# A cell, transferred 0.146 mol of electrons with a constant current of 1.24 A. How long, in hours,...

## Question:

A cell, transferred 0.146 mol of electrons with a constant current of 1.24 A. How long, in hours, did this take?

## Charge:

Charge is a fundamental property of particles. Charge as a property is defined as the ability of a particle to be affected by electric and magnetic fields. If we have a flowing charge distribution (that is, the particles are all moving continuously along a certain region), then if we average this flow over time, we get the electric current.

## Answer and Explanation:

Given:

- {eq}\displaystyle n = 0.146\ mol {/eq} is the amount of electrons

- {eq}\displaystyle I = 1.24\ A {/eq} is the current

Let us first determine the exact number of electrons we have. We multiply our amount of electrons by Avogadro's number:

{eq}\displaystyle N = (0.146\ mol)(6.022\ \times\ 10^{23}) {/eq}

We get:

{eq}\displaystyle N = 8.7921\ \times\ 10^{22} {/eq}

Now since an electron carries a constant charge of {eq}\displaystyle q_e = 1.602\ \times \10^{-19}\ C {/eq}, then we can express our electric current as:

{eq}\displaystyle I = \frac{Nq_e}{t} {/eq}

Let us isolate our time here:

{eq}\displaystyle t = \frac{Nq_e}{I} {/eq}

We substitute:

{eq}\displaystyle t = \frac{( 8.7921\ \times\ 10^{22})(1.602\ \times \10^{-19}\ C)}{1.24\ A} {/eq}

We get:

{eq}\displaystyle t = 1.1359\ \times\ 10^4\ s {/eq}

We now convert this to hours:

{eq}\displaystyle t = 1.1359\ \times\ 10^4\ s \left(\frac{1\ min}{60\ s} \right)\left(\frac{1\ h}{60\ min} \right) {/eq}

We cancel like units:

{eq}\displaystyle t = 1.1359\ \times\ 10^4\ \require{cancel}\cancel{s} \left(\frac{1\ \require{cancel}\cancel{min}}{60\ \require{cancel}\cancel{s}} \right)\left(\frac{1\ h}{60\ \require{cancel}\cancel{min}} \right) {/eq}

We thus obtain:

{eq}\displaystyle \boxed{t = 3.16\ h} {/eq}

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from CLEP Natural Sciences: Study Guide & Test Prep

Chapter 6 / Lesson 7