In a zinc manganese dioxide dry cell, the anode is madeup of `Zn` and cathode of carbon rod surrounded by a mixture of `MnO_(2)`, carbon, `NH_(4)Cl`, and `ZnCl_(2)` in aqueous base. The cathodic reaction is `: `
`MnO_(2)(s)+Zn^(2+)+2e^(-) rarr ZnMn_(2)O_(4)(s)`
`8.7g` of `MnO_(2)` is taken in the cathodic compartment. How many days will the dry cell continue to give a current of `9.65xx10^(-3)A` ? `(` Atomic weight of `Mn=55)(Mw` of `MnO_(2)=87g mol^(-1))`

Explanation

To solve the problem, we need to determine how long the zinc manganese dioxide dry cell will continue to provide a current of \(9.65 \times 10^{-3} \, \text{A}\) given that \(8.7 \, \text{g}\) of \(MnO_2\) is present in the cathodic compartment.

### Step-by-step Solution:

1. **Determine the equivalent weight of \(MnO_2\)**:
- The reaction provided is:
\[
MnO_2(s) + Zn^{2+} + 2e^- \rightarrow ZnMn_2O_4(s)
...