Chemistry Test-8

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Chemistry Test-8

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Weekly Quiz Competition

Question 1
4 / -1

The most dangerous pollutant for metals is _________________

A
Sulphur dioxide

B
Ammonia

C
Hydrogen sulphide

D
Hydrogen cyanide

Solution

Exposure to sulphur dioxide causes deterioration of fabrics (cotton, rayon), paper and leather. Being corrosive, it has damaging effects on metals, such as iron, steel, aluminium, and on building materials, such as limestone, marble, roof slate, etc. Marble structures, limestone and geological sources of ammonia act as a sink for atmospheric sulphur dioxide.
Question 2
4 / -1

Gay Lussac law gives relation between:

A
\(\mathrm{P}\) and \(\mathrm{V}\)

B
\(\mathrm{P}\) and \(\mathrm{T}\)

C
\(\mathrm{V}\) and\(\mathrm{T}\)

D
None of these

Solution

Gay Lussac's Law - states that the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature. If you heat a gas you give the molecules more energy so they move faster. This means more impacts on the walls of the container and an increase in the pressure.
Question 3
4 / -1

Calculate the heat required to make \(6.4 {Kg} {~CaC}_{2}\) from \({CaO}({s})\) and \({C}({s})\) from the reaction:

\({CaO}({s})+3 {C}({s}) \rightarrow\) \({CaC}_{2}({~s})+{CO}({g})\)

given that \(\Delta_{f} \mathrm{H^o}\left(\mathrm{CaC}_{2}\right)=-14.2\) kcal. \(\Delta \mathrm{f} \mathrm{H}^{\circ}(\mathrm{CO})=-26.4\) kcal.

A
\(5624{~kcal}\)

B
\(1.11 \times 10^{4}{~kcal}\)

C
\(86.24 \times 10^{3}{~kcal}\)

D
\(1100{~kcal}\)

Solution

We know,

\(n =\frac {Mass}{Molecular~weight}\) where \(n=\) No. of moles

For 1 mole of \(\mathrm{CaC}_{2}\)

\(\Delta \mathrm{H}=\Delta \mathrm{H}_{\mathrm{f}}(\mathrm{CaC})+\mathrm{H}_{\mathrm{f}}(\mathrm{CO})-\mathrm{H}_{\mathrm{f}}(\mathrm{CaO})\)

\(\Rightarrow -14.2-26.4+151.6=111.1 \mathrm{kcal}\)

\(\therefore\) For 100 moles, \(\Delta H=1.11 \times 10^{4}\) Kcal
Question 4
4 / -1

Which of the following is a balanced chemical equation of the reaction between carbon monoxide gas and hydrogen gas?

A
\(3\mathrm{CO ( g )+2 H _{2}( g ) \stackrel{340 atm }{\longrightarrow} CH _{3} OH (l)}\)

B
\(\mathrm{CO (g)+3 H _{2}( g ) \stackrel{340 atm}{\longrightarrow} CH _{3} OH (l)}\)

C
\(\mathrm{CO ( g )+2 H _{2}( g ) \stackrel{340 atm }{\longrightarrow} 2 CH _{3} OH (l)}\)

D
\(\mathrm{CO ( g )+2 H _{2}( g ) \stackrel{340 atm }{\longrightarrow} CH _{3} OH (l)}\)

Solution

The chemical equation of the reaction between carbon monoxide gas and hydrogen gas can be written as:

\(\mathrm{CO}+\mathrm{H}_{2} \rightarrow \mathrm{CH}_{3} \mathrm{OH}\)

The chemical formula of carbon monoxide is \(\mathrm{CO}\) and its physical state is gas, the chemical formulaof hydrogen is \(\mathrm{H}_{2}\) and its physical state is also gaseous. In the above reaction we can seethat the number of carbon and oxygen atoms are balanced on both sides. But the number of \(\mathrm{H}\) on the reactant side is \(2\) whereas the number of \(\mathrm{H}\) on the product side is \(4\). And hence, multiplying \(2\) as the coefficient of \(\mathrm{H_{2}}\), we will get a balanced chemical equation.

So, the balanced chemical equation is:

\(\mathrm{CO ( g )+2 H _{2}( g ) \stackrel{340 atm }{\longrightarrow} CH _{3} OH (l)}\)
Question 5
4 / -1

Choose the correct statement.

A
Non-metals have higher ionisation enthalpies and higher electronegativity then the metals

B
non metals and metalloids exist only in the p-block of the periodic table.

C
change of non-metallic to metallic character can be illustrated by the nature of oxide they form

D
all are correct,

Solution

The size of non-metals is small as compared to metals due to which electrons are more tightly held in non-metals.Non-Metals are electronegative elements and tend to share or accept electrons in a chemical reaction.Due to the electrons accepting nature of non-metals, they gains more electrons to complete their octet.

So, non-metals have high ionization enthalpy due to their small size and high electronegativity as compared to metal.
Question 6
4 / -1

X and Y are:

A
X- Natural rubber, Y- Gutta percha

B
X- Gutta percha , Y- Natural rubber

C
X- Natural rubber, Y- Natural rubber

D
X- Gutta percha, Y- Gutta percha

Solution

X and Y are Natural rubber and Gutta percha.
Question 7
4 / -1

Which one of the following statements regarding Henry's law is not correct?

A
The value of \(\mathrm{K}_{\mathrm{H}}\) increases with the function of the nature of the gas.

B
Higher the value of \(\mathrm{K}_{\mathrm{H}}\) at a given pressure, higher is the solubility of the gas in the liquids.

C
The partial pressure of the gas in vapour phase is proportional to the mole fraction of the gas in the solution.

D
Different gases have different \(\mathrm{K}_{\mathrm{H}}\) (Henry's law constant) values at the same temperature.

Solution

Different gases have different \(\mathrm{K}_{\mathrm{H}}\) values at the same temperature. Thissuggests that \(\mathrm{K}_{\mathrm{H}}\) is a function of the nature of the gas.

\(\therefore\) Statement (A) and (D) both are correct.

The most commonly used form of Henry's law states that "the partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution" and is expressed as:

\(\mathrm{p}=\mathrm{K}_{\mathrm{H}} \mathrm{X}\)

Here \(\mathrm{K}_{\mathrm{H}}\) is Henry's law constant.
Question 8
4 / -1

A photocell employs the photoelectric effect to convert:

A
Change in the frequency of light into a change in electric voltage

B
Change in the intensity of illumination into a change in photoelectric current

C
Change in the intensity of illumination into a change in the work function of the photocathode

D
Change in the frequency of light into a change in the electric current

Solution

A photocell employs the photoelectric effect to convert a change in the intensity of illumination into a change in photoelectric current.

In the photoelectric effect when monochromatic radiations of suitable frequency fall on the photo-sensitive plate called the cathode, the photoelectrons are emitted which get accelerated towards the anode. These electrons flow in the outer circuit resulting in the photoelectric current.

Using the incident radiations of a fixed frequency, it is found that the photoelectric current increases linearly with the intensity of incident light. Hence, a photocell employs the photoelectric effect to convert the change in the intensity of illumination into a change in photoelectric current.
Question 9
4 / -1

According to Charles law:

A
At constant pressure, volume of gas is proportional to its absolute, temperature.

B
At constant pressure, the volume of a gas is not proportional to its absolute temperature.

C
At constant gauge pressure, the molecular volume of a gas is proportional to its absolute temperature.

D
At constant volume, the absolute pressure is proportional to absolute temperature.

Solution

According to the statement of charles law,

For a given mass of a gas, at constant pressure, volume is proportional to absolute temperature of the gas.

Charles law deals with volume and temperature. While Boyle’s law deals with pressure and volume and Dalton’s law deals with total pressure and partial pressure. Use the ideal gas equation to get a relation of Charles law.

According to ideal gas equation,

PV = nRT

Where,

P = pressure

V = volume

N = number of moles

R = universal gas constant, whose value is 8.314mol⁻¹K⁻¹

T = absolute temperature

Therefore, we can write,

V=(nR/P)T

From the above equation we can conclude that volume is directly proportional to temperature.

Mathematically it is written as,

V∝T

And above equation is nothing but Charles law which states that “At a fixed pressure and for given mass of a gas, the volume of a gas is proportional to its absolute temperature”

So as we can see in options, option (A) is correct because the meaning of its statement is the same as that of Charles' law.

Do not get confused between Boyle’s law and Charles law. Boyle’s law states that “At constant temperature, the pressure exerted by a fixed mass of a gas is inversely proportional to volume occupied by gas.
Question 10
4 / -1

The number of water molecule(s) not coordinated to copper ion directly in \(\mathrm{CuSO}_{4} . 5 \mathrm{H}_{2} \mathrm{O}\) is:

A
\(4\)

B
\(3\)

C
\(1\)

D
\(2\)

Solution

\(1\) water molecule as shown in the diagram is not coordinated to copper ion directly.

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Re-Attempt Weekly Quiz Competition

Chemistry Test-8

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Chemistry Test-8

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SHARING IS CARING

Question 1

Sulphur dioxide

Ammonia

Hydrogen sulphide

Hydrogen cyanide

Solution

Question 2

\(\mathrm{P}\) and \(\mathrm{V}\)

\(\mathrm{P}\) and \(\mathrm{T}\)

\(\mathrm{V}\) and\(\mathrm{T}\)

None of these

Solution

Question 3

\(5624{~kcal}\)

\(1.11 \times 10^{4}{~kcal}\)

\(86.24 \times 10^{3}{~kcal}\)

\(1100{~kcal}\)

Solution

Question 4

\(3\mathrm{CO ( g )+2 H _{2}( g ) \stackrel{340 atm }{\longrightarrow} CH _{3} OH (l)}\)

\(\mathrm{CO (g)+3 H _{2}( g ) \stackrel{340 atm}{\longrightarrow} CH _{3} OH (l)}\)

\(\mathrm{CO ( g )+2 H _{2}( g ) \stackrel{340 atm }{\longrightarrow} 2 CH _{3} OH (l)}\)

\(\mathrm{CO ( g )+2 H _{2}( g ) \stackrel{340 atm }{\longrightarrow} CH _{3} OH (l)}\)

Solution

Question 5

Non-metals have higher ionisation enthalpies and higher electronegativity then the metals

non metals and metalloids exist only in the p-block of the periodic table.

change of non-metallic to metallic character can be illustrated by the nature of oxide they form

all are correct,

Solution

Question 6

X- Natural rubber, Y- Gutta percha

X- Gutta percha , Y- Natural rubber

X- Natural rubber, Y- Natural rubber

X- Gutta percha, Y- Gutta percha

Solution

Question 7

The value of \(\mathrm{K}_{\mathrm{H}}\) increases with the function of the nature of the gas.

Higher the value of \(\mathrm{K}_{\mathrm{H}}\) at a given pressure, higher is the solubility of the gas in the liquids.

The partial pressure of the gas in vapour phase is proportional to the mole fraction of the gas in the solution.

Different gases have different \(\mathrm{K}_{\mathrm{H}}\) (Henry's law constant) values at the same temperature.

Solution

Question 8

Change in the frequency of light into a change in electric voltage

Change in the intensity of illumination into a change in photoelectric current

Change in the intensity of illumination into a change in the work function of the photocathode

Change in the frequency of light into a change in the electric current

Solution

Question 9

At constant pressure, volume of gas is proportional to its absolute, temperature.

At constant pressure, the volume of a gas is not proportional to its absolute temperature.

At constant gauge pressure, the molecular volume of a gas is proportional to its absolute temperature.

At constant volume, the absolute pressure is proportional to absolute temperature.

Solution

Question 10

\(4\)

\(3\)

\(1\)

\(2\)

Solution

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