Chemistry Test

Result

Chemistry Test - 28

Score
-
out of -
Rank
-
out of -

TIME Taken - -

Weekly Quiz Competition

Question 1
1 / -0

The process of converting ammonia into nitrates is called:

A
Nitrogen-fixation

B
Ammonification

C
Denitrification

D
Nitrification
Solution
The process of conversion of ammonia into nitrate nitrogen is called nitrification. It is done in two steps- Nitrite formation, Nitrate formation. Nitrite formation: Ammonia are oxidized to nitrites by Nitrococcus, Nitrosomonas.

The nitrogen cycle has four steps:
- Nitrogen Fixation: Conversion of molecular nitrogen into the inorganic nitrogenous compound.
- Ammonification: Conversion of the dead organic nitrogenous compound into ammonia.
- Nitrification: Oxidation of ammonia into nitrates.
- Denitrification: Nitrites or nitrates converts back into molecular nitrogen.
Question 2
1 / -0

Rain is called acid rain when its pH is below:

A
7

B
6.5

C
6

D
4 to 5

Solution

Acid rain is rain that is acidic, meaning it has a very low pH. When sulfur dioxide and nitrogen oxides are released into the air, they react with water and oxygen and produces sulfuric and nitric acids. These acids then mix with water and result in acid rain. Normal rainwater is slightly acidic with a pH range of 5-6. When the pH level of rainwater falls below this range, it becomes acid rain. It can have harmful effects on plants, aquatic animals and infrastructure.
Question 3
1 / -0

Which of the following will give propyne on hydrolysis?

A
\(A l_{4} C_{3}\)

B
\(M g_{2} C_{3}\)

C
\(B_{4} C\)

D
\(L a_{4} C_{3}\)

Solution

Hydrolysis is a chemical reaction where the water molecules are added to the compound undergoing the reaction.

During hydrolysis, the water molecule undergoes decomposition to give \(H^{+}\) and \(O H^{-}\) ions respectively.

\(H O H \rightleftarrows H^{+}+O H^{-}\)

All the given options are examples of metal carbides.

Now, we know that carbides are compounds that have carbon and other elements which have less electronegative character than carbon atoms.

Metal carbides are compounds that have a transition metal ion and carbon atom as the constituents.

These structures are more of a complex type. Propyne is basically, \(C_{3} H_{4}\).

\(M g_{2} C_{3}\) is an ionic carbide and on hydrolysis with water yields magnesium hydroxide and propyne which is a three-carbon product.

\(\mathrm{Mg}_{2} \mathrm{C}_{3}+4 \mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{Mg}(\mathrm{OH})_{2}+\mathrm{C}_{3} \mathrm{H}_{4}\)

\(A l_{4} C_{3}\) is a metal carbide and on hydrolysis yields methane and aluminium hydroxide.

\(A l_{3} C_{4}+12 H O H \rightarrow 3 C H_{4}+4 A l(O H)_{3}\)

\(B_{4} C\) on hydrolysis gives boron oxide while carbon monoxide and hydrogen gases are evolved.

\(B_{4} C+7 H_{2} O \rightarrow 2 B_{2} O_{3}+C O+7 H_{2}\)

\(L a_{4} C_{3}\) on hydrolysis gives lanthanum hydroxide and methane.

\({La}_{4} {C}_{3}+{HOH} \rightarrow{La}(OH)_{3}+{CH}_{4}\)

Thus, from the above chemical reactions, it is clear that \(M g_{2} C_{3}\) is the correct option. As it gives propyne as a product.
Question 4
1 / -0

The freezing point of a \(4 \%\) aqueous solution of \(X\) is equal to the freezing point of \(12 \%\) aqueous solution of \(Y\). If the molecular weight of \(X\) is \(A\), then the molecular weight of \(Y\) is:

A
\(A\)

B
\(3A\)

C
\(4A\)

D
\(2A\)

Solution

Freezing point of \(4 \%\) aqueous solution of \(X=\) Freezing point of \(12 \%\) aqueous solution of \(Y\).
The molecular weight of \(X=A\)
We have to find the molecular weight of \(Y\).
So the freezing point of \(4 \%\) aqueous solution of \(X\) is equal to freezing point of \(12 \%\) aqueous solution of \(Y\), their depression in freezing point will be equal\(\triangle T_{f}=T_{f}^{0}-T_{f}\)
where, \(\triangle T_{f}=\) Depression in freezing point,\(T_{f}^{0}=\) Freezing point of pure solvent,\({T}_{{f}}=\) Freezing point of solution,

\(\Delta T_{f}\) is given as: \(\Delta T_{f}=i K_{f} m\),where, \({i}=\) van't Hoff factor,\({K}_{{f}}=\) Molal depression constant,\({m}=\) Molality

Taking W kg of solvent
Molarity \(=\) No. of moles of solute \(/\) Mass of solvent (in kg)
\(m_{z}=\frac{\left(4 / M_{z}\right)}{w}\)
Question 5
1 / -0

The molarity of a solution obtained by mixing \(750~ mL\) of \(0.5 M~ HCl\) with \(250 ~mL\) of \(2 M~ HCl\) will be:

A
\(1.00 M\)

B
\(1.75 M\)

C
\(0.975 M\)

D
\(0.875 M\)

Solution

\(750~ ml\) of \(0.5 M~ HCl\) have \(0.375\) moles.

\(250~ ml\) of \(2 M~ HCl\) have \(0.5\) moles.

Total moles \(=0.875~ moles\)

Total volume \(=750+250=1000~ ml =1~ L\)

Molarity \(=\frac{\text { Moles }}{\text { Volume }}=\frac{0.875}{1}=0.875~ M\)

Hence, the correct option is (A).
Question 6
1 / -0

Which of the following is the functional group of Amine?

A
Oxygen

B
Nitrogen

C
Helium

D
Barium

Solution

Amines:
- An amine is a functional group with a nitrogen atom having a lone pair.
- Basically, Amines are derived from ammonia \(\left( NH _3\right)\).
- Nitrogen has a valency of 5 , that's why it makes a trivalent with a lone pair.

Uses:

- Amines are used in water purification, medicine manufacturing, and insecticides, and pesticides.

- These are also used in the production of Amino acids.

- It is also used in pain-relieving medicines.

Types:

- Amines are generally of four types:

1. Primary Amines

2. Secondary Amines

3. Tertiary Amines

4. Cyclic Amines

- Amines can be obtained from Halogen alkanes.
Question 7
1 / -0

A substance forms Zwitter ion. It can have functional groups of:

A
\(-\mathrm{NH}_{2},-\mathrm{COOH}\)

B
\(-\mathrm{NH}_{2},-\mathrm{SO}_{3} \mathrm{H}\)

C
Both (A) and (B)

D
None of the above

Solution

Zwitter ion is a molecule or ion having separate positively and negatively charged groups. The amine functional group can take up hydrogen in solution and form \(-{NH_3^+}\) ion, and the carboxyl group can lose hydrogen to for \(COO^-\) and the sulphonyl group can also lose hydrogen to form \(SO_3^-\). Therefore the molecules having an \(NH_2\) group and either the \(-COOH\) group or \(-SO_3H\) group can form zwitter ion.
Question 8
1 / -0

In which of the following conditions, the potential for the following half-cell reaction is maximum?

\(2 \mathrm{H}^{+}+2 e \rightarrow \mathrm{H}_{2}\)

A
\(1.0 \mathrm{M~HCl}\)

B
A solution having \(p H=4\)

C
Pure water

D
\(1.0 \mathrm{M~NaOH}\)

Solution

In the given options,

(A) Concentration of \(H^{+}\)in \(1.0 \mathrm{M} \mathrm{~HCl}=1 M\)

(B) Since, \(p H\) of solution is 4 .

So, concentration of \(H^{+}=10^{-4} M\)

(C) \(p H\) of pure water \(=7\)

So, concetration of \(H^{+}=10^{-7} M\)

(D) Concentration of \(H^{+}\)in \(1.0 \mathrm{M~NaOH}\) is much less than the all the above options.

Since, concentration of \(H^{+}\)is highest in option (A). So, potential of option (A) is maximum.
Question 9
1 / -0

During discharging of a lead storage battery, which of the following is/are true?

A
\(\mathrm{PbSO}_{4}\) is formed at both electrodes

B
Density of electrolytic solution decreases

C
(A) and (B) both

D
\(\mathrm{H}_{2} \mathrm{SO}_{4}\) is produced

Solution

Oxidation at anode is represented by the half cell reaction:

\(\mathrm{Pb}(s)+\mathrm{SO}_{4}^{2-}(a q) \rightarrow \mathrm{PbSO}_{4}(s)+2 e^{-}\)

Reduction at cathode is represented by the half cell reaction:

\(\mathrm{PbO}_{2}(s)+4 H^{+}(a q)+S O_{4}^{2-}(a q)+2 e^{-} \rightarrow P b S O_{4}(s)+2 H_{2} O(l)\)

The overall cell reaction is:

\(\mathrm{Pb}(s)+\mathrm{PbO}_{2}(s)+2 \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow 2 \mathrm{PbSO}_{4}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)\)

During discharging of a lead storage battery, sulfuric acid is consumed and water is produced.

Lead sulphate is formed at both electrodes.

Since sulfuric acid is consumed, the density of the electrolytic solution decreases.
Question 10
1 / -0

From which of the following reaction primary amine is produced?

A
Reduction of Nitrile Compounds

B
Reduction of Amide Compounds

C
Hoffmann bromamide degradation reaction

D
Above all reactions

Solution

Formation of Primary Amines:
- Primary amines can be synthesized through various chemical reactions, each involving different starting materials and reagents.
- Some common methods for the preparation of primary amines include:
- Reduction of nitrile compounds
- Reduction of amide compounds
- Hoffmann bromamide degradation reaction
- Reduction of nitrile compounds:
\(R-C \equiv N+2 H_2 \rightarrow R-CH_2-NH_2\)
In this reaction, a nitrile \(( R - C \equiv N )\) is reduced to a primary amine \(\left( R - CH _2- NH _2\right)\) using hydrogen gas \(\left( H _2\right)\).
\(\underset{\text { Nitrile }}{ R - C \equiv N } \xrightarrow[\text { 2) } H ^{+}]{\text {3) } NaOH } \xrightarrow[\text { Amine }]{\text { 3) } LiAlH _4}\) R
Reduction of amide compounds:
\(R-CONH_2+4[H] \rightarrow R-CH_2-NH_2+H_2 O\)
In this reaction, an amide \(\left( R - CONH _2\right)\) is reduced to a primary amine \(\left( R - CH _2- NH _2\right)\) using a reducing agent such as lithium aluminium hydride \(\left(\right.\) LiAlH \(\left._4\right)\).
\(R ^{\stackrel{ O }{ }} NH _2 \xrightarrow[\text { 2) } H ^{+}]{\text {1) } LiAlH _4} R ^{\wedge} NH _2\)
Hoffmann bromamide degradation reaction:
\(R-CONH_2+Br_2+4 NaOH \rightarrow R-NH_2+Na_2 CO_3+2 NaBr+2 H_2 O\)
(In this reaction, an amide \(\left( R - CONH _2\right)\) reacts with bromine \(\left( Br _2\right)\) and sodium hydroxide \(( NaOH )\) to form a primary amine \(\left( R - NH _2\right)\), along with sodium carbonate \(\left( Na _2 CO _3\right)\), sodium bromide ( NaBr ), and water \(\left( H _2 O \right)\).
Therefore, primary amines can be produced from the reduction of nitrile compounds, the reduction of amide compounds, and the Hoffmann bromamide degradation reaction.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Chemistry Test - 28

Result

Chemistry Test - 28

Score

-

Rank

-

Question 1

Nitrogen-fixation

Ammonification

Denitrification

Nitrification

Solution

Question 2

7

6.5

6

4 to 5

Solution

Question 3

\(A l_{4} C_{3}\)

\(M g_{2} C_{3}\)

\(B_{4} C\)

\(L a_{4} C_{3}\)

Solution

Question 4

\(A\)

\(3A\)

\(4A\)

\(2A\)

Solution

Question 5

\(1.00 M\)

\(1.75 M\)

\(0.975 M\)

\(0.875 M\)

Solution

Question 6

Oxygen

Nitrogen

Helium

Barium

Solution

Question 7

\(-\mathrm{NH}_{2},-\mathrm{COOH}\)

\(-\mathrm{NH}_{2},-\mathrm{SO}_{3} \mathrm{H}\)

Both (A) and (B)

None of the above

Solution

Question 8

\(1.0 \mathrm{M~HCl}\)

A solution having \(p H=4\)

Pure water

\(1.0 \mathrm{M~NaOH}\)

Solution

Question 9

\(\mathrm{PbSO}_{4}\) is formed at both electrodes

Density of electrolytic solution decreases

(A) and (B) both

\(\mathrm{H}_{2} \mathrm{SO}_{4}\) is produced

Solution

Question 10

Reduction of Nitrile Compounds

Reduction of Amide Compounds

Hoffmann bromamide degradation reaction

Above all reactions

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.