Chemistry Test-39

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Chemistry Test-39

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Weekly Quiz Competition

Question 1
4 / -1

The ability of an ion to bring coagulation of a given colloidal solution depends on:

A
charge

B
sign of charge

C
magnitude of charge

D
both magnitude and charge

Solution

According to the Hardy - Schulz law, the greater the valency of the active ion, the greater is the power to cause coagulation. Active ion is responsible for coagulation.

The colloidal sols are stable by the presence of electric charges on the colloidal particles. It is because of the electric repulsion, that the particles do not come close to one another and coalesce. The removal of charge by any means will lead to the aggregation of particles and hence, precipitation immediately. The process by means of which the particles of the dispersed phase in a sol are precipitated is known as coagulation or flocculation.
Question 2
4 / -1

Which molecule/ion out of the following does not contain unpaired electrons?

A
\(N _{2}^{+}\)

B
\(O _{2}\)

C
\(O _{2}^{2-}\)

D
\(B _{2}\)

Solution
Question 3
4 / -1

Which of the following is not an antibiotic?

A
Sulphaguanidine

B
Penicillin

C
Streptomycin

D
Chloramphenicol

Solution

Sulphaguanidine is not an antibiotic.

Antibiotics are those drugs, which act against bacterial and viral infections e.g., Chloramphenicol Streptomycin, Penicillin, Tetracycline etc.

Sulphaguanidine is a Sulpha drug. It is a sulfonamide incorporating a guanidine moiety used to block the synthesis of folic acid; mostly used in veterinary medicine It has a role as an antiinfective agent. A sulfanilamide antimicrobial agent that is used to treat enteric infections., had been shown to have dramatic curative effects and had been used safely for some time in tablet and powder form.
Question 4
4 / -1

The standard heat of formation of \(\mathrm{U}_{3} \mathrm{O}_{8}\) is \(-853.5\) kcal/mole and the standard heat of the reaction \(3 \mathrm{UO}_{2}+\mathrm{O}_{2} \rightarrow \mathrm{U}_{3} \mathrm{O}_{8}\) is \(-76.01\) kcal.

The standard heat of formation of \(\mathrm{UO}_{2}\) is:

A
\(-1083 \mathrm{~kJ} \mathrm{~mole}^{-1}\)

B
\(-108.3 \mathrm{~kJ} \mathrm{~mole}^{-1}\)

C
\(-10.83 \mathrm{~kJ} \mathrm{~mole}^{-1}\)

D
\(-1.083 \mathrm{~kJ} \mathrm{~mole}^{-1}\)

Solution

\(3 U O_2+O_2 \rightarrow U_3 O_8\)

\(\Delta H^{\circ}=\Delta H^{\circ} f(U_3 O_8)-\Delta H^{\circ} f\left(O_{2}\right)-3 \Delta H^{\circ} f\left(U O_{2}\right)\)

\(\Rightarrow 3 \Delta H^{\circ} f\left(U O_{2}\right)=-\Delta H^{\circ}+\Delta H^{\circ} f(U_3 O_8)-\Delta H^{\circ} f (O)_{2} \)

\(=\frac{-(-76.0)+(-853.5)-0}{3}\)

or \(\Delta H^{\circ} f\left(UO_{2}\right) =-259.17\) kcal

\(-1083 \mathrm{~kJ} \mathrm{~mole}^{-1}\)
Question 5
4 / -1

Which one of the following element in period \(3\) has zero electron affinity?

A
Sulphur

B
Sodium

C
Neon

D
Chlorine

Solution

The element in period \(3\) which has zero electron affinity is neon.

The electron affinity of an element is defined as the ease with which it can accept an electron in the gaseous state.It is the relative electron-attracting tendency of an atom.It is the absolute electron attracting tendency of an isolated atom.

Nitrogen, oxygen, fluorine, and neon all lie along the same period \(2\). As we move from Nitrogen to Fluorine, the electron affinity increases as electron affinity increases along a period but is zero for halogens. So, neon has zero electron affinity.
Question 6
4 / -1

In the presence of peroxide, hydrogen chloride and hydrogen iodide do not give anti-Markovnikov's addition to alkenes because:

A
Both are highly ionic

B
One is oxidising and the other is reducing

C
One of the step is endothermic in both the cases

D
All the steps are exothermic in both cases

Solution

In the presence of peroxide, hydrogen chloride and hydrogen iodide do not give anti-Markovnikov's addition to alkenes because one of the step is endothermic in both the cases.

HI and HCl do not do this for energetic reasons. The addition of Cl and I radicals to alkanes in an anti-Markovnikov fashion is an endothermic reaction and therefore counterproductive.

Without the presence of peroxide, all hydrogen halides will add according to the Markovnikov rule.
Question 7
4 / -1

The volume of water to be added to \(100 \mathrm{~cm}^3\) of \(0.5 \mathrm{~N}~ \mathrm{H}_2 \mathrm{SO}_4\) to get decinormal concentration is:

A
\(100 \mathrm{~cm}^3\)

B
\(450 \mathrm{~cm}^3\)

C
\(500 \mathrm{~cm}^3\)

D
\(400 \mathrm{~cm}^3\)

Solution

Given,

\(\mathrm{N}_1=0.5 \mathrm{~N} \)

\(\mathrm{~N}_2=0.1 \mathrm{~N} \)

\(\mathrm{~V}_1=100 \mathrm{~cm}^3 \)

\(\mathrm{~V}_2=? \)

\(\therefore 0.5 \times 100=0.1 \times \mathrm{V}_2\)

\(\mathrm{V}_2=500 \mathrm{~cm}^3\) (Total final volume)

\(\therefore\) Volume to be added

\(=500-100=400 \mathrm{~cm}^3\)
Question 8
4 / -1

In which of the following reactions, \(K _{ p }\) is less than \(K _{ C }\) ?

A
\(N _{2} O _{4( g )} \rightleftharpoons 2 NO _{2( g )}\)

B
\(2 HI _{( g )} \rightleftharpoons H _{2( g )}+ I _{2( g )}\)

C
\(2 SO _{2( g )}+ O _{2( g )} \rightleftharpoons 2 SO _{3( g )}\)

D
\(N _{2( g )}+ O _{2( g )} \rightleftharpoons 2 NO _{( g )}\)

Solution
Question 9
4 / -1

Which of the following statement in not true?

A
Ionic bonds are non-directional while covalent bonds are directional.

B
Formation of \(\pi\) - bond shortens the distance between the two concerned atoms.

C
The ionic bond is possible between similar and dissimilar atoms.

D
Linear overlapping of atoms p- orbitals leads to a sigma bond.

Solution

The ionic bond is possible between similar and dissimilar atoms. This statement is not correct.

A covalent bond is formed by the overlap of half-filled atomic orbitals which have definite directions. So, covalent bonds are directional. In ionic compounds, each ion is surrounded by a number of oppositely charged ions and hence there is no definite direction. So, ionic bonds are non-directional. Thus, option (A) represents the correct statement.

In a pi bond, there is sideways overlapping which brings the nuclei closer and shortens the internuclear distance. Thus, option (B) also represents the correct statement.

Ionic bond or electrovalent is basically electrostatic force means attraction between positive and negative charge. That's why the ionic bond is only possible between different elements like metal and non-metal. Thus, option (C) represents the incorrect statement.

Sigma bond is formed by the linear or head to head or end on the overlapping of orbitals. Sigma bonds are strong bonds due to the maximum overlapping of orbitals.Thus, option (D)represents the correct statement.
Question 10
4 / -1

A certain substance answers Molisch's test as well as Benedict's test, but it does not answer Seliwanoff's test. Most probably, the compound under consideration is:

A
Protein

B
Sucrose

C
Maltose

D
Fructose

Solution

A certain substance answers Molisch's test as well as Benedict's test, but it does not answer Seliwanoff's test. Most probably, the compound under consideration is maltose.

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Chemistry Test-39

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Chemistry Test-39

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Question 1

charge

sign of charge

magnitude of charge

both magnitude and charge

Solution

Question 2

\(N _{2}^{+}\)

\(O _{2}\)

\(O _{2}^{2-}\)

\(B _{2}\)

Solution

Question 3

Sulphaguanidine

Penicillin

Streptomycin

Chloramphenicol

Solution

Question 4

\(-1083 \mathrm{~kJ} \mathrm{~mole}^{-1}\)

\(-108.3 \mathrm{~kJ} \mathrm{~mole}^{-1}\)

\(-10.83 \mathrm{~kJ} \mathrm{~mole}^{-1}\)

\(-1.083 \mathrm{~kJ} \mathrm{~mole}^{-1}\)

Solution

Question 5

Sulphur

Sodium

Neon

Chlorine

Solution

Question 6

Both are highly ionic

One is oxidising and the other is reducing

One of the step is endothermic in both the cases

All the steps are exothermic in both cases

Solution

Question 7

\(100 \mathrm{~cm}^3\)

\(450 \mathrm{~cm}^3\)

\(500 \mathrm{~cm}^3\)

\(400 \mathrm{~cm}^3\)

Solution

Question 8

\(N _{2} O _{4( g )} \rightleftharpoons 2 NO _{2( g )}\)

\(2 HI _{( g )} \rightleftharpoons H _{2( g )}+ I _{2( g )}\)

\(2 SO _{2( g )}+ O _{2( g )} \rightleftharpoons 2 SO _{3( g )}\)

\(N _{2( g )}+ O _{2( g )} \rightleftharpoons 2 NO _{( g )}\)

Solution

Question 9

Ionic bonds are non-directional while covalent bonds are directional.

Formation of \(\pi\) - bond shortens the distance between the two concerned atoms.

The ionic bond is possible between similar and dissimilar atoms.

Linear overlapping of atoms p- orbitals leads to a sigma bond.

Solution

Question 10

Protein

Sucrose

Maltose

Fructose

Solution

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