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Chapter 9: Problem 52

Why would it be unreasonable for an amendment to the Clean Air Act to call for 0% pollution emissions from cars with combustion engines?

Short Answer

Expert verified
It's unreasonable to expect 0% pollution emissions from cars with combustion engines because the process of burning fuel (combustion) inherently produces emission gases. With current technology, these emissions are unavoidable as long as fuel is being burned.

Step by step solution

01

Understanding How a Combustion Engine Works

Cars with combustion engines work by igniting a mixture of fuel and air, a process that inevitably leads to the generation of exhaust gases. These gases are commonly referred to as 'pollution emissions'.
02

Understanding the Process of Combustion

The process of combustion in engines takes in fuel (like petrol or diesel) and oxygen, then produces work (the power to drive your car), water, and pollutants such as carbon dioxide, carbon monoxide and nitrogen oxides. This implies that a certain level of emission is always present when using combustion engines, as the reaction itself unavoidably creates gases.
03

Linking the Combustion Process with Zero Pollution Emissions Policy

Expecting zero pollution emissions would mean a complete absence of these gases, which is simply not feasible with the current technology of combustion engines. As long as fuel is burned, pollution emissions will occur, making 0% pollution emissions from cars with combustion engines unreasonable and impracticable with present technology.

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Most popular questions from this chapter

Graphing Calculator Calculating Percentage Yield of a Chemical Reaction The graphing calculator can run a program that calculates the percentage yield of a chemical reaction when you enter the actual yield and the theoretical yield. Using an example in which the actual yield is 38.8 g and the theoretical yield is 53.2 g, you will calculate the percentage yield. First, the pro- gram will carry out the calculation. Then you can use it to make other calculations. Go to Appendix C. If you are using a TI-83 Plus, you can download the program YIELD and data and run the application as directed. If you are using another calculator, your teacher will provide you with keystrokes and data sets to use. After you have run the program, answer the questions. Note: all answers are written with three significant figures. \begin{equation}\begin{array}{l}{\text { a. What is the percentage yield when the }} \\ {\text { actual yield is } 27.3 \mathrm{g} \text { and the theoretical }} \\ {\text { yield is } 44.6 \mathrm{g} \text { ? }} \\ {\text { b. What is the percentage yield when the }} \\ {\text { actual yield is } 5.40 \mathrm{g} \text { and the theoretical }} \\ {\text { c. What actual yield theoretical yield pair }} \\ {\text { produced the largest percentage yield? }}\end{array}\end{equation}

The percentage yield of nitric acid is 95\(\% .\) If 9.88 \(\mathrm{kg}\) of nitrogen dioxide react, what mass of nitric acid is isolated? $$3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g)$$

What mass of oxygen is required to burn 688 g of octane, \(\mathrm{C}_{8} \mathrm{H}_{18},\) completely?

Define stoichiometry.

Calcium carbide, \(\mathrm{CaC}_{2},\) reacts with water to form acetylene. $$\mathrm{CaC}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{C}_{2} \mathrm{H}_{2}(g)+\mathrm{Ca}(\mathrm{OH})_{2}(s)$$ \begin{equation}\begin{array}{l}{\text { a. How many grams of water are needed to }} \\ {\text { react with } 485 \text { g of calcium carbide? }} \\\ {\text { b. How many grams of } \mathrm{CaC}_{2} \text { could make }} \\\ {\text { 23.6 g } \mathrm{C}_{2} \mathrm{H}_{2} \text { ? }} \\ {\text { c. If } 55.3 \mathrm{g} \mathrm{Ca}(\mathrm{OH})_{2} \text { are formed, how many }} \\ {\text { grams of water reacted? }}\end{array}\end{equation}
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