Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 105

Gold has a density of \(19.3 \mathrm{~g} / \mathrm{mL}\). Suppose you have \(100.0\) glonkins of gold. What volume in liters will the gold occupy? Here are some conversion factors to help you: \(0.911\) ounce per glonkin and \(28.35 \mathrm{~g}\) per ounce. Use unit analysis to calculate your answer, and show your work. Treat both conversion factors as exact.

Short Answer

Expert verified
The gold will occupy approximately \(0.137\) liters of volume.

Step by step solution

01

Use the conversion factor of \(0.911\) ounce per glonkin to convert the given gold mass of \(100.0\) glonkins to ounces: \(100.0\,\text{glonkins} \times \frac{0.911\,\text{ounce}}{1\,\text{glonkin}}\) #Step 2: Convert ounces to grams#:

Use the conversion factor of \(28.35\) grams per ounce to convert the mass in ounces to grams: \(100.0\,\text{glonkins} \times \frac{0.911\,\text{ounce}}{1\,\text{glonkin}} \times \frac{28.35\,\text{g}}{1\,\text{ounce}}\) #Step 3: Find the volume in milliliters#:
02

Divide the mass in grams by the density of gold (19.3 g/mL) to get the volume in milliliters: \(\frac{100.0\,\text{glonkins} \times \frac{0.911\,\text{ounce}}{1\,\text{glonkin}} \times \frac{28.35\,\text{g}}{1\,\text{ounce}}}{19.3\frac{\text{g}}{\text{mL}}}\) #Step 4: Convert volume in milliliters to liters#:

Use the conversion factor of \(1\) liter \(=1000\) milliliters to convert volume in milliliters to liters: \(\frac{100.0\,\text{glonkins} \times \frac{0.911\,\text{ounce}}{1\,\text{glonkin}} \times \frac{28.35\,\text{g}}{1\,\text{ounce}}}{19.3\frac{\text{g}}{\text{mL}}} \times \frac{1\,\text{L}}{1000\,\text{mL}}\) Now, calculate the volume of gold in liters: \(\frac{100.0\,\text{glonkins} \times 0.911\,\text{ounce} \times 28.35\,\text{g}}{19.3\,\text{g/mL} \times 1000\,\text{mL}} \approx 0.137\,\text{L}\) So, the gold will occupy approximately \(0.137\) liters of volume.

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

A train traveling at \(45.0\) miles \(/ \mathrm{h}\) has to make a trip of \(100.0\) miles. How many minutes will the trip take? Use unit analysis to calculate your answer, and show your work.

Using a bathroom scale, a tub, a sponge, and a measuring cup, explain how you would measure your own density.

A student reports a series of five length measurements that are accurate but not precise. Is it more likely that his laboratory technique is very good but the measuring instrument is bad, or that his laboratory technique is bad? Explain.

The density of an irregularly shaped object is determined by immersing the object in water. If the mass of the object is \(8.34 \mathrm{~g}\) and the water level rises from \(25.00 \mathrm{~mL}\) to \(28.10 \mathrm{~mL}\), what is the density of the object in grams per milliliter?

On a hot summer day, you want to cool two glasses of warm lemonade, but have no ice. Not wanting to wait until you can make some, you place two small metal blocks in the freezer. One block is pure iron, the other pure aluminum, and each has a mass of exactly \(50 \mathrm{~g}\). After both have cooled to \(-10^{\circ} \mathrm{C}\), you put them into separate glasses and add \(200 \mathrm{~mL}\) of warm lemonade to each. After a few minutes, both blocks have warmed up to \(+10{ }^{\circ} \mathrm{C}\). At this point, is the lemonade in one glass cooler than the lemonade in the other glass? If so, which is cooler and why? (Despite all the numerical information, you should be able to use specific heat values from Table \(2.5\) to answer without doing any calculations.)
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Analysis

Read Explanation

Chemistry Branches

Read Explanation

Nuclear Chemistry

Read Explanation

Physical 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