Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 8

Roughly speaking, the radius of an atom is about 10,000 times greater than that of its nucleus. If an atom were magnified so that the radius of its nucleus became \(10 \mathrm{~cm},\) what would be the radius of the atom in miles? \((1 \mathrm{mi}=1609 \mathrm{~m} .)\)

Short Answer

Expert verified
If magnified such that the radius of the nucleus became 10 cm, the radius of the atom would be approximately \(0.621\) miles.

Step by step solution

01

Understand the Ratio

The ratio of the radius of an atom to the radius of its nucleus is approximately 10,000:1. Meaning, the atom is around 10,000 times bigger than its nucleus.
02

Scale up the Nucleus

The problem states that the radius of the nucleus has been magnified to 10 cm. We need to use the same scale factor to find the radius of the atom.
03

Compute the Atom's Radius in cm

Use the ratio to compute the atom's radius. This means multiplying the given size of the nucleus \(10 cm\) by \(10,000\) which results in \((10 \times 10,000)=100,000 cm\)
04

Convert cm to m

The next step is to convert this value from centimeters to meters, as the final answer must be given in miles and the conversion rate from meters to miles is given. Since \(1 m = 100 cm\), hence the radius is \((100,000 \div 100) = 1,000 m\)
05

Convert m to mi

Lastly, convert the radius from meters to miles. The problem states that \(1 mi = 1609 m\). So, divide the radius in meters by the conversion factor \((1,000 \div 1609) = 0.621 miles\)

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

Which of the following compounds are likely to be ionic? Which are likely to be molecular? \(\mathrm{SiCl}_{4}\), LiF, \(\mathrm{BaCl}_{2}, \mathrm{~B}_{2} \mathrm{H}_{6}, \mathrm{KCl}, \mathrm{C}_{2} \mathrm{H}_{4}\)

Explain why the chemical formulas of ionic compounds are usually the same as their empirical formulas.

Name these compounds: (a) \(\mathrm{KClO},\) (b) \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\), (c) \(\mathrm{FeCl}_{2},\) (d) \(\mathrm{KMnO}_{4},\) (e) \(\mathrm{CsClO}_{3},\) (f) \(\mathrm{HIO}\) (g) \(\mathrm{FeO}\) (h) \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) (i) \(\mathrm{TiCl}_{4},\) (j) \(\mathrm{NaH},(\mathrm{k}) \mathrm{Li}_{3} \mathrm{~N},\) (l) \(\mathrm{Na}_{2} \mathrm{O}\) \((\mathrm{m}) \mathrm{Na}_{2} \mathrm{O}_{2},(\mathrm{n}) \mathrm{FeCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}\)

Define molecular formula and empirical formula. What are the similarities and differences between the empirical formula and molecular formula of a compound?

Which of these symbols provides more information about the atom: \({ }^{23}\) Na or \({ }_{11}\) Na? Explain.
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Reactions

Read Explanation

Nuclear Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Chemical Analysis

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