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wedge (eq) of 0.10N solution of CaI(2) i...

`wedge _(eq)` of `0.10N` solution of `CaI_(2)` is `100.0 Scm^(2)eq^(-1)` at `298 K.G^(**)` of the cell `=0.25 cm^(-1)`. How much current will flow potential difference between the electrode is `5V` ?

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To solve the problem step-by-step, we will follow the concepts of equivalent conductivity, resistance, and Ohm's law. ### Step 1: Understand the Given Data We are given: - Equivalent conductivity (λ) of 0.10 N CaI₂ = 100 cm²/eq - Cell constant (G*) = 0.25 cm⁻¹ - Potential difference (V) = 5 V ...
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