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Chemistry-

General

Easy

Question

Find out A:

${C u}_{2} S$
${C u F e S}_{2}$
${F e S}_{2}$
${F e}_{3} O_{4}$

The correct answer is: $C u F e S subscript 2 end subscript$

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ThenBis:

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Extractionofaluminiumcanbeunderstoodby:

Electrolytic reduction of $A l subscript 2 end subscript O subscript 3 end subscript$
Electrolytic:( $A l subscript 2 end subscript O subscript 3 end subscript$ +Cryolite+ $C a F subscript 2 end subscript$ )
Cathode: Carbon inside the Fe-container
Anode: Graphite rods
The molten electrolytes contain $N a to the power of plus end exponent comma A l to the power of 3 plus end exponent$ , and $C a to the power of 2 plus end exponent$ only Al gets deposited at cathode because:

Extractionofaluminiumcanbeunderstoodby:

Electrolytic reduction of $A l subscript 2 end subscript O subscript 3 end subscript$
Electrolytic:( $A l subscript 2 end subscript O subscript 3 end subscript$ +Cryolite+ $C a F subscript 2 end subscript$ )
Cathode: Carbon inside the Fe-container
Anode: Graphite rods
The molten electrolytes contain $N a to the power of plus end exponent comma A l to the power of 3 plus end exponent$ , and $C a to the power of 2 plus end exponent$ only Al gets deposited at cathode because:

chemistry-General

Extraction of aluminium can be understood by:

Electrolytic reduction of $A l subscript 2 end subscript O subscript 3 end subscript$
Electrolytic:( $A l subscript 2 end subscript O subscript 3 end subscript$ +Cryolite+ $C a F subscript 2 end subscript$ )
Cathode: Carbon inside the Fe-container
Anode: Graphite rods
The function of fluorspar( $C a F subscript 2 end subscript$ )is:

Extraction of aluminium can be understood by:

Electrolytic reduction of $A l subscript 2 end subscript O subscript 3 end subscript$
Electrolytic:( $A l subscript 2 end subscript O subscript 3 end subscript$ +Cryolite+ $C a F subscript 2 end subscript$ )
Cathode: Carbon inside the Fe-container
Anode: Graphite rods
The function of fluorspar( $C a F subscript 2 end subscript$ )is:

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In the following nuclear reaction $blank subscript 13 end subscript A l to the power of 27 end exponent plus blank subscript 2 end subscript H e to the power of 4 end exponent rightwards arrow blank subscript 15 end subscript P to the power of 30 end exponent plus X$ , X will be

In the following nuclear reaction $blank subscript 13 end subscript A l to the power of 27 end exponent plus blank subscript 2 end subscript H e to the power of 4 end exponent rightwards arrow blank subscript 15 end subscript P to the power of 30 end exponent plus X$ , X will be

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When $blank subscript 15 end subscript P to the power of 30 end exponent$ decays to become $blank subscript 14 end subscript S i to the power of 30 end exponent$ , the particle released is

When $blank subscript 15 end subscript P to the power of 30 end exponent$ decays to become $blank subscript 14 end subscript S i to the power of 30 end exponent$ , the particle released is

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The binding energies of a deutron and an $alpha$ -particle are 1.125, 7.2 MeV/nucleon respectively. The more stable of the two, is

The binding energies of a deutron and an $alpha$ -particle are 1.125, 7.2 MeV/nucleon respectively. The more stable of the two, is

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