Chapter 3 Problems and Answers

Ace your homework & exams now with Quizwiz!

36. The average length of an O-O single bond is 132 pm. The average length of an O=O double bond is 121 pm. What do you predict the O-O bond lengths will be in ozone? Will they all be the same? Explain your predictions.

Here are the resonance structures for ozone: Both contain 1 double bond (expected length of 121 pm) and 1 single bond (expected length of 132 pm). But in reality, the bonds are neither single nor double. Rather, the length of each bond is intermediate between the single and double bond lengths. A reasonable prediction would be 126 or 127 pm for both O-to-O bonds, midway between the two lengths.

46. Explain how the small changes in CL concentrations (measured in parts per billion) can cause the much larger changes in O3 concentrations (measured in parts per million).

Cl acts as a catalyst in the series of reactions in which stratospheric O3 molecules are changed to O2 molecules. As it is not consumed in the reaction, can continue to catalyze the breakdown of O3. (Not sure this is right)

7. Using the periodic table as guide, specify the number of protons and electrons in a neutral atom of each of these elements. a. oxygen (O) b. magnesium (Mg) c. Nitrogen (N) d. Sulfur (S)

a. A neutral atom of oxygen has 8 protons and 8 electrons. b. A neutral atom of nitrogen has 7 protons and 7 electrons. c. A neutral atom of magnesium has 12 protons and 12 electrons. d. A neutral atom of sulfur has 16 protons and 16 electrons.

1. How does ozone differ from oxygen in its chemical formula? In its properties?

The chemical formulas of ozone and oxygen are O 3 and O 2 , respectively. Both are gases, but they differ in their properties. Oxygen has no odor; ozone has a very sharp odor. ... These oxygen atoms then react with oxygen molecules to produce ozone.

33. What are some of the reasons that the solution to ozone depletion propose in this Sydney Harris cartoon will not work? (Illustration in book)

The solution proposed in the cartoon will not work for several reasons. One is that the amount of ozone required is too large to ship up to the stratosphere (and of course we don't have a series of freight planes heading up there anyhow). More importantly, if the mechanism for ozone destruction remains in place, any new ozone will be destroyed as well.

24. CFCs were used in hair sprays, refrigerators, air, conditioners, and plastic foams. Which properties of CFCs made then desirable for these uses?

We used CFC's because they are nontoxic, nonflammable, cheap, and very easy to get. These molecules have very strong bonds that make them almost indestructible.

57. The effect a chemical substance has on the ozone layer is measured by a value called its ozone-depleting potential, ODP. This is a numerical scale that estimates the lifetime potential stratospheric ozone that could be destroyed by a given mass of the substance. All values are relative to CFC-11, which has an ODP defined as equal to 1.0. Use those facts to answer these questions. a. What factors do you think will influence the ODP value for a chemical? Why? b. Most CFCs have ODP values ranging from 0.6 to 1.0. What range do you expect for HCFCs? Explain your reasoning. c. What ODP values do you expect for HFCs? Explain your reasoning.

a. Factors include (1) the reactivity of the compound, because this in turn affects the length of time it will remains in the atmosphere, and (2) the presence of Cl or Br in the compound, because and , formed in the stratosphere from chlorine- and bromine- containing compounds, deplete ozone. b. Most HCFCs, developed as replacements for CFCs, would be expected to have ODP values lower than those of the CFCs. HCFCs tend to be somewhat more reactive in the troposphere and thus do not accumulate in the stratosphere. Their ODP values range from 0.01 to 0.11. c. HFCs were also developed as CFC substitutes. Without chlorine or bromine present, they generally do not have the potential to deplete stratospheric ozone. Their ODP values are 0.0.

11. Assuming that the octet rule applies, draw the Lewis structure for each of these molecules. a.CCl4(carbon tetrachloride, a substance formerly used as a cleaning agent) b.H2O2(hydrogen peroxide, a mild disinfectant; the atoms are bonded in this order: H-O-O-H) c.H2S (hydrogen sulfide, a gas with the unpleasant odor of rotten eggs) d.N2(nitrogen gas, the major component of the atmosphere)e.HCN (hydrogen cyanide, a molecule found in space and a poisonous gas) f.N2O (nitrous oxide, "laughing gas";the atoms are bonded N-N-O) g.CS2(carbon disulfide, used to kill rodents; the atoms are bonded S-C-S)

a. There are4 + 4(7) = 32 outer electrons b. There are 2(1)+ 2(6)= 14 outer electrons. The Lewis structure is: c. There are 2(1)+ 6= 8 outer electrons. The Lewis structure is: d. There are 2(5) = 10 outer electrons. The Lewis structure is: e. There are 1 + 4 + 5 = 10 outer electrons. The Lewis structure is: f. There are 2(5) + 6 = 16 outer electrons. One possible Lewis structure is:or Other resonance structures are possible for N2O as well. g. There are 4 + 2(6) = 16 outer electrons. The Lewis structure is:

20. Ultraviolet radiation is categorized as UVA, UVB, or UVC. Arrange these types in order of increasing: a. wavelength b. potential for biological damage c. energy

a. UVC, UVB, UVA b. UVC, UVB, UVA (I'm not sure I have this right) c. UVC, UVB, UVA (Not sure if this is right either)

13. Consider these two waves representing different parts of the electromagnetic spectrum. How do they compare in terms of: (Illustration in book) a. wavelength b.frequency c. speed of travel

a. Wave 1 has longer wavelength than wave 2. b. Wave 1 has lower frequency than wave 2. c. Both waves travel at the same speed.


Related study sets

Abeka: Science Order and Design Reading Quiz N 7th Grade

View Set

UNIV 1001 UOP Graded Quiz Unit 3

View Set

Chp 10: Reshaping of Medieval Europe

View Set

Chapter 9. Games and Strategic Behavior

View Set