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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) At the beginning of 2007, the national debt was \(\$ 8.7\) trillion. How many stacks like the one described would be necessary to pay off this debt?

Short Answer

Expert verified
Approximately 113,105 quarters are needed to reach the height of the Washington Monument. The stack weighs about 641.2 kg and has a value of $28,276.25. About 308 million stacks of quarters would be necessary to pay off the given national debt.

Step by step solution

01

Find the number of quarters to reach the height

Firstly, we need to find the number of quarters to stack to reach the height of the Washington Monument (575 feet). To do this, we will convert the thickness of the quarter to feet and then divide the height of the Washington Monument by the thickness. Thickness of a quarter in feet is given as \(1.55 \times 10^{-3} m\). To convert meters to feet, we use the conversion factor 1 m = 3.28084 ft as follows: Thickness of a quarter in feet = \(1.55 \times 10^{-3} m \cdot 3.28084 ft/m = 5.0843 \times 10^{-3} ft\) Now, we will divide the height of the Washington Monument by the thickness of a quarter to get the number of quarters: Number of quarters = \(575 ft / 5.0843 \times 10^{-3} ft ≈ 113105\) Approximately 113,105 quarters are needed to reach the height of the Washington Monument.
02

Calculate the weight of the stack

To find the total weight of the stack, we will multiply the number of quarters (113,105) by the mass of one quarter (5.67 g) and then convert the mass to kilograms. Total mass of the stack = 113105 quarters × 5.67 g/quarter = 641204.35 g Now convert the mass to kilograms using the conversion factor 1 kg = 1000 g: Total mass of the stack = 641204.35 g × (1 kg / 1000 g) = 641.20435 kg So, this stack of quarters weighs about 641.2 kg.
03

Calculate the money value of the stack

To calculate the total value of this stack of quarters, we will multiply the number of quarters by the value of one quarter (0.25 USD). Value of the stack = 113105 quarters × 0.25 USD/quarter = 28276.25 USD The value of the stack is $28,276.25.
04

Find the number of stacks needed to pay off the national debt

To find how many of these stacks would be necessary to pay off the national debt of $8.7 trillion, we will divide the national debt by the value of one stack. Number of stacks = (8.7 × 10^12 USD) / 28276.25 USD ≈ 3.08 × 10^8 About 308 million stacks of quarters would be necessary to pay off the given national debt.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Unit Conversion in Chemical Problem Solving
Grasping the concept of unit conversion is fundamental in chemistry and other scientific disciplines. It is the process of converting one unit of measurement to another while maintaining the same quantity. A robust understanding of unit conversion is necessary because measurements can be provided in various units, and being versatile in calculations is a cornerstone in scientific literacy.

In our exercise, to tackle part (a), we convert the thickness of the US quarter from millimeters to feet. The conversion involved using the fact that 1 meter equals 3.28084 feet, along with the knowledge that there are 1,000 millimeters in a meter. This step bridges SI units common in science to imperial units that are used in everyday life in the United States. Often, problems will require multiple conversions, as seen when translating the mass of the quarters from grams to kilograms to determine weight.

Why is this important? Chemistry, alongside various scientific fields, operates on universal standards such as the metric system. Yet, in practical applications and understanding real-world problems like the one featured, conversions to local customary units such as feet and pounds might be necessary. Thus, being adept with unit conversion allows students to maneuver between different measurement systems and apply their scientific knowledge more broadly.
Calculating Mass and Weight in Scientific Contexts
Understanding the difference between mass and weight is crucial in scientific measurements. In our context, mass refers to the amount of matter in an object and is measured in grams or kilograms, while weight is the force exerted by gravity on that mass and is commonly measured in newtons or pounds.

The calculation in step 2 focuses on finding the mass of a stack of quarters by multiplying the mass of a single quarter by the total number of quarters. The next step would be converting this mass to weight if required, which involves multiplying the mass by the acceleration due to gravity (\(9.81 \text{m/s}^2\) on Earth).

How do we apply this in chemistry?

Quantifying mass is an everyday necessity in a chemistry lab for tasks such as weighing reactants or products. Precision in these measurements is key as it directly impacts the stoichiometry and the subsequent outcomes of chemical reactions. Conceptualizing the difference between the inert property of mass and the gravitational-dependent property of weight prepares students for advanced learning and problem-solving in not only chemistry but also in fields like physics and engineering.
Currency Value Calculation in Practical Applications
Currency value calculations are integral to a myriad of professional fields, including economics, business, and also in scientific industries. This operation is often simpler than it appears: it generally involves determining the total monetary value by multiplying the quantity of the currency unit by its individual value.

In our exercise, the value of the stack in part (c) was calculated by multiplying the number of quarters by the value of an individual quarter. This basic principle of multiplication gives us the total monetary worth of the stack. The apparent simplicity of currency calculations belies their importance; they are essential for budgeting in research projects, understanding the financial implications of scientific endeavors, and even in macroeconomic contexts like assessing national debts, as seen in part (d).

Comprehending these calculations ensures that students are not only scientifically literate but also possess practical skills that will benefit them in various aspects of life, from personal finance to understanding the larger economic picture that shapes global scientific funding and policymaking.

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Most popular questions from this chapter

A beaker contains a clear, colorless liquid. If it is water, how could you determine whether it contained dissolved table salt? Do not taste it!

(a) To identify a liquid substance, a student determined its density. Using a graduated cylinder, she measured out a 45-mLsample of the substance. She then measured the mass of the sample, finding that it weighed \(38.5 \mathrm{~g}\) She knew that the substance had to be either isopropyl alcohol (density \(0.785 \mathrm{~g} / \mathrm{mL}\) ) or toluene (density \(0.866 / \mathrm{mL})\). What are the calculated density and the probable identity of the substance? (b) An experiment requires \(45.0 \mathrm{~g}\) of ethylene glycol, a liquid whose density is \(1.114 \mathrm{~g} / \mathrm{mL}\). Rather than weigh the sample on a balance, a chemist chooses to dispense the liquid using a graduated cylinder. What volume of the liquid should he use? (c) A cubic piece of metal measures \(5.00 \mathrm{~cm}\) on each edge. If the metal is nickel, whose density is \(8.90 \mathrm{~g} / \mathrm{cm}^{3}\), what is the mass of the cube?

A sample of ascorbic acid (vitamin C) is synthesized in the laboratory. It contains \(1.50 \mathrm{~g}\) of carbon and \(2.00 \mathrm{~g}\) of oxygen. Another sample of ascorbic acid isolated from citrus fruits contains \(6.35 \mathrm{~g}\) of carbon. How many grams of oxygen does it contain? Which law are you assuming in answering this question?

Suppose you decide to define your own temperature scale using the freezing point \(\left(-11.5^{\circ} \mathrm{C}\right)\) and boiling point \(\left(197.6^{\circ} \mathrm{C}\right)\) of ethylene glycol. If you set the freezing point as \(0^{\circ} \mathrm{G}\) and the boiling point as \(100^{\circ} \mathrm{G}\), what is the freezing point of water on this new scale?

Carry out the following conversions: (a) \(0.105\) in. to \(\mathrm{mm}\), (b) \(0.650\) qt to \(\mathrm{mL}\), (c) \(8.75 \mu \mathrm{m} / \mathrm{s}\) to \(\mathrm{km} / \mathrm{hr}\), (d) \(1.955 \mathrm{~m}^{3}\) to \(\mathrm{yd}^{3}\), (e) \(\$ 3.99 / \mathrm{lb}\) to dollars per \(\mathrm{kg}\), (f) \(8.75 \mathrm{lb} / \mathrm{ft}^{3}\) to \(\mathrm{g} / \mathrm{mL}\).

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