Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 1: Problem 65

The US quarter has a mass of \(5.67 \mathrm{~g}\) and is approximately \(1.55 \mathrm{~mm}\) thick. (a) How many quarters would have to be stacked to reach \(575 \mathrm{ft},\) the height of the Washington Monument? (b) How much would this stack weigh? (c) How much money would this stack contain? (d) The US National Debt Clock showed the outstanding public debt to be $$\$ 11,687,233,914,811.11$$ on August \(19,2009 .\) How many stacks like the one described would be necessary to pay off this debt?

Short Answer

Expert verified
To answer the given question: (a) Approximately 112,501 quarters would need to be stacked to reach the height of the Washington Monument. (b) The total weight of this stack would be about 1,468.37 pounds. (c) This stack would have a total value of \(\$28,125.25\). (d) To pay off the debt, about 415,492 stacks like the one described would be required.

Step by step solution

01

Calculate the number of quarters needed to reach 575 ft

First, we need to convert the thickness of a quarter, which is given in mm, to ft: \(1.55~mm * (1~in/(25.4~mm))(1~ft/(12~in))\) Now, we divide the height of the Washington Monument (575 ft) by the calculated thickness in ft to find the number of quarters needed. \(Number~of~Quarters = \frac{Height~of~Washington~Monument}{Thickness~of~a~Quarter~in~ft}\)
02

Calculate the total weight of the stack

Since the mass of one quarter is given in grams, we will convert it to pounds by using the conversion factor. Then, we'll multiply the number of quarters from Step 1 by the mass of one quarter to find the total weight of the stack. \(5.67~g * (1~lb)/(453.592~g)\) \(Total~Weight = Number~of~Quarters * Weight~of~a~Quarte_r~in~lb\)
03

Calculate the total value of the stack

Each quarter is worth \(0.25. We'll multiply the number of quarters from Step 1 by \)0.25 to find the total value of the stack. \(Total~Value = Number~of~Quarters * 0.25\)
04

Calculate the number of stacks required to pay off the debt

We'll divide the outstanding public debt by the total value of one stack (calculated in Step 3) to find the number of stacks required to pay off the public debt. \(Number~of~Stacks = \frac{Outstanding~Public~Debt}{Total~Value~of~One~Stack}\)

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 thief plans to steal a gold sphere with a radius of \(28.9 \mathrm{~cm}\) from a museum. If the gold has a density of \(19.3 \mathrm{~g} / \mathrm{cm}^{3},\) what is the mass of the sphere in pounds? [The volume of a sphere is \(\left.V=(4 / 3) \pi r^{3} .\right]\) Is the thief likely to be able to walk off with the gold sphere unassisted?

Show the steps to convert the speed of sound, 344 meters per second, into miles per hour. [Section 1.6]

(a) After the label fell off a bottle containing a clear liquid believed to be benzene, a chemist measured the density of the liquid to verify its identity. A 25.0 -mL portion of the liquid had a mass of \(21.95 \mathrm{~g}\). A chemistry handbook lists the density of benzene at \(15^{\circ} \mathrm{C}\) as \(0.8787 \mathrm{~g} / \mathrm{mL}\). Is the calculated density in agreement with the tabulated value? (b) An experiment requires \(15.0 \mathrm{~g}\) of cyclohexane, whose density at \(25^{\circ} \mathrm{C}\) is \(0.7781 \mathrm{~g} / \mathrm{mL}\). What volume of cyclohexane should be used? (c) A spherical ball of lead has a diameter of \(5.0 \mathrm{~cm}\). What is the mass of the sphere if lead has a density of \(11.34 \mathrm{~g} / \mathrm{cm}^{3}\) ? (The volume of a sphere is \((4 / 3) \pi r^{3}\) where \(r\) is the radius.)

The density of air at ordinary atmospheric pressure and \(25^{\circ} \mathrm{C}\) is \(1.19 \mathrm{~g} / \mathrm{L}\). What is the mass, in kilograms, of the air in a room that measures \(14.5 \mathrm{ft} \times 16.5 \mathrm{ft} \times 8.0 \mathrm{ft} ?\)

What type of quantity (for example, length, volume, density) do the following units indicate: (a) \(\mathrm{mL},(\mathbf{b}) \mathrm{cm}^{2},(\mathrm{c}) \mathrm{mm}^{3}\) (d) \(\mathrm{mg} / \mathrm{L},(\mathrm{e}) \mathrm{ps},(\mathrm{f}) \mathrm{nm},(\mathrm{g}) \mathrm{K} ?\)
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Making Measurements

Read Explanation

Nuclear Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Reactions

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