330174 The density of Cu is \({\rm{8}}{\rm{.94}}\,\,{\rm{c}}{{\rm{m}}^{{\rm{ - 3}}}}\) The quantity of electricity needed to plate an area \({\rm{10}}\,\,{\rm{cm \times 10}}\,\,{\rm{cm}}\) to a thickness of \({\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}\,\,{\rm{cm}}\) using \({\rm{CuS}}{{\rm{O}}_{\rm{4}}}\) solution would be
330176
Match Column I (Conversion) with Column II (Number of Faraday required).
Column I
Column II
A
\({\rm{1 mol}}\,{{\rm{H}}_{{\rm{2}}\,}}{\rm{O}}\,\,{\rm{to}}\,\,{{\rm{O}}_{\rm{2}}}\)
P
3F
B
\({\rm{1 mol}}\,{\rm{MnO}}_4^ - \,\,{\rm{to}}\,\,{\rm{M}}{{\rm{n}}^{{\rm{2 + }}}}\)
Q
2F
C
\[\begin{array}{l}
{\rm{1}}{\rm{.5 \,mol}}\,{\rm{of}}\,{\rm{Ca}}\,{\rm{from}}\,\,\\
{\rm{molten}}\,{\rm{CaC}}{{\rm{l}}_{\rm{2}}}
\end{array}\]
R
1F
D
\({\rm{1 mol}}\,{\rm{of}}\,{\rm{FeO}}\,\,{\rm{to}}\,{\rm{F}}{{\rm{e}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}\)
S
5F
330174 The density of Cu is \({\rm{8}}{\rm{.94}}\,\,{\rm{c}}{{\rm{m}}^{{\rm{ - 3}}}}\) The quantity of electricity needed to plate an area \({\rm{10}}\,\,{\rm{cm \times 10}}\,\,{\rm{cm}}\) to a thickness of \({\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}\,\,{\rm{cm}}\) using \({\rm{CuS}}{{\rm{O}}_{\rm{4}}}\) solution would be
330176
Match Column I (Conversion) with Column II (Number of Faraday required).
Column I
Column II
A
\({\rm{1 mol}}\,{{\rm{H}}_{{\rm{2}}\,}}{\rm{O}}\,\,{\rm{to}}\,\,{{\rm{O}}_{\rm{2}}}\)
P
3F
B
\({\rm{1 mol}}\,{\rm{MnO}}_4^ - \,\,{\rm{to}}\,\,{\rm{M}}{{\rm{n}}^{{\rm{2 + }}}}\)
Q
2F
C
\[\begin{array}{l}
{\rm{1}}{\rm{.5 \,mol}}\,{\rm{of}}\,{\rm{Ca}}\,{\rm{from}}\,\,\\
{\rm{molten}}\,{\rm{CaC}}{{\rm{l}}_{\rm{2}}}
\end{array}\]
R
1F
D
\({\rm{1 mol}}\,{\rm{of}}\,{\rm{FeO}}\,\,{\rm{to}}\,{\rm{F}}{{\rm{e}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}\)
S
5F
330174 The density of Cu is \({\rm{8}}{\rm{.94}}\,\,{\rm{c}}{{\rm{m}}^{{\rm{ - 3}}}}\) The quantity of electricity needed to plate an area \({\rm{10}}\,\,{\rm{cm \times 10}}\,\,{\rm{cm}}\) to a thickness of \({\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}\,\,{\rm{cm}}\) using \({\rm{CuS}}{{\rm{O}}_{\rm{4}}}\) solution would be
330176
Match Column I (Conversion) with Column II (Number of Faraday required).
Column I
Column II
A
\({\rm{1 mol}}\,{{\rm{H}}_{{\rm{2}}\,}}{\rm{O}}\,\,{\rm{to}}\,\,{{\rm{O}}_{\rm{2}}}\)
P
3F
B
\({\rm{1 mol}}\,{\rm{MnO}}_4^ - \,\,{\rm{to}}\,\,{\rm{M}}{{\rm{n}}^{{\rm{2 + }}}}\)
Q
2F
C
\[\begin{array}{l}
{\rm{1}}{\rm{.5 \,mol}}\,{\rm{of}}\,{\rm{Ca}}\,{\rm{from}}\,\,\\
{\rm{molten}}\,{\rm{CaC}}{{\rm{l}}_{\rm{2}}}
\end{array}\]
R
1F
D
\({\rm{1 mol}}\,{\rm{of}}\,{\rm{FeO}}\,\,{\rm{to}}\,{\rm{F}}{{\rm{e}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}\)
S
5F
330174 The density of Cu is \({\rm{8}}{\rm{.94}}\,\,{\rm{c}}{{\rm{m}}^{{\rm{ - 3}}}}\) The quantity of electricity needed to plate an area \({\rm{10}}\,\,{\rm{cm \times 10}}\,\,{\rm{cm}}\) to a thickness of \({\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}\,\,{\rm{cm}}\) using \({\rm{CuS}}{{\rm{O}}_{\rm{4}}}\) solution would be
330176
Match Column I (Conversion) with Column II (Number of Faraday required).
Column I
Column II
A
\({\rm{1 mol}}\,{{\rm{H}}_{{\rm{2}}\,}}{\rm{O}}\,\,{\rm{to}}\,\,{{\rm{O}}_{\rm{2}}}\)
P
3F
B
\({\rm{1 mol}}\,{\rm{MnO}}_4^ - \,\,{\rm{to}}\,\,{\rm{M}}{{\rm{n}}^{{\rm{2 + }}}}\)
Q
2F
C
\[\begin{array}{l}
{\rm{1}}{\rm{.5 \,mol}}\,{\rm{of}}\,{\rm{Ca}}\,{\rm{from}}\,\,\\
{\rm{molten}}\,{\rm{CaC}}{{\rm{l}}_{\rm{2}}}
\end{array}\]
R
1F
D
\({\rm{1 mol}}\,{\rm{of}}\,{\rm{FeO}}\,\,{\rm{to}}\,{\rm{F}}{{\rm{e}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}\)
S
5F