1
MHT CET 2023 11th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

Identify nonbenzenoid aromatic compound from following.

A
Aniline
B
Tropone
C
Naphthalene
D
Phenol
2
MHT CET 2023 11th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

Methyl propanoate on hydrolysis with dil $$\mathrm{NaOH}$$ forms a salt which on further acidification with conc. $$\mathrm{HCl}$$ forms _________.

A
$$\mathrm{CH}_3-\mathrm{COOH}$$
B
MHT CET 2023 11th May Evening Shift Chemistry - Carboxylic Acids and Its Derivatives Question 16 English Option 2
C
MHT CET 2023 11th May Evening Shift Chemistry - Carboxylic Acids and Its Derivatives Question 16 English Option 3
D
MHT CET 2023 11th May Evening Shift Chemistry - Carboxylic Acids and Its Derivatives Question 16 English Option 4
3
MHT CET 2023 11th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

Identify the product obtained in the following reaction.

$$\left(\mathrm{CH}_3 \mathrm{CO}\right)_2 \mathrm{O} \stackrel{\mathrm{H}_2 \mathrm{O}}{\longrightarrow} \text { Product }$$

A
$$\mathrm{CH}_3 \mathrm{COCH}_3$$
B
MHT CET 2023 11th May Evening Shift Chemistry - Carboxylic Acids and Its Derivatives Question 15 English Option 2
C
$$\mathrm{CH}_3-\mathrm{OH}$$
D
$$\mathrm{CH}_3 \mathrm{COOH}$$
4
MHT CET 2023 11th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

Identify the expression for average rate for following reaction.

$$\mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{NH}_{3(\mathrm{~g})}$$

A
$$\frac{-\Delta\left[\mathrm{N}_2\right]}{\Delta \mathrm{t}}=-\frac{1}{3} \frac{\Delta\left[\mathrm{H}_2\right]}{\Delta \mathrm{t}}=\frac{1}{2} \frac{\Delta\left[\mathrm{NH}_3\right]}{\Delta \mathrm{t}}$$
B
$$-\frac{1}{3} \frac{\Delta\left[\mathrm{N}_2\right]}{\Delta \mathrm{t}}=\frac{\Delta\left[\mathrm{H}_2\right]}{\Delta \mathrm{t}}=\frac{1}{2} \frac{\Delta\left[\mathrm{NH}_3\right]}{\Delta \mathrm{t}}$$
C
$$\frac{-\Delta\left[\mathrm{N}_2\right]}{\Delta \mathrm{t}}=\frac{-\Delta\left[\mathrm{H}_2\right]}{\Delta \mathrm{t}}=\frac{\Delta\left[\mathrm{NH}_3\right]}{\Delta \mathrm{t}}$$
D
$$-\frac{1}{2} \frac{\Delta\left[\mathrm{N}_2\right]}{\Delta \mathrm{t}}=\frac{-\Delta\left[\mathrm{H}_2\right]}{\Delta \mathrm{t}}=\frac{1}{3} \frac{\Delta\left[\mathrm{NH}_3\right]}{\Delta \mathrm{t}}$$
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