Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 1: Problem 39

In the year \(2013,\) an estimated amount of 36 billion metrictons ( 1 metric ton \(=1000 \mathrm{kg}\) ) of carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) was emitted worldwide due to fossil fuel combustion and cement production. Express this mass of \(\mathrm{CO}_{2}\) in grams without exponential notation, using an appropriate metric prefix.

Short Answer

Expert verified
The mass of CO₂ emitted worldwide in 2013 is approximately 36 Teragrams (Tg).

Step by step solution

01

Convert metric tons to grams

First, we need to convert the given mass from metric tons to grams. We are given that 1 metric ton is equal to 1000 kg. Furthermore, we know that 1 kg contains 1000 grams. We have: Mass in metric tons = 36 billion metric tons To convert this mass to grams, we multiply it by 1000 (to convert to kg) and again by 1000 (to convert to grams): 36,000,000,000 metric tons * 1000 kg/metric ton * 1000 g/kg = 36,000,000,000,000 g
02

Determine the appropriate metric prefix for the value in grams

Now we have the mass of CO₂ in grams, and we need to express this using an appropriate metric prefix. We have the mass in grams as 36,000,000,000,000 g. To express this using a suitable metric prefix, we need to rewrite this as a multiple of a power of 10. 36,000,000,000,000 g can be written as \(3.6 * 10^{13}\) g.
03

Express mass in grams using the appropriate metric prefix

Finally, let's express the mass of CO₂ in grams using the appropriate metric prefix associated with a power of 10. The closest metric prefix for \(10^{13}\) is Tera (T) which represents \(10^{12}\). So, we have the mass of CO₂ as \(3.6 * 10^{13}\) g, and we can divide this value by \(10^{12}\) to get the mass in Teragrams. \(3.6 * \frac{10^{13}}{10^{12}} = 3.6 * 10^1\) Teragrams. Thus, the mass of CO₂ emitted worldwide in 2013 is approximately 36 Teragrams (Tg).

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

Label each of the following as either a physical process or a chemical process: (a) rusting of a metal can, (b) boiling a cup of water, (c) pulverizing an aspirin, (d) digesting a candy bar, (e) exploding of nitroglyerin.

In the United States, water used for irrigation is measured in acre-feet. An acre-foot of water covers an acre to a depth of exactly 1 ft. An acre is 4840 yd. An acre-foot is enough water to supply two typical households for 1.00 yr. (a) If desalinated water costs \(\$ 1950\) per acre-foot, how much does desalinated water cost per liter? (b) How much would it cost one household per day if it were the only source of water?

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. $$ \begin{array}{l}{\text { (a) } 320.5-(6104.5 / 2.3)} \\ {\text { (b) }\left[\left(285.3 \times 10^{5}\right)-\left(1.200 \times 10^{3}\right)\right] \times 2.8954} \\ {\text { (c) }(0.0045 \times 20,000.0)+(2813 \times 12)} \\ {\text { (a) } 863 \times[1255-(3.45 \times 108)]}\end{array} $$

Two students determine the percentage of lead in a sample as a laboratory exercise. The true percentage is 22.52\(\%\) . The students' results for three determinations are as follows: $$\begin{array}{l}{\text { (1) } 22.52,22.48,22.54} \\ {\text { (2) } 22.64,22.58,22.62}\end{array}$$ (a) Calculate the average percentage for each set of data and state which set is the more accurate based on the average. (b) Precision can be judged by examining the average of the deviations from the average value for that data set. (Calculate the average value for each data set; then calculate the average value of the absolute deviations of each measurement from the average.) Which set is more precise?
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Kinetics

Read Explanation

Chemical Analysis

Read Explanation

Making Measurements

Read Explanation

Nuclear Chemistry

Read Explanation

Inorganic Chemistry

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