7.10: Exercises
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Writing and Balancing Chemical Equations
1. What does it mean to say an equation is balanced? Why is it important for an equation to be balanced?
2. Consider molecular, complete ionic, and net ionic equations.
(a) What is the difference between these types of equations?
(b) In what circumstance would the complete and net ionic equations for a reaction be identical?
3. Balance the following equations:
(a) \(\mathrm{PCl}_5(s)+\mathrm{H}_2 \mathrm{O}(l) \longrightarrow \mathrm{POCl}_3(l)+\mathrm{HCl}(a q)\)
(b) \(\mathrm{Cu}(s)+\mathrm{HNO}_3(a q) \longrightarrow \mathrm{Cu}\left(\mathrm{NO}_3\right)_2(a q)+\mathrm{H}_2 \mathrm{O}(l)+\mathrm{NO}(g)\)
(c) \(\mathrm{H}_2(g)+\mathrm{I}_2(s) \longrightarrow \mathrm{HI}(s)\)
(d) \(\mathrm{Fe}(s)+\mathrm{O}_2(g) \longrightarrow \mathrm{Fe}_2 \mathrm{O}_3(s)\)
(e) \(\mathrm{Na}(s)+\mathrm{H}_2 \mathrm{O}(l) \longrightarrow \mathrm{NaOH}(a q)+\mathrm{H}_2(g)\)
(f) \(\left(\mathrm{NH}_4\right)_2 \mathrm{Cr}_2 \mathrm{O}_7(s) \longrightarrow \mathrm{Cr}_2 \mathrm{O}_3(s)+\mathrm{N}_2(g)+\mathrm{H}_2 \mathrm{O}(g)\)
(g) \(\mathrm{P}_4(s)+\mathrm{Cl}_2(g) \longrightarrow \mathrm{PCl}_3(l)\)
(h) \(\mathrm{PtCl}_4(s) \longrightarrow \operatorname{Pt}(s)+\mathrm{Cl}_2(g)\)
4. Balance the following equations
(a) \(\mathrm{Ag}(s)+\mathrm{H}_2 \mathrm{~S}(g)+\mathrm{O}_2(g) \longrightarrow \mathrm{Ag}_2 \mathrm{~S}(s)+\mathrm{H}_2 \mathrm{O}(l)\)
(b) \(\mathrm{P}_4(s)+\mathrm{O}_2(g) \longrightarrow \mathrm{P}_4 \mathrm{O}_{10}(s)\)
(c) \(\mathrm{Pb}(s)+\mathrm{H}_2 \mathrm{O}(l)+\mathrm{O}_2(g) \longrightarrow \mathrm{Pb}(\mathrm{OH})_2(s)\)
(d) \(\mathrm{Fe}(s)+\mathrm{H}_2 \mathrm{O}(l) \longrightarrow \mathrm{Fe}_3 \mathrm{O}_4(s)+\mathrm{H}_2(g)\)
(e) \(\mathrm{Sc}_2 \mathrm{O}_3(s)+\mathrm{SO}_3(l) \longrightarrow \mathrm{Sc}_2\left(\mathrm{SO}_4\right)_3(s)\)
(f) \(\mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2(a q)+\mathrm{H}_3 \mathrm{PO}_4(a q) \longrightarrow \mathrm{Ca}\left(\mathrm{H}_2 \mathrm{PO}_4\right)_2(a q)\)
(g) \(\mathrm{Al}(s)+\mathrm{H}_2 \mathrm{SO}_4(a q) \longrightarrow \mathrm{Al}_2\left(\mathrm{SO}_4\right)_3(s)+\mathrm{H}_2(g)\)
(h) \(\mathrm{TiCl}_4(s)+\mathrm{H}_2 \mathrm{O}(g) \longrightarrow \mathrm{TiO}_2(s)+\mathrm{HCl}(g)\)
5. Write a balanced molecular equation describing each of the following chemical reactions.
(a) Solid calcium carbonate is heated and decomposes to solid calcium oxide and carbon dioxide gas.
(b) Gaseous butane, \(\mathrm{C}_4 \mathrm{H}_{10}\), reacts with diatomic oxygen gas to yield gaseous carbon dioxide and water vapor.
(c) Aqueous solutions of magnesium chloride and sodium hydroxide react to produce solid magnesium hydroxide and aqueous sodium chloride.
(d) Water vapor reacts with sodium metal to produce solid sodium hydroxide and hydrogen gas.
6. Write a balanced equation describing each of the following chemical reactions.
(a) Solid potassium chlorate, \(\mathrm{KClO}_3\), decomposes to form solid potassium chloride and diatomic oxygen gas.
(b) Solid aluminum metal reacts with solid diatomic iodine to form solid \(\mathrm{Al}_2 \mathrm{I}_6\).
(c) When solid sodium chloride is added to aqueous sulfuric acid, hydrogen chloride gas and aqueous sodium sulfate are produced.
(d) Aqueous solutions of phosphoric acid and potassium hydroxide react to produce aqueous potassium dihydrogen phosphate and liquid water.
7. Colorful fireworks often involve the decomposition of barium nitrate and potassium chlorate and the reaction of the metals magnesium, aluminum, and iron with oxygen.
(a) Write the formulas of barium nitrate and potassium chlorate.
(b) The decomposition of solid potassium chlorate leads to the formation of solid potassium chloride and diatomic oxygen gas. Write an equation for the reaction.
(c) The decomposition of solid barium nitrate leads to the formation of solid barium oxide, diatomic nitrogen gas, and diatomic oxygen gas. Write an equation for the reaction.
(d) Write separate equations for the reactions of the solid metals magnesium, aluminum, and iron with diatomic oxygen gas to yield the corresponding metal oxides. (Assume the iron oxide contains \(\mathrm{Fe}^{3+}\) ions.)
8. Fill in the blank with a single chemical formula for a covalent compound that will balance the equation:
9. Aqueous hydrogen fluoride (hydrofluoric acid) is used to etch glass and to analyze minerals for their silicon content. Hydrogen fluoride will also react with sand (silicon dioxide).
- Write an equation for the reaction of solid silicon dioxide with hydrofluoric acid to yield gaseous silicon tetrafluoride and liquid water.
- The mineral fluorite (calcium fluoride) occurs extensively in Illinois. Solid calcium fluoride can also be prepared by the reaction of aqueous solutions of calcium chloride and sodium fluoride, yielding aqueous sodium chloride as the other product. Write complete and net ionic equations for this reaction.
- The first step is the decomposition of solid calcium carbonate from seashells to form solid calcium oxide and gaseous carbon dioxide.
- The second step is the formation of solid calcium hydroxide as the only product from the reaction of the solid calcium oxide with liquid water.
- Solid calcium hydroxide is then added to the seawater, reacting with dissolved magnesium chloride to yield solid magnesium hydroxide and aqueous calcium chloride.
- The solid magnesium hydroxide is added to a hydrochloric acid solution, producing dissolved magnesium chloride and liquid water.
- Finally, the magnesium chloride is melted and electrolyzed to yield liquid magnesium metal and diatomic chlorine gas.
11. From the balanced molecular equations, write the complete ionic and net ionic equations for the following:
(a) \(\mathrm{K}_2 \mathrm{C}_2 \mathrm{O}_4(a q)+\mathrm{Ba}(\mathrm{OH})_2(a q) \longrightarrow 2 \mathrm{KOH}(a q)+\mathrm{BaC}_2 \mathrm{O}_4(s)\)
(b) \(\mathrm{Pb}\left(\mathrm{NO}_3\right)_2(a q)+\mathrm{H}_2 \mathrm{SO}_4(a q) \longrightarrow \mathrm{PbSO}_4(s)+2 \mathrm{HNO}_3(a q)\)
(c) \(\mathrm{CaCO}_3(s)+\mathrm{H}_2 \mathrm{SO}_4(a q) \longrightarrow \mathrm{CaSO}_4(s)+\mathrm{CO}_2(g)+\mathrm{H}_2 \mathrm{O}(l)\)
7.2 Classifying Chemical Reactions
12. Use the following equations to answer the next four questions:
i. \(\mathrm{H}_2 \mathrm{O}(s) \longrightarrow \mathrm{H}_2 \mathrm{O}(l)\)
ii. \(\mathrm{Na}^{+}(a q)+\mathrm{Cl}^{-}(\)aq \()+\mathrm{Ag}^{+}(a q)+\mathrm{NO}_3{ }^{-}(a q) \longrightarrow \mathrm{AgCl}(s)+\mathrm{Na}^{+}(a q)+\mathrm{NO}_3{ }^{-}(a q)\)
iii. \(\mathrm{CH}_3 \mathrm{OH}(g)+\mathrm{O}_2(g) \longrightarrow \mathrm{CO}_2(g)+\mathrm{H}_2 \mathrm{O}(g)\)
iv. \(2 \mathrm{H}_2 \mathrm{O}(l) \longrightarrow 2 \mathrm{H}_2(g)+\mathrm{O}_2(g)\)
v. \(\mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{H}_2 \mathrm{O}(l)\)
(a) Which equation describes a physical change?
(b) Which equation identifies the reactants and products of a combustion reaction?
(c) Which equation is not balanced?
(d) Which is a net ionic equation?
13. Indicate what type, or types, of reaction each of the following represents:
(a) \(\mathrm{Ca}(s)+\mathrm{Br}_2(l) \longrightarrow \mathrm{CaBr}_2(s)\)
(b) \(\mathrm{Ca}(\mathrm{OH})_2(a q)+2 \mathrm{HBr}(a q) \longrightarrow \mathrm{CaBr}_2(a q)+2 \mathrm{H}_2 \mathrm{O}(l)\)
(c) \(\mathrm{C}_6 \mathrm{H}_{12}(l)+9 \mathrm{O}_2(g) \longrightarrow 6 \mathrm{CO}_2(g)+6 \mathrm{H}_2 \mathrm{O}(g)\)
14. Indicate what type, or types, of reaction each of the following represents:
(a) \(\mathrm{H}_2 \mathrm{O}(g)+\mathrm{C}(s) \longrightarrow \mathrm{CO}(g)+\mathrm{H}_2(g)\)
(b) \(2 \mathrm{KClO}_3(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_2(g)\)
(c) \(\mathrm{Al}(\mathrm{OH})_3(a q)+3 \mathrm{HCl}(a q) \longrightarrow \mathrm{AlCl}_3(a q)+3 \mathrm{H}_2 \mathrm{O}(l)\)
(d) \(\mathrm{Pb}\left(\mathrm{NO}_3\right)_2(a q)+\mathrm{H}_2 \mathrm{SO}_4(a q) \longrightarrow \mathrm{PbSO}_4(s)+2 \mathrm{HNO}_3(a q)\)
15. Silver can be separated from gold because silver dissolves in nitric acid while gold does not. Is the dissolution of silver in nitric acid an acid-base reaction or an oxidation-reduction reaction? Explain your answer.
16. Determine the oxidation states of the elements in the following compounds:
(a) NaI
(b) \(\mathrm{GdCl}_3\)
(c) \(\mathrm{LiNO}_3\)
(d) \(\mathrm{H}_2 \mathrm{Se}\)
(e) \(\mathrm{Mg}_2 \mathrm{Si}\)
(f) \(\mathrm{RbO}_2\), rubidium superoxide
(g) HF
17. Determine the oxidation states of the elements in the compounds listed. None of the oxygen-containing compounds are peroxides or superoxides.
(a) \(\mathrm{H}_3 \mathrm{PO}_4\)
(b) \(\mathrm{Al}(\mathrm{OH})_3\)
(c) \(\mathrm{SeO}_2\)
(d) \(\mathrm{KNO}_2\)
(e) \(\mathrm{In}_2 \mathrm{~S}_3\)
(f) \(\mathrm{P}_4 \mathrm{O}_6\)
18. Determine the oxidation states of the elements in the compounds listed. None of the oxygen-containing compounds are peroxides or superoxides.
(a) \(\mathrm{H}_2 \mathrm{SO}_4\)
(b) \(\mathrm{Ca}(\mathrm{OH})_2\)
(c) BrOH
(d) \(\mathrm{ClNO}_2\)
(e) \(\mathrm{TiCl}_4\)
(f) NaH
19. Classify the following as acid-base reactions or oxidation-reduction reactions:
(a) \(\mathrm{Na}_2 \mathrm{~S}(a q)+2 \mathrm{HCl}(a q) \longrightarrow 2 \mathrm{NaCl}(a q)+\mathrm{H}_2 \mathrm{~S}(g)\)
(b) \(2 \mathrm{Na}(s)+2 \mathrm{HCl}(a q) \longrightarrow 2 \mathrm{NaCl}(a q)+\mathrm{H}_2(g)\)
(c) \(\mathrm{Mg}(s)+\mathrm{Cl}_2(g) \longrightarrow \mathrm{MgCl}_2(s)\)
(d) \(\mathrm{MgO}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{MgCl}_2(a q)+\mathrm{H}_2 \mathrm{O}(l)\)
(e) \(\mathrm{K}_3 \mathrm{P}(s)+2 \mathrm{O}_2(g) \longrightarrow \mathrm{K}_3 \mathrm{PO}_4(s)\)
(f) \(3 \mathrm{KOH}(a q)+\mathrm{H}_3 \mathrm{PO}_4(a q) \longrightarrow \mathrm{K}_3 \mathrm{PO}_4(a q)+3 \mathrm{H}_2 \mathrm{O}(l)\)
20. Identify the atoms that are oxidized and reduced, the change in oxidation state for each, and the oxidizing and reducing agents in each of the following equations:
(a) \(\mathrm{Mg}(s)+\mathrm{NiCl}_2(a q) \longrightarrow \mathrm{MgCl}_2(a q)+\mathrm{Ni}(s)\)
(b) \(\mathrm{PCl}_3(l)+\mathrm{Cl}_2(g) \longrightarrow \mathrm{PCl}_5(s)\)
(c) \(\mathrm{C}_2 \mathrm{H}_4(g)+3 \mathrm{O}_2(g) \longrightarrow 2 \mathrm{CO}_2(g)+2 \mathrm{H}_2 \mathrm{O}(g)\)
(d) \(\mathrm{Zn}(s)+\mathrm{H}_2 \mathrm{SO}_4(a q) \longrightarrow \mathrm{ZnSO}_4(a q)+\mathrm{H}_2(g)\)
(e) \(2 \mathrm{~K}_2 \mathrm{~S}_2 \mathrm{O}_3(s)+\mathrm{I}_2(s) \longrightarrow \mathrm{K}_2 \mathrm{~S}_4 \mathrm{O}_6(s)+2 \mathrm{KI}(s)\)
(f) \(3 \mathrm{Cu}(s)+8 \mathrm{HNO}_3(a q) \longrightarrow 3 \mathrm{Cu}\left(\mathrm{NO}_3\right)_2(a q)+2 \mathrm{NO}(g)+4 \mathrm{H}_2 \mathrm{O}(l)\)
21. Complete and balance the following acid-base equations:
- HCl gas reacts with solid Ca(OH) 2 ( s ).
- A solution of Sr(OH) 2 is added to a solution of HNO 3 .
- A solution of HClO 4 is added to a solution of LiOH.
- Aqueous H 2 SO 4 reacts with NaOH.
- Ba(OH) 2 reacts with HF gas.
23. Complete and balance the following oxidation-reduction reactions, which give the highest possible oxidation state for the oxidized atoms.
(a) \(\mathrm{Al}(s)+\mathrm{F}_2(g)\) \(\qquad\)
(b) \(\mathrm{Al}(s)+\mathrm{CuBr}_2(a q)\)
\(\longrightarrow\) (single displacement)
(c) \(\mathrm{P}_4(s)+\mathrm{O}_2(g)\) \(\qquad\)
(d) \(\mathrm{Ca}(s)+\mathrm{H}_2 \mathrm{O}(l) \longrightarrow\) (products are a strong base and a diatomic gas)
24. Complete and balance the following oxidation-reduction reactions, which give the highest possible oxidation state for the oxidized atoms.
(a) \(\mathrm{K}(s)+\mathrm{H}_2 \mathrm{O}(l)\) \(\qquad\) \(\rightarrow\)
(b) \(\mathrm{Ba}(s)+\operatorname{HBr}(a q)\) \(\qquad\) \(\rightarrow\)
(c) \(\operatorname{Sn}(s)+\mathrm{I}_2(s)\) \(\qquad\) \(\rightarrow\)
25. Complete and balance the equations for the following acid-base neutralization reactions. If water is used as a solvent, write the reactants and products as aqueous ions. In some cases, there may be more than one correct answer, depending on the amounts of reactants used.
(a) \(\mathrm{Mg}(\mathrm{OH})_2(s)+\mathrm{HClO}_4(a q)\) \(\qquad\)
(b) \(\mathrm{SO}_3(g)+\mathrm{H}_2 \mathrm{O}(l)\) \(\qquad\) (assume an excess of water and that the product dissolves)
(c) \(\mathrm{SrO}(s)+\mathrm{H}_2 \mathrm{SO}_4(l)\) \(\qquad\)
26. When heated to \(700-800^{\circ} \mathrm{C}\), diamonds, which are pure carbon, are oxidized by atmospheric oxygen. (They burn!) Write the balanced equation for this reaction.
27. The military has experimented with lasers that produce very intense light when fluorine combines explosively with hydrogen. What is the balanced equation for this reaction?
28. Write the molecular, total ionic, and net ionic equations for the following reactions:
(a) \(\mathrm{Ca}(\mathrm{OH})_2(a q)+\mathrm{HC}_2 \mathrm{H}_3 \mathrm{O}_2(a q)\) \(\qquad\)
(b) \(\mathrm{H}_3 \mathrm{PO}_4(a q)+\mathrm{CaCl}_2(a q)\) \(\qquad\)
29. Great Lakes Chemical Company produces bromine, \(\mathrm{Br}_2\), from bromide salts such as NaBr , in Arkansas brine by treating the brine with chlorine gas. Write a balanced equation for the reaction of NaBr with \(\mathrm{Cl}_2\).23. Complete and balance the following oxidation-reduction reactions, which give the highest possible oxidation state for the oxidized atoms.
31.Lithium hydroxide may be used to absorb carbon dioxide in enclosed environments, such as manned spacecraft and submarines. Write an equation for the reaction that involves 2 mol of LiOH per 1 mol of CO 2 . (Hint: Water is one of the products.)
33. Complete and balance the equations of the following reactions, each of which could be used to remove hydrogen sulfide from natural gas:
(a) \(\mathrm{Ca}(\mathrm{OH})_2(s)+\mathrm{H}_2 \mathrm{~S}(g)\) \(\qquad\)
(b) \(\mathrm{Na}_2 \mathrm{CO}_3(a q)+\mathrm{H}_2 \mathrm{~S}(g) \longrightarrow\)
35. Write balanced chemical equations for the reactions used to prepare each of the following compounds from the given starting material(s). In some cases, additional reactants may be required.
- solid ammonium nitrate from gaseous molecular nitrogen via a two-step process (first reduce the nitrogen to ammonia, then neutralize the ammonia with an appropriate acid)
- gaseous hydrogen bromide from liquid molecular bromine via a one-step redox reaction
- gaseous H 2 S from solid Zn and S via a two-step process (first a redox reaction between the starting materials, then reaction of the product with a strong acid)
37. Complete and balance each of the following half-reactions (steps \(2-5\) in half-reaction method):
(a) \(\mathrm{Sn}^{4+}(a q) \longrightarrow \mathrm{Sn}^{2+}(a q)\)
(b) \(\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}(a q) \longrightarrow \mathrm{Ag}(s)+\mathrm{NH}_3(a q)\)
(c) \(\mathrm{Hg}_2 \mathrm{Cl}_2(s) \longrightarrow \mathrm{Hg}(l)+\mathrm{Cl}^{-}(a q)\)
(d) \(\mathrm{H}_2 \mathrm{O}(l) \longrightarrow \mathrm{O}_2(g)\) (in acidic solution)
(e) \(\mathrm{IO}_3{ }^{-}(a q) \longrightarrow \mathrm{I}_2(s)\) (in basic solution)
(f) \(\mathrm{SO}_3{ }^{2-}(a q) \longrightarrow \mathrm{SO}_4{ }^{2-}(a q)\) (in acidic solution)
(g) \(\mathrm{MnO}_4^{-}(a q) \longrightarrow \mathrm{Mn}^{2+}(a q)\) (in acidic solution)
(h) \(\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{ClO}_3{ }^{-}(a q)\) (in basic solution)
38. Complete and balance each of the following half-reactions (steps \(2-5\) in half-reaction method):
(a) \(\mathrm{Cr}^{2+}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)\)
(b) \(\mathrm{Hg}(l)+\mathrm{Br}^{-}(a q) \longrightarrow \mathrm{HgBr}_4{ }^{2-}(a q)\)
(c) \(\mathrm{ZnS}(s) \longrightarrow \mathrm{Zn}(s)+\mathrm{S}^{2-}(a q)\)
(d) \(\mathrm{H}_2(g) \longrightarrow \mathrm{H}_2 \mathrm{O}(l)\) (in basic solution)
(e) \(\mathrm{H}_2(g) \longrightarrow \mathrm{H}_3 \mathrm{O}^{+}(a q)\) (in acidic solution)
(f) \(\mathrm{NO}_3{ }^{-}(a q) \longrightarrow \mathrm{HNO}_2(a q)\) (in acidic solution)
(g) \(\mathrm{MnO}_2(s) \longrightarrow \mathrm{MnO}_4{ }^{-}(a q)\) (in basic solution)
(h) \(\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{ClO}_3^{-}(a q)\) (in acidic solution)
39. Balance each of the following equations according to the half-reaction method:
(a) \(\mathrm{Sn}^{2+}(a q)+\mathrm{Cu}^{2+}(a q) \longrightarrow \mathrm{Sn}^{4+}(a q)+\mathrm{Cu}^{+}(a q)\)
(b) \(\mathrm{H}_2 \mathrm{~S}(g)+\mathrm{Hg}_2{ }^{2+}(a q) \longrightarrow \mathrm{H} g(l)+\mathrm{S}(s)\) (in acid)
(c) \(\mathrm{CN}^{-}(a q)+\mathrm{ClO}_2(a q) \longrightarrow \mathrm{CNO}^{-}(a q)+\mathrm{Cl}^{-}(a q)\) (in acid)
(d) \(\mathrm{Fe}^{2+}(a q)+\mathrm{Ce}^{4+}(a q) \longrightarrow \mathrm{Fe}^{3+}(a q)+\mathrm{Ce}^{3+}(a q)\)
(e) \(\mathrm{HBrO}(a q) \longrightarrow \mathrm{Br}^{-}(a q)+\mathrm{O}_2(g)\) (in acid)
Reaction Stoichiometry
42. Write the balanced equation, then outline the steps necessary to determine the information requested in each of the following:
(a) The number of moles and the mass of chlorine, \(\mathrm{Cl}_2\), required to react with 10.0 g of sodium metal, Na , to produce sodium chloride, NaCl .
(b) The number of moles and the mass of oxygen formed by the decomposition of 1.252 g of mercury(II) oxide.
(c) The number of moles and the mass of sodium nitrate, \(\mathrm{NaNO}_3\), required to produce 128 g of oxygen. ( \(\mathrm{NaNO}_2\) is the other product.)
(d) The number of moles and the mass of carbon dioxide formed by the combustion of 20.0 kg of carbon in an excess of oxygen.
(e) The number of moles and the mass of copper(II) carbonate needed to produce 1.500 kg of copper(II) oxide. \(\left(\mathrm{CO}_2\right.\) is the other product. \()\)
The number of moles and the mass of
43. Determine the number of moles and the mass requested for each reaction in Exercise 42.
44. Write the balanced equation, then outline the steps necessary to determine the information requested in each of the following:
(a) The number of moles and the mass of Mg required to react with 5.00 g of HCl and produce \(\mathrm{MgCl}_2\) and \(\mathrm{H}_2\).
(b) The number of moles and the mass of oxygen formed by the decomposition of 1.252 g of silver(l) oxide.
(c) The number of moles and the mass of magnesium carbonate, \(\mathrm{MgCO}_3\), required to produce 283 g of carbon dioxide. ( MgO is the other product.)
(d) The number of moles and the mass of water formed by the combustion of 20.0 kg of acetylene, \(\mathrm{C}_2 \mathrm{H}_2\), in an excess of oxygen.
(e) The number of moles and the mass of barium peroxide, \(\mathrm{BaO}_2\), needed to produce 2.500 kg of barium oxide, \(\mathrm{BaO}\left(\mathrm{O}_2\right.\) is the other product. \()\)
(f)
45. Determine the number of moles and the mass requested for each reaction in Exercise 4.44.
46. H 2 is produced by the reaction of 118.5 mL of a 0.8775-M solution of H 3 PO 4 according to the following equation:
- Outline the steps necessary to determine the number of moles and mass of H 2 .
- Perform the calculations outlined.
- Outline the steps necessary to determine the number of moles and mass of gallium chloride.
- Perform the calculations outlined.
- How many molecules of I 2 are produced?
- What mass of I 2 is produced?
- How many molecules of Zn(CN) 2 are produced by the reaction of 35.27 g of K[Ag(CN) 2 ]?
- What mass of Zn(CN) 2 is produced?
51. Carborundum is silicon carbide, SiC, a very hard material used as an abrasive on sandpaper and in other applications. It is prepared by the reaction of pure sand, SiO 2 , with carbon at high temperature. Carbon monoxide, CO, is the other product of this reaction. Write the balanced equation for the reaction, and calculate how much SiO 2 is required to produce 3.00 kg of SiC.
52. Automotive air bags inflate when a sample of sodium azide, NaN 3 , is very rapidly decomposed.
What mass of sodium azide is required to produce 2.6 ft 3 (73.6 L) of nitrogen gas with a density of 1.25 g/L?
54. In an accident, a solution containing 2.5 kg of nitric acid was spilled. Two kilograms of Na 2 CO 3 was quickly spread on the area and CO 2 was released by the reaction. Was sufficient Na 2 CO 3 used to neutralize all of the acid?
56. What volume of 0.750 M hydrochloric acid solution can be prepared from the HCl produced by the reaction of 25.0 g of NaCl with excess sulfuric acid?
58. A mordant is a substance that combines with a dye to produce a stable fixed color in a dyed fabric. Calcium acetate is used as a mordant. It is prepared by the reaction of acetic acid with calcium hydroxide.
What mass of Ca(OH) 2 is required to react with the acetic acid in 25.0 mL of a solution having a density of 1.065 g/mL and containing 58.0% acetic acid by mass?
59. The toxic pigment called white lead, Pb 3 (OH) 2 (CO 3 ) 2 , has been replaced in white paints by rutile, TiO 2 . How much rutile (g) can be prepared from 379 g of an ore that contains 88.3% ilmenite (FeTiO 3 ) by mass?Reaction Yields
60. The following quantities are placed in a container: 1.5 10 24 atoms of hydrogen, 1.0 mol of sulfur, and 88.0 g of diatomic oxygen.
- What is the total mass in grams for the collection of all three elements?
- What is the total number of moles of atoms for the three elements?
- If the mixture of the three elements formed a compound with molecules that contain two hydrogen atoms, one sulfur atom, and four oxygen atoms, which substance is consumed first?
- How many atoms of each remaining element would remain unreacted in the change described in (c)?
62. Which of the postulates of Dalton's atomic theory explains why we can calculate a theoretical yield for a chemical reaction?
64. A sample of 0.53 g of carbon dioxide was obtained by heating 1.31 g of calcium carbonate. What is the percent yield for this reaction?
66. Citric acid, C 6 H 8 O 7 , a component of jams, jellies, and fruity soft drinks, is prepared industrially via fermentation of sucrose by the mold Aspergillus niger . The equation representing this reaction is
What mass of citric acid is produced from exactly 1 metric ton (1.000 10 3 kg) of sucrose if the yield is 92.30%?
67. Toluene, C 6 H 5 CH 3 , is oxidized by air under carefully controlled conditions to benzoic acid, C 6 H 5 CO 2 H, which is used to prepare the food preservative sodium benzoate, C 6 H 5 CO 2 Na. What is the percent yield of a reaction that converts 1.000 kg of toluene to 1.21 kg of benzoic acid?68. In a laboratory experiment, the reaction of 3.0 mol of H 2 with 2.0 mol of I 2 produced 1.0 mol of HI. Determine the theoretical yield in grams and the percent yield for this reaction.
69. Outline the steps needed to solve the following problem, then do the calculations. Ether, (C 2 H 5 ) 2 O, which was originally used as an anesthetic but has been replaced by safer and more effective medications, is prepared by the reaction of ethanol with sulfuric acid.
2C 2 H 5 OH + H 2 SO 4 ⟶ (C 2 H 5 ) 2 O + H 2 SO 4 ·H 2 O
What is the percent yield of ether if 1.17 L (d = 0.7134 g/mL) is isolated from the reaction of 1.500 L of C
2
H
5
OH
(d = 0.7894 g/mL)?
71. Outline the steps needed to determine the limiting reactant when 0.50 mol of Cr and 0.75 mol of H 3 PO 4 react according to the following chemical equation.
Determine the limiting reactant.
73. Uranium can be isolated from its ores by dissolving it as UO 2 (NO 3 ) 2 , then separating it as solid UO 2 (C 2 O 4 )·3H 2 O. Addition of 0.4031 g of sodium oxalate, Na 2 C 2 O 4 , to a solution containing 1.481 g of uranyl nitrate, UO 2 (NO 3 ) 2 , yields 1.073 g of solid UO 2 (C 2 O 4 )·3H 2 O.
Na 2 C 2 O 4 + UO 2 (NO 3 ) 2 + 3H 2 O ⟶ UO 2 (C 2 O 4 )·3H 2 O + 2NaNO 3
Determine the limiting reactant and the percent yield of this reaction.
75. How many molecules of the sweetener saccharin can be prepared from 30 C atoms, 25 H atoms, 12 O atoms, 8 S atoms, and 14 N atoms?
- What is the limiting reactant when 0.200 mol of P 4 and 0.200 mol of O 2 react according to
- Calculate the percent yield if 10.0 g of P 4 O 10 is isolated from the reaction.
Quantitative Chemical Analysis
78. What volume of 0.0105-M HBr solution is required to titrate 125 mL of a 0.0100- M Ca(OH) 2 solution?
80. What is the concentration of NaCl in a solution if titration of 15.00 mL of the solution with 0.2503 M AgNO 3 requires 20.22 mL of the AgNO 3 solution to reach the end point?
81. In a common medical laboratory determination of the concentration of free chloride ion in blood serum, a serum sample is titrated with a Hg(NO 3 ) 2 solution.
What is the Cl − concentration in a 0.25-mL sample of normal serum that requires 1.46 mL of 8.25 10 −4 M Hg(NO 3 ) 2 ( aq ) to reach the end point?
83. A sample of gallium bromide, GaBr 3 , weighing 0.165 g was dissolved in water and treated with silver nitrate, AgNO 3 , resulting in the precipitation of 0.299 g AgBr. Use these data to compute the %Ga (by mass) GaBr 3 .
85. A 0.025-g sample of a compound composed of boron and hydrogen, with a molecular mass of ~28 amu, burns spontaneously when exposed to air, producing 0.063 g of B 2 O 3 . What are the empirical and molecular formulas of the compound?
87. What volume of 0.600 M HCl is required to react completely with 2.50 g of sodium hydrogen carbonate?
89. What volume of a 0.3300- M solution of sodium hydroxide would be required to titrate 15.00 mL of 0.1500 M oxalic acid?
91. A sample of solid calcium hydroxide, Ca(OH) 2 , is allowed to stand in water until a saturated solution is formed. A titration of 75.00 mL of this solution with 5.00 10 −2 M HCl requires 36.6 mL of the acid to reach the end point. What is the molarity?
94. Potassium hydrogen phthalate, KHC 8 H 4 O 4 , or KHP, is used in many laboratories, including general chemistry laboratories, to standardize solutions of base. KHP is one of only a few stable solid acids that can be dried by warming and weighed. A 0.3420-g sample of KHC 8 H 4 O 4 reacts with 35.73 mL of a NaOH solution in a titration. What is the molar concentration of the NaOH?


13.6: Applications of Acid-Base Equilibria
Make sure you thoroughly understand the following essential concepts:
- Describe the major features of the global carbon cycle , and outline the role of acid-base processes in both the slow and fast parts of the cycle.
- Describe the principal acid-base reactions in aerobic respiration .
- Define acid rain , and explain its origins. Outline its major effects on the environment.
Acid-base reactions pervade every aspect of industrial-, physiological-, and environmental chemistry. In this unit we touch on a few highlights that anyone who studies or practices chemical science should be aware of.
Some applications of Buffers
Buffer solutions and the buffering effect they produce are extremely important in many practical applications of chemistry. The reason for this is that many chemical processes are quite sensitive to the pH; the extent of the reaction, its rate, and even the nature of the products can be altered if the pH is allowed to change. Such a change will tend to occur, for example, when the reaction in question, or an unrelated parallel reaction, consumes or releases hydrogen- or hydroxide ions.
Buffers in biochemistry and physiology
Many reactions that take place in living organisms fall into this category. Most biochemical processes are catalyzed by enzymes whose activities are highly dependent on the pH; if the local pH deviates too far from the optimum value, the enzyme may become permanently deactivated.
- For example, the oxidation of glucose to carbon dioxide and water — the fundamental energy-producing process of aerobic metabolism — releases H + ions. The multiple steps of this process are catalyzed by enzymes whose activity is strongly affected by the pH. If the region of the cell where these reactions take place were not buffered, the very act of living would soon reduce the local pH below the range in which the enzymes are active, halting energy production and killing the cell.
- Blood is strongly buffered (mainly by bicarbonate) to maintain its pH at 7.4±0.3; pH values below 7.0 or above 7.8 cause death within minutes.
- The interiors of most cells are buffered near pH 7.0 (by phosphates and proteins)
Buffers in industry
Buffers are employed in a wide variety of laboratory procedures and industrial processes:
- solutions for the calibration of pH meters
- growth of bacteria in culture media
- fermentation processes, including winemaking and the brewing of beer
- controlling the colors of dyes used in coloring fabrics
- shampoos (maintaining pH at or below 7 to conteract the alkalinity of soaps that can cause irritation)
- other personal care products such as baby lotions (keeping the pH around 6 to inhibit the growth of bacteria)
- pH control in printing inks to assure their proper penetration into the paper
Buffers in the environment
- Natural waters (lakes and streams) able to support aquatic life are buffered by the sediments they are in contact with. This buffering is not always able to offset the effects of acid rain.
- Seawater is buffered mainly by bicarbonate and borates. This allows the oceans to absorb about half of the CO 2 emitted to the atmosphere by human activities.
- Soils, in order to remain productive must maintain a near-neutral pH, not only for plants but also for microorganisms which fix- and recycle nitrogen. Soils rendered highly acidic by acid rain can keep essential nutrients such as phosphates in forms such as insoluble Ca 3 (PO 4 ) 2 that prevents their uptake by plants.
Acid-base chemistry in physiology
Acid-base chemistry plays a crucial role in physiology, both at the level of the individual cell and of the total organism. The reasons for this are twofold:
- Many of the major chemical components of an organism can themselves act as acids and/or bases. Thus proteins contain both acidic and basic groups, so that their shapes and their functional activities are highly dependent on the pH.
- Virtually all important metabolic processes involve the uptake or release of hydrogen ions. The very act of being alive tends to change the surrounding pH (usually reducing it); this will eventually kill the organism in the absence of buDering mechanisms.
About two-thirds of the weight of an adult human consists of water. About two-thirds of this water is located within cells, while the remaining third consists of extracellular water, mostly interstial fluid that bathes the cells, and the blood plasma. The latter, amounting to about 5% of body weight (about 5 L in the adult), serves as a supporting fluid for the blood cells and acts as a means of transporting chemicals between cells and the external environment. It is basically a 0.15 M solution of NaCl containing smaller amounts of other electrolytes, the most important of which are HCO 3 – and protein anions.
Respiration, the most important physiological activity of a cell, is an acid-producing process. Carbohydrate substances are broken down into carbon dioxide, and thus carbonic acid:
\[C(H_2O)_{n} + O_2 → CO_2 + n H_2O\]
Maintenance of acid-base balance
It is remarkable that the pH of most cellular fluids can be kept within such a narrow range, given the large number of processes that tend to upset it. This is due to the exquisite balance between a large number of interlinked processes operating at many different levels. Acid-base balance in the body is maintained by two general mechanisms: selective excretion of acids or bases, and by the buffering action of weak acid-base systems in body fluids.
-
Over a 24-hour period, the adult human eliminates the equivalent of about 20-40 moles of H
+
by way of the lungs in the form of CO
2
. In addition, the kidneys excrete perhaps 5% of this amount of acid, mostly in the form of H
2
PO
4
–
and
NH 4 + . Owing to their electric charges, these two species are closely linked to salt balance with ions such as Na + or K + and Cl – . - The major buffering system in the body is the carbonate system, which exists mainly in the form of HCO 3 – at normal physiological pH. Secondary buffering action comes from phosphates, from proteins and other weak organic acids, and (within blood cells), the hemoglobin.
There is a huge industry aimed at the notoriously science-challenged "alternative health" market that promotes worthless nostrums such as "ionized water" that are claimed to restore or preserve "the body's acid-base balance". They commonly point out that most foods are "acidic", (i.e., they are metabolized to H 2 CO 3 ), but they never explain that most of this acid is almot immediately exhaled in the form of CO 2 . The implication is that our health is being ruined by the resulting "acidification" of the body; some further imply this can be a cause of cancer and other assorted ailments.
 Those who, like you, have studied Chemistry, should consider that they have a social duty to counter this kind of predatory and deceptive pseudoscience.
Disturbances of acid-base balance
Deviations of the blood plasma pH from its normal value of 7.4 by more than about ±.1 can be very serious. These conditions are known medically as acidosis and alkalosis . They can be caused by metabolic disturbances such as diabetes and by kidney failure in which excretion of H 2 PO 4 – , for example, is inhibited.
Numerous other processes lead to temporary unbalances. Thus hyperventilation, which can result from emotional upset, leads to above-normal loss of CO 2 , and thus to alkalosis. Similarly, hypoventilation can act as a compensatory mechanism for acidosis. On the other hand, retention of CO 2 caused by bronchopneumonia, for example, can give rise to acidosis. Acidosis can also result from diarrhea (loss of alkaline fluid from the intestine), while loss of gastric contents by vomiting promotes alkalosis.
Acid rain
The atmosphere is naturally acidic
The natural, unpolluted atmosphere receives acidic, basic, and neutral substances from natural sources (volcanic emissions, salt spray, windblown dust and microbial metabolism) as well as from pollution (Figure 11.6.X). These react in a kind of gigantic acid-base titration to give a solution in which hydrogen ions must predominate to maintain charge balance (indicated by the identical widths of the bar graphs at the bottom labeled "anions" and "cations").
Carbon dioxide, however, is the major source of acidity in the unpolluted atmosphere. As will be explained farther on, CO 2 makes up 0.032 vol-% of dry air, and dissolves in water to form carbonic acid:
\[CO_{2(g) }+ H_2O(l) \rightleftharpoons H_2CO_{3(aq)}\]
Thus all rain is “acidic” in the sense that it contains dissolved CO 2 which will reduce its pH to 5.7. The term acid rain is therefore taken to mean rain whose pH is controlled by pollutants which can lower the pH into the range of 3-4, resulting in severe damage to the environment.
Origin of acid rain
Acid rain originates from emissions of SO 2 and various oxides of nitrogen (known collectively as "NO x ") that are formed during combustion processes, especially those associated with the burning of coal. Incineration of wastes, industrial operations such as smelting, and forest fires are other combustion-related sources; these also release particulate material into the atmosphere which plays an important role in concentrating and distributing these acid-forming substances.
Origins of NO x
Nitrogen oxides are formed naturally by soil bacteria acting on nitrate ions, an essential plant nutrient and itself a product of natural nitrogen fixation. Ammonia, a product of bacterial decomposition of organic matter, is eventually transformed into NO 3 – by bacteria into nitrogen oxides.
The quantity of fixed nitrogen produced industrially to support intensive agriculture now exceeds the amount formed naturally, and has become the major source of anthropogenic nitrogen oxides.
- Atmospheric nitrogen can be thermally decomposed to its various oxides NO x when subjected to high temperatures; one minor natural source is lightning. More significantly, these oxides are also formed when "clean" fuels such as natural gas (mainly methane, CH 4 ) is burnt in air.
- NO x can also react with other atmospheric pollutants to produce photochemical smog .
Formation of acids
The gases noted above react with atmospheric oxygen, with each other, and with particulate materials to form sulfuric, nitric, and hydrochloric acids.
- Most of the H 2 SO 4 comes from the photooxidation of SO 2 released from the burning of fossil fuels and from industrial operations such as smelting.
- Nitric acid similarly results from photooxidation of nitrogen oxides. Large quantities of these oxides are formed in high-temperature combustion processes in gasoline- and jet engines, and in the turbine engines of fossil-fueled electric power plants.
- Hydrochloric acid is formed naturally through the reaction of sea salt aerosols with atmospheric H 2 SO 4 and HNO 3 . The salt particles form when ocean waves break the surface and some of the water evaporates before the spray settles back to the surface. On a much smaller scale, fossil fuel combustion and waste incineration are believed to the principal human-caused sources of HCl.
Wet and dry deposition
There are two basic mechanisms by which acidic material is transported through the environment. The acid that dissolves in the water droplets that form clouds and precipitation, and is eventually incorporated into fog and snowflakes, undergoes wet deposition ; this is ordinarily what is referred to as "acid rain". Because it is widely dispersed, often high in the atmosphere, it can travel very large distances from the original source.
About 40% of the acidic substances introduced into the atmosphere becomes adsorbed onto particulate matter, which can be soot, smoke, windblown dust, and salt particles formed naturally from sea spray. This spreads less widely (typically around 30 km from the source), and falls onto surfaces by " dry deposition ". Once the acid-laden particles land on surfaces, ordinary rain, fog, and dew release the acidic components, often in considerably more concentrated form than occurs through dry deposition alone.
Effects on soils
Because soils support the growth of plants and of the soil microorganisms that are essential agents in the recycling of dead plant materials, acid rain has a indirect but profound effect on soil health and plant growth.
Soils containing alkaline components (most commonly limestone CaCO 3 and other insoluble carbonates) can neutralize added acid and mitigate its effects. But soils in high mountain regions tend to be thin and unable to provide adequate buffering capacity. The same is true in almost half of Canada, in which granite rock of the Canadian Shield is very close to the surface; the eastern provinces of the country are strongly impacted by acid rain.
- Most soils also contain clay and humic substances that bind and retain ions such as Ca 2 + , Mg 2 + and K + which are essential plant nutrients. Added H + binds even more strongly to these substances, displacing the nutrient ions from the upper layers. The overall effect is to leach these ions to greater depths where they may be inaccessible to plants, or to wash them away into ground water.
- The SO 4 2 – component of acid rain converts some nutrient cations into insoluble sulfates, reducing their availability to plants.
- Clays, which are complex aluminosilicates, gradually break down, releasing aluminum ions which normally form insoluble Al(OH) 3 . Addition of H + dissolves this hydroxide, raising the concentration of Al(H 2 O) 6 3+ to levels that can be toxic to plants.
- Many toxic heavy metals such as Pb, Cd and Cr are present in trace amounts in soils but are tied up in the form of insoluble salts. Acid rain can mobilize the ions of these metals, much as happens with aluminum.
Effects on plants
↓ Click on image to enlarge
↓ Click on image to enlarge
The effects on soils noted above affect plants most strongly. However, direct impingment of acidic rain and fog on leaves has other effects which can be especially serious when air pollutants such as SO 2 are present.
- Acid deposition on leaves and needles tends to weaken them by eroding away their protective waxy coatings. This often results in the development of brown spots that interfere with photosynthesis.
- Forests in mountain regions are frequently bathed in clouds and fog which can be even more acidic than normal precipitation, exacerbating the above problem. In addition, the leaves and needles tend to accumulate the fog aerosol into larger droplets. These eventually fall to the forest floor, magnifying the effects on soils described in the previous section.
- Food crops tend to be less affected by acid rain where good farming practices such as the addition of fertilizers to replace depleted nutrients and the addition of limestone to raise the soil pH.
Effects on lakes and aquatic ecosystems
Lakes are subject not only to wet and dry acidic deposition, but also by the water they receive from streams and surface runoff. Any toxic elements released by the action of acid rain on soils and sediments can thus be conveyed to, and concentrated within lakes and the streams that empty them.
↓ Click on image to enlarge
↓ Click on image to enlarge
Lakes in poorly-buffered areas such as are found in alpine regions (western North America, Colorado and much of Switzerland) or on the Canadian Shield and in the Adirondacs and Appalachians) are highly sensitive to acid deposition.
Severely acidified lakes (such as the one depicted at the above right) can be so devoid of life, including algae, that the water appears to be perfectly clear and bright blue in color.
↓ Click on image to enlarge
Aquatic organisms are generally adapted to "ordinary" pH conditions of 6-8, but vary greatly in their tolerance of low pH. As pH declines below 6, the diversity of aquatic animals, plants, and microorganisms diminishes.
- Some plant species such as sphagnum mosses and certain filamentous algaes that thrive at very low pH can proliferate in acidified lakes, producing thick mats that seal off oxygen and thus inhibit the decay of litter on the lake floor.
- Acid deposited onto winter snow is released during spring melting and can cause rapid drops of pH in poorly-buffered lakes that receive waters from these sources. This "spring acid shock" can seriously affect viability of organisms such as fish, amphibians, and insects that lay their eggs in the water which hatch in the spring. The hatchlings are often unable to adapt to the rapid change, and end up deformed or killed.
-
Even those aquatic species able to survive at in low-pH waters can be indirectly affected if their food supply is pH-limited. For example, mayflies and some other insects, which, which are important food sources for frogs, cannot survive below pH 5.5.
↓ Click on image to enlarge
- Fish are severely affected by the aluminum that is released from the action of acid on sediments; it causes a coating of mucus to form on the gills, impeding absorption of oxygen. This has led to the extinction of some species from affected lakes.
- Low pH increases the solubility of calcium salts, making it more difficult for bone to form in developing embryos of fish and amphibians.
- Low pH can also make the eggs of aquatic organism thicker and more difficult for embryos to penetrate, thus delaying hatching. When the embryos continue to grow within the confined space of the egg, their spines can be deformed, interfering with their viability when they finally hatch.
Effects on buildings and statues
Acid rain is not new!
"It has often been observed that the stones and bricks of buildings, especially under projecting parts, crumble more readily in large towns where coal is burnt... I was led to attribute this effect to the slow but constant action of acid rain." Robert Angus Smith, 1856
Acid deposition most strongly affects heritage buildings made of limestone and similar carbonate-containing stone materials.
The principal chemical process involves the reaction of sulfuric acid with calcium carbonate:
\[\ce{CO3(s) + H2SO4 → CaSO4(s) + CO2}.\]
The acid first erodes and breaks up the surface of the stone. As the hydrated calcium sulfate (gypsum) forms, it picks up iron and other components of the stone and forms an unsightly black coating. Some of this gradually blisters off, exposing yet more stone suface. The gypsum crystals can sometimes grow into the stone, further undermining the surface.
Very old structures such as the Taj Mahal, Notre Dame, the Colosseum and Westminster Abbey have all been affected.
↓ Click on image to enlarge
Statues and monuments, including those made from marble, are also susceptible to erosion and damage from acid deposition.
Modern buildings are less affected, although acid rain can erode some painted surfaces.
The Geochemical Carbonate system
The carbonate system, consisting as it does of a soluble gas CO 2 , soluble ions HCO 3 – and CO 3 2– , and sparingly soluble carbonate salts, spans all four realms of nature: the atmosphere, hydrosphere (mainly the oceans), the lithosphere, and the biosphere. And owing to the acid-base equilibria that govern the transformations between these carbonate species, carbon is readily transported between these geochemical reservoirs. This chapter presents an overview of carbon geochemistry.
Distribution of carbon on Earth
Carbon is the fourth most abundant element in the universe. Within the Earth's crust it ranks 15th, mostly in the form of carbonates in limestones and dolomites. Kerogens, which are fossilized plant-derived carbon mostly in the form of oil shale, constitute another major repository of terrestrial carbon.
| Source | relative to atmosphere | |
|---|---|---|
|
carbonate rock
|
carbonate rock
|
28,500
|
|
fossil carbon
|
572
|
10,600
|
|
land - organic carbon
|
0.065
|
1.22
|
|
ocean
|
3.2
|
61.8
|
|
atmospheric CO
2
|
0.0535
|
1.0
|
The geochemical carbon cycle
The carbonate system encompasses virtually all of the environmental compartments– the atmosphere, hydrosphere, biosphere, and major parts of the lithosphere. The complementary processes of photosynthesis and respiration drive a global cycle in which carbon passes slowly between the atmosphere and the lithosphere, and more rapidly between the atmosphere and the oceans. Thus the "carbon cycle" can be divided into "fast" and "slow" parts, operating roughly on annual and geological time scales.
Figure (\PageIndex{2}\): Image by David Bise, Pennsylvania State University; see here for a full discussion of this diagram.
Another excellent depiction of the carbon cycle, by the U.S. Dept of Energy. ( See description ):
Carbon Cycle
Carbon dioxide in the atmosphere
CO 2 has probably always been present in the atmosphere in the relatively small absolute amounts now observed. Precambrian limestones, possibly formed by reactions with rock-forming silicates, e.g.
\[\ce{CaSiO_3 + CO_2 → CaCO_3 + SiO_2} \label{4-1}\]
have likely had a moderating influence on the CO 2 abundance throughout geological time.
The volume-percent of CO 2 in dry air is 0.032%, leading to a partial pressure of 3 × 10 –4 (10 3.5 ) atm. In a crowded and poorly-ventilated room, P CO2 can be as high as 100 × 10 –4 atm. About 54 × 10 14 moles per year of CO 2 is taken from the atmosphere by photosynthesis, divided about equally between land and sea. Of this, all except 0.05% is returned by respiration (mostly by microorganisms); the remainder leaks into the slow, sedimentary part of the geochemical cycle where it can remain for thousands to millions of years. Trends in the growth of atmospheric CO 2 for the past five years can be seen on this US NOAA page .
Since the advent of large-scale industrialization around 1860, the amount of CO 2 in the atmosphere has been increasing. Most of this has been due to fossil-fuel combustion; in 1966, about 3.6 × 10 15 g of C was released to the atmosphere; this is about 12 times greater than the estimated natural removal of carbon into sediments. The large-scale destruction of tropical forests, which has accelerated greatly in recent years, is believed to exacerbate this effect by removing a temporary sink for CO 2 .
About 30-50% of the CO
2
released into the atmosphere by combustion remains there; the remainder enters the hydrosphere and biosphere. The oceans have a large absorptive capacity for CO
2
by virtue of its transformation into bicarbonate and carbonate in a slightly alkaline aqueous medium, and they contain about 60 times as much inorganic carbon as is in the atmosphere. However, effcient transfer takes place only into the topmost
(100 m) wind-mixed layer, which contains only about one atmosphere equivalent of CO
2
; mixing time into the deeper parts of the ocean is of the order of 1000 years. For this reason, only about ten percent of added CO
2
is taken up by the oceans.
Most of the carbon in the oceans is in the form of bicarbonate, as would be expected from the pH which ranges between 7.8 and 8.2. In addition to atmospheric CO 2 , there is a carbonate input to the ocean from streams. This is mostly in the form of HCO 3 – which derives from the weathering of rocks and terrestrial carbonate sediments, and the acid-base reaction
\[H_2CO_3 + CO_3^{2–} → 2 HCO_3^– \label{4-2}\]
which can be considered to be the source of bicarbonate in seawater. In this sense, the ocean can be considered the site of a gigantic acid-base titration in which atmospheric acids (mainly CO 2 but also SO 2 , HCl, and other acids of volcanic origin) react with bases that originate from rocks and are introduced through carbonate-bearing streams or in windblown dust.
Carbon in the hydrosphere
Dissolution of CO 2 in water
Carbon dioxide is slightly soluble in water:
| °C |
0
|
4
|
10
|
20
|
|---|---|---|---|---|
| mol L –1 |
0.077
|
0.066
|
0.054
|
0.039
|
Henry's law is followed up to a CO 2 pressure of about 5 atm:
\[[\ce{CO2 (aq)}] = 0.032 P_{\ce{CO_2}} \label{4-3}\]
“Dissolved carbon dioxide” consists mostly of the hydrated oxide CO2(aq) together with a small proportion of carbonic acid:
\[[\ce{CO2 (aq)}] = 650 [\ce{H2CO3}] \label{4-4}\]
The acid dissociation constant \(K_{a1}\) that is commonly quoted for "H 2 CO 3 " is really a composite equilibrium constant that includes the above equilibrium. This means that "pure" H 2 CO 3 (which cannot be isolated) is a considerably stronger acid than is usually appreciated.
Distribution of carbonate species in aqueous solutions
Water exposed to the atmosphere with P CO2 = 10 3.5 atm will take up carbon dioxide, which becomes distributed between the three carbonate species CO 2 , HCO 3 – and CO 3 2– in proportions that depend on K 1 and K 2 and on the pH. The "total dissolved carbon" is given by the mass balance
\[C_t = [H_2CO_3] + [HCO_3^–] + [CO_3^{2–}] \label{4-5}\]
The distribution of these species as a function of pH can best be seen by constructing a log C -pH diagram for C t = 10 –5 M .
This Sillén plot is drawn for two different concentrations of the carbonate system. The lower one, in heavier lines, is for a 10 –5 M solution, corresponding roughly to atmospheric CO 2 in equilibrium with pure water. The upper plot, for a 10 –3 M solution, is representative of many natural waters such as lakes and streams where the water is in contact with sediments.
It is important to note that this diagram applies only to a system in which C t is constant. In a solution that is open to the atmosphere, this will not be the case at high pH values where the concentration of CO 2 is appreciable. Under these conditions this ion will react with H 2 CO 3 and the solution will absorb CO 2 from the atmosphere, eventually resulting in the formation of a solid carbonate precipitate.
The lower of the two above plots can be used to predict the pH of 10^{5} M solutions of carbon dioxide, sodium bicarbonate, and sodium carbonate in pure water. The pH values are estimated by using the mass and charge balance conditions for each solution as noted below.
The reasoning leading to these calculations is explained in the discussion of the 10 –3 M carbonate system in a previous chapter.
- Solution of CO 2 or H 2 CO 3
- [H + ] = [OH – ] + [HCO 3 – ] + 2 [CO 3 2– ](4-6)
- which, since the solution will be acidic, can be simplified to
- [H + ] = [HCO 3 – ] (point )(4-7)
- Solution of NaHCO 3
- [H + ] + [H 2 CO 3 ] = [CO 3 2– ] + [OH](4-8)
- or
- [H + ] = [HCO 3 – ] (point ) (4-9)
- Solution of Na 2 CO 3
- [H + ] + 2 [H 2 CO 3 ] + [CO 3 2– ] = [OH – ](4-10)
- or
- [HCO 3 – ] = [OH – ] (point ) (4-11)
Carbon in the oceans
Natural waters aquire carbon from sediments they are in contact with, and of course also from the atmosphere. The pH is an important factor here; CO 2 and solid carbonates are more soluble at high pH, and pH also controls the distribution of carbon species, as is seen in the plot just above.
| rain | river/lake water | sea water | |
|---|---|---|---|
| ppm of carbon |
1.2
|
35
|
140
|
| pH (unpolluted) |
5.6
|
6.5 - 8.5
|
7.5 - 8.4
|
At the slightly alkaline pH of most bodies of water (of which the oceans constitute 97% of the earth's surface waters), bicarbonate is the principal dissolved carbon species. The quantity of organic carbon is fairly small.
| CO 2 (aq) | HCO 3 – | CO 3 2– | dead org. | living org. |
|---|---|---|---|---|
|
0.18
|
2.6
|
0.33
|
0.23
|
0.05
|


14.4: Transposons Since McClintock
Transposons exist everywhere we look in prokaryotes and account for much of eukaryotic repetitive DNA. As such, they can be a large proportion of eukaryotic genomes, including some that no longer even transpose. Transposons were once considered useless or junk DNA , with no obvious function…, or selfish genes with no other purpose than selfreplication. But in light of some new evidence, perhaps not!
As you will see, mechanisms of transposition share many features with DNA replication, recombination and repair, and even viral infection. As you study these mechanisms, keep in the back of your mind that transposition is often triggered by cellular stress.
A. Insertion Sequences (IS Elements)
Bacterial IS elements were the first mobile elements described after those in maize. As we’ll see, they share some structural features of eukaryotic transposons. Discovered in the late 1960s, many have been identified (IS1, IS2…, IS10 etc.). Some are inserted into well-known genes (e.g., those of the lac operon), but most are not, likely because there is little ‘extra’ DNA in the compact bacterial genome. Without extra non-coding DNA to buffer against damaging mutations, few bacterial cells would live to tell a tale of transposition! It should surprise us that IS elements can be made to transpose in the lab, but are generally silent in nature.
Members of the IS element family vary in length from about 750 to 1425 bp. Within this stretch of DNA lie transposase and resolvase genes whose products are necessary for mobility. At either end of the IS element are inverted repeats , and when found in either genomic or plasmid DNA, the IS sequence itself is flanked by direct repeats of host genome or plasmid DNA that result from the mechanism of transposition. Again, because of their compact genomes, bacteria can only tolerate low copy numbers of IS elements in their genome or on plasmids (less than ten copies and as few as one!). A typical IS element is illustrated below.
B. Composite Transposons: Tn Elements
If a pair of IS elements should lie close to each other, separated by a short stretch of genomic or plasmid DNA, they can transpose together, carrying the DNA between them as part of a composite transposon , or Tn element . If some of the DNA between IS elements in a Tn element contains antibiotic resistance genes, its transposition can carry and spread these genes to other DNA in the cell. Tn elements (like IS elements) are present in low copy number. A generic Tn element is drawn below.
Antibiotic resistance genes have the medical community worried; their spread has led to antibiotic-resistant pathogens that cause diseases that are increasingly hard and even impossible to treat. Earlier we saw genetic ‘transformation’ of streptococcal cells that pick up virulence genes in DNA from dead cell. We routinely transform cells with plasmids as part of recombinant DNA experiments. But bacteria can transfer plasmid DNA between themselves quite naturally. During bacterial conjugation, an F ( fertility ) plasmid normally transfers DNA between compatible bacterial mating types (review bacterial conjugation elsewhere in this text for more details). An F plasmid containing a Tn element harboring an antibiotic resistance gene can thus is passed from donor to recipient during conjugation. The Tn element can transpose into to the recipient bacterial genome. In this way, transposition is a major pathway for the transfer and spread of antibiotic resistance.
C. Complex Transposons
Bacterial Complex Transposons also contain other genes in addition to those required for mobility. Some complex transposons resemble a bacteriophage, or as in the case of phage Mu , actually are phage! In fact, Mu can function either as an infectious phage that reproduces in an infected cell, or as a transposon in the bacterial genome. Transposon genes in Mu phage are illustrated below.
After infecting a bacterium, Mu can enter the lytic phase of its life cycle, replicating its DNA, producing and ultimately releasing new infectious phage ‘particles’ by lysing the host bacterial cell. Alternatively, like other phage, Mu can undergo lysogeny , inserting its DNA into the host cell chromosome. Integrated copies of Mu might excise and re-enter the lytic phase to produce more phage, particularly if some environmental stress threatens host bacterial survival. But, a third lifestyle choice, transposition, is available to Mu once the phage integrates into the bacterial chromosome. The three lifestyle options for Mu phage are illustrated in the next few pages.
Lytic and lysogenic lifestyle options for Mu phage are shown below.
Mu phage DNA can act as a transposable element while in the lysogenic pathway, as shown below
As we turn to a description of eukaryotic transposons, look for similarities to bacterial IS and Tn elements.
D. Overview of Eukaryotic Transposable Elements
There are two classes transposons in eukaryotes:
Class I (Retrotransposons) move/’jump’ by transcription of RNA at one locus, followed by reverse transcription and integration of the cDNA back into genomic DNA at a different location. Retrotransposons may be derived from (or be the source of) retroviruses since active retroviruses excise from and integrate into DNA much like retrotransposons. Retroposons are a sub-class of retrotransposons (see below).
Class II (DNA Transposons) move by one of two mechanisms. In the cut-&-paste pathway, the transposon leaves one locus and integrates at another. In the replicative pathway, the original transposon remains in place while new copies are mobile. The table below shows the distribution and proportion of genomes represented by different classes/types of transposable elements.
The table confirms that bacteria contain few transposons, while eukaryotes vary widely in transposon load (transposons as a percentage of genomic DNA), from as low as 4% to more than 70%.
The table below summarizes transposable elements by class, sub-type, size, genomic distribution, mechanism of transposition, etc.
Between the two tables above, we can conclude the following:
- Transposon load is not correlated with evolutionary complexity of organisms.
- Shared Transposons have different evolutionary histories in different organisms.
- Where transposons remain active, they continue to shape genomic landscapes, especially in organisms with a high transposons load.
We will revisit some of these conclusions later, after looking at the structure and mechanism of mobility of different transposable elements.
E. The Structure of Eukaryotic DNA (Class II) Transposons
Active eukaryotic DNA transposons share structural features with bacterial mobile elements, including genes required for transposition, flanking inverted repeats and flanking insertion-site direct repeats of host cell DNA. The characteristic structure of a eukaryotic DNA transposon is shown below.
Class II transposons can ‘jump’ by cut-and-paste or replicative mechanisms. Cut-and-paste transposition removes a copy from one location and moves ( transposes ) it to another location. As its name suggests, replicative transposition leaves a copy of the original transposon in place while inserting a new copy elsewhere in the genome. Transposition by the cut-and-paste mechanism is diagrammed below.
Note that after transcription of the transposase gene, the enzyme nicks the DNA and trims the 3’OH ends to create a staggered cut to excise the transposon. The transposase actually brings the transposon ends together during the cut step and mediates its insertion at a new DNA site. After ligation of the 3’OH ends of the transposon to the 5’OH at the insertion site, replication replaces the missing bases, generating the direct repeats of host cell genomic DNA at the insertion site. A final ligation step completes transposition.
In the replicative transposition , the transposon also nicks and trims the DNA at its source (original) insertion site. But, unlike the cut and paste mechanism, the source transposon is not excised.
Details of the replicative mechanism of transposition is summarized below.
After nicking the 3’ ends of the transposon at the insertion site, transposase holds the transposon ends together while catalyzing a hydrolytic attack of DNA at a new insertion site. This is followed by priming of transposon strand replication from the 3’OH ends of the insertion site DNA strands. A cointegrate structure forms in which each transposon copy has been made by semi-conservative replication. The cointegrate is resolved by one of two recombinational mechanisms. The result leaves copies of the transposon at both the original site and the new insertion site.
Let’s compare and contrast the features of cut-&-paste and replicative DNA transposition. The common features are that:
- Transposon-encoded transposase binds, brings transposon ends together and catalyzes single-stranded cleavage (hydrolysis) leaving ‘staggered ends’.
- Transposase holds the transposon ends together for the remaining steps.
The differences between the two mechanisms are that in cut & paste transposition, the transposon is completely excised and then transposed to a new site in genomic DNA. In contrast, after single stranded cleavage in replicative transposition, transposase-bound free 3’ ends of the transposon hydrolyze both strands of stranded DNA at a new insertion site. After ligation of the 3’ ends of transposon strands to 5’ ends of cut genomic DNA insertion-site ends, the remaining 3’ ends of the insertion site DNA ends prime replication of the transposon, forming the cointegrate , which is followed by its resolution by one of two recombination pathways.
F. The Structures of Eukaryotic RNA (Class I) Transposons
Like DNA transposons, all RNA transposons leave insert-site footprints, i.e., direct repeats of genomic DNA flanking the element. Unlike DNA transposons, active eukaryotic Class I transposons move via an RNA intermediate. Also unlike DNA transposons, they lack terminal inverted repeats.
The mobility of the RNA intermediate of all retrotransposons requires a promoter that recognizes a reverse transcriptase enzyme as well as endonuclease and integrase enzymes (to be described below). Autonomous Class I RNA transposons include LTR retrotransposons (e.g., the yeast Ty element) as well as Non-LTR retrotransposons ). The latter include the autonomous LINEs ( L ong I nterspersed N uclear E lements). The autonomous LTR and Non-LTR LINEs contain and express genes needed for enzymes required for transposition. On the other hand, SINEs (a sub-class of Non-LTR retrotransposons) lack genes for enzymes required for transposition and therefore can’t transpose independently. Thus, they are non-autonomous retrotransposons that rely on “true” (autonomous) retrotransposon activity for mobility. SINEs are sometimes called retroposons to distinguish them from the autonomous retrotransposons.
249 Introduction to Features of Retrotransposition
Next we take a closer look at Retrotransposon structures and the genes and enzyme activities required for retrotransposition
1. LTR retrotransposons: The Yeast Ty Element
The Ty transposon harbors several genes needed for transposition. These include:
- the Gag gene that encodes group-specific antigen , a protein that forms a viruslike particle that will contain reverse-transcribed transposon DNA,
- the RT gene that encodes the reverse transcriptase that will make reversetranscribed copies of retrotransposon transcript RNAs.
- the Prt gene that encodes a protease that will break down the virus-like particle as the retrotransposon enters the nucleus.
- the Int gene that encodes the integrase required for integration of the retrotransposon into a genomic DNA insertion site.
A representative Ty element is shown below as it would exist integrated into yeast genomic DNA.
In fact, many of the events in Ty transposition occur in the cytoplasmic “virus-like particle” in yeast cells. To see more, click here . Note that the Pol region in the illustration above consists of overlapping open reading frames (ORFs) encoding the Prt, RT and Int genes. The ready-to-move transposon consists only of the region of DNA symbolized in yellow.
2. Non-LTR Retrotransposons: LINEs
LINEs ( L ong I nterspersed N uclear E lements) also encode enzymes needed for transposition and like other transposons, generate target-site direct repeats flanking the inserted element. But they do not have the long terminal repeats! Instead, their ORFs (genes) are flanked by 5’ and a 3’ untranslated regions ( UTRs ).
The structure of the human L1 Line is drawn below.
The 5’ UTR contains a promoter from which cellular RNA polymerase II can transcribe the downstream genes (see the Transcription chapter). The second of these (ORF2) encodes the reverse transcriptase and an integrase activity essential for transposition of the LINE. All Class I (RNA-intermediate) autonomous transposons share the following features:
a) a Promoter in the 5’ UTR from which they can be transcribed.
b) a Reverse Transcriptase that generates a cDNA copy of the transposable element.
c) RNAse H (an endonuclease) that degrades that transcript after reverse transcription.
d) Integrase (like a transposase) that catalyzes insertion of the retrotransposon copy at insertion sites.
3. Non-LTR SINE retrotransposons
Non-LTR SINE retrotransposons typically lack genes, but their non-genic DNA is nonetheless flanked by 5’ and 3’ UTRs. RNA polymerase III, which also transcribes transfer RNAs, also transcribes SINEs. However, to transpose, they rely on the concurrent activity of a Non-LTR transposon (a LINE) to provide the requisite enzymatic activities.
A typical SINE (e.g., the Alu element) is shown below.
G. Mechanisms of Retrotransposition
There are two mechanisms of retrotransposition: Extrachromosomally Primed Retrotransposition (LTR retrotransposons for example) and Insertion Target-Site Primed Retrotransposition (non-LTR Retrotransposons like LINEs and SINEs). These will be considered next.
1. Extrachromosomally Primed Retrotransposition (e.g., of a LINE)
As its name suggests, in extrachromosomally primed retrotransposition , a circular reverse transcript of the retrotransposon attacks, nicks and integrates into a genomic insertion site. In this mechanism, reverse transcriptase creates a cDNA copy of a transcribed retro-element. Integrase/endonuclease then binds the cDNA copy, holding the ends together, in effect circularizing it. This isolable ribonucleoprotein resembles an intasome , a structure similar to the nucleoprotein complex that catalyzes the integration of retroviral cDNAs during lysogeny.
Extrachromosomally primed retrotransposition is illustrated below.
The three-dimensional structure of a retroviral intasome interacting with DNA and nucleosomes was recently determined (for more, see Retroviral Intasome 3D Structure ). In this form, the retrotransposon attacks DNA at an insertion site, creating staggered ends. After insertion, the gaps in the DNA are filled in. Ligation seals the retrotransposon in its new location, creating direct insertion site repeats.
2. Target-Site Primed SINE Retrotransposition (e.g., of a SINE)
A key feature of target-site primed retrotransposition ( retroposition ) is the absence of an integrase-bound, circular double-stranded reverse transcript. In SINE transposition, RNA polymerase III (the same enzyme that catalyzes tRNA and 5S rRNA transcription) transcribes the SINE. If a LINE is concurrently transcribed, its enzymes will be made. When its integrase-endonuclease catalyzes hydrolysis of one strand of DNA at a new insertion site, the 3’OH end of this strand can prime reverse transcription of the one SINE cDNA strand by the LINE reverse transcriptase . After hydrolysis of the second target site DNA strand, its 3’-OH end primes replication of the second strand of the SINE cDNA. Integrase completes insertion of the copy-SINE in its new genomic location. The target-site primed retrotransposition mechanism of retrotransposition is illustrated below.


 In a university environment, faculty frequently add to the mix, though some argue that they fit into one of the first three categories. When these people invest money in the early stages of your company, it’s called a seed round of investment. Seed money helps pay for the business plan and the prototype, and supports you while you find additional management talent and secure the next round of financing.
Angel investors
In the wake of friends and family come angel investors. As we mentioned previously, angel investors are affluent individuals who provide capital for a start-up, usually in exchange for ownership equity. The friends and family round of financing comes from your own personal relationships with people. But angel investors are typically not people you know. They’re people who are judging you based on the business concept, the team and the opportunity for their capital to propel the company to the next level.
Building your credibility
To secure angel financing, you need to make a credible case for your business. You may not have tremendous successes under your belt, but if your business concept is worth anything, and if your entrepreneurial skills are up to the task, you’ll find a way to assemble a team of advisors who have more experience and credibility than you do, and will let you borrow a piece of their reputation as currency when you present your business to the angels.
Typically, a group of angel investors contribute less than $1 million. Often that amount is significantly less with an unproven team and an early stage concept. Your task is to make that money work. You need to hit such significant milestones with that capital that you can use your accomplishments to secure the next round of money.
Venture capital
How do I apply? People often assume that venture capital funding is akin to a bank loan process. Recently, online firms have made moves to democratize the venture capital process, but a fair number of these have failed. In reality, securing venture capital remains a complicated process without a specific formula for success.Venture capital access is a networking game. It is not democratic, and because of several factors, it is inherently elitist. If you live in a major metropolitan area, you have better access to VC resources (though www.villageventures.com) is aiming to change that. Unfortunately, if you look like the VCs you’re targeting (who are frequently white, male, Ivy Leaguers), that may also up your chances of success.
That said, VCs are also hungry. Once you’re in the door, your presentation will stand or fall on its own merits. Alternative VC sources and investor networks are a good resource if you don’t fit the above profile. A source for venture capital for women is www.springboard2000.org. Entrepreneurs from the Indian-American community should investigate www.indianceo.com. If you’re affiliated with a college or university, visit www.universityangels.com.
Here’s a general framework for thinking about your business, the type of capital that is appropriate, and some of the key criteria that may help secure the jet fuel of a high growth business. (Of course you have to do something productive with the money and provide a return to your investor.) Here are some basic questions that an investor asks to assess a business.
Is the management team proven? Does its experience reflect its competence? If it’s not proven, what are its attributes? How can the missing elements be filled with additional seasoned managers? How committed is this team to the business?
How big is the market for this product or service? Typically, an equity investor wants to see a market for a technology product that is, in total, greater than $1 billion. A good market is one in which customers feel real pain, where they will readily adopt your product because of an immediate and urgent need.
What does the competition look like? Is this company the first to market? Is it defining the market? Will it be able to quickly gain market share?
Is the product/technology difficult to replicate? Does it have some proprietary position (meaning patents or other intellectual property protection)? Will it take the competition a long time to replicate what the company does?
Does the business plan present a credible story that suggests that the company can forecast results? The financial plan is a financial expression of your business strategy. It shows the interrelation of timelines, functions, and hires, and reflects a detailed understanding of the business.
Capital intensity. Is the business appropriately scaled to the money that might be available? Your business can’t require too much money to be successful.
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Playing the money game
The father of venture capital: J.H. Whitney
Advice for before you go to the VC
From "What to Know Before You Go to the VCs" by Joseph Bartlett, Fast Company, November, 2000.
The oyster and the pearl
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7.13: La donna e mobile
As you read this page on this aria from Verdi’s Rigoletto, pay particular attention to the information on the theme and the form of the piece.
Introduction
“ La donna è mobile ” ( The woman is fickle ) is the Duke of Mantua’s canzone from the beginning of act 3 of Giuseppe Verdi’s opera Rigoletto (1851). The inherent irony is that the Duke, a callous playboy, is the one who is mobile (“inconstant”). Its reprise towards the end of the opera is chilling, as Rigoletto realizes from the sound of the Duke’s lively voice coming from within the tavern (offstage), that the body in the sack over which he has grimly triumphed is not that of the Duke after all: Rigoletto had paid Sparafucile, an assassin, to kill the Duke but Sparafucile deceived him by killing Gilda, Rigoletto’s beloved daughter, instead.
The canzone is famous as a showcase for tenors. Raffaele Mirate’s performance of the bravura aria at the opera’s 1851 premiere was hailed as the highlight of the evening. Before its first public performance (in Venice), it was rehearsed under tight secrecy: a necessary precaution, because it proved to be catchy and soon after its first public performance every gondolier in Venice was singing it.
The Music
The almost comical-sounding theme of “ La donna è mobile ” is introduced immediately, and runs as illustrated (transposed from the original key of B major). The theme is repeated several times in the approximately two to three minutes it takes to perform the aria, but with the important—and obvious—omission of the last bar. This has the effect of driving the music forward as it creates the impression of being incomplete and unresolved, which it is, ending not on the tonic or dominant but on the submediant. Once the Duke has finished singing, however, the theme is once again repeated; but this time it includes the last, and conclusive, bar and finally resolving to the tonic. The song is strophic in form with an orchestral ritornello.
Libretto
| Italian | Prosaic translation | Poetic translation |
|---|---|---|
|
1.
La donna è mobile
Sempre un amabile,
Refrain
2. È sempre misero
Pur mai non sentesi
Refrain
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Woman is flighty.
Always a lovely,
Refrain
Always miserable
Yet one never feels
Refrain
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Plume in the summerwind
Thus heart of womankind
Refrain
Sorrow and misery
Yet all felicity
Refrain
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Contributors and Attributions
- Authored by : Elliott Jones. Provided by : Santa Ana College. Located at : http://www.sac.edu . License : CC BY: Attribution


3.S: Basic Concepts of Chemical Bonding (Summary)
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8.1: CHEMICAL BONDS , LEWIS SYMBOLS AND THE OCTET RULE
Lewis dot symbols can be used to predict the number of bonds formed by most elements in their compounds. Lewis electron dot symbols, which consist of the chemical symbol for an element surrounded by dots that represent its valence electrons, grouped into pairs often placed above, below, and to the left and right of the symbol. The structures reflect the fact that the elements in period 2 and beyond tend to gain, lose, or share electrons to reach a total of 8 valence electrons in their compounds.
- Ionic bond : bond formed on the basis of electrostatic forces that exist between oppositely charged ions. The ions are formed from atoms by transfer of one or more electrons
- Covalent bond : bond formed between two or more atoms by a sharing of electrons
- Metallic bond : bonding in which the bonding electrons are relatively free to move throughout the 3D structure
- Electron dot symbols : aka Lewis symbols ; simple and convenient way of showing the valence electrons of atoms and keeping tack of them in the course of bond formation
- The number of valence electrons of any representative element is the same as the column number of the element in the periodic table
- Octet rule : atoms tend to lose or gain electrons until they are surrounded by 8 valence electrons
8.2: IONIC BONDING
The amount of energy needed to separate a gaseous ion pair is its bond energy. Forming ionic compounds is usually extremely exothermic. The strength of the electrostatic attraction between ions with opposite charges is directly proportional to the magnitude of the charges on the ions and inversely proportional to the internuclear distance.
Energies of Ionic Bond Formation
The formation of ionic compounds is very exothermic
Removing an electron from an atom, such as Na, is endothermic because energy needs to be used to overcome the attractive forces within the atom. Adding an electron is the opposite process and releases lots of energy
The principal reason that ionic compounds are stable is the attraction between ions of unlike charge. This attraction draws the ions together, releasing energy and causing the ions to form a solid array (lattice)
Lattice energy : energy required to separate completely a mole of a solid ionic compounds into its gaseous ions
Large values of lattice energy mean that the ions are strongly attracted to one another
Energy released by the attraction between the ions of unlike charges more than makes up for the endothermic nature of ionization energies, making the formation of ionic compounds an exothermic process
\[E = k \dfrac{Q_1Q_2}{ d} \nonumber \]
- E = potential energy of two interacting charged particles
- Q 1 and Q 2 = charges on the particles
- D= distance between the particles
- K = constant; 8.99 X 10 9 Jm/C 2
For a given arrangement of ions, the lattice energy increases as the charges of ions increase and as their radii decrease. The magnitude of lattice energies depends primarily on the ionic charges because ionic radii do not vary over a wide range.
Electron Configurations of Ions
Many ions tend to have noble gas electron configurations. This is why Na can have a +1 charge, but not a +2 one. Once an ion has reached noble gas configuration, it wants to stay there.
- Na: 1s 2 2s 2 2p 6 3s 1 = [Ne] 3s 1
- Na + : 1s 2 2s 2 2p 6 = [Ne]
- Na 2 + : 1s 2 2s 2 2p 5
Similarly, addition of electrons to nonmetals is either exothermic or slightly endothermic as long as electrons are being added to the valence shell. Further addition of electrons requires tremendous amounts of energy; more than is available form the lattice energy
- Cl: 1s 2 2s 2 2p 6 3s 2 3p 5 = [Ne] 3s 2 3p 5
- Cl: 1s 2 2s 2 2p 6 3s 2 3p 6 = [Ar]
- Cl 2 : 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 = [Ar]
The lattice energies of ionic compounds are generally large enough to compensate for the loss of up to only 3 electrons from atoms. Thus we find cations only having charges of +1, +2, or +3.
Because most transition metals have more than 3 electrons beyond a noble gas core, attainment of a noble gas configuration for these ions is usually impossible.
When a positive ion is formed from an atom, electrons are always lost first from the subshell with the largest value of n . Thus, a transition metal always loses the outer s electrons before it loses electrons from the underlying d subshell.
8.3 Sizes of Ions
Sizes of ions are important in determining both the way in which the ions pack in a solid and the lattice energy of the solid. It is also a major factor governing the properties of ions in solution
The size of an atom depends on its nuclear charge, the number of electrons it possesses, and the orbitals in which the outer-shell electrons reside
Positive ions are formed by removing 1 or more electrons from the outermost region of the atom. Thus, the formation of a cation not only vacates the most spatially extended orbitals, it also decreases the total electron-electron repulsions. Hence, cations are smaller than the original atoms from which they came.
The opposite happens when speaking of negative ions. An added electron increases electron-electron repulsions and causes the electrons to spread out more in space.
For ions of the same charge, size increases as we go down a group
8.3: COVALENT BONDING
The strength of a covalent bond depends on the overlap between the valence orbitals of the bonded atoms. Bond order is the number of electron pairs that hold two atoms together. Single bonds have a bond order of one, and multiple bonds with bond orders of two (a double bond) and three (a triple bond) are quite common. In closely related compounds with bonds between the same kinds of atoms, the bond with the highest bond order is both the shortest and the strongest.
Ionic substances are usually brittle with high melting points. They are usually crystalline, meaning that they have flat surfaces that make characteristic angles with one another.
- Covalent bond : chemical bond formed by sharing a pair of electrons
- Lewis structure : structure that represents bonding using dots for unpaired electrons and lines for bonds
For nonmetals, the number of valence electrons is the same as the group number
Knowing this, we can predict that an element in Group 7A would need one covalent bond in order to get an octet, an element in Group 6A would need two, and so on.
Multiple Bonds
- Single bond : sharing of one pair of electrons, one covalent bond
- Double bond : two shared electrons
Distance between bonded atoms decreases as the number of shared electron pairs increases
8.4: BOND POLARITY AND ELECRONEGATIVITY
Bond polarity and ionic character increase with an increasing difference in electronegativity. The electronegativity (χ) of an element is the relative ability of an atom to attract electrons to itself in a chemical compound and increases diagonally from the lower left of the periodic table to the upper right. The Pauling electronegativity scale is based on measurements of the strengths of covalent bonds between different atoms, whereas the Mulliken electronegativity of an element is the average
- Bond polarity : measure of how equally the electrons are shared between the atoms in a chemical bond
- Nonpolar bond : one in which the electrons are shared equally between the two atoms
- Polar covalent bond : one of the atoms exerts a greater attraction for the electron than the other
Electronegativity
Used to estimate whether a bond will be nonpolar, polar covalent, or ionic
Electronegativity : ability of an atom in a molecule to attract electrons to itself
An atom with a very negative electron affinity and high ionization energy will both attract electrons from other atoms and resist having its electrons attracted away; it will be highly electronegative
Highest electronegativity = 4.0 (Fluorine), lowest = 0.7 (Cesium)
Electronegativity increases form left to right, and usually decreases with increasing atomic number in any one group
Electronegativity and Bond Polarity
Differences in electronegativities:
Nonpolar = 0 – 0.4
Polar covalent = 0.4 – 1.6
Ionic = > 1.6 (> 50% = ionic)
δ + and δ : "delta sign"; represent partial positive and negative charge. The atom with the δ is the more electronegative one
8.5: DRAWING LEWIS STRUCTURES
Lewis dot symbols provide a simple rationalization of why elements form compounds with the observed stoichiometries. A plot of the overall energy of a covalent bond as a function of internuclear distance is identical to a plot of an ionic pair because both result from attractive and repulsive forces between charged entities. Lewis structures are an attempt to rationalize why certain stoichiometries are commonly observed for the elements of particular families.
- Sum the valence electrons from all atoms. For an anion, add an electron to the total negative charge. For a cation, subtract an electron.
- Write the symbols for the atoms to show which atoms are attached to which, and connect them with a single bond.
- Complete the octets of the atoms bonded to the central atom.
- Place any leftover electrons on the central atom, even if doing so results in more than an octet
- If there are not enough electrons to give the central atom an octet, try multiple bonds
8.6: RESONANCE STRUCTURES
Some molecules have two or more chemically equivalent Lewis electron structures, called resonance structures. Resonance is a mental exercise and method within the Valence Bond Theory of bonding that describes the delocalization of electrons within molecules. These structures are written with a double-headed arrow between them, indicating that none of the Lewis structures accurately describes the bonding but that the actual structure is an average of the individual resonance structures.
Resonance structures ( resonance forms) are individual Lewis structures in cases where two or more Lewis structures are equally good descriptions of a single molecule. If a molecule (or ion) has two or more resonance structures, the molecule is a blend of these structures. The molecule does not oscillate rapidly between two or more different forms.
8.7: EXCEPTIONS TO THE OCTET RULE
Following the Octet Rule for Lewis Dot Structures leads to the most accurate depictions of stable molecular and atomic structures and because of this we always want to use the octet rule when drawing Lewis Dot Structures. There are three exceptions: (1) When there are an odd number of valence electrons, (2) When there are too few valence electrons, and (3) when there are too many valence electrons
- Molecules with an odd number of electrons
- Molecules in which an atom has less than an octet
- Molecules in which an atom has more than an octet
Odd Number of Electrons
In a few molecules, such as ClO 2 , NO , and NO 2 , the number of electrons is odd. In NO for example, there are 5+6 = 11 valence electron. Hence, complete pairing of these electrons is impossible and an octet around each atom cannot be achieved.
Less Than an Octet
Second type of exception occurs when there are fewer than eight electrons around an atom in a molecule or ion. Relatively rare situation; most often encountered in compounds of Boron and Beryllium. For example, let’s consider Boron Trifluoride, BF 3
There are 6 electrons around the Boron atom. We can form a double bond between Boron and any of the 3 Fluorine atoms (3 possible resonance structures)
However, by doing so, we forced a Fluorine atom to share additional electrons with Boron. This would make the F atom to have a +1 charge, and the Boron atom to have a –1 charge, which is extremely unfavorable.
We then conclude that the structures containing the double bonds are less important than the one illustrated on the right. Since in this case Boron has only 6 valence electrons, it will react violently with molecules that have an unshared pair of electrons.
More Than an Octet
The octet rule works as well as it does because the representative elements usually employ only an ns and three np valence shell orbitals in bonding, and these hold eight electrons.
Because elements of the second period have only 2s and 2p orbitals, they cannot have more than an octet of electrons in their valence shells. However, from the third period on, the elements have unfilled nd orbitals that can be used in bonding.
Size also plays an important role in determining whether an atom can accommodate more than eight electrons. The larger the central atom, the larger the number of atoms that can surround it. The size of the surrounding atoms is also important. Expanded valence shells occur most often when the central atom is bonded to the smallest and most electronegative atoms.
8.8: STRENGTHS OF COVALENT BONDS
Bond order is the number of electron pairs that hold two atoms together. Single bonds have a bond order of one, and multiple bonds with bond orders of two (a double bond) and three (a triple bond) are quite common. The bond with the highest bond order is both the shortest and the strongest. In bonds with the same bond order between different atoms, trends are observed that, with few exceptions, result in the strongest single bonds being formed between the smallest atoms.
Bonddissociation energy : aka bond energy ; enthalpy change, ΔH, required to break a particular bond in a mole of gaseous substance.
For polyatomic molecules, we must often utilize average bond energies.
Bond energy is always positive, the greater the bond energy, the stronger the bond
A molecule with strong bonds generally has less tendency to undergo chemical change than does one with weak bonds
Bond Energies and the Enthalpy of Reactions
ΔH = Σ (bond energies of bonds broken) – Σ (bond energies of bonds formed)
If ΔH > 0, the reaction is endothermic
If ΔH < 0, the reaction is exothermic
Cl – Cl (g) + H – CH 3 (g) → H – Cl (g) + Cl – CH 3 (g)
Bonds broken: 1 mol Cl – Cl, 1 mol C – H
Bonds made: 1 mol H – Cl, 1 mol C – Cl
\[∆H = [D (Cl – Cl) + D(C – H)] [D (H – Cl) + D (Cl – Cl)] \nonumber \]
= (242 kJ + 413 kJ) – (431kJ + 328 kJ)
= 104 kJ
Bond Strength and Bond Length
As the number of bonds between a given element increase, the bond energy increases and the bond length decreases. Hence, the atoms are held more tightly and closely together. In general, as the number of bonds between two atoms increases, the bond grows shorter and stronger.
8.10 Oxidation Numbers
Oxidation Numbers : aka Oxidation states ; a positive or negative whole number assigned to an element in a molecule or ion on the basis of a set of normal rules; to some degree it reflects the positive or negative character of an atom
Oxidation numbers do NOT correspond to real charges on the atoms, EXCEPT in the special case of simple ionic substances
- The oxidation form of an element in its elemental form is zero .
- The oxidation number of a monoatomic ion is the same as its charge. For example, the oxidation number of sodium in Na + is +1, and that of sulfur in S 2 is –2
- In binary compounds (those with two different elements), the element with greater electronegativity is assigned a negative oxidation number equal to its charge in simple ionic compounds the element. For example, consider the oxidation state of Cl in PCl 3 . Cl is more electronegative than P. In its simple ionic compounds, Cl appears as the ion Cl. Thus, in PCl 3 , Cl is assigned an oxidation number of –1.
- The sum of the oxidation numbers equals zero for an electrically neutral compound and equals the overall charge of the ionic species. For example, PCl 3 is a neutral molecule. Thus, the sum of oxidation number of the P and Cl atoms must equal zero Because the oxidation number of each Cl in this compound is –1 (rule 3), the oxidation number of P must be +3.
Group 1A elements are +1, Group 2A elements are +2, and Aluminum is +3.
The most electronegative element, F, is always found in the –1 oxidation state. Oxygen is usually in the –2 state, however, it can be –1 in peroxides.
Hydrogen has an oxidation number of +1 when it is bonded to a more electronegative element (most nonmetals), and of –1 when bonded to less electronegative elements (most metals)
Oxidation Numbers and Nomenclature
Name of the less electronegative element is given first, followed by the name of the more electronegative element modified to have an –ide ending
Compounds of metals in higher oxidation states tend to be molecular rather than ionic


1.1.2: Union, Intersection, and Complement
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Commonly sets interact.  At this party, two sets are being combined, though it might turn out that there are some friends that were in both sets.
Union, Interection, and Complement
The union of two sets contains all the elements contained in either set (or both sets).
The union is notated \(A \cup B\)
More formally, \(x \in A \cup B\) if \(x \in A\) or \(x \in B\) (or both)
The intersection of two sets contains only the elements that are in both sets.
The intersection is notated \(A \cap B\)
More formally, \(x \in A \cap B\) if \(x \in A\) and \(x \in B\)
The complement of a set A contains everything that is not in the set A .
The complement is notated \(A\), or \(A^{c}\), or sometimes \(\sim A\).
Example 5
Consider the sets:
\(\quad A=\{\text { red, green, blue }\} \quad B=\{\text { red, yellow, orange }\} \quad C=\{\text { red, orange, yellow, green, blue, purple }\}\)
- Find \(A \cup B\)
- Find \(A \cap B\)
- Find \(A^{c} \cap C\)
Solution
a) The union contains all the elements in either set: \(A \cup B=\{\text { red, green, blue, yellow, orange }\}\)
Notice we only list red once.
b) The intersection contains all the elements in both sets: \(A \cap B=\{\text { red }\}\)
c) Here we're looking for all the elements that are not in set \(A\) and are also in \(C\).
\(A^{c} \cap C=\{\text { orange, yellow, purple }\}\)
Try it Now 2
Using the sets from the previous example, find \(A \cup C\) and \(B^{c} \cap A\)
- Answer
-
\(A \cup C=\{\text { red, orange, yellow, green, blue purple }\}\)
\(B^{c} \cap A=\{\text { green, blue }\}\)
Notice that in the example above, it would be hard to just ask for \(A^{c},\) since everything from the color fuchsia to puppies and peanut butter are included in the complement of the set. For this reason, complements are usually only used with intersections, or when we have a universal set in place.
Universal Set
A universal set is a set that contains all the elements we are interested in. This would have to be defined by the context.
A complement is relative to the universal set, so \(A^{C}\) contains all the elements in the universal set that are not in \(A\).
Example 6
- If we were discussing searching for books, the universal set might be all the books in the library.
- If we were grouping your Facebook friends, the universal set would be all your Facebook friends.
- If you were working with sets of numbers, the universal set might be all whole numbers, all integers, or all real numbers
Example 7
Suppose the universal set is \(U=\) all whole numbers from \(1\) to \(9 .\) If \(A=\{1,2,4\}\), then
\(A^{c}=\{3,5,6,7,8,9\}\)
As we saw earlier with the expression \(A^{c} \cap C,\) set operations can be grouped together. Grouping symbols can be used like they are with arithmetic - to force an order of operations.
Example 8
Suppose
\(H=\{\text { cat, dog, rabbit, mouse }\}, F=\{\text { dog, cow, duck, pig, rabbit }\} \quad W=\{\text { duck, rabbit, deer, frog, mouse }\}\)
- Find \((H \cap P) \cup W\)
- Find \(H \cap(F \cup W)\)
- Find \((H \cap P) \cap W\)
Solution
a) We start with the intersection: \(H \cap F=\{\text { dog, rabbit }\}\)
Now we union that result with \(W:(H \cap F) \cup W=\{\text{dog, duck, rabbit, deer, frog, mouse }\}\)
b) We start with the union: \(F \cup W=\{\text{dog, cow, rabbit, duck, pig, deer, frog, mouse }\}\)
Now we intersect that result with \(H: H \cap(F \cup W)=\{\text { dog, rabbit, mouse }\}\)
c) We start with the intersection: \(H \cap F=\{\mathrm{dog}, \text { rabbit }\}\)
Now we want to find the elements of \(W\) that are not in \(\mathrm{H} \cap F\)
\((H \cap P)^{c} \cap W=\{\text { duck, deer, frog, mouse }\}\)