General Principles and Processes of Isolation of Elements Test

Result

General Principles and Processes of Isolation of Elements Test - 24

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

According to Ellingham diagram, the oxidation reaction of carbon to carbon monoxide may be used to reduce which one of the following oxides at the lowest temperature?

A
$$Al_2O_3$$

B
$$Cu_2O$$

C
$$MgO$$

D
$$ZnO$$

Solution

$$Cu_2O$$ is least stable among all so the oxidation reaction of carbon to carbon monoxide may be used to reduce this at the lowest temperature.
Question 2
1 / -0

Directions For Questions

The Ellingham diagram for zinc, magnesium and carbon converting into corresponding oxides is shown below :

...view full instructions

At what temperature, zinc and carbon have equal affinity for oxygen?

A
$$1000^{\small\circ}C$$

B
$$1500^{\small\circ}C$$

C
$$500^{\small\circ}C$$

D
$$1200^{\small\circ}C$$

Solution

As both oxides met at $$1000^oC$$ in graph so at this temperature both C and Zn have equal affinity for oxygen.
Question 3
1 / -0

Froth flotation process used for the concentration of sulfide ore _______________.

A
is based on the difference in wettability of different minerals

B
uses sodium ethly xanthate, $$C_{2}H_{5}OCS_{2}Na$$ as collector

C
uses $$NaCN$$ as depressant in the mixture of ZnS and PbS when ZnS forms soluble complex and Pbs forms froth

D
all of the above statements are correct

Solution

(A) Froth floatation process used for the concentration of sulphide ore is based on the difference in wettability of different minerals. The metallic sulphide particles of ore are preferentially wetted by oil and the gangue particles by water.
(B) Froth floatation process used for the concentration of sulphide ore uses sodium ethyl xanthate, $$C_{2}H_{5}OCS_{2}Na$$ as collector. Other examples of collector include ethyl xanthate and potassium ethyl xanthate. They attach themselves to the grains of mineral by polar groups so that minerals become water repellent and pass on into the froth.
(C) Froth floatation process used for the concentration of sulphide ore uses NaCN as depressant in the mixture of ZnS and PbS when ZnS forms soluble complex and Pbs forms froth. Other depressant used is KCN.
Depressants depress the floating property of one of the components of the ore and help in the separation of different minerals present in the same ore.
Thus, all of the above statements are correct.
Question 4
1 / -0

Directions For Questions

For a spontaneous reaction, the free energy change must be negative.

$$\Delta G=\Delta H- T $$ $$\Delta S$$

$$\Delta$$$$H$$ is the enthalpy change during the reaction. $$T$$ is the absolute temperature, and $$\Delta$$$$S$$ is the change in entropy during the reaction. Consider a reaction such as the formation of an oxide.

$$M+O_{2}$$ $$\rightarrow$$ $$MO$$

Dioxygen is used up in the course of this reaction. Gases have a more random structure (less ordered) than liquid or solids. Consequently, gases have a higher entropy than liquids and solids. In this reaction, $$S$$ (entropy or randomness) decreases, hence $$\Delta$$S is negative. Thus, if the temperature is raised then $$T\Delta$$S becomes more negative. Since $$T\Delta$$S is subtracted in the equation, then $$\Delta$$G becomes negative.

Thus, the free energy change increases with the increase in temperature. The free energy changes that occur when one mole of common reactant (in this case dioxygen) is used may be plotted graphically against temperature for a number of reactions of metals to their oxides. The following plot is called an Ellingham diagram for metal oxide. Understanding of Ellingham diagram is extremely important for the efficient extraction of metals

...view full instructions

Free energy change of Hg and Mg for the conversion to oxides the slope of $$\Delta G$$ vs. T has been changes above the boiling points of the given metal because -

A
Above the boiling point of the metal, the entropy of metal become more

B
Above the boiling point of the metal, the entropy is constant

C
Above the boiling point of the metal the entropy is decreased

D
All of these
Question 5
1 / -0

Select the correct statement.

A
During the decomposition of an oxide into oxygen and metal vapour, entropy increases.

B
Decomposition of an oxide is an endothermic change.

C
To make $$\triangle G^{0}$$ negative, temperature should be high enough so that $$T$$ $$\triangle$$ $$S^{0}$$> $$\triangle$$ $$H^{0}$$.

D
All are correct answers.

Solution

During the decomposition of an oxide into oxygen and metal vapour entropy increases as gas is produced in the reaction.

Decomposition of an oxide is an endothermic change. This is because it is reverse of combustion reaction and all combustion reactions are exothermic in nature.

Since both $$\Delta H $$ and $$\Delta S $$ are positive, the temperature should be high enough so that $$T\Delta S >\Delta H $$,

Thus, all are correct answers.
Question 6
1 / -0

Directions For Questions

For a spontaneous reaction, the free energy change must be negative.

$$\Delta G=\Delta H- T $$ $$\Delta S$$

$$\Delta$$$$H$$ is the enthalpy change during the reaction. $$T$$ is the absolute temperature, and $$\Delta$$$$S$$ is the change in entropy during the reaction. Consider a reaction such as the formation of an oxide.

$$M+O_{2}$$ $$\rightarrow$$ $$MO$$

Dioxygen is used up in the course of this reaction. Gases have a more random structure (less ordered) than liquid or solids. Consequently, gases have a higher entropy than liquids and solids. In this reaction, $$S$$ (entropy or randomness) decreases, hence $$\Delta$$S is negative. Thus, if the temperature is raised then $$T\Delta$$S becomes more negative. Since $$T\Delta$$S is subtracted in the equation, then $$\Delta$$G becomes negative.

Thus, the free energy change increases with the increase in temperature. The free energy changes that occur when one mole of common reactant (in this case dioxygen) is used may be plotted graphically against temperature for a number of reactions of metals to their oxides. The following plot is called an Ellingham diagram for metal oxide. Understanding of Ellingham diagram is extremely important for the efficient extraction of metals

...view full instructions

For the conversion of Ca(s) to CaO(s) which of the following represent the $$\Delta G$$ vs. T

A

B

C

D
Question 7
1 / -0

Sulphide ores are common for metals:

A
Ag, Cu and Pb

B
Ag, Cu and Sn

C
Ag, Mg and Pb

D
none
Solution
Correct Answer: Option A
Explanation:

Sulfide ores are common for the metals:
- Copper: Copper glance $$ Cu_{2}S$$ and Copper pyrite $$ CuFeS_{2}$$
- Lead: Galena $$PbS$$
- Mercury: Cinnabar $$HgS$$
- Iron: Iron Pyrite $$FeS_{2}$$
- Silver: Argentite $$Ag_{2}S$$
Question 8
1 / -0

Gold is extracted by hydrometallurgical process, based on its property:

A
of being electropositive

B
of being less reactive

C
to form complex which are water soluble

D
None of these

Solution

Hydrometallurgy is a method for obtaining metals from their ores.

It is a technique within the field of extractive metallurgy involving the use of aqueous chemistry for the recovery of metals from ores, concentrates, and recycled or residual materials.

Gold is extracted by hydrometallurgical process, based on its property to form complexes which are water soluble.

Option C is correct.
Question 9
1 / -0

The oxide of metal $${R}$$ can be reduced by the metal $${P}$$ and metal $${R}$$ can reduce the oxide of metal $${Q}$$. Then the decreasing order of the reactivity of metal $${P}$$, $${Q}$$ and $${R}$$ with oxygen is:

A
$$P>Q>R$$

B
$$P>R>Q$$

C
$$R>P>Q$$

D
$$Q>P>R$$
Solution
Correct Answer: Option $$B$$.

Explanation:
A more reactive metal can reduce the oxide of a less reactive metal.
Since $$P$$ can reduce the oxide of $$R$$, $$P$$ is more reactive than $$R$$.
Since $$R$$ can reduce the oxide of $$Q$$, $$R$$ is more reactive than $$Q$$.
So, the order of reactivity will be $$P>R>Q$$
Question 10
1 / -0

Method used for obtained highly pure silicon, used as a semiconductor material, is

A
oxidation

B
electrochemical

C
crystallization

D
zone refining

Solution

High purity silicon is required for the semiconductor industry and the needed levels cannot be reached by chemical methods alone. So for this;
A standard technique for purifying semiconductors (zone refining process) is used. In this, a horizontal semi-cylinder of material is held in a boat made from an unreactive material such as quartz and a narrow molten zone is passed from one end to the other of the charge. At the end of the pass the heater is rapidly moved to the beginning of the charge and the process is repeated. Impurities are dissolved in the molten zone and preferentially segregated to the first to freeze or last to freeze ends of the charge.