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- PrepTest 146
- Section 4, Question 26
By 1970 it was well established that ultraviolet(442 more words not shown)
Based on the passage, the(27 more words not shown)
- Atesting to see(14 more words not shown)
- Btesting to see(17 more words not shown)
- Ctesting the chemical(9 more words not shown)
- Dtesting to see(12 more words not shown)
- Etesting the chemical(15 more words not shown)
Reading the passage
Would the replacement damage ozone? That is the question our test needs to help answer. It is not enough to discover that the replacement reacts with something or breaks apart under ultraviolet light.
What we’re looking for
Remember the full CFC mechanism, too: CFCs release chlorine when they break down, and the chlorine then attacks ozone. We need to consider both the replacement chemical and the substances it might release.
Working the answers
- A
The words “other than ozone” take the test away from our target. A reaction with another form of oxygen does not tell us whether the replacement would damage the ozone layer.
- B
Reactions with unspecified chemicals in the lower atmosphere do not directly answer whether the replacement or its products would attack stratospheric ozone.
- C
Reacting with chlorine is different from reacting like chlorine. The harmful behavior we are investigating is chlorine’s destruction of ozone, not whether another chemical reacts with chlorine itself.
- DCorrect
We want to know whether the replacement or its constituent elements share chlorine’s relevant chemical properties—especially its ability to react with and destroy ozone. Including the constituents matters because a chemical might release a damaging substance after it breaks down, just as CFCs release chlorine.
- E
Ultraviolet breakdown is part of the CFC story, so this test is relevant to one step. But breakdown alone does not establish ozone damage. The components released might be harmless to ozone. D gets closer to the property that makes the released chlorine destructive.
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