Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 73

Determine the only possible \(2+\) ion for which the following two conditions are both satisfied: \(\bullet\) The net ionic charge is one-tenth the nuclear charge. \(\bullet\) The number of neutrons is four more than the number of electrons.

Short Answer

Expert verified
The only possible \(2+\) ion for which both conditions are satisfied is an atomic ion with 20 protons (nuclear charge), 18 electrons, and 22 neutrons. This corresponds to the isotope of Calcium, represented as \(^{42}Ca^{2+}\).

Step by step solution

01

Interpret the conditions into equations

The conditions given can be expressed as follows: 1. The net ionic charge is one-tenth the nuclear charge. The net ionic charge (nuclear charge - number of electrons) is \(2+\), and the nuclear charge is the number of protons (let's denote this as \(p\)). Therefore, this condition can be written as \(2 = p/10\). 2. The number of neutrons is four more than the number of electrons. Let's denote the number of neutrons as \(n\) and the number of electrons as \(e\). This can be written as \(n = e + 4\).
02

Solve the first equation

Solving the first equation \(2 = p/10\) for \(p\), we find that \(p = 20\). So, the ion has 20 protons.
03

Determine the number of electrons

Knowing that it's a \(2+\) ion, it means that the ion has lost 2 electrons. So, the number of electrons (\(e\)) would be 20 (the number of protons) - 2 (due to the \(2+\) charge), which gives us 18 electrons.
04

Solve the second equation

Now, substituting the number of electrons into the second equation \(n = e + 4\), we get \(n = 18 + 4 = 22\). So, the ion has 22 neutrons.

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

Before \(1961,\) the standard for atomic masses was the isotope \(^{16} \mathrm{O},\) to which physicists assigned a value of exactly \(16 .\) At the same time, chemists assigned a value of exactly 16 to the naturally occurring mixture of the isotopes \(^{16} \mathrm{O},^{17} \mathrm{O},\) and \(^{18} \mathrm{O}\). Would you expect atomic masses listed in a 60 -year-old text to be the same, generally higher, or generally lower than in this text? Explain.

Refer to the periodic table inside the front cover and identify (a) the element that is in group 14 and the fourth period (b) one element similar to and one unlike sulfur (c) the alkali metal in the fifth period (d) the halogen element in the sixth period

Appendix E describes a useful study aid known as concept mapping. Using the method presented in Appendix \(\mathrm{E}\), construct a concept map illustrating the different concepts in Sections \(2-7\) and \(2-8\).

Americium-241 is a radioactive isotope that is used in high-precision gas and smoke detectors. How many neutrons, protons, and electrons are there in an atom of americium-241?

A particular silver solder (used in the electronics industry to join electrical components) is to have the atom ratio of \(5.00 \mathrm{Ag} / 4.00 \mathrm{Cu} / 1.00 \mathrm{Zn}\). What masses of the three metals must be melted together to prepare \(1.00 \mathrm{kg}\) of the solder?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Chemical Reactions

Read Explanation

Physical Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Ionic and Molecular Compounds

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