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In the year 2007 , an estimated amount of 31 billion tons 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 carbon dioxide emitted in 2007 can be expressed as 31 Peta grams of CO₂, or 31 Pg of CO₂.

Step by step solution

01

Convert tons to grams

To convert tons to grams, we know that one metric ton equals 1,000,000 grams. So, we will multiply the given value by the conversion factor: \(31 \: billion \: tons * 1,000,000 \, grams/ton\)
02

Calculate the result

Now, we'll compute the value in grams: \(31,000,000,000 * 1,000,000 = 31,000,000,000,000,000 \: grams\)
03

Choose an appropriate metric prefix

The available metric prefixes are: - Kilo (k) for \(10^3\) - Mega (M) for \(10^6\) - Giga (G) for \(10^9\) - Tera (T) for \(10^{12}\) - Peta (P) for \(10^{15}\) - Exa (E) for \(10^{18}\) In our case, the mass of carbon dioxide is \(31,000,000,000,000,000 \: grams\), which can be written as \(31 * 10^{15} grams\) which corresponds to the Peta (P) prefix.
04

Write the final answer

So, the mass of carbon dioxide emitted in 2007 can be expressed as: 31 Peta grams of CO₂, or 31 Pg of CO₂.

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Key Concepts

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

Carbon Dioxide Emissions
When discussing climate change and environmental impact, carbon dioxide emissions are a frequently cited concern. Carbon dioxide (CO₂) is a greenhouse gas that contributes to the warming of our planet. It’s produced naturally through processes like respiration and volcanic eruptions, but a significant increase in CO₂ levels has been attributed to human activities, particularly the combustion of fossil fuels and industrial processes like cement production.

The year 2007 saw an immense amount of CO₂ being released into the atmosphere, mainly due to these human activities, totaling an estimated 31 billion tons. To grasp the enormity of this number, it helps to convert these emissions into a unit like grams which is more commonly used in chemistry for measuring smaller masses. Understanding the scale of these emissions in terms familiar to us aids in comprehending the vast impact that these actions have on our environment.
Conversion of Mass Units
The conversion of mass units from tons to grams is a fundamental concept in chemistry and other sciences when discussing quantities at vastly different scales. The provided exercise shows an enormous figure - 31 billion tons, which is not easy to visualize or utilize in scientific calculations. Bringing the mass down to the gram scale makes it more fathomable.

One metric ton is equivalent to 1,000,000 grams. Therefore, to convert from tons to grams, you multiply the number of tons by the conversion factor of 1,000,000. In the context of the exercise:
  • 31 billion tons translates to \(31 \times 10^9 \) tons,
  • Multiplying by the conversion factor gives us \(31 \times 10^9 \) tons \(\times \) 1,000,000 grams/ton = \(31 \times 10^{15}\) grams.
Through this process, we're able to communicate such large figures in a more manageable form, using units like grams, that are standardized in scientific documentation and familiar within the framework of chemistry.
Scientific Notation
Handling extremely large or small numbers in science is facilitated by scientific notation, a method of expression that uses powers of ten to simplify numbers. This representation is crucial for scientists and students alike when calculating or reading extensive quantities.

In scientific notation, a number is represented as a product of a coefficient (between 1 and 10) and a power of ten. For example, \(\text{31 billion} = 31 \times 10^9\). The exercise reveals the importance of this method: \(31 \times 10^15\) grams is a simplified way of expressing 31 quadrillion grams. This approach not only prevents confusion and reduces the likelihood of errors but also makes comparisons and calculations with large numbers more efficient.

Additionally, matching a number with an appropriate metric prefix can further simplify expression. In our case, the prefix is 'Peta,' which stands for a quadrillion or \(10^{15}\) units. Therefore, expressing the CO₂ emissions as 31 Pg (Peta grams) conveys the magnitude of the number succinctly and accurately within the context of chemistry and environmental science.

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Most popular questions from this chapter

A package of aluminum foil contains \(50 \mathrm{ft}^{2}\) of foil, which weighs approximately 8.0 oz. Aluminum has a density of \(2.70 \mathrm{~g} / \mathrm{cm}^{3}\). What is the approximate thickness of the foil in millimeters?

(a) A sample of tetrachloroethylene, a liquid used in dry cleaning that is being phased out because of its potential to cause cancer, has a mass of \(40.55 \mathrm{~g}\) and a volume of \(25.0 \mathrm{~mL}\) at \(25^{\circ} \mathrm{C}\). What is its density at this temperature? Will tetrachloroethylene float on water? (Materials that are less dense than water will float.) (b) Carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) is a gas at room temperature and pressure. However, carbon dioxide can be put under pressure to become a "supercritical fluid" that is a much safer drycleaning agent than tetrachloroethylene. At a certain pressure, the density of supercritical \(\mathrm{CO}_{2}\) is \(0.469 \mathrm{~g} / \mathrm{cm}^{3} .\) What is the mass of a \(25.0-\mathrm{mL}\) sample of supercritical \(\mathrm{CO}_{2}\) at this pressure?

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: 1\. 22.52,22.48,22.54 2\. 22.64,22.58,22.62 (a) Calculate the average percentage for each set of data, and state which set is the more accurate based on the average. \((\mathbf{b})\) 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?

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

What is the number of significant figures in each of the following measured quantities? (a) \(601 \mathrm{~kg}\), (b) \(0.054 \mathrm{~s}\), (c) \(6.3050 \mathrm{~cm}\), (d) \(0.0105 \mathrm{~L}\) (e) \(7.0500 \times 10^{-3} \mathrm{~m}^{3}\) (f) \(400 \mathrm{~g}\).

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