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(B) The reduction reaction at the cathode is: $Ca^{2+} + 2e^- \longrightarrow Ca_{(s)}$.From the reaction,the $n$-factor $(nf)$ for calcium is $2$.The

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$0.04$

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(B) The reduction reaction at the cathode is: $Ca^{2+} + 2e^- \longrightarrow Ca_{(s)}$.From the reaction,the $n$-factor $(nf)$ for calcium is $2$.The molar mass of calcium is $40 \ g/mol$.The number of moles of calcium deposited is: $	ext{moles} = \frac{0.8 \ g}{40 \ g/mol} = 0.02 \ mol$.The number of equivalents of calcium is: $	ext{equivalents} = 	ext{moles} 	imes nf = 0.02 	imes 2 = 0.04 \ eq$.According to Faraday's law of electrolysis,$1 \ Faraday$ of electricity is required to deposit $1 \ equivalent$ of a substance.Therefore,to deposit $0.04 \ equivalents$ of calcium,$0.04 \ F$ of electricity is required.

How many Faraday of electricity is required to deposit $0.8 \ g$ of calcium at cathode by the electrolysis of $CaCl_2$ (in $F$)?

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