A) \[{{600}^{o}}C\]
B) \[{{C}_{p}}=5.006+0.0026T\]
C) \[138\text{ }cal/deg\]
D) \[13.75\text{ }cal/deg\]
Correct Answer: B
Solution :
Volume of Ag deposited \[I<IV<II<III\] Mass of Ag deposited = Volume \[\text{I}<III<II<IV\] density Hence, number of Ag atoms deposited \[I>III>IV>II\] \[FeC{{r}_{2}}{{O}_{4}}\xrightarrow{I}N{{a}_{2}}Cr{{O}_{4}}\xrightarrow{II}C{{r}_{2}}{{O}_{3}}\xrightarrow{III}Cr\] \[NaOH/air\]
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