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A coin dealer offers to sell you an ancient gold coin that is \(2.2 \mathrm{~cm}\) in diameter and \(3.0 \mathrm{~mm}\) in thickness. (a) The density of gold is \(19.3 \mathrm{~g} / \mathrm{cm}^{3} .\) How much should the coin weigh if it is pure gold? (b) If gold sells for \(\$ 640\) per troy ounce, how much is the gold content worth? \((1\) troy ounce \(=31.1 \mathrm{~g})\)

Short Answer

Expert verified
(a) The coin should weigh approximately \(21.82\mathrm{~g}\) if it's made of pure gold. (b) The gold content in the coin is worth approximately $449.28.

Step by step solution

01

Find the volume of the coin

To find the volume of the coin, we'll use the formula for the volume of a cylinder, since the coin is a cylinder-shaped object. The formula is: Volume = \(Ï€r^{2}h\) where r is the radius, and h is the height or thickness of the coin. We're given the diameter of the coin (2.2 cm), so we can find the radius by dividing the diameter by 2.
02

Calculate the radius of the coin

Radius = Diameter / 2 Radius = \(2.2\mathrm{~cm} /2\) Radius = \(1.1\mathrm{~cm}\) Now that we have the radius, we can find the volume of the coin.
03

Calculate the volume of the coin

Volume = \(π(1.1\mathrm{~cm})^{2}(3.0\mathrm{~mm})\) We need to convert the thickness from millimeters to centimeters. \(1 \mathrm{cm} = 10 \mathrm{mm}\) So, \(3.0\mathrm{~mm} = 0.3\mathrm{~cm}\) Volume = \(π(1.1\mathrm{~cm})^{2}(0.3\mathrm{~cm})\) Volume ≈ \(1.131\mathrm{~cm}^{3}\)
04

Calculate the mass of the coin

Now that we have the volume, we can use the density formula to find the mass of the coin. Density = Mass / Volume We need to rearrange the formula to solve for Mass: Mass = Density × Volume We're given the density of gold (19.3 g/cm³): Mass = \(19.3\mathrm{~g/cm}^{3} × 1.131\mathrm{~cm}^{3}\) Mass ≈ \(21.82\mathrm{~g}\)
05

Convert mass to troy ounces

We're given the conversion factor (1 troy ounce = 31.1 g): Troy ounces = Mass (g) / 31.1 g Troy ounces ≈ \(21.82\mathrm{~g} / 31.1\mathrm{~g}\) Troy ounces ≈ \(0.702\)
06

Calculate the worth of the gold content

We're given the price of gold ($640 per troy ounce): Worth = Price per troy ounce × Number of troy ounces Worth = \(\$640 × 0.702\) Worth ≈ \(\$449.28\) Now we have the answers: (a) The coin should weigh approximately 21.82 g if it's made of pure gold. (b) The gold content in the coin is worth approximately $449.28.

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Key Concepts

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

Cylinder Volume
When calculating the volume of an object shaped like a cylinder, such as a coin, you need to apply the cylinder volume formula:
  • Volume = \( \pi r^2 h \)
Here, \(r\) represents the radius of the base, and \(h\) stands for the height or thickness.
To find the radius, remember that it is half the diameter. For this particular coin:
  • Diameter = 2.2 cm, hence Radius = \(2.2 / 2\), which equals 1.1 cm.
  • The thickness originally given in mm, needs conversion to cm as volume requires consistent units.
  • Thus, 3.0 mm becomes 0.3 cm.
Substituting these values into the formula, you arrive at the volume:
\[ Volume = \pi (1.1\, \text{cm})^2 (0.3\, \text{cm}) \approx 1.131\, \text{cm}^3 \]
Gold Density
Density is the mass per unit volume of a substance, measured in g/cm³ for solids. Gold, known for its high density, has a value of 19.3 g/cm³.
Understanding gold's density is essential for determining how much a given volume will weigh. In this exercise, the density formula is rearranged to find mass:
  • Density \(= \frac{\text{Mass}}{\text{Volume}}\)
  • Mass \(= \text{Density} \times \text{Volume}\)
After calculating the coin's volume (1.131 cm³), use the density of gold to find its mass:
\[ \text{Mass} = 19.3\, \text{g/cm}^3 \times 1.131\, \text{cm}^3 \approx 21.82\, \text{g} \]
Mass Calculation
Mass calculation involves using density and volume to determine how heavy an object is, using the formula derived from the definition of density. For this gold coin exercise:
  • Ensure units are consistent: Use cm and grams.
  • Multiply density by volume: \(19.3\, \text{g/cm}^3 \times 1.131\, \text{cm}^3 \).
  • This gives a precise mass of approximately 21.82 grams for the pure gold coin.
Mass calculation is crucial for verifying the purity of the coin and understanding its value.
Troy Ounce Conversion
In precious metals like gold, the weight unit 'troy ounce' is prevalent. It differs slightly from regular ounces, equaling 31.1 grams. To convert between grams and troy ounces, follow this simple calculation:
  • Troy ounces \(= \frac{\text{Mass (g)}}{31.1\, \text{g}}\)
  • For our coin mass of 21.82 g, this gives \(\approx 0.702\) troy ounces.
Once transformed to troy ounces, it becomes easy to assess the coin's financial worth by multiplying the number of troy ounces by the market price per troy ounce:
  • Price per troy ounce: \\(640
  • Total worth: \(\\)640 \times 0.702 \approx \$449.28\)
This conversion simplifies the process of evaluating and trading gold based on commonly accepted standards.

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Most popular questions from this chapter

(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 \(25.0-\mathrm{mL}\) 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 \(\left(\frac{4}{3}\right) \pi r^{3}\) where \(r\) is the radius.)

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