Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 1: Problem 32

Use appropriate metric prefixes to write the following measurements without use of exponents: (a) \(7.29 \times 10^{6} \mathrm{~g}\) (b) \(6.1 \times 10^{-10} \mathrm{~m}\) (c) \(1.828 \times 10^{-3} \mathrm{~s}\) (d) \(3.523 \times 10^{9} \mathrm{~m}^{3}\) (g) \(3.552 \times 10^{12} \mathrm{~L}\) (e) $9.62 \times 10^{2} \mathrm{~m} / \mathrm{s}(\mathbf{f}) 8.923 \times 10^{-12} \mathrm{~kg}$

Short Answer

Expert verified
(a) \(7.29 \mathrm{~Mg}\) (b) \(610.0 \mathrm{~nm}\) (c) \(1.828 \mathrm{~ms}\) (d) \(3.523 \mathrm{~Gm}^{3}\) (g) \(3.552 \mathrm{~TL}\) (e) \(0.962 \mathrm{~km/s}\) (f) \(8.923 \mathrm{~pg}\)

Step by step solution

01

(a) Convert g to appropriate metric prefix

To convert \(7.29 \times 10^{6} \mathrm{~g}\), we will use the mega (M) prefix which corresponds to 10^6. Divide by 10^6 and add the prefix: \[ 7.29 \times 10^{6} \mathrm{~g} = 7.29 \mathrm{~Mg} \]
02

(b) Convert m to appropriate metric prefix

To convert \(6.1 \times 10^{-10} \mathrm{~m}\), we will use the nano (n) prefix which corresponds to 10^-9. Multiply by 10^9 to remove the -10 exponent and add the prefix: \[ 6.1 \times 10^{-10} \mathrm{~m} = 610.0 \mathrm{~nm} \]
03

(c) Convert s to appropriate metric prefix

To convert \(1.828 \times 10^{-3} \mathrm{~s}\), we will use the milli (m) prefix which corresponds to 10^-3. Divide by 10^-3 and add the prefix: \[ 1.828 \times 10^{-3} \mathrm{~s} = 1.828 \mathrm{~ms} \]
04

(d) Convert \(m^3\) to appropriate metric prefix

To convert \(3.523 \times 10^{9} \mathrm{~m}^{3}\), we will use the giga (G) prefix which corresponds to 10^9. Divide by 10^9 and add the prefix: \[ 3.523 \times 10^{9} \mathrm{~m}^{3} = 3.523 \mathrm{~Gm}^{3} \]
05

(g) Convert L to appropriate metric prefix

To convert \(3.552 \times 10^{12} \mathrm{~L}\), we will use the tera (T) prefix which corresponds to 10^12. Divide by 10^12 and add the prefix: \[ 3.552 \times 10^{12} \mathrm{~L} = 3.552 \mathrm{~TL} \]
06

(e) Convert \(\mathrm{m/s}\) to appropriate metric prefix

To convert \(9.62 \times 10^{2} \mathrm{~m/s}\), we will use the kilo (k) prefix which corresponds to 10^3. Divide by 10^3 with adjustment to maintain the same value and add the prefix: \[ 9.62 \times 10^{2} \mathrm{~m/s} = 0.962 \mathrm{~km/s} \]
07

(f) Convert kg to appropriate metric prefix

To convert \(8.923 \times 10^{-12} \mathrm{~kg}\), we will use the pico (p) prefix which corresponds to 10^-12. Divide by 10^-12 and add the prefix: \[ 8.923 \times 10^{-12} \mathrm{~kg} = 8.923 \mathrm{~pg} \]

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

(a) The diameter of Earth at the equator is \(12756.27 \mathrm{~km}\). Round this number to three significant figures and express it in standard exponential notation. (b) The circumference of Earth through the poles is \(40,008 \mathrm{~km}\). Round this number to four significant figures and express it in standard exponential notation.

The total rate at which power is used by humans worldwide is approximately 15 TW (terawatts). The solar flux averaged over the sunlit half of Earth is $680 \mathrm{~W} / \mathrm{m}^{2}$ (assuming no clouds). The area of Earth's disc as seen from the Sun is \(1.28 \times 10^{14} \mathrm{~m}^{2}\). The surface area of Earth is approximately 197,000,000 square miles. How much of Earth's surface would we need to cover with solar energy collectors to power the planet for use by all humans? Assume that the solar energy collectors can convert only \(10 \%\) of the available sunlight into useful power.

(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?

Silicon for computer chips is grown in large cylinders called "boules" that are \(300 \mathrm{~mm}\) in diameter and \(2 \mathrm{~m}\) in length, as shown. The density of silicon is \(2.33 \mathrm{~g} / \mathrm{cm}^{3}\). Silicon wafers for making integrated circuits are sliced from a \(2.0-\mathrm{m}\) boule and are typically \(0.75 \mathrm{~mm}\) thick and \(300 \mathrm{~mm}\) in diameter. (a) How many wafers can be cut from a single boule? (b) What is the mass of a silicon wafer? (The volume of a cylinder is given by \(\pi r^{2} h,\) where \(r\) is the radius and \(h\) is its height.)

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.
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Inorganic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Chemical Reactions

Read Explanation

Physical Chemistry

Read Explanation

Making Measurements

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.

Sign-up for free