Compounds Containing Nitrogen Question 135 English Option 1
B
JEE Main 2020 (Online) 9th January Evening Slot Chemistry - Compounds Containing Nitrogen Question 135 English Option 2
C
JEE Main 2020 (Online) 9th January Evening Slot Chemistry - Compounds Containing Nitrogen Question 135 English Option 3
D
JEE Main 2020 (Online) 9th January Evening Slot Chemistry - Compounds Containing Nitrogen Question 135 English Option 4
3
JEE Main 2020 (Online) 9th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A, B and C are three biomolecules. The results of the tests performed on them are given below:

$$ \begin{array}{|l|l|l|l|} \hline & \begin{array}{l} \text { Molisch's } \\ \text { Test } \end{array} & \begin{array}{l} \text { Barfoed } \\ \text { Test } \end{array} & \begin{array}{l} \text { Biuret } \\ \text { Test } \end{array} \\ \hline \text { A } & \text { Positive } & \text { Negative } & \text { Negative } \\ \hline \text { B } & \text { Positive } & \text { Positive } & \text { Negative } \\ \hline \text { C } & \text { Negative } & \text { Negative } & \text { Positive } \\ \hline \end{array} $$

A, B and C are respectively :
A
A = Lactose, B = Glucose, C = Albumin
B
A = Lactose, B = Fructose, C = Alanine
C
A = Lactose, B = Glucose, C = Alanine
D
A = Glucose, B = Fructose, C = Albumin
4
JEE Main 2020 (Online) 9th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The reaction of H3N3B3Cl3 (A) with LiBH4 in tetrahydrofuran gives inorganic benzene (B). Further, the reaction of (A) with (C) leads to H3N3B3(Me)3. Compounds (B) and (C) respectively, are :
A
Borazine and MeBr
B
Diborane and MeMgBr
C
Boron nitride and MeBr
D
Borazine and MeMgBr
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