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Chapter 1: Problem 86

A \(10.0 \mathrm{~g}\) block of gold is hammered into a thin gold sheet which has an area of \(150 \mathrm{~cm}^{2}\). Given the density of gold is \(19.3 \mathrm{~g} / \mathrm{cm}^{3}\), what is the approximate thickness of the gold sheet in millimeters?

Short Answer

Expert verified
The approximate thickness of the gold sheet is \(0.03448 \mathrm{~mm}\).

Step by step solution

01

Understand the problem

We need to find the thickness of a gold sheet given its mass, area, and the density of gold. We know that the volume of an object is related to its mass by the density. Furthermore, we understand that the volume of a rectangular object is the product of its length, width, and height (or thickness in this case).
02

Write the volume formula for the gold sheet

Since the gold sheet can be considered a rectangular object, its volume can be expressed as: \( V = A \times h\) Where \(V\) is the volume, \(A\) is the area of the gold sheet, and \(h\) is its thickness (height).
03

Express the volume of the gold sheet using its mass and density

The volume of an object is related to its mass and density by: \(V = \frac{m}{\rho}\) Where \(m\) is the mass of the gold sheet, and \(\rho\) is the density of gold.
04

Equate the two expressions for the volume of the gold sheet

From Steps 2 and 3, we equate the expressions for the volume of the gold sheet: \(A \times h = \frac{m}{\rho}\)
05

Solve for the thickness of the gold sheet

We rearrange the equation from Step 4 to isolate the thickness (height) of the gold sheet: \(h = \frac{m}{A \times \rho}\)
06

Substitute the given values in the formula and compute the thickness of the gold sheet

We are given the mass of the gold sheet as \(10.0 \mathrm{~g}\), its area as \(150 \mathrm{~cm}^{2}\), and the density of gold as \(19.3 \mathrm{~g} / \mathrm{cm}^{3}\). Plugging these values into the equation from Step 5: \(h = \frac{(10.0 \mathrm{~g})}{(150 \mathrm{~cm}^{2}) \times (19.3 \mathrm{~g} / \mathrm{~cm}^{3})}\) Now, perform the calculation: \(h \approx \frac{10}{(150\times19.3)} \approx 0.003448 \mathrm{~cm}\)
07

Convert the thickness of the gold sheet from centimeters to millimeters

Since we're asked for the thickness in millimeters, we need to convert it from centimeters. We know that there are 10 millimeters in 1 centimeter, so: \(h \approx 0.003448 \mathrm{~cm} \times \frac{10 \mathrm{~mm}}{1 \mathrm{~cm}} \approx 0.03448 \mathrm{~mm}\) The approximate thickness of the gold sheet is \(0.03448 \mathrm{~mm}\).

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Key Concepts

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

Volume
Volume is an essential concept in working out density-related problems. It refers to the amount of space that an object occupies. For geometric shapes like cubes or rectangular prisms, volume is calculated by multiplying length, width, and height. However, for this problem, where the gold sheet is extremely thin, volume can be expressed as the area of the sheet multiplied by its thickness.To calculate the volume of our gold sheet, we use the formula:- \[ V = A \times h \] - Where:\( V \) is the volume, - \( A \) is the area of the gold sheet, - \( h \) is the thickness (or height).Understanding these relationships helps serve as the foundation to find the thickness when given area and mass, using the concept of volume.
Mass
Mass is a measure of the amount of matter in an object, often weighed in grams or kilograms. In problems involving density, mass plays a crucial role as it helps express the relationship between an object's volume and its density.In this particular exercise, the gold sheet has a mass of \( 10.0 \, \text{g} \). The equation linking mass, density, and volume is:- \[ V = \frac{m}{\rho} \] - Where: - \( m \) represents mass, - \( \rho \) symbolizes density, - \( V \) represents volume.So, knowing the mass helps us to calculate volume when the density of the material, like gold in this example, is provided.
Area
Area is a measure of the total surface of a two-dimensional shape. For a rectangle or a sheet, it is calculated by multiplying its length and width. In our gold sheet problem, we are given an area of \( 150 \, \text{cm}^2 \).The area helps determine how much space the surface of the gold sheet occupies, and it is pivotal to finding the thickness when combined with volume. By utilizing the area, we can express the volume equation as a function of thickness:
  • \( V = A \times h \)
  • Using the density equation, we equate to solve for \( h \):
  • \( A \times h = \frac{m}{\rho} \)
Using the area information in this way simplifies the task of finding the unknown thickness in density-related problems.

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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 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.)

Indicate the number of significant figures in each of the following measured quantities: (a) \(62.65 \mathrm{~km} / \mathrm{hr}\), (b) \(78.00 \mathrm{~K}\), (c) \(36.9 \mathrm{~mL}\) (d) \(250 \mathrm{~mm}\), (e) 89.2 metric tons, (f) \(6.4224 \times 10^{2} \mathrm{~m}^{3}\)

The U.S. 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 U.S. National Debt Clock showed the outstanding public debt to be \(\$ 16,213,166,914,811\) on October \(28,2012 .\) How many stacks like the one described would be necessary to pay off this debt?

Give the chemical symbol or name for the following elements, as appropriate: (a) helium, (b) platinum, (c) cobalt, (d) tin, (e) silver, (f) \(\mathrm{Sb},(\mathbf{g}) \mathrm{Pb}\) (h) Br, (i) \(V\), \((\mathbf{j}) \mathrm{Hg}\).

Indicate which of the following are exact numbers: (a) the mass of a 945-mL can of coffee, \((\mathbf{b})\) the number of students in your chemistry class, \((\mathbf{c})\) the temperature of the surface of the Sun, \((\mathbf{d})\) the mass of a postage stamp, \((\mathbf{e})\) the number of milliliters in a cubic meter of water, (f) the average height of NBA basketball players.
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