Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Q17.2E (page 957)

For the scenario in the previous question, how many electrons moved through the circuit.

Short Answer

Expert verified

Number of electrons is \({\rm{3}}{\rm{.28 \times 1}}{{\rm{0}}^{{\rm{22}}}}\)

Step by step solution

01

Formula of number of electron

The charge of 1 electron is equal to\({\bf{1}}{\bf{.6022 \times 1}}{{\bf{0}}^{{\bf{ - 19}}}}{\bf{C}}\). Thus, in 1C, the number of electrons will be:

\(\begin{array}{l}{\bf{1C = }}\frac{{\bf{1}}}{{{\bf{1}}{\bf{.6022 \times 1}}{{\bf{0}}^{{\bf{ - 19}}}}}}{\bf{ }}\\{\bf{electrons = 6}}{\bf{.2414 \times 1}}{{\bf{0}}^{{\bf{18}}}}{\bf{ electrons}}\end{array}\)

02

Determine the number of electrons

For 5250 C of charge, the number of electrons can be calculated as follows:

\({\rm{1C}} \to {\rm{6}}{\rm{.2414 \times 1}}{{\rm{0}}^{{\rm{18}}}}{\rm{ electrons }}\)

Thus, \(5250{\rm{C}} \to (5250)\left( {6.2414 \times {{10}^{18}}{\rm{ electrons }}} \right) = 3.28 \times {10^{22}}{\rm{ electrons }}\)

Therefore, the number of electrons are \(3.28 \times 1{0^{22}}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

What value of Q for the previous concentration cell would result in a voltage of 0.10 V? If the concentration of zinc ion at the cathode was 0.50 M, what was the concentration at the anode?

Determine \({\bf{\Delta G}}\) and \({\bf{\Delta G}}^\circ \) for each of the reactions in the previous problem.

From the information provided, use cell notation to describe the following systems:

(a) In one half-cell, a solution of \({\rm{Pt}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}\)forms Pt metal, while in the other half-cell, Cu metal goes into a \({\rm{Cu}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}\)solution with all solute concentrations 1M.

(b) The cathode consists of a gold electrode in a \(0.55{\rm{MAu}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_3}\) solution and the anode is a magnesium electrode in \(0.75{\rm{MMg}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}\) solution.

(c) One half-cell consists of a silver electrode in a \(1{\rm{MAgN}}{{\rm{O}}_3}\) solution, and in the other half-cell, a copper electrode in \(1M{\rm{Cu}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}\) is oxidized.

Why is a salt bridge necessary in galvanic cells like the one in Figure 17.4?

An active (metal) electrode was found to gain mass as the oxidation-reduction reaction was allowed to proceed. Was the electrode part of the anode or cathode? Explain.
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Analysis

Read Explanation

Inorganic Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free