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) \(2.3 \times 10^{-10} \mathrm{L}\) ,(b) \(4.7 \times 10^{-6} \mathrm{g},\) (c) \(1.85 \times 10^{-12} \mathrm{m},\) (d) \(16.7 \times 10^{6} \mathrm{s}\) (e) \(15.7 \times 10^{3} \mathrm{g},(\mathrm{f}) 1.34 \times 10^{-3} \mathrm{m},(\mathrm{g}) 1.84 \times 10^{2} \mathrm{cm}\)

Short Answer

Expert verified
(a) \(230 \,\mathrm{nL}\), (b) \(4.7 \,\mathrm{µg}\), (c) \(1.85 \,\mathrm{pm}\), (d) \(16.7 \,\mathrm{Ms}\), (e) \(15.7 \,\mathrm{kg}\), (f) \(1.34 \,\mathrm{mm}\), (g) \(1.84 \,\mathrm{hm}\)

Step by step solution

01

a)

Convert \(2.3 \times 10^{-10} \mathrm{L}\) to an equivalent value using appropriate metric prefix. Since the exponent is \(-10\), we will use nano (n) prefix: \[2.3 \times 10^{-10} \mathrm{L} = 230 \,\mathrm{nL}\]
02

b)

Convert \(4.7 \times 10^{-6} \mathrm{g}\) to an equivalent value using appropriate metric prefix. Since the exponent is \(-6\), we will use micro (µ) prefix: \[4.7 \times 10^{-6} \mathrm{g} = 4.7 \,\mathrm{µg}\]
03

c)

Convert \(1.85 \times 10^{-12} \mathrm{m}\) to an equivalent value using appropriate metric prefix. Since the exponent is \(-12\), we will use pico (p) prefix: \[1.85 \times 10^{-12} \mathrm{m} = 1.85 \,\mathrm{pm}\]
04

d)

Convert \(16.7 \times 10^{6} \mathrm{s}\) to an equivalent value using appropriate metric prefix. Since the exponent is \(6\), we will use Mega (M) prefix: \[16.7 \times 10^{6} \mathrm{s} =16.7 \,\mathrm{Ms}\]
05

e)

Convert \(15.7 \times 10^{3} \mathrm{g}\) to an equivalent value using appropriate metric prefix. Since the exponent is \(3\), we will use kilo (k) prefix: \[15.7 \times 10^{3} \mathrm{g} = 15.7 \,\mathrm{kg}\]
06

f)

Convert \(1.34 \times 10^{-3} \mathrm{m}\) to an equivalent value using appropriate metric prefix. Since the exponent is \(-3\), we will use milli (m) prefix: \[1.34 \times 10^{-3} \mathrm{m} = 1.34 \,\mathrm{mm}\]
07

g)

Convert \(1.84 \times 10^{2} \mathrm{cm}\) to an equivalent value using appropriate metric prefix. Since the exponent is \(2\), we will use hecto (h) prefix. Note that 1 hectometer (hm) = 10000 centimeters (cm): \[1.84 \times 10^{2} \mathrm{cm} = 1.84 \,\mathrm{hm}\]

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!

Key Concepts

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

Scientific Notation
When dealing with very large or very small numbers, scientists and engineers use scientific notation to make these numbers more manageable. Scientific notation expresses numbers as a product of two parts: a coefficient and a power of ten. For example, the number 5000 can be written in scientific notation as \(5 \times 10^3\). This format is particularly helpful in simplifying complex calculations, as it allows for straightforward multiplication and division.

Numbers smaller than one are also easily represented in scientific notation. For instance, 0.00012 becomes \(1.2 \times 10^{-4}\). Each step in the exponent represents a shift of the decimal point by one place. Using scientific notation aligns with our decimal system, making it a universal tool for scientists around the world in all fields ranging from physics to economics.
Unit Conversion
Unit conversion is a critical technique in many scientific calculations, allowing for comparison and computation across different measurement systems. Unit conversion involves the use of multiplication or division to convert one unit to another while maintaining the same quantity. This is often achieved by utilising conversion factors that are based on relationships between units. For example, knowing that 1 kilometer is equal to 1000 meters, one can convert 5 kilometers to meters by multiplying 5 by 1000, yielding 5000 meters.

To ensure accuracy, the unit you are converting to should cancel out the unit you are converting from. When working with scientific notation, this process also means adjusting the exponent as needed to reflect the new unit's scale.
Metric System
The metric system is an internationally agreed-upon decimal system of measurement. It is used almost universally around the world, which simplifies international communication and trade. One of the system's key features is its use of metric prefixes, which indicate a multiple of the base unit. These prefixes serve as shorthand to express large or small quantities.

Common metric prefixes include:
  • Kilo (k): Represents 1,000 units of the base
  • Mega (M): Equals 1,000,000 units of the base
  • Micro (µ): Means one millionth of the base
  • Nano (n): Signifies one billionth of the base
  • Pico (p): Indicates one trillionth of the base
Instead of writing out the whole number, such as 0.000001 meters, we can use the metric prefix to simplify it to 1 micrometer (1 µm). This simplicity is one of the reasons why the metric system is preferred in science and engineering.

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

Carry out the following operations and express the answers with the appropriate number of significant figures. $$ \begin{array}{ll}{\text { (a) } 14.3505+2.65} & {\text { (b) } 952.7-140.7389} \\\ {\text { (c) }\left(3.29 \times 10^{4}\right)(0.2501)} & {\text { (d) } 0.0588 / 0.677}\end{array} $$

(a) To identify a liquid substance, a student determined its density. Using a graduated cylinder, she measured out a 45 -mL. sample of the substance. She then measured the mass of the sample, finding that it weighed 38.5 \(\mathrm{g}\) . She knew that the substance had to be either isopropylalcohol (density 0.785 \(\mathrm{g} / \mathrm{mL}\) )or toluene (density 0.866 \(\mathrm{g} / \mathrm{mL} ) .\) What are the calculated density and the probable identity of the substance? (b) An experiment requires 45.0 \(\mathrm{g}\) of ethylene glycol, a liquid whose density is 1.114 \(\mathrm{g} / \mathrm{mL}\) . Rather than weigh the sample on a balance, a chemist chooses to dispense the liquid using a graduated cylin-der. What volume of the liquid should he use? (c) Is a graduated cylinder such as that shown in Figure 1.21 likely to afford the accuracy of measurement needed? (d) A cubic piece of metal measures 5.00 \(\mathrm{cm}\) on each edge. If the metal is nickel, whose density is \(8.90 \mathrm{g} / \mathrm{cm}^{3},\) what is the mass of the cube?

A copper refinery produces a copper ingot weighing 150 \(\mathrm{lb}\) . If the copper is drawn into wire whose diameter is 7.50 \(\mathrm{mm}\) , how many feet of copper can be obtained from the ingot? The density of copper is 8.94 \(\mathrm{g} / \mathrm{cm}^{3} .\) (Assume that the wire is a cylinder whose volume \(V=\pi r^{2} h,\) where ris its radius and \(h\) is its height or length.)

Carry out the following operations and express the answer with the appropriate number of significant figures. $$ \begin{array}{l}{\text { (a) } 320.5-(6104.5 / 2.3)} \\ {\text { (b) }\left[\left(285.3 \times 10^{5}\right)-\left(1.200 \times 10^{3}\right)\right] \times 2.8954} \\ {\text { (c) }(0.0045 \times 20,000.0)+(2813 \times 12)} \\ {\text { (a) } 863 \times[1255-(3.45 \times 108)]}\end{array} $$

The distance from Earth to the Moon is approximately \(240,000 \mathrm{mi.}\) (a) What is this distance in meters? (b) The peregrine falcon has been measured as traveling up to 350 \(\mathrm{km} /\) hr in a dive. If this falcon could fly to the Moon at this speed, how many seconds would it take? (c) The speed of light is \(3.00 \times 10^{8} \mathrm{m} / \mathrm{s} .\) How long does it take for light to travel from Earth to the Moon and back again? (\boldsymbol{d} ) ~ E a r t h ~ t r a v e l s ~ a r o u n d ~the Sun at an average speed of 29.783 \(\mathrm{km} / \mathrm{s} .\) Convert this speed to miles per hour.
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

The Earths Atmosphere

Read Explanation

Making Measurements

Read Explanation

Kinetics

Read Explanation

Chemistry Branches

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