Chapter 5.3, Problem 12P

To determine

To Find: How much silver will be deposited by a current of $12.0$ amperes in one hour.

Answer to Problem 12P

For 12 Amperes of Current $48.384$ g of silver in one hour.

Explanation of Solution

Given information:

For $1$ Ampere of current $1.12\times {10}^{-3}$ g silver will be deposited in one second.

For 1 second

  $1$ Ampere $\to 1.12\times {10}^{-3}$ g of silver

Multiplying with $12$ on both sides

We get

  $\Rightarrow 1\times 12$ Ampere $\to 1.12\times {10}^{-3}\times 12$ g of silver

  $\Rightarrow 12$ Ampere $\to 13.44\times {10}^{-3}$ g of silver

Therefore,

For 1 second for 12 Ampere of current $13.44\times {10}^{-3}$ g of silver.

For 12 amperes

  $1$ second $\to 13.44\times {10}^{-3}$ g of silver.

Multiplying with $3600$ on both sides

We get

  $\Rightarrow 1\times 3600$ seconds $\to 13.44\times {10}^{-3}\times 3600$ g of silver.

  $\Rightarrow 3600$ seconds $\to 48.384$ g of silver.

Therefore,

For 12 Amperes of Current $48.384$ g of silver in one hour.