1
COMEDK 2024 Evening Shift
MCQ (Single Correct Answer)
+1
-0

A current of 3.0A is passed through 750 ml of 0.45 M solution of CuSO$$_4$$ for 2 hours with a current efficiency of 90%. If the volume of the solution is assumed to remain constant, what would be the final molarity of CuSO$$_4$$ solution?

A
0.296
B
0.4
C
0.237
D
0.316
2
COMEDK 2024 Evening Shift
MCQ (Single Correct Answer)
+1
-0

Match the Vitamins given in Column I with the diseases caused by their deficiency as given in Column II.

Vitamin Deficiency
(A) K (P) Cheilosis
(B) B$$_12$$ (Q) Osteomalacia
(C) B$$_2$$ (R) Permicious Anaemia
(D) D (S) Haemophilia

A
$$ A-S \quad B-R \quad C-P \quad D-Q $$
B
$$ A-Q \quad B-P \quad C-S \quad D-R $$
C
$$ \mathrm{A}-\mathrm{S} \quad \mathrm{B}-\mathrm{R} \quad \mathrm{C}-\mathrm{Q} \quad \mathrm{D}-\mathrm{P} $$
D
$$ \mathrm{A}-\mathrm{P} \quad \mathrm{B}-\mathrm{S} \quad \mathrm{C}-\mathrm{Q} \quad \mathrm{D}-\mathrm{R} $$
3
COMEDK 2024 Evening Shift
MCQ (Single Correct Answer)
+1
-0

Which one of the following is the correct order of reagents to be used to convent [A] to [X] ?

COMEDK 2024 Evening Shift Chemistry - Aldehyde and Ketone Question 1 English

$$ \text { Reagents: } \mathrm{PCC}, \mathrm{PCl}_5, \mathrm{LiAlH}_4, \frac{\mathrm{KCN}}{\mathrm{H}_3 \mathrm{O}^{+}} $$

A
$$ \frac{K C N}{H_3 \mathrm{O}^{+}} \quad \mathrm{PCl}_5 \quad \mathrm{PCC} \quad \mathrm{LiAlH}_4 $$
B
$$ \mathrm{PCl}_5, \quad \frac{K C N}{H_3 \mathrm{O}^{+}} \quad \mathrm{LiAlH}_4 \quad \mathrm{PCC} $$
C
$$ \mathrm{LiAlH}_4, \frac{K C N}{H_3 \mathrm{O}^{+}} \quad \mathrm{PCl}_5 \quad \mathrm{PCC} $$
D
$$ \mathrm{PCC} \quad \mathrm{PCl}_5 \quad \frac{K C N}{\mathrm{H}_3 \mathrm{O}^{+}} \quad \mathrm{LiAlH}_4 $$
4
COMEDK 2024 Evening Shift
MCQ (Single Correct Answer)
+1
-0

Arrange the following redox couples in the increasing order of their reducing strength:

$$\begin{array}{ll} {[\mathrm{A}]=\mathrm{Cu} / \mathrm{Cu}^{2+}} & \mathrm{E}^0=-0.34 \mathrm{~V} \\ {[\mathrm{~B}]=\mathrm{Ag} / \mathrm{Ag}^{+}} & \mathrm{E}^0=-0.8 \mathrm{~V} \\ {[\mathrm{C}]=\mathrm{Ca} / \mathrm{Ca}^{2+}} & \mathrm{E}^0=+2.87 \mathrm{~V} \\ {[\mathrm{D}]=\mathrm{Cr} / \mathrm{Cr}^{3+}} & \mathrm{E}^0=+0.74 \mathrm{~V} \end{array}$$

A
B < A < D < C
B
A < C < B < D
C
C < D < A < B
D
D < A < C < B
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