▶ 실전 연습 해답 ▶실전 연습 해답 제2장 제3장 제4장 사이_줌달의대학기초화학_해답.indd 1 2015. 2. 16. 오후 2:02 1 1 2 2 줌달의 대학기초화학 제5장 제6장 사이_줌달의대학기초화학_해답.indd 2 2015. 2. 16. 오후 2:02 줌달의 대학기초화학 실전 연습 해답 3 3 제7장 사이_줌달의대학기초화학_해답.indd 3 2015. 2. 16. 오후 2:02 실전 연습 해답 4 4 줌달의 대학기초화학 제8장 사이_줌달의대학기초화학_해답.indd 4 2015. 2. 16. 오후 2:02 줌달의 대학기초화학 실전 연습 해답 5 5 제9장 사이_줌달의대학기초화학_해답.indd 5 2015. 2. 16. 오후 2:02 실전 연습 해답 6 6 줌달의 대학기초화학 사이_줌달의대학기초화학_해답.indd 6 2015. 2. 16. 오후 2:02 줌달의 대학기초화학 실전 연습 해답 7 7 사이_줌달의대학기초화학_해답.indd 7 2015. 2. 16. 오후 2:02 실전 연습 해답 8 8 줌달의 대학기초화학 제10장 제11장 사이_줌달의대학기초화학_해답.indd 8 2015. 2. 16. 오후 2:02 줌달의 대학기초화학 실전 연습 해답 9 9 사이_줌달의대학기초화학_해답.indd 9 2015. 2. 16. 오후 2:02 실전 연습 해답 10 10 줌달의 대학기초화학 제12장 사이_줌달의대학기초화학_해답.indd 10 2015. 2. 16. 오후 2:02 줌달의 대학기초화학 실전 연습 해답 11 11 사이_줌달의대학기초화학_해답.indd 11 2015. 2. 16. 오후 2:02 실전 연습 해답 12 12 줌달의 대학기초화학 제13장 제14장 사이_줌달의대학기초화학_해답.indd 12 2015. 2. 16. 오후 2:02 줌달의 대학기초화학 실전 연습 해답 13 13 사이_줌달의대학기초화학_해답.indd 13 2015. 2. 16. 오후 2:02 실전 연습 해답 14 14 줌달의 대학기초화학 제15장 제16장 사이_줌달의대학기초화학_해답.indd 14 2015. 2. 16. 오후 2:02 줌달의 대학기초화학 제17장 실전 연습 해답 15 15 사이_줌달의대학기초화학_해답.indd 15 2015. 2. 16. 오후 2:02 실전 연습 해답 16 16 줌달의 대학기초화학 제18장 제19장 사이_줌달의대학기초화학_해답.indd 16 2015. 2. 16. 오후 2:02 줌달의 대학기초화학 ▶ 질문과 문제의 선택된 번호 정답 제1장 제3장 4. Answers will depend on the student’s choices. 5. Recognize the problem and state it clearly; propose possible solutions or explanations; decide which solution/explanation is best through experments. 7. Answers will depend on student responses. A quantitative ob- servation must include a number, such as “There are three win- dows in this room.” A qualitative observation could include something like “The chair is blue.” 8. A natural law is a summary of observed, measurable behavior that occurs repeatedly and consistently. A theory is an at- tempt to explain such behavior. 9. Chemistry is not just a set of facts that have to be memorized. To be successful in chemistry, you have to be able to apply what you have learned to new situations, new phenomena, new experiments. Rather than just learning a list of facts or studying someone else’s solution to a problem, your instruc- tor hopes you will learn how to solve problems yourself, so that you will be able to apply what you have learned in future circumstances. 제2장 4. time 5. (a) 1022; 7. the woman 9. (a) centimeter; 10. uncertainty 13. (a) one; ber); (d) two; 14. (a) 9.96 3 1021; (d) 4.00 3 1029; 15. (a) 8.8 3 1024; (d) 1.00 3 103 16. (a) 5.16; (d) 2423 17. (a) one; 18. (a) 2.045; 20. 1000 mL (d) 3.8418 3 1027 1 L 1000 mL 1 L ; (b) 106; (c) 109; (d) 1021; (e) 1023; (f) 1026 (b) meter; (c) kilometer (b) infinite (definition); (c) infinite (fixed num- (e) two (b) 4.40 3 103; (e) 8.42 3 1022 (b) 9.375 3 104; (c) 8.22 3 1021; (c) 8.97 3 1021; (b) 2423 (2.423 3 103); (c) 0.516 (5.16 3 1021); (b) four; (c) two; (b) 3.8 3 103; (d) three (c) 5.19 3 1025; 21. (a) 2.44 yd; (b) 42.2 m; (c) 115 in.; (d) 2238 cm; (e) 648.1 mi; (h) 5.01 3 104 cm . 23. 1 3 1028 cm; 4 3 1029 in.; 0.1 nm (g) 0.0362 km; (f ) 716.9 km; 24. freezing/melting 28. (a) 63 K; 29. (a) 173 °F; 31. g/cm3 (g/mL) 34. (a) 22 g/cm3; (d) 2.1 3 1025 g/cm3 26. 273 27. Fahrenheit (F) (b) 2 °C; (c) 505 °C; (d) 1051 K (b) 104 °F; (c) 2459 °F; (d) 90. °F 33. copper (b) 0.034 g/cm3; (c) 0.962 g/cm3; 35. 0.91 L (two significant figures) 36. 11.7 mL 37. (a) 966 g; (b) 394 g; (c) 567 g; (d) 135 g 2. forces 4. liquids 6. gaseous 9. Because gases are mostly empty space, they can be compressed easily to smaller volumes. In solids and liquids, most of the sample’s bulk volume is filled with the molecules, leaving lit- tle empty space. 11. chemical 13. malleable; ductile 14. (a) physical; (e) physical; (i) physical; (b) chemical; (f ) physical; (j) physical; (c) chemical; (g) chemical; (k) chemical (d) chemical; (h) physical; 16. Compounds consist of two or more elements combined to- gether chemically in a fixed composition, no matter what their source may be. For example, water on earth consists of molecules containing one oxygen atom and two hydrogen atoms. Water on Mars (or any other planet) has the same composition. 19. Assuming the magnesium and sulfur had been measured out in exactly the correct ratio for complete reaction, what would remain after heating would be a pure compound. If there were an excess of either magnesium or sulfur, however, the material left after reaction would be a mixture of the compound and the excess reagent. 21. Heterogeneous mixtures: salad dressing, jelly beans, the change in my pocket; solutions: window cleaner, shampoo, rubbing alcohol 22. (a) primarily a pure compound, but fillers and anticaking (c) mixture; (b) mixture; agents may have been added; (d) pure substance 23. (a) homogeneous; (d) heterogeneous; (c) heterogeneous; (b) heterogeneous; (e) homogeneous 25. Consider a mixture of salt (sodium chloride) and sand. Salt is soluble in water; sand is not. The mixture is added to water and stirred to dissolve the salt, and is then filtered. The salt so- lution passes through the filter; the sand remains on the fil- ter. The water can then be evaporated from the salt. 제4장 2. Robert Boyle 4. 115 elements are known; 88 occur naturally; 27 are man- made. Table 4.1 lists the most common elements on the earth. 6. The symbols for these elements are based upon their names in (d) 9; (e) 13; (f) 12; (g) 6; other languages. 7. (a) 8; (h) 11; (b) 5; (i) 7; (c) 2; (j) 1 lium, No; hafnium, Hf 8. praseodymium, Pr; lawrencium, Lr; californium, Cf; nobe- 10. (a) Elements are made of tiny particles called atoms. (b) All atoms of a given element are identical; (c) The atoms of a given element are different from those of any other element; (d) A given compound always has the same numbers and types of atoms; (e) Atoms are neither created nor destroyed in chemical processes. A chemical reaction simply changes the way the atoms are grouped together. 12. (a) PbO2; (e) Na2CO3; (b) CoCl3; (f) CaH2 (c) C6H12O6; (d) Al2O3; 17 사이_줌달의대학기초화학_해답.indd 17 2015. 2. 16. 오후 2:02 18 23. 13. (a) False; Rutherford’s bombardment experiments with metal foil suggested that the a particles were being de- flected by coming near a dense, positively charged atomic nucleus; (b) False; the proton and the electron have op- posite charges, but the mass of the electron is much smaller than the mass of the proton; (c) True 15. The protons and neutrons are found in the nucleus. The pro- tons are positively charged; the neutrons have no charge. The protons and neutrons each weigh approximately the same. 17. neutron; electron 18. The electrons; outside the nucleus 20. mass 22. Atoms of the same element (atoms with the same number of protons in the nucleus) may have different numbers of neutrons, and so will have different masses. 24. 25. (a) 19 protons, 20 neutrons, 19 electrons; (b) 24 protons, 29 neutrons, 24 electrons; (c) 34 protons, 50 neutrons, 34 electrons; (d) 33 protons, 43 neutrons, 33 electrons; (e) 36 protons, 55 neutrons, 36 protons; (f) 27 protons, 32 neutrons, 27 electrons 26. 29. Metallic elements are found toward the left and bottom of the periodic table; there are far more metallic elements than nonmetals. 31. nonmetallic gaseous elements: oxygen, nitrogen, fluorine, chlorine, hydrogen, and the noble gases; There are no metal- lic gaseous elements at room conditions 33. A metalloid is an element that has some properties common to both metallic and nonmetallic elements. The metalloids are found in the “stair-step” region marked on most periodic tables. 35. (a) fluorine, chlorine, bromine, iodine, astatine; (b) lithium, sodium, potassium, rubidium, cesium, francium; (c) beryllium, magnesium, calcium, strontium, barium, radium; (d) helium, neon, argon, krypton, xenon, radon 36. 38. These elements are found uncombined in nature and do not readily react with other elements. Although these elements were once thought to form no compounds, this now has been shown to be untrue. 39. diatomic gases: H2, N2, O2, F2, Cl2; monatomic gases: He, Ne, Kr, 42. diamond Xe, Rn, Ar 40. chlorine 46. 31 51. (a) 10; (g) 23; 48. -ide (b) 22; (h) 2 53. (a) two electrons gained; 44. electrons 50. nonmetallic (c) 10; (d) 10; (e) 23; (f) 54; (c) three electrons lost; electron lost; (f) two electrons lost (b) three electrons gained; (d) two electrons lost; (e) one (f) Se22 (c) At2; (e) Cs1; (b) Ra21; (d) no ion; 54. (a) P32; 55. Sodium chloride is an ionic compound, consisting of Na1 and Cl2 ions. When NaCl is dissolved in water, these ions are set free and can move independently to conduct the electric cur- rent. Sugar crystals, although they may appear similar visu- ally, contain no ions. When sugar is dissolved in water, it dis- solves as uncharged molecules. No electrically charged species are present in a sugar solution to carry the electric current. 57. (a) Cr2S3; (b) CrO; (c) AlF3; (d) Al2O3; (e) AlP; (f ) Li3N 제5장 2. A binary chemical compound contains only two elements; the major types are ionic (compounds of a metal and a non- metal) and nonionic or molecular (compounds between two nonmetals). Answers depend on student responses. 4. Some substances do not contain molecules; the formula we write reflects only the relative number of each type of atom present. (d) magnesium sulfide; (b) zinc chloride; (c) cesium (e) aluminum iodide; (f ) (g) beryllium fluoride; 5. Roman numeral 6. (a) potassium bromide; oxide; magnesium bromide; (h) barium hydride (b) BaH2; 8. (a) Ag2S; 9. (a) copper(II) chloride; (c) Al2O3; (d) MgF2; (e) correct (b) copper(I) iodide; (c) manganese(II) bromide; (e) chromium(III) chloride; (d) chromium(II) iodide; (f ) mercury(II) oxide 10. (a) cupric iodide; (b) mercurous bromide; (c) chromous bromide; (e) cobaltic oxide; 11. (a) xenon difluoride; (d) cobaltous oxide; (f ) stannous chloride (b) diboron trisulfide; (c) dichlorine hept(a)oxide; (e) nitrogen monoxide; (d) silicon tetrabromide; (f) sulfur trioxide (b) aluminum sulfide; (d) calcium phosphide; (c) diphospho- (e) krypton penta- 12. (a) barium nitride; rus trisulfide; fluoride; 13. (a) barium fluoride; (f ) copper(I) selenide/cuprous selenide (b) radium oxide; (c) dinitrogen (d) rubidium oxide; (e) diarsenic pent(a)oxide; (f ) oxide; calcium nitride 15. An oxyanion is a polyatomic ion containing a given element and one or more oxygen atoms. The oxyanions of chlorine and bromine are given below: (b) NO2 2; 2; 17. (a) NO3 18. (a) ammonium ion; (c) sulfate ion; (e) perchlorate ion; 1; (d) CN2 (c) NH4 (b) dihydrogen phosphate ion; (d) hydrogen sulfite ion (bisulfite ion); (f ) iodate ion 사이_줌달의대학기초화학_해답.indd 18 2015. 2. 16. 오후 2:02 줌달의 대학기초화학 19. (a) ammonium acetate; (b) lithium perchlorate; (c) sodium hydrogen sulfate; (e) calcium chlorate; 21. (a) hypochlorous acid; (d) hypoiodous acid; acid; (g) hydroselenic acid; (d) gold(III) carbonate; (f ) hydrogen peroxide (b) sulfurous acid; (e) perbromic acid; (c) bromic acid; (f ) hydrosulfuric (h) phosphorous acid 42. (a) lithium nitrate; (b) chromium(III) carbonate/chromic carbonate; (c) copper(II) carbonate/cupric carbonate; (d) copper (I) selenide/cuprous selenide; (e) manganese(IV) sulfate; (f) magnesium nitrite (b) NO2; (c) N2O4; (g) OCl2 (e) PBr3; (d) SF6; (e) Fe2(SO4)3; (b) Ca(H2PO4)2; (f) CoCO3; (c) NH4ClO4; (g) Ni(OH)2; (d) NaMnO4; (h) ZnCrO4 24. (a) N2O; (f) CI4; 25. (a) BaSO3; 26. (a) HCN; (b) HNO3; (c) H2SO4; (d) H3PO4; (e) HClO or HOCl; (f ) HBr; 27. (a) Mg(HSO4)2; (e) Sr(NO3)2; (i) Na2HPO4; (b) CsClO4; (f) Sn(C2H3O2)4; (j) Li2O2; (g) HBrO2; (h) HF (d) H2Te(aq); (c) FeO; (g) MnSO4; (h) N2O4; (k) HNO2; (l) Co(NO3)3 제6장 2. Most of these products contain a peroxide, which decom- poses and releases oxygen gas. 3. Bubbling takes place as the hydrogen peroxide chemically decomposes into water and oxygen gas. 7. water 8. H2O2(aq) S H2( g) 1 O2( g) 9. AgNO3(aq) 1 HCl(aq) S AgCl(s) 1 HNO3(aq); Pb(NO3)2(aq) 1 HCl(aq) S PbCl2(s) 1 HNO3(aq) 10. C3H8( g) 1 O2( g) S CO2( g) 1 H2O( g); C3H8( g) 1 O2( g) S CO( g) 1 H2O( g) 11. CaCO3(s) 1 HCl(aq) S CaCl2(aq) 1 H2O(l) 1 CO2( g) 12. SiO2(s) 1 C(s) S Si(s) 1 CO( g) 13. H2S( g) 1 O2( g) S SO2( g) 1 H2O( g) 14. SO2( g) 1 H2O(l) S H2SO3(aq); SO3( g) 1 H2O(l) S H2SO4(aq) 15. NO( g) 1 O3( g) S NO2( g) 1 O2( g) 16. NH3( g) 1 HNO3(aq) S NH4NO3(aq) 17. Xe( g) 1 F2( g) S XeF4(s) 18. NH4Cl(s) 1 NaOH(s) heat NH3( g) 1 H2O( g) 1 NaCl(s) ¡ 20. whole numbers 21. (a) 2Al(s) 1 3CuO(s) S Al2O3(s) 1 3Cu(l); (c) Xe( g) 1 3F2( g) S XeF6(s); (b) S8(s) 1 24F2( g) S 8SF6( g); (d) NH4Cl( g) 1 KOH(s) S NH3( g) 1 H2O( g) 1 KCl(s); (e) SiC(s) 1 2Cl2( g) S SiCl4(l) 1 C(s); 2KOH(aq); (h) 8H2S( g) 1 8Cl2( g) S S8(s) 1 16HCl( g) (g) N2O5( g) 1 H2O(l) S 2HNO3(aq); (f ) K2O(s) 1 H2O(l) S 22. (a) Na2SO4(aq) 1 CaCl2(aq) S CaSO4(s) 1 2NaCl(aq); (b) 3Fe(s) 1 4H2O( g) S Fe3O4(s) 1 4H2( g); (c) Ca(OH)2(aq) 1 2HCl(aq) S CaCl2(aq) 1 2H2O(l); (d) Br2( g) 1 2H2O(l) 1 SO2( g) S 2HBr(aq) 1 H2SO4(aq); (e) 3NaOH(s) 1 H3PO4(aq) S Na3PO4(aq) 1 3H2O(l); (f ) 2NaNO3(s) S 2NaNO2(s) 1 O2( g); 2H2O(l) S 4NaOH(aq) 1 O2( g); 2Si2S4(s) (h) 4Si(s) 1 S8(s) S ( g) 2Na2O2(s) 1 23. (a) 4NaCl(s) 1 2SO2( g) 1 2H2O( g) 1 O2( g) S 2Na2SO4(s) 1 (b) 3Br2(l) 1 I2(s) S 2IBr2(s); 4HCl( g); (c) Ca(s) 1 2H2O(g) S Ca(OH)2(aq) 1 H2(g); 3H2O(g) S B2O3(s) 1 6HF( g); SOCl2(l) 1 Cl2O(g); (g) Mg(s) 1 CuO(s) S MgO(s) 1 Cu(l); 4H2( g) S 3Fe(l) 1 4H2O( g) (e) SO2( g) 1 2Cl2( g) S (f) Li2O(s) 1 H2O(l) S 2LiOH(aq); (h) Fe3O4(s) 1 (d) 2BF3(g) 1 24. (a) Ba(NO3)2(aq) 1 Na2CrO4(aq) S BaCrO4(s) 1 2NaNO3(aq); (b) PbCl2(aq) 1 K2SO4(aq) S PbSO4(s) 1 2KCl(aq); (c) C2H5OH(l) 1 3O2( g) S 2CO2( g) 1 3H2O(l ); (d) CaC2(s) 1 2H2O(l ) S Ca(OH)2(s) 1 C2H2( g); 19 (e) Sr(s) 1 2HNO3(aq) S Sr(NO3)2(aq) 1 H2( g); (f ) BaO2(s) 1 H2SO4(aq) S BaSO4(s) 1 H2O2(aq); (g) 2AsI3(s) S 2As(s) 1 3I2(s); 2CuI(s) 1 I2(s) 1 2K2SO4(aq) (h) 2CuSO4(aq) 1 4KI(s) S 제7장 2. Driving forces are types of changes in a system that pull a re- action in the direction of product formation; driving forces in- clude formation of a solid, formation of water, formation of a gas, and transfer of electrons. 4. The net charge of a precipitate must be zero. The total number of positive charges equals the total number of negative charges. 7. The simplest evidence is that solutions of ionic substances conduct electricity. 9. b, c, f, h 10. (a) Rule 5; 11. (a) MnCO3, Rule 6; (b) Rule 6; (c) Rule 6; (d) Rule 6 (b) CaSO4, Rule 4; (d) no precipitate, most sodium and nitrate salts are soluble; (f ) BaSO4, Rule 4 (e) Ni(OH)2, Rule 5; (c) Hg2Cl2, Rule 3; 12. (a) Na2S(aq) 1 CuCl2(aq) S CuS(s) 1 2NaCl(aq); (b) K3PO4(aq) 1 AlCl3(aq) S AlPO4(s) 1 3KCl(aq); (c) H2SO4(aq) 1 BaCl2(aq) S BaSO4(s) 1 2HCl(aq); (d) 3NaOH(aq) 1 FeCl3(aq) S Fe(OH)3(s) 1 3NaCl(aq); (e) 2NaCl(aq) 1 Hg2(NO3)2(aq) S Hg2Cl2(s) 1 2NaNO3(aq); (f ) 3K2CO3(aq) 1 2Cr(C2H3O2)3(aq) S Cr2(CO3)3(s) 1 6KC2H3O2(aq) 13. (a) CaCl2(aq) 1 2AgNO3(aq) S Ca(NO3)2(aq) 1 2AgCl(s); (b) 2AgNO3(aq) 1 K2CrO4(aq) S Ag2CrO4(s) 1 2KNO3(aq); (c) BaCl2(aq) 1 K2SO4(aq) S BaSO4(s) 1 2KCl(aq) 14. (a) CaCl2(aq) 1 2AgC2H3O2(aq) S 2AgCl(s) 1 Ca(C2H3O2)2(aq); Ba(OH)2(s) 1 2NH4NO3(aq); NiCO3(s) 1 2NaCl(aq) (b) Ba(NO3)2(aq) 1 2NH4OH(aq) S (c) NiCl2(aq) 1 Na2CO3(aq) S 16. Spectator ions are ions that remain in solution during a precip- itation/double-displacement reaction. For example, in the re- action BaCl2(aq) 1 K2SO4(aq) S BaSO4(s) 1 2KCl(aq), the K1 and Cl2 ions are spectator ions. 17. (a) Ca21(aq) 1 SO4 22(aq) S CaSO4(s); (b) Ni21(aq) 1 2OH2(aq) S Ni(OH)2(s); (c) 2Fe31(aq) 1 3S2(aq) S Fe2S3(s) 18. Ag1(aq) 1 Cl2(aq) S AgCl(s); Pb21(aq) 1 2Cl2(aq) S PbCl2(s); Hg2 21(aq) 1 2Cl2(aq) S Hg2Cl2(s) 20. The strong bases are those hydroxide compounds that disso- ciate fully when dissolved in water. The strong bases that are highly soluble in water (NaOH, KOH) are also strong electrolytes. 23. A salt is the ionic product remaining in solution when an acid neutralizes a base. For example, in the reaction HCl(aq) 1 NaOH(aq) S NaCl(aq) 1 H2O(l), sodium chloride is the salt produced by the neutralization reaction. 25. RbOH(s) S Rb1(aq) 1 OH2(aq); CsOH(s) S Cs1(aq) 1 OH2(aq) 27. (a) HCl(aq) 1 KOH(aq) S H2O(l) 1 KCl(aq); (b) HClO4(aq) 1 NaOH(aq) S NaClO4(aq) 1 H2O(l ); (c) CsOH(aq) 1 HNO3(aq) S CsNO3(aq) 1 H2O(l); (d) 2KOH(aq) 1 H2SO4(aq) S 2H2O(l) 1 K2SO4(aq) 29. Answer depends on student choice of example: Na(s) 1 Cl2(g) S 2NaCl(s) is an example. 30. The metal loses electrons, the nonmetal gains electrons. 31. Each magnesium atom would lose two electrons. Each oxygen atom would gain two electrons (so the O2 molecule would gain four electrons). Two magnesium atoms would be re- quired to react with each O2 molecule. Magnesium ions are charged 21, oxide ions are charged 22. 32. Each potassium atom loses one electron. The sulfur atom 사이_줌달의대학기초화학_해답.indd 19 2015. 2. 16. 오후 2:02 질문과 문제의 선택된 번호 정답 20 gains two electrons. So two potassium atoms are required to react with one sulfur atom. 2 3 (K S K1 1 e2) S 1 2e2 S S22 33. (a) P4(s) 1 5O2(g) S P4O10(s); (b) MgO(s) 1 C(s) S Mg(s) 1 CO(g); (d) Co(s) 1 2HCl(aq) S CoCl2(aq) 1 H2(g) (c) Sr(s) 1 2H2O(l) S Sr(OH)2(aq) 1 H2(g); 35. (a) oxidation–reduction; (b) oxidation–reduction; (c) acid–base; (e) precipitation; (g) oxidation–reduction; (i) acid–base (d) acid–base, precipitation; (f ) precipitation; (h) oxidation–reduction; 37. oxidation–reduction 39. A decomposition reaction is one in which a given com- pound is broken down into simpler compounds or con- stituent elements. The reactions CaCO3(s) S CaO(s) 1 CO2( g) and 2HgO(s) S 2Hg(l) 1 O2(g) represent decomposition reac- tions. Such reactions often may be classified in other ways. For example, the reaction of HgO(s) is also an oxidation–reduc- tion reaction. 41. (a) C3H8(g) 1 5O2(g) S 3CO2(g) 1 4H2O(g); (b) C2H4(g) 1 3O2(g) S 2CO2(g) 1 2H2O(g); (c) 2C8H18(l) 1 25O2(g) S 16CO2(g) 1 18H2O(g) 42. (a) 8Fe(s) 1 S8(s) S 8FeS(s); (b) 4Co(s) 1 3O2(g) S 2Co2O3(s); (c) Cl2O7(g) 1 H2O(l) S 2HClO4(aq) 43. (a) 2NI3(s) S N2( g) 1 3I2(s); (b) BaCO3(s) S BaO(s) 1 (c) C6H12O6(s) S 6C(s) 1 6H2O( g); CO2( g); (d) Cu(NH3)4SO4(s) S CuSO4(s) 1 4NH3(g); Na3N(s) 1 4N2( g) (e) 3NaN3(s) S 제8장 2. 307 corks; 116 stoppers; 613 corks; 2640 g (2.64 3 103 g) 4. The average atomic mass takes into account the various iso- topes of an element and the relative abundances in which those isotopes are found. 5. (a) one; 6. A sample containing 35 tin atoms would weigh 4155 amu; (b) five; (c) ten; (d) 50; (e) ten 2967.5 amu of tin would represent 25 tin atoms. 7. 3 3 Avogadro’s number (3 3 6.022 3 1023 5 1.807 3 1024) 8. 32.00 g 9. 177 g 10. 2.326 3 10223 g 11. 0.50 mol O 12. (a) 3.500 mol of F atoms; (c) 3.000 mol Si; (f ) 0.5000 mol Mo (b) 2.000 mmol Hg; (d) 0.2500 mol Pt; (e) 100.0 mol Mg; 13. (a) 2.34 3 1021 g; (d) 2.80 3 1025g; 14. (a) 9.77 3 103 amu; (b) 0.252 g; (e) 0.413 g; (c) 1.10 3 106 g; (f ) 105 g (b) 1.62 3 10220 g; (c) 9.77 3 103 g; (d) 2.56 3 1026 atoms; (f) 18.5 mol Na; (g) 449 g Mg (e) 1.11 3 1025 atoms; 16. The molar mass is calculated by summing the individual atomic masses of the atoms in the formula. 17. (a) carbon monoxide, 28.01 g; (b) sodium carbonate, 105.99 g; (d) hydrogen iodide, 127.9 g; (c) iron(III) nitrate/ferric nitrate, 241.88 g; (e) sulfur trioxide, 80.07 g (b) sodium phosphate, 18. (a) aluminum fluoride, 83.98 g; (c) magnesium carbonate, 84.32 g; 163.94 g; hydrogen carbonate/lithium bicarbonate, 67.96 g; (e) chromium(III) oxide/chromic oxide, 152 g (d) lithium 19. (a) 8.92 3 1024 mol NaCl; (c) 0.0392 mol Al2O3; (e) 1.69 3 1023 mol Li2CO3; 20. (a) 3.55 3 1025 mol; (d) 0.0697 mol; (e) 467 mol (b) 0.125 mol MgCO3; (d) 0.151 mol Fe2O3; (f) 40.3 mol Fe (b) 5.26 mol; (c) 2704 mol; 21. (a) 0.132 g; (d) 0.633 g; 22. (a) 0.313 g; (b) 5.31 3 1024 g; (e) 0.115 g; (b) 139 g; (f) 112 g (c) 1.67 g; (d) 11.3 g; (c) 3.70 3 104 g; (e) 4.66 3 104 g; (f) 4.45 g 23. (a) 3.84 3 1024 molecules; (c) 8.76 3 1016 molecules; (e) 4.03 3 1022 molecules 24. (a) 0.0141 mol S; (d) 0.0127 mol S 26. less 27. (a) 88.82% Cu, 11.18% O; (c) 77.73% Fe, 22.27% O; (e) 46.68% N, 53.32% O; (b) 1.37 3 1023 molecules; (d) 1.58 3 1018 molecules; (b) 0.0159 mol S; (c) 0.0258 mol S; (b) 79.89% Cu, 20.11% O; (d) 69.94% Fe, 30.06% O; (f ) 30.45% N, 69.55% O 28. (a) 28.45% Cu; (d) 34.43% Fe; (g) 88.12% Sn; 29. (a) 34.43% Fe; (d) 11.92% N; (g) 30.45% N; (b) 44.29% Cu; (e) 18.85% Co; (h) 78.77% Sn (b) 29.63% O; (e) 93.10% Ag; (h) 43.66% Mn (c) 44.06% Fe; (f) 13.40% Co; (c) 92.25% C; (f) 45.39% Co; 33. a, c 34. BaH2 35. CaO2Cl2/Ca(OCl)2 36. N2H8CO3 [actually (NH4)2CO3] 37. Co2S3 38. AlF3 39. AlF3 40. Li3N 41. Al2S3O12 [actually Al2(SO4)3] 42. PCl3, PCl5 44. molar mass 46. C4H10O2 제9장 2. The coefficients of the balanced chemical equation indicate the relative numbers of molecules (or moles) of each reactant that combine, as well as the number of molecules (or moles) of each product formed. 3. Balanced chemical equations tell us in what molar ratios sub- stances combine to form products, not in what mass propor- tions they combine. 4. (a) (NH4)2CO3(s) S 2NH3(g) 1 CO2(g) 1 H2O(g). One formula unit of solid ammonium carbonate decomposes to produce two molecules of ammonia gas, one molecule of carbon dioxide gas, and one molecule of water vapor. One mole of solid ammonium carbonate decomposes into two moles of gaseous ammonia, one mole of carbon dioxide gas, and one mole of water vapor. (b) 6Mg(s) 1 P4(s) S 2Mg3P2(s). Six atoms of magnesium metal react with one molecule of solid phosphorus (P4) to make two formula units of solid mag- nesium phosphide. Six moles of magnesium metal react with one mole of solid phosphorus (P4) to produce two moles of solid magnesium phosphide. (c) 4Si(s) 1 S8(s) S 2Si2S4(l). Four atoms of solid silicon react with one molecule of solid sulfur (S8) to form two molecules of liquid disilicon tetrasul- fide. Four moles of solid silicon react with one mole of solid sulfur (S8) to form two moles of liquid disilicon tetrasulfide. (d) C2H5OH(l) 1 3O2(g) S 2CO2(g) 1 3H2O(g). One molecule of liquid ethanol burns with three molecules of oxygen gas to produce two molecules of carbon dioxide gas and three molecules of water vapor. One mole of liquid ethanol burns with three moles of oxygen gas to produce two moles of gaseous carbon dioxide and three moles of water vapor. 사이_줌달의대학기초화학_해답.indd 20 2015. 2. 16. 오후 2:02 줌달의 대학기초화학 5. Balanced chemical equations tell us in what molar ratios sub- stances combine to form products, not in what mass propor- tions they combine. How could 2 g of reactant produce a to- tal of 3 g of products? 6. S(s) 1 2H2SO4(l) S 3SO2(g) 1 2H2O(l); for SO2, ; for H2O, for H2SO4, 7. (a) 17.5 mol/18 mol; (d) 3.75/3.8 mol (b) 25 mol; (c) 5.0 mol; 8. (a) 0.50 mol NH4Cl (27 g); (b) 0.13 mol CS2 (9.5 g); 0.25 mol H2S (8.5 g); (c) 0.50 mol H3PO3 (41 g); 1.5 mol HCl (55 g); (d) 0.50 mol NaHCO3 (42 g) (b) 0.938 mol Se; (c) 0.625 mol CH3CHO; 9. (a) 0.469 mol O2; (d) 1.25 mol Fe 11. (a) 3.19 3 1025 mol; (d) 0.0256 mol; (b) 8.07 3 1026 mol; (c) 11.5 mol; (e) 0.138 mol 12. (a) 119 g; (b) 678 g; (c) 0.0438 g; (d) 256 g; (e) 0.0206 g; (f ) 170. g; (g) 2.11 3 1024 g 13. (a) 0.326 mol; (d) 0.0448 mol (b) 0.202 mol; (c) 0.124 mol; 14. (a) 1.38 g B, 14.0 g HCl; (c) 35.9 g Cu, 6.04 g SO2; (b) 13.5 g Cu2O, 6.04 g SO2; (d) 29.0 g CaSiO3, 11.0 g CO2 15. 3.82 g H2SO4 16. 0.959 g Na2CO3 17. 2.68 g ethyl alcohol 18. 0.443 g NH3 19. 8.62 kg Hg 20. 0.501 g C 21. 2.07 g MgO 22. To determine the limiting reactant, first calculate the number of moles of each reactant present. Then determine how these numbers of moles correspond to the stoichiometric ratio indi- cated by the balanced chemical equation for the reaction. For each reactant, use the stoichiometric ratios from the balanced chemical equation to calculate how much of the other reac- tants would be required to react completely. 24. (a) H2SO4 is limiting, 4.90 g SO2, 0.918 g H2O; limiting, 6.30 g Mn(SO4)2, 0.918 g H2O; 6.67 g SO2, 1.88 g H2O; 2.09 g Al(NO3)3 (b) H2SO4 is (c) O2 is limiting, (d) AgNO3 is limiting, 3.18 g Ag, 25. (a) O2 is limiting, 0.458 g CO2; (c) MnO2 is limiting, 0.207 g H2O; (b) CO2 is limiting, 0.409 g (d) I2 is limiting, H2O; 1.28 g ICl 26. (a) CO is limiting reactant; 11.4 mg CH3OH; (b) I2 is limit- ing reactant; 10.7 mg AlI3; 12.4 mg CaBr2; 2.23 mg H2O; tant; 15.0 mg CrPO4; 0.309 mg H2 (c) HBr is limiting reactant; (d) H3PO4 is limiting reac- 27. 136 g urea 28. 1.79 g Fe2O3 29. 0.627 g CuI; 0.691 g KI3; 0.573 g K2SO4 30. 0.67 kg SiC 32. theoretical yield, 1.16 g; percent yield, 94.0% 33. 2LiOH(s) 1 CO2(g) S Li2CO3(s) 1 H2O(g). 142 g of CO2 can be ultimately absorbed; 102 g is 71.8% of the canister’s capacity. 34. theoretical, 2.72 g BaSO4; percent, 74.3% 제10장 2. potential 4. The total energy of the universe is constant. Energy cannot be created or destroyed, but can only be converted from one 21 form to another. 6. Ball A initially possesses potential energy by virtue of its posi- tion at the top of the hill. As ball A rolls down the hill, its po- tential energy is converted to kinetic energy and frictional (heat) energy. When ball A reaches the bottom of the hill and hits ball B, it transfers its kinetic energy to ball B. Ball A then has only the potential energy corresponding to its new posi- tion. 8. The hot tea is at a higher temperature, which means the par- ticles in the hot tea have higher average kinetic energies. When the tea spills on the skin, energy flows from the hot tea to the skin, until the tea and skin are at the same temperature. This sudden inflow of energy causes the burn. 10. Temperature is the concept by which we express the thermal energy contained in a sample. We cannot measure the mo- tions of the particles/kinetic energy in a sample of matter di- rectly. We know, however, that if two objects are at different temperatures, the one with the higher temperature has mole- cules that have higher average kinetic energies than the mol- ecules of the object at the lower temperature. 12. When the chemical system evolves energy, the energy evolved from the reacting chemicals is transferred to the sur- roundings. 14. exactly equal to 17. losing 16. internal 18. gaining 20. 21. 6540 J 5 6.54 kJ 23. (a) 243 kJ to three significant figures; (c) 0.000251 kJ; (d) 0.4503 kJ (b) 0.004184 kJ; 26. 89 cal 29. The specific heat is 0.89 J/g °C, so the element is most likely 28. 0.057 cal/g °C 27. 29° C aluminum. (b) 2148 kJ; (c) 1296 kJ/mol (b) 2445 kJ/mol dioxygen 31. (a) 29.23 kJ; 32. (a) 2445 kJ/mol water; 33. 2220 kJ 34. 2233 kJ 35. Once everything in the universe is at the same temperature, no further thermodynamic work can be done. Even though the total energy of the universe will be the same, the energy will have been dispersed evenly, making it effectively useless. 38. Tetraethyl lead was used as an additive for gasoline to pro- mote smoother running of engines. It is no longer widely used because of concerns about the lead being released into the environment as the leaded gasoline burns. 39. The greenhouse effect is a warming effect due to the presence of gases in the atmosphere that absorb infrared radiation that has reached the earth from the sun; the gases do not allow the energy to pass back into space. A limited greenhouse effect is desirable because it moderates the temperature changes in the atmosphere that would otherwise be more drastic between daytime when the sun is shining and nighttime. Having too high a concentration of greenhouse gases, however, will ele- vate the temperature of the earth too much, affecting climate, crops, the polar ice caps, the temperatures of the oceans, and so on. Carbon dioxide produced by combustion reactions is our greatest concern as a greenhouse gas. 41. If a proposed reaction involves either or both of those phe- nomena, the reaction will tend to be favorable. 44. The molecules in liquid water are moving around freely and are therefore more “disordered” than when the molecules are held rigidly in a solid lattice in ice. The entropy increases dur- ing melting. 사이_줌달의대학기초화학_해답.indd 21 2015. 2. 16. 오후 2:02 질문과 문제의 선택된 번호 정답 (c) F2 (the electrons are pulled in closer to the nucleus by the pos- itive charge). (b) F2; 31. (a) I; 33. When atoms form covalent bonds, they try to attain a valence-electron configuration similar to that of the follow- ing noble gas element. When the elements in the first few horizontal rows of the periodic table form covalent bonds, they attempt to achieve the configurations of the noble gases helium (two valence electrons, duet rule) and neon and argon (eight valence electrons, octet rule). 37. (a) 8; (b) 16; (c) 26; (d) 26 22 제11장 (d) P H (H and P 40. 2. bond energy 4. covalent 5. In H2 and HF, the bonding is covalent in nature, with an electron pair being shared between the atoms. In H2, the two atoms are identical and so the sharing is equal; in HF, the two atoms are different and so the bonding is polar covalent. Both of these are in marked contrast to the situation in NaF: NaF is an ionic compound, and an electron is completely transferred from sodium to fluorine, thereby producing the separate ions. 7. The difference in electronegativity between the atoms in the bond 8. (a) I is most electronegative, Rb is least electronegative; (b) Mg is most electronegative, Ca and Sr have similar electronegativities; least electronegative (c) Br is most electronegative, K is 9. (a) ionic; (b) polar covalent (c) covalent (d) Cl (c) Cl; (b) Ba–Cl; (b) neither; 10. c and d 11. (a) F; 12. (a) Ca–Cl; (c) Fe–I; 14. The presence of strong bond dipoles and a large overall dipole moment makes water a very polar substance. Properties of wa- ter that are dependent on its dipole moment involve its freez- ing point, melting point, vapor pressure, and ability to dis- solve many substances. (b) Cl; (d) Be–F 15. (a) H; 16. (a) d1P S Fd2; (c) I (c) d1P S Cd2; (b) d1P S Od2; have essentially the same electronegativities) (c) d1S S Nd2; (b) d1N S Od2; 17. (a) d1H S Cd2; (d) d1C S Nd2; 19. preceding 21. Atoms in covalent molecules gain a configuration like that of a noble gas by sharing one or more pairs of electrons be- tween atoms: such shared pairs of electrons “belong” to each of the bonding atoms at the same time. In ionic bonding, one atom completely donates one or more electrons to an- other atom, and then the resulting ions behave indepen- dently of one another (they are not “attached” to one an- other, although they are mutually attracted). (b) Cs1 [Xe]; (c) P32 [Ar]; 22. (a) Br– [Kr]; 24. (a) Na2S; (b) BaSe; 25. (a) S21 [Kr]; O22 [Ne]; (d) Li3N; (c) MgBr2; (b) Ca21 [Ar]; H– [He]; (d) S2– [Ar] (e) KH (c) K1 [Ar]; P3– [Ar]; (d) Ba21 [Xe]; Se2– [Kr] 27. An ionic solid such as NaCl consists of an array of alternat- ing positively and negatively charged ions: that is, each pos- itive ion has as its nearest neighbors a group of negativeions, and each negative ion has a group of positive ions sur- rounding it. In most ionic solids, the ions are packed as tightly as possible. 29. In forming an anion, an atom gains additional electrons in its outermost (valence) shell. Having additional electrons in the valence shell increases the repulsive forces between elec-trons, and the outermost shell becomes larger to accommo-date this. 30. (a) F2 is larger than Li1. The F2 ion has a filled n 5 2 shell. A lithium atom has lost the electron from its n 5 2 shell, leav- ing the n 5 1 shell as its outermost. (b) Cl2 is larger than Na1, since its valence electrons are in the n 5 3 shell (Na1 has lost its 3s electron). (c) Ca is larger than Ca21. Positive ions are always smaller than the atoms from which they are formed. (d) I2 is larger. Both Cs1 and I2 have the same elec- tron configuration (isoelectronic with Xe) and have their va- lence electrons in the same shell. However, Cs1 has two more positive charges in its nucleus than does I2; this charge causes the n 5 5 shell of Cs1 to be smaller than that of I2 36. 38. 39. 41. 42. 사이_줌달의대학기초화학_해답.indd 22 2015. 2. 16. 오후 2:03 줌달의 대학기초화학 23 14. Charles’s law indicates that an ideal gas decreases by 1/273 of its volume for every Celsius degree its temperature is low- ered. This means an ideal gas would approach a volume of zero at 2273 °C. 15. V 5 bT; V1/T1 5 V2/T2 16. 35.4 mL 17. (a) 80.2 mL; (c) 208 mL (2.1 3 102 mL) (b) 277 °C (196 K); 18. (a) 35.4 K 5 2238 °C; (b) 0 mL (absolute zero; a real gas would condense to a solid or liquid); (c) 40.5 mL 19. If the temperature is decreased by a factor of 2, the volume will also decrease by a factor of 2. The new volume of the sample will be half its original volume. 20. 90 °C, 124 mL; 80 °C, 121 mL; 70 °C, 117 mL; 60 °C, 113 mL; 50 °C, 110. mL; 40 °C, 107 mL; 30 °C, 103 mL; 20 °C, 99.8 mL 21. V 5 an; V1/n1 5 V2/n2 22. 5.60 L 23. 80.1 L 25. Real gases behave most ideally at relatively high tempera- tures and relatively low pressures. We usually assume that a real gas’s behavior approaches ideal behavior if the temper- ature is over 0 °C (273 K) and the pressure is 1 atm or lower. 27. For an ideal gas, PV 5 nRT is true under any conditions. Con- sider a particular sample of gas (n remains constant) at a par- ticular fixed pressure (P remains constant). Suppose that at temperature T1 the volume of the gas sample is V1. For this set of conditions, the ideal gas equation would be given by PV1 5 nRT1. If the temperature of the gas sample changes to a new temperature, T2, then the volume of the gas sample changes to a new volume, V2. For this new set of conditions, the ideal gas equation would be given by PV2 5 nRT2. If we make a ratio of these two expressions for the ideal gas equation for this gas sample, and cancel out terms that are constant for this situation (P, n, and R), we get which can be rearranged to the familiar form of Charles’s law, 44. The geometric structure of NH3 is that of a trigonal pyra- mid. The nitrogen atom of NH3 is surrounded by four elec- tron pairs (three are bonding, one is a lone pair). The HONOH bond angle is somewhat less than 109.5° (because of the presence of the lone pair). 47. The general molecular structure of a molecule is determined by how many electron pairs surround the central atom in the molecule, and by which of those pairs are used for bond- ing to the other atoms of the molecule. 49. In NF3, the nitrogen atom has four pairs of valence electrons; in BF3, only three pairs of valence electrons surround the boron atom. The nonbonding pair on nitrogen in NF3 pushes the three F atoms out of the plane of the N atom. 50. (a) four electron pairs arranged in a tetrahedral arrange- ment with some distortion due to the nonbonding pair; (b) four electron pairs in a tetrahedral arrangement; (c) four electron pairs in a tetrahedral arrangement 51. (a) tetrahedral; (b) bent(nonlinear); 52. (a) basically tetrahedral arrangement of the oxygens around (c) tetrahedral the phosphorus; (c) trigonal pyramid (b) tetrahedral; 53. (a) approximately tetrahedral (a little less than 109.5°); (b) approximately tetrahedral (a little less than 109.5°); (c) tetrahedral (109.5°); (d) trigonal planar (120°) because of the double bond 54. The ethylene molecule contains a double bond between the carbon atoms. This makes the molecule planar (flat), with HOCOH and HOCOC bond angles of approximately 120°. The 1,2-dibromoethane molecule would not be planar, how- ever. Each carbon would h0ave four bonding pairs of elec- trons around it, and so the orientation around each carbon atom would be basically tetrahedral with bond angles of ap- proximately 109.5° (assuming all bonds are similar). 제12장 2. Solids are rigid and incompressible and have definite shapes and volumes. Liquids are less rigid than solids; although they have definite volumes, liquids take the shape of their con- tainers Gases have no fixed volume or shape; they take the volume and shape of their container and are affected more by changes in their pressure and temperature than are solids or liquids. 3. Pressure units include mm Hg, torr, pascals, and psi. The unit “mm Hg” is derived from the barometer, because in a traditional mercury barometer, we measure the height of the mercury column (in millimeters) above the reservoir of mercury. 4. (a) 0.980 atm; (b) 1.01 atm; (c) 0.916 atm; (d) 3.41 3 1026 atm 5. (a) 792 mm Hg; (d) 8.18 mm Hg 6. (a) 2.07 3 103 kPa; (d) 87.9 kPa (b) 714 mm Hg; (c) 746 mm Hg; (b) 106 kPa; (c) 1.10 3 103 kPa; 7. Additional mercury increases the pressure on the gas sam- ple, causing the volume of the gas upon which the pressure is exerted to decrease (Boyle’s law). 9. PV 5 k; P1V1 5 P2V2 10. (a) 4.69 atm; 11. (a) 146 mL; 12. 0.520 L 13. 27.2 atm (b) 1.90 3 104 mm Hg; (b) 0.354 L; (c) 687 mm Hg (c) 0.270 atm (c) 334 K 33. 340 atm (b) 3.56 atm 5 2.70 3 103 mm Hg; 28. (a) 5.02 L; 29. 2428 K/2.43 3 103 K 30. 131 atm 31. 0.150 atm; 0.163 atm 32. 238 K/235 °C 34. 0.332 atm; 0.346 atm 36. As a gas is bubbled through water, the bubbles of gas be- come saturated with water vapor, thus forming a gaseous mix- ture. The total pressure for a sample of gas that has been col- lected by bubbling through water is made up of two components: the pressure of the sample gas and the pressure of water vapor. The partial pressure of the gas equals the total pressure of the sample minus the vapor pressure of water. 35. 150. atm 37. 0.314 atm 38. 0.0984 mol; 0.0984 mol 39. 3.07 atm 40. Phydrogen 5 0.990 atm; 9.55 3 1023 mol H2; 0.625 g Zn 42. pressure 43. If the temperature of a sample of gas is increased, the aver- age kinetic energy of the particles of gas increases. This means that the speeds of the particles increase. If the parti- cles have a higher speed, they hit the walls of the container more frequently and with greater force, thereby increasing the pressure. 44. STP 5 0 °C, 1 atm pressure. These conditions were chosen because they are easy to attain and reproduce experimentally. 사이_줌달의대학기초화학_해답.indd 23 2015. 2. 16. 오후 2:03 질문과 문제의 선택된 번호 정답 24 The barometric pressure within a laboratory will usually be near 1 atm, and 0 °C can be attained with a simple ice bath. 45. 1.93 L 46. 0.941 L 47. 0.941 L; 0.870 L 48. 5.03 L (dry volume) 49. 52.7 L 50. 184 mL 51. 40.5 L; PHe 5 0.864 atm; PNe 5 0.136 atm 52. 1.72 L 53. 0.365 g 제13장 3. Because it requires so much more energy to vaporize water than to melt ice, this suggests that the gaseous state is signifi- cantly different from the liquid state, but that the liquid and solid states are relatively similar. 4. See Figure 13.2. 6. When a solid is heated, the molecules begin to vibrate/move more quickly. When enough energy has been added to over- come the intermolecular forces that hold the molecules in a crystal lattice, the solid melts. As the liquid is heated, the mol- ecules begin to move more quickly and more randomly. When enough energy has been added, molecules having sufficient kinetic energy will begin to escape from the surface of the liq- uid. Once the pressure of vapor coming from the liquid is equal to the pressure above the liquid, the liquid boils. Only intermolecular forces need to be overcome in this process: no chemical bonds are broken. 8. intramolecular; intermolecular 9. fusion; vaporization 10. (a) more energy is required to separate the molecules of a liq- uid into the freely moving and widely separated molecules of a (c) 4.22 kJ; vapor/gas; (b) 0.113 kJ; (d) 24.22 kJ 11. 8.35 kJ; 84.4 kJ; 23.1 kJ 12. 107 kJ/mol; 39.5 kJ; 1.07 3 103 kJ 16. The hydrogen bonding that can exist when H is bonded to O (or N or F) is an additional intermolecular force, which means additional energy must be added to separate the molecules during boiling. 17. London dispersion forces are instantaneous dipole forces that arise when the electron cloud of an atom is momentarily dis- torted by a nearby dipole, temporarily separating the centers of positive and negative charge in the atom. 18. (a) London dispersion forces; (c) London dispersion forces (b) hydrogen bonding; (d) dipole–dipole forces 20. For a homogeneous mixture to form, the forces between mol- ecules of the two substances being mixed must be at least com- parable in magnitude to the intermolecular forces within each separate substance. In the case of a water–ethanol mixture, the forces that exist when water and ethanol are mixed are stronger than water–water or ethanol–ethanol forces in the separate substances. Ethanol and water molecules can ap- proach one another more closely in the mixture than either substance’s molecules could approach a like molecule in the separate substances. Strong hydrogen bonding occurs in both ethanol and water. 23. (a) HF: Although both substances are capable of hydrogen bonding, water has two OOH bonds that can be involved in hydrogen bonding versus only one FOH bond in HF; (b) CH3OCH3: Because no H is attached to the O atom, no hydrogen bonding can exist. Thus, the molecule should be relatively more volatile than CH3CH2OH even though it con- tains the same number of atoms of each element; (c) CH3SH: Hydrogen bonding is not as important for a SOH bond (because S has a lower electronegativity than O). Since there is relatively little hydrogen bonding, CH3SH is more volatile than CH3OH. 24. Both substances have the same molar mass. Ethyl alcohol contains a hydrogen atom directly bonded to an oxygen atom, however. Therefore, hydrogen bonding can exist in ethyl alcohol, whereas only weak dipole–dipole forces exist in dimethyl ether. Dimethyl ether is more volatile; ethyl alcohol has a higher boiling point. 26. Ionic solids have positive and negative ions as their funda- mental particles; a simple example is sodium chloride, in which Na1 and Cl2 ions are held together by strong electro- static forces. Molecular solids have molecules as their funda- mental particles, with the molecules being held together in the crystal by dipole–dipole forces, hydrogen-bonding forces, or London dispersion forces (depending on the identity of the substance); simple examples of molecular solids include ice (H2O) and ordinary table sugar (sucrose). Atomic solids have simple atoms as their fundamental particles, with the atoms being held together in the crystal by either covalent bonding (as in graphite or diamond) or metallic bonding (as in copper or other metals). 28. Ionic solids consist of a crystal lattice of positively and neg- atively charged ions. A given ion is surrounded by several ions of the opposite charge, all of which electrostatically at- tract it strongly. This pattern repeats itself throughout the crystal. The existence of these strong electrostatic forces throughout the crystal means a great deal of energy must be applied to overcome the forces and melt the solid. 29. Ice contains nonlinear, highly polar water molecules, with ex- tensive, strong hydrogen bonding. Dry ice consists of linear, nonpolar molecules, and only very weak intermolecular forces are possible. 30. Although ions exist in both the solid and liquid states, in the solid state the ions are rigidly held in place in the crystal lat- tice and cannot move so as to conduct an electric current. 제14장 2. A nonhomogeneous mixture may differ in composition in various places in the mixture, whereas a solution (a homo- geneous mixture) has the same composition throughout. Examples of nonhomogeneous mixtures include spaghetti sauce, a jar of jelly beans, and a mixture of salt and sugar. 5. “Like dissolves like.” The hydrocarbons in oil have inter- molecular forces that are very different from those in water, so the oil spreads out rather than dissolving in the water. 8. unsaturated 10. large 12. 100. 13. (a) 0.0224%; 14. (a) 3.11 g NaCl, 121.89 g water; (b) 18.3%; (c) 0.223%; (d) 18.3% (c) 62.1 g NaCl, 937.9 g water; (b) 1.74 mg NaCl, 33.46 g (d) 0.0292 g NaCl, water; 29.17 g water (b) 1.0 M; 17. 7.81 g KBr 18. approximately 71 g 20. 0.110 mol; 0.220 mol 22. b 23. (a) 2.0 M; 24. (a) 1.08 M; 25. 0.464 M 26. 0.0902 M 27. 0.619 M 28. (a) 0.0133 mol, 0.838 g; (c) 0.00505 mol, 0.490 g; (b) 1.08 M; 29. (a) 25.6 g; (b) 901 g; (c) 1.3 g; (d) 7.59 g (b) 2.34 mol, 39.9 g; (d) 0.0299 mol, 1.09 g (c) 0.67 M; (d) 0.50 M (c) 0.0108 M; (d) 0.108 M 사이_줌달의대학기초화학_해답.indd 24 2015. 2. 16. 오후 2:03 줌달의 대학기초화학 (b) 2.42 mL; (c) 50.1 mL; (d) 1.22 L (b) 3.37 3 1023 N; (b) 0.0104 N; (c) 13.3 N (c) 1.63 N 23. (a) [H1] 5 2.4 3 10211 M, basic; (b) [H1] 5 1.3 3 1026 M, acidic; (c) [H1] 5 9.6 3 10213 M, basic; (d) [H1] 5 1.5 3 1028 M, basic 31. (a) 4.60 3 1023 mol Al31, 1.38 3 1022 mol Cl2; (b) 1.70 mol Na1, 0.568 mol PO4 Cu21, 4.38 3 1023 mol Cl2; 7.91 3 1025 mol OH2 32; (c) 2.19 3 1023 mol (d) 3.96 3 1025 mol Ca21, (b) 1.69 M; (c) 0.0426 M; (d) 0.625 M 32. 1.33 g 34. half 36. (a) 0.0717 M; 38. 0.13 M 39. 10.3 mL 40. 31.2 mL 41. 0.523 g 42. 0.300 g 44. 1.8 3 1024 M 45. (a) 63.0 mL; 47. 1 N 49. (a) 0.277 N; 50. (a) 0.134 N; 52. 22.2 mL, 11.1 mL 53. 0.05583 M, 0.1117 N 제15장 2. acid; base 4. A conjugate acid–base pair differs by one hydrogen ion, H1. For example, HC2H3O2 (acetic acid) differs from its conju- gate base, C2H3O2 2(acetate ion), by a single H1 ion. HC2H3O2(aq) C2H3O2 (aq) 1 H (aq) 6. When an acid is dissolved in water, the hydronium ion (H3O1) is formed. The hydronium ion is the conjugate acid of water (H2O). 7. (a) a conjugate pair (HSO4 2 is the acid, SO4 22 is the base); (b) a conjugate pair (HBr is the acid, Br2 is the base; they dif- fer by one proton); is the and also the conjugate base of conjugate acid of HPO4 32; (d) not a con- H3PO4; HPO4 2 is the jugate pair (NO3 conjugate base of HNO2) is the conjugate base of HNO3; NO2 (c) not a conjugate pair (H2PO4 is the conjugate acid of PO4 32 2 2 2 8. (a) NH3(aq)(base) 1 H2O(l)(acid) NH4 1(aq)(acid) 1(aq)(acid) 1 H2O(l)(base) NH3(aq)(base) 2(aq)(base) 1 H2O(l)(acid) S NH3(aq)(acid) 1 OH2(aq)(base); (b) NH4 1 H3O1(aq)(acid); (c) NH4 1 OH2(aq)(base) (b) HF; (b) HSO3 9. (a) HBrO3; 10. (a) BrO2; 11. (a) CN2 1 H2O (c) HSO3 (c) SO3 2; 22; HCN 1 OH2; 2; (d) H2SO3 (d) CH3NH2 22 1 H2O (b) CO3 (c) H3PO4 1 H2O 2 1 H2O (d) NH2 2 1 OH2; 22 1 H3O1; HCO3 H2PO4 NH3 1 OH2 13. If an acid is weak in aqueous solution, it does not easily transfer protons to water (and does not fully ionize). If an acid does not lose protons easily, then the acid’s anion must strongly attract protons. 14. A strong acid loses its protons easily and fully ionizes in wa- ter; the acid’s conjugate base is poor at attracting and hold- ing protons and is a relatively weak base. A weak acid resists loss of its protons and does not ionize to a great extent in water; the acid’s conjugate base attracts and holds protons tightly and is a relatively strong base. 16. H2SO4 (sulfuric): H2SO4 1 H2O S HSO4 2 1 H3O1; HCl (hydrochloric): HCl 1 H2O S Cl2 1 H3O1; HNO3 (nitric): HNO3 1 H2O S NO3 HClO4 (perchloric): HClO4 1 H2O S ClO4 21 H3O1; 2 1 H3O1 18. An oxyacid is an acid containing a particular element that is 25 bonded to one or more oxygen atoms. HNO3, H2SO4, and HClO4 are oxyacids. HCl, HF, and HBr are not oxyacids. (b) HBr is a strong acid; 21. HCO3 (d) HC2H3O2 is a weak acid 19. (a) HSO4 is a moderately strong acid; (c) HCN is a weak acid; 2 can behave as an acid if it reacts with a substance 2 itself. For that more strongly gains protons than does HCO3 2 would behave as an acid when reacting example, HCO3 2(aq) 1 with hydroxide ion (a much stronger base): HCO3 OH2(aq) S CO3 2 would behave as a base when reacted with a substance that 2 itself. For ex- more readily loses protons than does HCO3 2 would behave as a base when reacting with ample, HCO3 2(aq) 1 hydrochloric acid (a much stronger acid): HCO3 HCl(aq) S H2CO3(aq) 1 Cl2(aq). H2PO4 22 2 1 H3O1 S H3PO4 1 H2O. 1 H2O and H2PO4 22(aq) 1 H2O(l ). On the other hand, HCO3 2 1 OH2 S HPO4 24. (a) 7.5 3 10213 M, acidic; (b) 1.4 3 1028 M, acidic; (c) 2.5 3 1026 M, basic; 25. (a) [OH2] 5 0.105 M; (c) [OH ] 5 8.41 3 1022 M (d) 2.5 3 1022 M, basic (b) [OH2] 5 5.22 3 1025 M; 27. Answers will depend on student choices. 29. The pH of a solution is defined as the negative of the loga- rithm of the hydrogen ion concentration, pH 5 2log[H1]. Mathematically, the negative sign in the definition means the pH decreases as the hydrogen ion concentration increases. 30. (a) pH 5 3.000 (acidic); (c) pH 5 10.037 (basic); (b) pH 5 3.660 (acidic); (d) pH 5 6.327 (acidic) 31. (a) 1.983, acidic; (b) 12.324, basic; (c) 11.368, basic; (d) 3.989, acidic 32. (a) pOH 5 6.55, basic; (c) pOH 5 0.85, basic; (b) pOH 5 12.11, acidic; (d) pOH 5 8.45, acidic 33. (a) pH 5 1.719, [OH2] 5 5.2 3 10213 M; (b) pH 5 6.316, [OH2] 5 2.1 3 1028 M; (c) pH 5 10.050, [OH2] 5 1.1 3 1024 M; (d) pH 5 4.212, [OH2] 5 1.6 3 10210 M 34. (a) [H1] 5 9.1 3 10210 M; (c) [H1] 5 5.0 3 10210 M; 35. (a) [OH2] 5 1.0 3 10213 M; (c) [OH2] 5 6.0 3 10212 M; 36. (a) pH 5 2.69; (d) pH 5 6.17 (b) pH 5 9.86; (b) [H1] 5 6.8 3 1026 M; (d) [H1] 5 2.6 3 10211 M (b) [OH2] 5 9.8 3 1022 M; (d) [OH2] 5 3.1 3 1022 M (c) pH 5 3.003; 38. The solution contains water molecules, H3O1 ions (protons), 2 ions. Because HNO3 is a strong acid that is com- and NO3 pletely ionized in water, no HNO3 molecules are present. 39. (a) pH 5 2.917; (b) pH 5 3.701; (c) pH 5 4.300; (d) pH 5 2.983 41. A buffered solution consists of a mixture of a weak acid and its conjugate base; one example of a buffered solution is a mixture of acetic acid (HC2H3O2) and sodium acetate (NaC2H3O2). 43. CH3COO2 1 HCl S CH3COOH 1 Cl2; CH3COOH 1 NaOH S H2O 1 NaCH3COO 제16장 1. The carbon–oxygen bonds two carbon monoxide molecules and the oxygen–oxygen bond in an oxygen gas molecule must break, and the carbon–oxygen bonds in two carbon dioxide molecules must form. in 2. Ea represents the activation energy for the reaction, which is the minimum energy needed for the reaction to be able to occur. 4. Enzymes are biochemical catalysts that speed up the com- 사이_줌달의대학기초화학_해답.indd 25 2015. 2. 16. 오후 2:03 질문과 문제의 선택된 번호 정답 plicated reactions that would be too slow to sustain life at normal body temperatures. 38. Ksp 5 1.9 3 1024; 10. g/L 70. 4 3 10217 M, 4 3 10215 g/L 5. A state of equilibrium is attained when two opposing processes are exactly balanced. The development of a vapor pressure above a liquid in a closed container is an example of a physical equilibrium. Any chemical reaction that ap- pears to “stop” before completion is an example of a chem- ical equilibrium. 8. The two curves come together when a state of chemical equilibrium has been reached, after which point the forward and reverse reactions are occurring at the same rate so that there is no further net change in concentration. 제17장 26 10. 11. 15. 16. 12. 9.4 3 10220 13. 1.2 3 1010 17. [CO2] increases; K does not change 19. If heat is applied to an endothermic reaction (the temper- ature is raised), the equilibrium is shifted to the right. More product will be present at equilibrium than if the temperature had not been increased. The value of K in- creases. (b) no change (S is solid); (c) shift right; 20. (a) shift right; (d) no change (d) shift right 21. (a) no change (B is solid); (b) shift right; (c) shift left; (c) no; (b) yes; 22. (a) no; (d) yes 23. For an endothermic reaction, an increase in temperature will shift the position of equilibrium to the right (toward products). 25. A small equilibrium constant implies that not much product forms before equilibrium is reached. The reaction would not be a good source of the products unless Le Châtelier’s prin- ciple can be used to force the reaction to the right. 26. 1.26 3 103 27. 1.06 3 1021 28. [O2( g)] 5 8.0 3 1022 M 29. 5.4 3 1024 M 32. only the temperature 33. (a) Ni(OH)2(s) 2. Oxidation is a loss of one or more electrons by an atom or ion. Reduction is the gaining of one or more electrons by an atom or ion. Equations depend on student responses. 3. (a) Potassium is oxidized, oxygen is reduced; oxidized, chlorine is reduced; rine is reduced; (d) carbon is oxidized, oxygen is reduced (b) iodine is (c) cobalt is oxidized, chlo- 4. (a) sulfur is oxidized, oxygen is reduced; (b) phosphorus is oxidized, oxygen is reduced; (c) hydrogen is oxidized, carbon is reduced; (d) boron is oxidized, hydrogen is reduced 6. Oxidation numbers represent a “relative charge” one atom has compared to another in a compound. In an element, all the atoms are equivalent. 8. Because fluorine is the most electronegative element, its ox- idation state is always negative relative to other elements; because fluorine gains only one electron to complete its out- ermost shell, its oxidation number in compounds is always 21. The other halogen elements are almost always more electronegative than the atoms to which they bond, and al- most always have -1 oxidation numbers. However, in an in- terhalogen compound involving fluorine and some other halogen, since fluorine is the most electronegative element of all, the other halogens in the compound will have posi- tive oxidation states relative to fluorine. 10. 3– 11. (a) Cr, 13; Cl, 21; (b) Cu, 11; O, 22; (c) Cu, 12; O, 22; (d) 0 12. (a) Al, 13; P, 15; O, 22; (c) Ba, 12; C, 14; O, 22; (b) Mn, 14; O, 22; (d) Cl, 11; F, 21 13. (a) H, 11; P, 15; O, 22; (c) H, 11; N, 15; O, 22; 14. (a) K, 11; Cl, 15; O, 22; (d) Na, 11; I, 15; O, 22 (b) H, 11; Br, 11; O, 22; (d) H, 11; Cl, 17; O, 22 (b) 0; (c) C, 12; O, 22; 15. (a) Fe, 13; O, 22; (b) Al, 13; C, 14; O, 22; (c) Ba, 12; Cr, 16; O, 22; (d) Ca, 12; H, 11; C, 14; O, 22 17. Electrons are negative; when an atom gains electrons, it gains one negative charge for each electron gained. For ex- ample, in the reduction reaction Cl 1 e2 S Cl2, the oxida- tion state of chlorine decreases from 0 to 21 as the electron is gained. 19. An oxidizing agent oxidizes another species by gaining the electrons lost by the other species; therefore, an oxidizing agent itself decreases in oxidation state. A reducing agent in- creases its oxidation state when acting on another atom or molecule. 21. (a) manganese is oxidized, hydrogen is reduced; (b) sulfur is oxidized, oxygen is reduced; (c) aluminum is oxidized, hydrogen is reduced; (d) nitrogen is oxidized, oxygen is reduced 22. (a) carbon is oxidized, chlorine is reduced; (b) carbon is oxidized, oxygen is reduced; (c) phosphorus is oxidized, chlorine is reduced; (d) calcium is oxidized, hydrogen is reduced Ni21(aq) 1 2OH2(aq), Ksp 5 (c) Hg(OH)2(s) (b) Cr2S3(s) [Ni21(aq)][OH2(aq)]2; Ksp 5 [Cr31(aq)]2[S22(aq)]3; 2OH2(aq), Ksp 5 [Hg21(aq)][OH2(aq)]2; 2Ag1(aq) 1 CO3 34. 6.5 3 1025 M; 0.021 g/L 35. 7.4 3 1024 g/L 36. Ksp 5 2.3 3 10247 37. Ksp 5 1.23 3 10215 22 (aq), Ksp 5 [Ag1(aq)]2[CO3 2Cr31(aq) 1 3S22(aq), Hg21(aq) 1 (d) Ag2CO3(s) 22(aq)] 23. Iron is reduced [13 in Fe2O3(s), 0 in Fe(l )]; carbon is oxidized [12 in CO( g), 14 in CO2( g)]. Fe2O3(s) is the oxidizing agent; CO( g) is the reducing agent. 24. (a) chlorine is reduced, iodine is oxidized; chlorine is the ox- idizing agent, iodide ion is the reducing agent; (b) iron is reduced, iodine is oxidized; iron(III) is the oxidizing agent, iodide ion is the reducing agent; (c) copper is reduced, io- dine is oxidized; copper(II) is the oxidizing agent, iodide ion is the reducing agent 사이_줌달의대학기초화학_해답.indd 26 2015. 2. 16. 오후 2:03 줌달의 대학기초화학 26. Under ordinary conditions it is impossible to have “free” electrons that are not part of some atom, ion, or molecule. Thus, the total number of electrons lost by the species being oxidized must equal the total number of electrons gained by the species being reduced. 27. (a) 2Cl2(aq) S Cl2(g) 1 2e2; (c) Fe(s) S Fe31(aq) 1 3e2; 28. (a) 8e2 1 10H1(aq) 1 NO3 (b) Fe21(aq) S Fe31(aq) 1 e2; (d) Cu21(aq) 1 e2 S Cu1(aq) 1(aq) 1 3H2O(l); 2(aq) S NH4 (b) 2e2 1 2H1(aq) 1 C2N2(g) S 2HCN(aq); (c) 6e2 1 6H1(aq) 1 ClO3 (d) 2e2 1 4H1(aq) 1 MnO2(s) S Mn21(aq) 1 2H2O(l) 2(aq) S Cl2(aq) 1 3H2O(l); 29. (a) 2Al 1 6H1 S 2Al31 1 3H2; 3S 1 2NO 1 4H2O; 2H1 1 10HCl; 1 2 1 H2O 30. Cu(s) 1 2HNO3(aq) 1 2H1(aq) S Cu21(aq) 1 2NO2(g) 1 2H2O(l); Mg(s) 1 2HNO3(aq) S Mg(NO3)2(aq) 1 H2(g) (b) 8H1 1 2NO3 (c) 6H2O 1 I2 1 5Cl2 S 2IO3 1 S22 S 1 AsO3 (d) 2H1 1 AsO4 2 1 3S22 S 2 2 31. A salt bridge typically consists of a U-shaped tube filled with an inert electrolyte (one involving ions that are not part of the oxidation–reduction reaction). A salt bridge completes the electrical circuit in a cell. Any method that al- lows transfer of charge without allowing bulk mixing of the solutions may be used (another common method is to set up one half-cell in a porous cup, which is then placed in the beaker containing the second half-cell). 33. Reduction takes place at the cathode and oxidation takes place at the anode. 34. Pb21(aq) ion is reduced; Zn(s) is oxidized. The anode reaction is Zn(s) S Zn21(aq) 1 2e2. The cathode reaction is Pb21(aq) 1 2e2 S Pb(s). 36. Cd 1 2OH2 S Cd(OH)2 1 2e2 (oxidation); NiO2 1 2H2O 1 2e2 S Ni(OH)2 1 2OH2 (reduction) 38. Aluminum is a very reactive metal when freshly isolated in the pure state. Upon standing for even a relatively short pe- riod of time, aluminum metal forms a thin coating of Al2O3 on its surface from reaction with atmospheric oxygen. This Al2O3 coating is much less reactive than the metal and pro- tects the metal’s surface from further attack. 40. The magnesium is used for cathodic protection of the steel pipeline. Magnesium is more reactive than iron, and will be oxidized in preference to the iron of the pipeline. 42. The main recharging reaction for the lead storage battery is 2PbSO4(s) 1 2H2O(l ) S Pb(s) 1 PbO2(s) 1 2H2SO4(aq). A ma- jor side reaction is the electrolysis of water, 2H2O(l) S 2H2(g) 1 O2(g), which produces an explosive mixture of hy- drogen and oxygen that accounts for many accidents during the recharging of such batteries. 제18장 2. The radius of a typical atomic nucleus is on the order of 10213 cm, which is roughly 100,000 times smaller than the radius of an atom overall. 4. The mass number represents the total number of protons and neutrons in a nucleus. 5. The atomic number (Z) is written as a left subscript, while the mass number (A) is written as a left superscript. That is, 27 the general symbol for a nuclide is A zX. As an example, con- sider the isotope of oxygen with 8 protons and 8 neutrons: its symbol would be 16 8O. 7. a neutron 10. Gamma rays are high-energy photons of electromagnetic ra- diation; they are not normally considered to be particles. When a nucleus produces only gamma radiation, the atomic number and mass number of the nucleus do not change. 12. Electron capture occurs when one of the inner-orbital elec- 14. 7 trons is pulled into, and becomes part of, the nucleus. 3Li and 6 3Li; since there is a much larger abundance of 7 average mass number will be closer to 7. (c) neutron; (b) positron; 16. (a) electron; 3Li, the (b) 14 (d) proton 0e; 17. (a) 21 18. (a) 21 8 6Rn; 8 19. (a) 13 7 5Cs S 21 5 6 (c) 21 4Po S 21 8 0e (c) 21 7N; 1e (positron); (b) 3 6Ba; 5At (b) 0 0e 1 13 7 5 0e 1 21 6 8 (c) 1H S 21 0e 1 3 2He; 23. 24. The half-life of a nucleus is the time required for one-half of the original sample of nuclei to decay. A given isotope of an element always has the same half-life, although different iso- topes of the same element may have greatly different half- lives. Nuclei of different elements have different half-lives. 4Ra is the 2 6Ra is the most stable (longest half-life); 2 2 25. 2 8 8 8 8 “hottest” (shortest half-life). 28. After four half-lives, a little over 6 mg 29. For an administered dose of 100 mg, 0.39 mg remains after 2 days. The fraction remaining is 0.39/100 5 0.0039; on a percentage basis, less than 0.4% of the original radioisotope remains. 31. Carbon-14 is produced in the upper atmosphere by the bombardment of nitrogen with neutrons from space: 33. We assume that the concentration of C-14 in the atmosphere is effectively constant. A living organism is constantly re- plenishing C-14 through the processes of either metabolism (sugars ingested in foods contain C-14) or photosynthesis (carbon dioxide contains C-14). When a plant dies, it no longer replenishes itself with C-14 from the atmosphere. As the C-14 undergoes radioactive decay, its amount decreases with time. 35. (a) thyroid gland; (b) heart muscle; (c) bones, heart, liver, lungs; (d) circulatory system 37. fission, fusion, fusion, fission 39. 41. A critical mass of a fissionable material is the amount needed to provide a high enough internal neutron flux to sustain the chain reaction (production of enough neutrons to cause the continuous fission of further material). A sam- ple with less than a critical mass is still radioactive, but can- not sustain a chain reaction. 43. An actual nuclear explosion, of the type produced by a nu- clear weapon, cannot occur in a nuclear reactor because the concentration of the fissionable materials is not sufficient to form a supercritical mass. 46. protons (hydrogen), helium 47. Somatic damage is damage directly to the organism itself, causing nearly immediate sickness or death to the organism. Genetic damage is damage to the genetic machinery of the organism, which will be manifested in future generations of offspring. 49. Gamma rays penetrate long distances, but seldom cause ion- ization of biological molecules. Because they are much heav- ier, although less penetrating, alpha particles ionize biologi- cal molecules very effectively and leave a dense trail of 사이_줌달의대학기초화학_해답.indd 27 2015. 2. 16. 오후 2:03 질문과 문제의 선택된 번호 정답 28 제19장 damage in the organism. Isotopes that decay by releasing al- pha particles can be ingested or breathed into the body, where the damage from the alpha particles will be more acute. C3H8( g) 1 5O2( g) S 3CO2( g) 1 4H2O( g) 21. (a) 2C6H14(l ) 1 19O2( g) S 12CO2( g) 1 14H2O( g); (b) CH4( g) 1 Cl2( g) S CH3Cl(l ) 1 HCl( g); (c) CHCl3(l ) 1 Cl2( g) S CCl4(l ) 1 HCl( g) 23. An alkyne is a hydrocarbon containing a carbon – carbon triple bond. The general formula is CnH2n22. 25. (a) 2-chloro-1-butene; (c) 3-chloro-2-butene; (b) 3-chloro-1-butene; (d) 1-chloro-2-butene 26. The carbon skeletons are 2. A given carbon atom can be attached to a maximum of four other atoms. Carbon atoms have four valence electrons. By making four bonds, carbon atoms exactly complete their va- lence octet. 4. A triple bond represents the sharing of six electrons (three pairs of electrons). The simplest example of an organic mole- cule containing a triple bond is acetylene, 6. 7. (a) CH3OCH2OCH2OCH2OCH2OCH2OCH3; (b) CH3OCH2OCH2OCH2OCH3; (c) CH3OCH2OCH3; (d) CH3OCH2OCH2OCH2OCH2OCH2OCH2OCH3 8. (a) CH3OCH2OCH2OCH2OCH2OCH3; (b) CH3OCH2OCH2OCH2OCH2OCH2OCH2O CH2OCH2OCH2OCH2OCH3; (c) CH3OCH2OCH2OCH3; (d) CH3OCH2OCH2OCH2OCH2OCH2OCH2OCH3 10. A branched alkane contains one or more shorter carbon-atom chains attached to the side of the main (longest) carbon-atom chain. The simplest branched alkane is 2-methylpropane, 12. Structures depend on student choices. 15. Multiple substituents are listed in alphabetical order, disre- garding any prefix. 16. (a) 2,3,4-trimethylpentane; (c) 3,4-dimethylhexane; (b) 2,3-dimethylpentane; (d) 4,5-dimethyloctane 17. 18. 19. Tetraethyl lead was added to gasoline to prevent “knocking” of high-efficiency automobile engines. Its use is being discontinued because of the danger to the environment from the lead in this substance. 20. Combustion reactions are used as a source of heat and light: 40. 27. For benzene, a set of equivalent Lewis structures can be drawn, with each structure differing only in the location of the three double bonds in the ring. Experimentally, benzene does not demonstrate the chemical properties expected for molecules having any double bonds. 29. ortho-: adjacent substituents (1,2-); meta-: two substituents with one unsubstituted carbon atom between them (1,3-); para-: two substituents with two unsubstituted carbon atoms between them (1,4-) 30. (a) anthracene; (b) 1,3,5-trimethylbenzene; (c) 1,4-dinitrobenzene; methylbenzene) (d) 4-bromotoluene (4-bromo-1- (d) ethers (c) ketones; (b) aldehydes; 32. (a) organic acids; 33. Primary alcohols have one hydrocarbon fragment (alkyl group) bonded to the carbon atom where the OOH group is attached. Secondary alcohols have two alkyl groups attached, and tertiary alcohols contain three alkyl groups. Examples are ethanol (primary) 2-propanol (secondary) 2-methyl-2-propanol (tertiary) 35. The yeast necessary for the fermentation process are killed if the concentration of ethanol is greater than 13%. More con- centrated ethanol solutions are most commonly made by distillation. 36. Methanol (CH3OH): starting material for synthesis of acetic acid and many plastics; ethylene glycol (CH2OHOCH2OH): automobile antifreeze; isopropyl alcohol (2-propanol, CH3O CH(OH)OCH3): rubbing alcohol 38. (a) CH3OCH2OCH2OCOOH; (b) CH3OCH2OC(PO)OCH3 사이_줌달의대학기초화학_해답.indd 28 2015. 2. 16. 오후 2:03 줌달의 대학기초화학 29 48. In addition polymerization, the monomer units add to- gether to form the polymer, with no other products. Poly- ethylene and polytetrafluoroethylene (Teflon) are examples. 50. A polyester is formed from the reaction of a dialcohol (two OOH groups) with a diacid (two OCOOH groups). One OOH group of the alcohol forms an ester linkage with one of the OCOOH groups of the acid. The resulting dimer pos- sesses an OOH and a OCOOH group, so the dimer can un- dergo further esterification reactions. Dacron is an example. 42. (a) CH3OCH2OCH2OCHO; (b) CH3OCH2OCOOH; (c) 80. 82. 사이_줌달의대학기초화학_해답.indd 29 2015. 2. 16. 오후 2:03 질문과 문제의 선택된 번호 정답