In a certain electrolysis experiment,0.650 g | vedclass

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(A) According to Faraday's second law of electrolysis,when cells are connected in series,the mass of substances deposited is proportional to their equ

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$0.635$

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(A) According to Faraday's second law of electrolysis,when cells are connected in series,the mass of substances deposited is proportional to their equivalent masses: $\frac{W_{Zn}}{E_{Zn}} = \frac{W_{Cu}}{E_{Cu}}$Equivalent mass of $Zn = \frac{	ext{Molar mass}}{	ext{Valency factor}} = \frac{65}{2} = 32.5 \ g \ eq^{-1}$Equivalent mass of $Cu = \frac{	ext{Molar mass}}{	ext{Valency factor}} = \frac{63.5}{2} = 31.75 \ g \ eq^{-1}$Substituting the values: $\frac{0.650}{32.5} = \frac{W_{Cu}}{31.75}$$W_{Cu} = \frac{0.650 	imes 31.75}{32.5} = 0.635 \ g$

In a certain electrolysis experiment,$0.650 \ g$ of zinc is deposited in a cell having $ZnSO_4$ solution. Calculate the mass of $Cu$ deposited in another cell having $CuSO_4$ solution arranged in series with the first cell (in $g$)? (Molar mass of $Zn = 65 \ g \ mol^{-1}$,$Cu = 63.5 \ g \ mol^{-1}$)

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