Chemistry Test

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Chemistry Test - 8

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Weekly Quiz Competition

Question 1
4 / -1

Which of the following will not act as an antacid?

A
Sodium hydrogen carbonate

B
Magnesium hydroxide

C
Sodium carbonate

D
Aluminum carbonate

Solution

An antacid is a substance that neutralizes stomach acidity and is used to relieve heartburn, indigestion, or an upset stomach. Some antacids have been used in the treatment of constipation and diarrhea. Currently marketed antacids contain salts of aluminum, calcium, magnesium, or sodium.

Sodium carbonate is a chemical compound with the molecular formula Na₂CO₃. It's commonly referred to as washing soda and is used in cleaning products, glass production, as a food additive, and more.
Question 2
4 / -1

Which of the following combination doesn't evolve \(\mathrm{Cl}_{2}\) gas?

A

\(\mathrm{HCl}(\) aq\()+\mathrm{KMnO}_{4}\)

B

\(\mathrm{HCl}+\mathrm{MnO}_{2}\)

C

\(\mathrm{HCl}+\mathrm{I}_{2}\)

D

\(\mathrm{HCl}+\mathrm{F}_{2}\)

Solution

Here \(\mathrm{HCl}+\mathrm{I}_{2}\) combination doesn't evolve \(\mathrm{Cl}_{2}\) gas.

\(\mathrm{HCl}+\mathrm{I}_{2} \rightarrow\) No reaction

\(8 \mathrm{HCl}+\mathrm{KMnO}_{4} \rightarrow \mathrm{MnCl}_{2}+\frac{5}{2} \mathrm{Cl}_{2} \uparrow\) \(+\mathrm{KCl}+4 \mathrm{H}_{2} \mathrm{O}\)

\(4 \mathrm{HCl}+\mathrm{MnO}_{2} \rightarrow \mathrm{MnCl}_{2}+\mathrm{Cl}_{2} \uparrow+2 \mathrm{H}_{2} \mathrm{O}\)

\(\mathrm{HCl}+\mathrm{F}_{2} \rightarrow \mathrm{HF}+\mathrm{Cl}_{2} \uparrow\)
Question 3
4 / -1

The increasing order of the following compounds towards HCN addition is:

A
\(\mathrm{(iii) < (iv) <(i)<(ii)}\)

B
\(\mathrm{(iii) <(i)<(iv)<(ii)}\)

C
\(\mathrm{(i) <(iii)<(iv)<(ii)}\)

D
\(\mathrm{(iii) <(iv)<(ii)<(i)}\)

Solution

In \(\mathrm{HCN}, \mathrm{CN}^{-}\)acts as a nucleophile, attack first that -CHO group which has a maximum positive charge. The magnitude of the (+ve) charge increases by \(-\mathrm{M}\) and \(-I\) group. So reactivity order will be:
Question 4
4 / -1

In context with the transition elements, which of the following statements is incorrect?

A
In addition to the normal oxidation states, the zero oxidation state is also shown by these elements in complexes.

B
In the highest oxidation states, the transition metals show basic character and form cationic complexes.

C
In the highest oxidation states of the first five transition elements (\(\mathrm{Sc}\) to \(\mathrm{Mn}\)), all the \(\mathrm{4s}\) and \(\mathrm{3d}\) electrons are used for bonding.

D
Once the \(\mathrm{d^{5}}\) configuration is exceeded, the tendency to involve all the \(\mathrm{3 \mathrm{d}}\)-electrons in bonding decreases.
Solution
Transition metals do not show any basic character and from cations complexes in higher oxidation states.
- In the higher oxidation states, the transition metals show acidic nature and form anionic complexes.
- In higher oxidation states metal oxides are more covalent in nature as the polarizing power of the metal ion proportional to its oxidation states. So, their oxides are acidic in nature.
Question 5
4 / -1

The process of converting alkyl halides to alcohols involve:

A
Addition reaction

B
Rearrangement reaction

C
Substitution reaction

D
Dehydrohalogenation reaction

Solution

Reaction with Alkyl halides:

Alkyl halides goes for two reactions - Substitution and elimination reaction.

In a substitution reaction, the X group in R-X is replaced by a different group i.e.,

\({R} .{X} \rightarrow {R}-{OH}+{X}^{+}\)

In the elimination reaction, the elements of H-X are eliminated from R-X; the product is very often alkene.
Question 6
4 / -1

When 32.25 g of ethyl chloride (atomic mass of chlorine is 35.5 g) is subjected to dehydrohalogenation reaction, the yield of the alkene formed is 50%. The mass of the ethene formed is:

A
7 g

B
28 g

C
64.5 g

D
56 g

Solution

Molar mass of ethyl chloride

=2×12+5×1+1×35.5

=24+9+35.5=64.5 gmol⁻¹

Molar mass of ethene =2+12+4×1

=28 gmol⁻¹

Actual yield = 50% of 28 g = 14 g

64.5 g of ethyl chloride on dehydrohalogenation gives 14 g ethene

32.25 g of ethyl chloride on dehydrohalogenation will give ethene = 7 g
Question 7
4 / -1

The expression to compute pH of NH₄CI solution is:

A
\(\mathrm{pH}=\frac{1}{2} \mathrm{pK}_{w}^{0}-\frac{1}{2} \mathrm{pK}_{\mathrm{b}}^{0}\left(\mathrm{NH}_{4} \mathrm{OH}\right)-\frac{1}{2} \log ([\mathrm{salt}] / \mathrm{M})\)

B
\(\mathrm{pH}=\frac{1}{2} \mathrm{pK}_{w}^{0}+\frac{1}{2} \mathrm{pK}_{\mathrm{b}}^{0}\left(\mathrm{NH}_{4} \mathrm{OH}\right)-\frac{1}{2} \log ([s a l t] / \mathrm{M})\)

C
\(\mathrm{pH}=\frac{1}{2} \mathrm{pK}_{w}^{0}+\frac{1}{2} \mathrm{pK}_{\mathrm{b}}^{0}\left(\mathrm{NH}_{4} \mathrm{OH}\right)+\frac{1}{2} \log ([\mathrm{salt}] / \mathrm{M})\)

D
\(p H=\frac{1}{2} p_{b}^{0}\left(\mathrm{NH}_{4} \mathrm{OH}\right)-\frac{1}{2} \mathrm{pK}_{w}^{0}+\frac{1}{2} \log ([s a l t] / \mathrm{M})\)

Solution

The expression to compute pH of NH₄CI solution is:\(\mathrm{pH}=\frac{1}{2} \mathrm{pK}_{w}^{0}-\frac{1}{2} \mathrm{pK}_{\mathrm{b}}^{0}\left(\mathrm{NH}_{4} \mathrm{OH}\right)-\frac{1}{2} \log ([\mathrm{salt}] / \mathrm{M})\)

NH₄Cl is a salt and it is formed from weak base NH₄OH and strong acid HCl (via neutralization reaction)

NH₄OH +HCl —-> NH4Cl

Now, since HCl is a strong acid it will dissociate completely as well as rapidly and gives H+ n Cl-

And NH₄OH is a weak base, it will dissociate slowly.

So, at a particular time, H+ in the solution will be more as compared to NH₄+ n OH^-. Thus, solution will be acidic at that particular time.

Now if you want to know the pH value, you can use the following equation :

pH =0.5 {pKa -pKb -logC}
Question 8
4 / -1

Which of the following will have a net dipole moment?

A
NO₃^-

B
CCl₄

C
BeF₂

D
SO₂

Solution

SO₂:

In Sulphur dioxide, the dipole actas from sulphur towards oxygen and from sulphur towards the lone pair of electrons.

The net dipole is acting downwards and the net dipole is thus not zero.
Question 9
4 / -1

Which of the following is the decreasing order of oxidizing power of halogens?

A
I > Br > Cl > F

B
I > Cl > Br > F

C
Cl > F > Br > I

D
F > Cl > Br > I

Solution

Halogens have a strong tendency to accept electrons.

Halogens accept electrons from other metals and become uni negative.

The other substance is oxidized.

The substances that accept electrons from others are termed as oxidizing agents. Thus, halogens act as good oxidizing agents.

The oxidizing powers of halogens are measured in terms of their electrode potentials.

The electrode potentials of halogens decrease in the order:

\(\mathrm{F}_{2}(+2.87 \mathrm{~V})>\mathrm{Cl}_{2}(+1.36 \mathrm{~V})>\mathrm{Br}_{2}(+1.09 \mathrm{~V})>\mathrm{I}_{2}(+0.54 \mathrm{~V})\)

Higher electrode potential means a stronger oxidizing agent.

Thus, the decreasing order of oxidizing power of halogens is F > Cl > Br > I.
Question 10
4 / -1

For the process to occur under adiabatic conditions, the correct conditionis:

A
\(\mathrm{\Delta T=0}\)

B
\(\mathrm{\Delta p=0}\)

C
\(\mathrm{q=0}\)

D
\(\mathrm{w=0}\)

Solution

For the process to occur under adiabatic conditions, the correct condition is\(\mathrm{q=0}\).

When a thermodynamic system undergoes a change in such a way that no exchange of heat takes place between Systems and surroundings, the process is known as an adiabatic process. During the adiabatic process, there is no exchange of heat that takes place between the system and the surroundings.

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Chemistry Test - 8

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Chemistry Test - 8

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Question 1

Sodium hydrogen carbonate

Magnesium hydroxide

Sodium carbonate

Aluminum carbonate

Solution

Question 2

\(\mathrm{HCl}(\) aq\()+\mathrm{KMnO}_{4}\)

\(\mathrm{HCl}+\mathrm{MnO}_{2}\)

\(\mathrm{HCl}+\mathrm{I}_{2}\)

\(\mathrm{HCl}+\mathrm{F}_{2}\)

Solution

Question 3

\(\mathrm{(iii) < (iv) <(i)<(ii)}\)

\(\mathrm{(iii) <(i)<(iv)<(ii)}\)

\(\mathrm{(i) <(iii)<(iv)<(ii)}\)

\(\mathrm{(iii) <(iv)<(ii)<(i)}\)

Solution

Question 4

In addition to the normal oxidation states, the zero oxidation state is also shown by these elements in complexes.

In the highest oxidation states, the transition metals show basic character and form cationic complexes.

In the highest oxidation states of the first five transition elements (\(\mathrm{Sc}\) to \(\mathrm{Mn}\)), all the \(\mathrm{4s}\) and \(\mathrm{3d}\) electrons are used for bonding.

Once the \(\mathrm{d^{5}}\) configuration is exceeded, the tendency to involve all the \(\mathrm{3 \mathrm{d}}\)-electrons in bonding decreases.

Solution

Question 5

Addition reaction

Rearrangement reaction

Substitution reaction

Dehydrohalogenation reaction

Solution

Question 6

7 g

28 g

64.5 g

56 g

Solution

Question 7

\(\mathrm{pH}=\frac{1}{2} \mathrm{pK}_{w}^{0}-\frac{1}{2} \mathrm{pK}_{\mathrm{b}}^{0}\left(\mathrm{NH}_{4} \mathrm{OH}\right)-\frac{1}{2} \log ([\mathrm{salt}] / \mathrm{M})\)

\(\mathrm{pH}=\frac{1}{2} \mathrm{pK}_{w}^{0}+\frac{1}{2} \mathrm{pK}_{\mathrm{b}}^{0}\left(\mathrm{NH}_{4} \mathrm{OH}\right)-\frac{1}{2} \log ([s a l t] / \mathrm{M})\)

\(\mathrm{pH}=\frac{1}{2} \mathrm{pK}_{w}^{0}+\frac{1}{2} \mathrm{pK}_{\mathrm{b}}^{0}\left(\mathrm{NH}_{4} \mathrm{OH}\right)+\frac{1}{2} \log ([\mathrm{salt}] / \mathrm{M})\)

\(p H=\frac{1}{2} p_{b}^{0}\left(\mathrm{NH}_{4} \mathrm{OH}\right)-\frac{1}{2} \mathrm{pK}_{w}^{0}+\frac{1}{2} \log ([s a l t] / \mathrm{M})\)

Solution

Question 8

NO3-

CCl4

BeF2

SO2

Solution

Question 9

I > Br > Cl > F

I > Cl > Br > F

Cl > F > Br > I

F > Cl > Br > I

Solution

Question 10

\(\mathrm{\Delta T=0}\)

\(\mathrm{\Delta p=0}\)

\(\mathrm{q=0}\)

\(\mathrm{w=0}\)

Solution

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NO₃^-

CCl₄

BeF₂

SO₂