Chemistry Test

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Chemistry Test - 20

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Weekly Quiz Competition

Question 1
1 / -0

The rate of diffusion of hydrogen and chlorine are in the ratio of 6:1. The density of chlorine as compared to that of hydrogen will be:

A
3

B
6

C
18

D
36

Solution

\(\frac{r_{H_{2}}}{r_{C l_{2}}}=\frac{6}{1}=\sqrt{\frac{d_{C l_{2}}}{d_{H_{2}}}}\)If density of hydrogen is 1, then density of chlorine is 36.
Question 2
1 / -0

Match the following:

A
A-5,B-4,C-3,D-1

B
A-4,B-3,C-2,D-1

C
A-3,B-5,C-2,D-4

D
A-2,B-5,C-1,D-3

Solution

\(R=8.314 J K^{-1}\) mole \(^{-1} \rightarrow\) S.I unit \(V=22.414\) litre \(\rightarrow\) Gram molar volume at STP \(P=760 \mathrm{mm}\) of \(H g, T=273 K \rightarrow\) STP condition \(R=0.8314 \times 10^{8}\) ergs \(k^{-1}\) mole \(^{-1} \rightarrow\) CGS units
Question 3
1 / -0

A
2

B
3

C
4

D
6

Solution

This is the type of reaction where none of the atoms involved undergo a
change in their oxidation states.

A simple way to find the n-factor of \(N a_{3} P O_{4}\) in such reactions would be to find the amount of cationic charge replaced per molecule. \(\ln 2 N a_{3} P O_{4},\) we have \(6 N a^{+}\)
So, \(N a_{3} P O_{4}\) has \(3 N a^{+}\)
\(N a^{+}\) can be cosider to be similar to \(H^{+}\) in \(H_{3} P O_{4}\) (just an analogy)
Question 4
1 / -0

The IUPAC name of the following compound is?

A
2-Methyl-3,2-methylethylpentane

B
3-Ethyl-2,3-dimethylpentane

C
3,2-Methylethyl-2-methylpentane

D
None of these

Solution

Decide parent chain.

Numbering so that more substituted substituent will get the lowest

number.

Five carbons \(\Rightarrow\) word root \(=\) pent single bonds \(=\) ane(suffix)
\(\rightarrow\) substituent at \(C_{2}\) and \(C_{3}\)
\(\Rightarrow 3\) -ethyl-2, 3-dimethylpentane(alphabetically)
Therefore,
(b) 3 -Ethyl-2,3-dimethylpentane
Question 5
1 / -0

Which of the following is not an isomer of butanal ?

A
2-Butanone

B
2-Methyl propanal

C
2-Butanol

D
None of these

Solution

Aldehydes and ketones can be functional isomers. But alcohols are not isomeric with aldehydes.

You can just write down the structures of each and count the number of carbons, hydrogens, and oxygens to be certain.
Question 6
1 / -0

The polymer used as electrodes in batteries is

A
polythene

B
polypropene

C
polyacetylene

D
poly vinyl chloride

Solution

A solid-state lithium/iodine battery has been formed by directly contacting metallic lithium with iodine-doped polyacetylene
Question 7
1 / -0

Calculate \(\Delta H / k J\) for the following reaction using the listed standard enthalpy of reaction data.
\[
\begin{array}{l}
2 N_{2}(g)+5 O_{2}(g) \longrightarrow 2 N_{2} O_{5}(s) \\
N_{2}(g)+3 O_{2}(g)+H_{2}(g) \longrightarrow 2 H N O_{3}(a q) ; \Delta H / k J=-414.0 \\
N_{2} O_{5}(s)+H_{2} O(I) \longrightarrow 2 H N O_{3}(a q) ; \Delta H / k J=-86.0 \\
2 H_{2}(g)+O_{2}(g) \longrightarrow 2 H_{2} O(I) ; \Delta H / k J=-571.6
\end{array}
\]

A
-84.4

B
-243.6

C
-71.2

D
-121.8

Solution

\(2 N_{2}(g)+5 O_{2}(g) \longrightarrow 2 N_{2} O_{5}(s)\)
\(N_{2}(g)+3 O_{2}(g)+H_{2}(g) \longrightarrow 2 H N O_{3}(a q) ; \Delta H / k J=-414.0\)
\(N_{2} O_{5}(s)+H_{2} O(l) \longrightarrow 2 H N O_{3}(a q) ; \Delta H / k J=-86.0\)
\(2 H_{2}(g)+O_{2}(g) \longrightarrow 2 H_{2} O(l) ; \Delta H / k J=-571.6\)
Applying Hess's law,
\(\Delta H_{r}=[2(-414)+2(86)+571.6]\)
\(\Delta H_{r}=-84.4 k J\)
Question 8
1 / -0

Consider the following changes.
\[
\begin{array}{ll}
M(s) \rightarrow M(g) \ldots \ldots(i) & \\
M(s) \rightarrow M^{2+}(g)+2 e^{-} & \ldots \ldots(i i) \\
M(g) \rightarrow M^{+}(g)+e^{-} & \ldots \ldots(i i i) \\
M^{+}(g) \rightarrow M^{2+}(g)^{+} e^{-} & \ldots \ldots(i v) \\
M(g) \rightarrow M^{2+}(g)+2 e^{-} & \ldots \ldots(v)
\end{array}
\]
The second ionization energy of \(M\) can be calculated from the energy values associated with:

A
1 + 3 + 4

B
2 - 1 + 3

C
1 + 5

D
5 - 3

Solution

Second ionization energy is amount of energy required to take out an electron from the monopositive cation. Hence, \(M(g) \rightarrow M^{2+}+2 e^{-}-M(g) \rightarrow M^{+}+e^{-}\)
\(\Longrightarrow M^{+} \rightarrow M^{2+}+e^{-}\)
Question 9
1 / -0

Select correct statement about valence-bond approach.

A
Each bond is formed by maximum overlap for its maximum stability.

B
It represents localised electron model of bonding.

C
Most of the electrons retain the same orbital locations as in a separated atoms.

D
All the above are correct statements.

Solution

The correct statement about valence-bond approach are:

(A) Each bond is formed by maximum overlap for its maximum stability. Higher is the overlap of the participating bonds, higher is the strength of bond formed and higher is the stability.

(B) It represents localized electron model of bonding. The most of the electron density of molecules is localized in the region between two nuclei.

(C) Most of the electrons retain the same orbital locations as in a separated atoms.

Thus, the location of the non bonding electrons in the valence shell and the electrons present in the inner shells is unaffected.
Question 10
1 / -0

The formation of the oxide ion \(O^{2-}(g)\) requires first an exothermic and then an endothermic step as shown below.
\(O_{(g)}+e^{-}=O^{-}(g), \Delta H^{0}=-142 k J m o l^{-1}\)
\(O^{-}(g)+e^{-}=O^{2-}(g), \Delta H^{0}=+844 k J m o l^{-1}\)
This is because :

A
O⁻ ion will tend to resist the addition of another electron

B
oxygen has high electron affinity

C
oxygen is more electronegative

D
O⁻ion has comparatively larger size than oxygen atom

Solution

O⁻ ion tends to resist the addition of another electron due to large repulsion of e⁻in small sized O⁻ ion. Thus, energy is required to overcome this force of repulsion.

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Chemistry Test - 20

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Chemistry Test - 20

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Question 1

3

6

18

36

Solution

Question 2

A-5,B-4,C-3,D-1

A-4,B-3,C-2,D-1

A-3,B-5,C-2,D-4

A-2,B-5,C-1,D-3

Solution

Question 3

2

3

4

6

Solution

Question 4

2-Methyl-3,2-methylethylpentane

3-Ethyl-2,3-dimethylpentane

3,2-Methylethyl-2-methylpentane

None of these

Solution

Question 5

2-Butanone

2-Methyl propanal

2-Butanol

None of these

Solution

Question 6

polythene

polypropene

polyacetylene

poly vinyl chloride

Solution

Question 7

-84.4

-243.6

-71.2

-121.8

Solution

Question 8

1 + 3 + 4

2 - 1 + 3

1 + 5

5 - 3

Solution

Question 9

Each bond is formed by maximum overlap for its maximum stability.

It represents localised electron model of bonding.

Most of the electrons retain the same orbital locations as in a separated atoms.

All the above are correct statements.

Solution

Question 10

O− ion will tend to resist the addition of another electron

oxygen has high electron affinity

oxygen is more electronegative

O− ion has comparatively larger size than oxygen atom

Solution

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O⁻ ion will tend to resist the addition of another electron

O⁻ion has comparatively larger size than oxygen atom