A) 7720 sq
B) 120 min
C) 772 s
D) 77.2 s
Correct Answer: A
Solution :
Volume of Ag metal required to coat the coffee tray \[=\,\,\,30\times 15\times 2\times 0.1 =90 c{{m}^{3}}\] \[\operatorname{Mass} of Ag required =\,\,\frac{Volume\,\,of\,\,Ag}{Density\,\,of\,\,Ag}\] \[=\,\,\,\frac{90}{10.5}=8.57\,\,g\] Gram equivalent of \[\operatorname{Ag} = \,\frac{8.57}{108} = 0.08\] \[Number of Faraday passed = 0.08\] Number of Coulomb?s passed \[= \,0.08 \times 96500\] \[=\,\,\,7720 C\]\[\operatorname{time} = 7720 s\]You need to login to perform this action.
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