Chemistry Test

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Chemistry Test - 12

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Weekly Quiz Competition

Question 1
1 / -0

One mole of calcium phosphide on reaction with an excess of water gives:

A
one mole of phosphine

B
two moles of phosphoric acid

C
two moles of phosphine

D
one mole of phosphorous oxide

Solution

The reaction is as follows:
\(\mathrm{Ca}_{3} \mathrm{P}_{2}+3 \mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{PH}_{3}+3 \mathrm{CaO}\)
Hence, one mole of calcium phosphide on reaction with an excess of water gives 2 moles of phosphine.
Question 2
1 / -0

In the reaction given below:
\(3 \mathrm{Br} 2+6 \mathrm{CO}_{2}-3+3 \mathrm{H} 2 \mathrm{O} \rightarrow 5 \mathrm{Br} \ominus+\mathrm{BrO}\ominus+_{3}+6 \mathrm{HCO}\ominus_{3}\)

A
Br2 is oxidised and CO₂₋₃is reduced

B
Br₂is reduced and H₂O is oxidised

C
Br₂is neither reduced nor oxidised

D
Br₂ is both reduced and oxidised

Solution

Disproportionation is a specific type of redox reaction in which a species is simultaneously reduced and oxidised to form two different products.

\(3 \mathrm{Br} 2+6 \mathrm{CO}_{2}-3+3 \mathrm{H} 2 \mathrm{O} \rightarrow 5 \mathrm{Br} \ominus+\mathrm{BrO}\ominus+_{3}+6 \mathrm{HCO}\ominus_{3}\)
The oxidation state of \(\mathrm{C}\) in both \(\mathrm{CO}_{2}-3\) and \(\mathrm{HCO} \ominus 3\) does not change. Hence, here Br2 is oxidised as well as reduced.
Question 3
1 / -0

The volume of the water needed to dissolve 1g of BaSO₄BaSO₄ (Ksp=1.1×10−10 at 25oC is:

A
820 L

B
410 L

C
205 L

D
None of these

Solution

Solubility of \(\mathrm{BaSO}_{4}=\mathrm{Ksp}-\mathrm{V}\)
\(=1.1 \times 10-10-1.05 \times 10-5 \mathrm{M}\)
\(\therefore\) mass of \(\mathrm{BaSO}[4=1.05 \times 10-5 \times 223\)
or \(\mathrm{wBaSO}_{4}=244.27 \times 10-5 \mathrm{g} /\) litre
\(\therefore\) Volume of water needed to dissolve \(1 \mathrm{g}\) BaSO 4 is equal to \(1244.27 \times 10-5=410\) litres
Question 4
1 / -0

The number of possible structures of amine (C₇H₉N) having one benzene ring is:

A
5

B
3

C
4

D
6

Solution

Amine with molecular formula C7H9N has five isomers.
Question 5
1 / -0

Determine the percentage comosition of calcium in calcium hydroxide, Ca(OH)₂. (1 mol =74g)

A
40%

B
43%

C
54%

D
69%

Solution

The atomic mass of Ca is 40.1
\(\mathrm{g} / \mathrm{mol}\)
The molar mass of \(\mathrm{Ca}(\mathrm{OH}) 2=74 \mathrm{g} / \mathrm{mol}\)
I mole of \(\mathrm{Ca}(\mathrm{OH}) 2=74 \mathrm{g}\)
\(\mathrm{Ca}(\mathrm{OH}) 2=40.1 \mathrm{g} \mathrm{Ca}\)
The percentage composition of calcium in calcium hydroxide
\(=40.1 \times 10074=54 \%\)
Question 6
1 / -0

Which atom that has the largest second ionization energy?

A
Mg

B
Cl

C
S

D
Na

Solution

Sodium, Na atoms have the largest second ionization energy.

When sodium loses one electron, it forms Na+ ion which has stable noble gas configuration of Ne.

When Na+ ions loses an electron, this stable configuration of Ne is broken. This requires high amount of energy. Hence, second ionization energy of sodium is largest.
Question 7
1 / -0

Which of the following would not react with benzene sulphonyl chloride in aq. NaOH ?

A
aniline

B
methylamine

C
N,N-dimethyl aniline

D
N-methyl aniline

Solution

N,N-dimethyl aniline does not contain H attached to nitrogen.

Hence, it would not react with benzene sulphonyl chloride in aqueous NaOH.
Question 8
1 / -0

Four lowest energy levels of H atoms are shown in the figure The number of emission lines could be :

A
3

B
4

C
5

D
6

Solution

From four lowest energy levels of H atoms, the number of emission lines could be 6.

n = 4

The number of emission lines =n(n−1)²=4(4−1)²=6.
Question 9
1 / -0

When glycerol is treated with excess of HI, it produces ?

A
allyl iodide

B
propene

C
glycerol triiodide

D
2-iodopropane

Solution

\(\mathrm{CH}_{2} \mathrm{OH}-\mathrm{CHOH}-\mathrm{CH}_{2} \mathrm{OH} \rightarrow\) excess \(\mathrm{HICH}_{2}=\mathrm{CH}-\mathrm{CH}_{3}\)
At first, Glycerol is reduced to form Propene via an unstable intermediate ( 1,2,3 triiodopropane). This propene then undergoes addition reaction with excess HI to form Isopropyl iodide or 2 -Iodopropane. \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{3} \rightarrow \mathrm{HICH}_{3}-\mathrm{CHI}-\mathrm{CH}_{3}\)
Hence, Option D is correct.
Question 10
1 / -0

Which of the following compounds contains S=O as well as S=S bonds?

A
Sulphuric acid

B
Thiosulphuric acid

C
Sulphurous acid

D
Thiosulphurous acid

Solution

Thiosulphuric acid contains both \(\mathrm{S}=\mathrm{O}\) as well as \(\mathrm{S}=\mathrm{S}\) bond.

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Chemistry Test - 12

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Chemistry Test - 12

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Question 1

one mole of phosphine

two moles of phosphoric acid

two moles of phosphine

one mole of phosphorous oxide

Solution

Question 2

Br2 is oxidised and CO2−3is reduced

Br2is reduced and H2O is oxidised

Br2 is neither reduced nor oxidised

Br2 is both reduced and oxidised

Solution

Question 3

820 L

410 L

205 L

None of these

Solution

Question 4

5

3

4

6

Solution

Question 5

40%

43%

54%

69%

Solution

Question 6

Mg

Cl

S

Na

Solution

Question 7

aniline

methylamine

N,N-dimethyl aniline

N-methyl aniline

Solution

Question 8

3

4

5

6

Solution

Question 9

allyl iodide

propene

glycerol triiodide

2-iodopropane

Solution

Question 10

Sulphuric acid

Thiosulphuric acid

Sulphurous acid

Thiosulphurous acid

Solution

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Br2 is oxidised and CO₂₋₃is reduced

Br₂is reduced and H₂O is oxidised

Br₂is neither reduced nor oxidised

Br₂ is both reduced and oxidised