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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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Most popular questions from this chapter

Water has a density of \(0.997 \mathrm{~g} / \mathrm{cm}^{3}\) at $25^{\circ} \mathrm{C}\(; ice has a density of \)0.917 \mathrm{~g} / \mathrm{cm}^{3}$ at \(-10^{\circ} \mathrm{C}\). (a) If a soft-drink bottle whose volume is $1.50 \mathrm{~L}\( is completely filled with water and then frozen to \)-10^{\circ} \mathrm{C},$ what volume does the ice occupy? (b) Can the ice be contained within the bottle?

Suppose you decide to define your own temperature scale with units of \(\mathrm{O}\), using the freezing point \(\left(13^{\circ} \mathrm{C}\right)\) and boiling point \(\left(360^{\circ} \mathrm{C}\right)\) of oleic acid, the main component of olive oil. If you set the freezing point of oleic acid as \(0^{\circ} \mathrm{O}\) and the boiling point as \(100^{\circ} \mathrm{O},\) what is the freezing point of water on this new scale?

Label each of the following as either a physical process or a (a) crushing a metal can, \((\mathbf{b})\) production chemical process: of urine in the kidneys, \((\mathbf{c})\) melting a piece of chocolate, \((\mathbf{d})\) burning fossil fuel, \((\mathbf{e})\) discharging a battery.

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

(a) The speed of light in a vacuum is $2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}$. Calculate its speed in miles per hour. (b) The Sears Tower in Chicago is \(1454 \mathrm{ft}\) tall. Calculate its height in meters. \((\mathbf{c})\) The Vehicle Assembly Building at the Kennedy Space Center in Florida has a volume of \(3,666,500 \mathrm{~m}^{3}\). Convert this volume to liters and express the result in standard exponential notation. (d) An individual suffering from a high cholesterol level in her blood has $242 \mathrm{mg}\( of cholesterol per \)100 \mathrm{~mL}$ of blood. If the total blood volume of the individual is \(5.2 \mathrm{~L}\), how many grams of total blood cholesterol does the individual's body contain?
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