/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 86 A \(10.0 \mathrm{~g}\) block of ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

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 thickness of the gold sheet is approximately 0.0345 mm.

Step by step solution

01

Find the Volume of the Gold

Since we know the mass of the gold and its density, we can find the volume of the gold block using the formula for density: \(\text{Density} = \frac{\text{Mass}}{\text{Volume}}\). Rearranging the formula gives us \(\text{Volume} = \frac{\text{Mass}}{\text{Density}}\). Substituting the given values, we have: \(\text{Volume} = \frac{10.0 \, \text{g}}{19.3 \, \text{g/cm}^3} \approx 0.518 \, \text{cm}^3\).
02

Relate Volume, Area, and Thickness

The volume of a sheet can also be expressed as the product of its area and thickness using the formula: \(\text{Volume} = \text{Area} \times \text{Thickness}\). Since we have the area and the volume, we can rearrange this formula to find the thickness: \(\text{Thickness} = \frac{\text{Volume}}{\text{Area}}\).
03

Calculate the Thickness in Centimeters

Using the formula from the previous step, plug in the values for volume and area to find the thickness in centimeters: \(\text{Thickness} = \frac{0.518 \, \text{cm}^3}{150 \, \text{cm}^2} \approx 0.00345 \, \text{cm}\).
04

Convert Thickness to Millimeters

Remember that 1 centimeter is equivalent to 10 millimeters. Therefore, multiply the thickness in centimeters by 10 to convert it to millimeters: \(0.00345 \, \text{cm} \times 10 = 0.0345 \, \text{mm}\).

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Volume Calculation
To determine the volume of an object, especially when its mass and density are known, we can use the handy density formula: \(\text{Density} = \frac{\text{Mass}}{\text{Volume}}\). Rearranging this equation gives us the formula for volume: \(\text{Volume} = \frac{\text{Mass}}{\text{Density}}\). This is particularly useful when dealing with solids like gold or other metals.

For example, if you have a block of gold weighing 10 grams and the density of gold is known to be \(19.3 \, \text{g/cm}^3\), you can easily substitute these values into the formula to get the volume of the gold block:

  • Mass = 10.0 g
  • Density = 19.3 g/cm\(^3\)
  • Volume = \(\frac{10.0 \, \text{g}}{19.3 \, \text{g/cm}^3} \approx 0.518 \, \text{cm}^3\)

This calculation reveals that this particular block of gold takes up approximately 0.518 cubic centimeters of space.
Area and Thickness Relationship
Once you have the volume of an object, like a thin sheet, understanding how area and thickness relate is vital for further calculations. For objects like a metal sheet, the volume can also be expressed as the multiplication of its area and its thickness:

\[\text{Volume} = \text{Area} \times \text{Thickness}\]

In this formula, if you know the volume and the area, you can rearrange the equation to solve for thickness:

\[\text{Thickness} = \frac{\text{Volume}}{\text{Area}}\]

Suppose you have a gold sheet with an area of 150 cm\(^2\) and a volume of 0.518 cm\(^3\). Using the formula for thickness:

  • Area = 150 cm\(^2\)
  • Volume = 0.518 cm\(^3\)
  • Thickness = \(\frac{0.518 \, \text{cm}^3}{150 \, \text{cm}^2} \approx 0.00345 \, \text{cm}\)

This means the gold sheet would be approximately 0.00345 cm thick.
Unit Conversion
When we perform calculations in science and engineering, unit conversion is often necessary to express results in the desired units. In this exercise, converting the thickness from centimeters to millimeters provides a more intuitive measurement for thin materials.

Remember, the relationship between centimeters and millimeters is straightforward: 1 cm equals 10 mm.

For the calculated thickness of 0.00345 cm, we convert it to millimeters by multiplying:

  • Thickness in cm: 0.00345 cm
  • Conversion factor: 10 mm/cm
  • Thickness in mm: \(0.00345 \, \text{cm} \times 10 = 0.0345 \, \text{mm}\)

This converts it to 0.0345 mm, offering a clearer perspective on just how thin the gold sheet truly is.

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

For each of the following processes, does the potential energy of the object(s) increase or decrease? (a) The charge of two oppositely charged particles is increased. (b) \(\mathrm{H}_{2} \mathrm{O}\) molecule is split into two oppositely charged ions, \(\mathrm{H}^{+}\) and \(\mathrm{OH}^{-}\). (c) A person skydives from a height of 600 meters.

What type of quantity (for example, length, volume, density) do the following units indicate? (a) \(\mathrm{m}^{3}\), (b) \(\mathrm{ns}\), (c) \(\mathrm{mm}\), (d) \(\mathrm{g} / \mathrm{dm}^{3}\), (e) \({ }^{\circ} \mathrm{C},(\mathbf{f}) \mathrm{ms}^{-1},(\mathbf{g}) \mathrm{Pa} .\)

Identify each of the following as measurements of length, area, volume, mass, density, time, or temperature: (a) \(25 \mathrm{ps}\), (b) \(374.2 \mathrm{mg}\), (c) \(77 \mathrm{~K}\), (d) \(100,000 \mathrm{~km}^{2}\), (e) \(1.06 \mu \mathrm{m}\), (f) \(16 \mathrm{nm}^{2},(\mathbf{g})-78^{\circ} \mathrm{C},(\mathbf{h}) 2.56 \mathrm{~g} / \mathrm{cm}^{3}\), (i) \(28 \mathrm{~cm}^{3}\).

What is the number of significant figures in each of the following measured quantities? (a) \(902.5 \mathrm{~kg}\), (b) \(3 \times 10^{-6} \mathrm{~m}\) (c) \(0.0096 \mathrm{~L}\), (d) \(2.94 \times 10^{3} \mathrm{~m}^{2}\) (e) \(92.03 \mathrm{~km}\) (f) \(782.234 \mathrm{~g}\).

(a) How many liters of wine can be held in a wine barrel whose capacity is 31 gal? \((\mathbf{b})\) The recommended adult dose of Elixophyllin \(^{\circ}\), a drug used to treat asthma, is \(6 \mathrm{mg} / \mathrm{kg}\) of body mass. Calculate the dose in milligrams for a \(185-\mathrm{lb}\) person. (c) If an automobile is able to travel \(400 \mathrm{~km}\) on \(47.3 \mathrm{~L}\) of gasoline, what is the gas mileage in miles per gallon? (d) When the coffee is brewed according to directions, a pound of coffee beans yields 50 cups of coffee \((4\) cups \(=1 \mathrm{qt}) .\) How many \(\mathrm{kg}\) of coffee are required to produce 200 cups of coffee?
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Organic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Chemistry Branches

Read Explanation

The Earths Atmosphere

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.