Solutions to Self Check Exercises

Transcription

1 hapter elf hek Exerise elf hek Exerise. a. hree signifiant figures. he leading zeros (to the left of the 1) do not ount, but the trailing zeros do. b. ive signifiant figures. he one aptive zero and the two trailing zeros all ount.. his is an exat number obtained by ounting the ars. It has an unlimited number of signifiant figures. elf hek Exerise.3 a imiting b ; 6.7 imiting imiting elf hek Exerise elf hek Exerise.5 5 mi h yd 1 mi elf hek Exerise qt qt 3 1 m yd 3 1 km 1000 m 5 36 km h he best way to solve this problem is to onvert 17 K to elsius degrees. o do this we will use the formula 5 K 73. In this ase 5 K o 17 K 5 101, whih is a lower temperature than 75. hus 17 K is older than 75. elf hek Exerise.7 he problem is 41 5?. Using the formula we have ? hat is, A elf hek Exerise. his problem an be stated as 39 5?. Using the formula we have in this ase 5? 5 hat is, elf hek Exerise We obtain the density of the leaner by dividing its mass by its volume. Density 5 mass volume 5.1 g 35. m g /m his density identifies the liquid as isopropyl alohol. hapter 3 elf hek Exerise 3.1 Items (a) and () are physial properties. When the solid gallium melts, it forms liquid gallium. here is no hange in omposition. Items (b) and (d) reflet the ability to hange omposition and are thus hemial properties. tatement (b) means that platinum does not reat with oxygen to form some new substane. tatement (d) means that opper does reat in the air to form a new substane, whih is green. elf hek Exerise 3. a. Milk turns sour beause new substanes are formed. his is a hemial hange. b. Melting the wax is a physial hange (a hange of state). When the wax burns, new substanes are formed. his is a hemial hange. elf hek Exerise 3.3 a. Maple syrup is a homogeneous mixture of sugar and other substanes dispersed uniformly in water. b. elium and oxygen form a homogeneous mixture.. il and vinegar salad dressing is a heterogeneous mixture. (Note the two distint layers the next time you look at a bottle of dressing.) d. ommon salt is a pure substane (sodium hloride), so it always has the same omposition. (Note that other substanes suh as iodine are often added to ommerial preparations of table salt, whih is mostly sodium hloride. hus ommerial table salt is a homogeneous mixture.) 99030_3_ans-se_A0-A6.indd

2 A9 hapter 4 elf hek Exerise 4.1 a. P 4 10 b. U 6. All 3 elf hek Exerise 4. In the symbol 90 3r, the number 3 is the atomi number, whih represents the number of protons in the nuleus of a strontium atom. Beause the atom is neutral overall, it must also have 3 eletrons. he number 90 (the mass number) represents the number of protons plus the number of neutrons. hus the number of neutrons is A Z elf hek Exerise 4.3 he atom 01 0g has 0 protons, 0 eletrons, and neutrons. elf hek Exerise 4.4 he atomi number for phosphorus is 15 and the mass number is hus the symbol for the atom is 3 15P. elf hek Exerise 4.5 Atomi Metal or amily Element ymbol Number Nonmetal Name a. argon Ar 1 nonmetal noble gas b. hlorine l 17 nonmetal halogen. barium Ba 56 metal alkaline earth metal d. esium s 55 metal alkali metal elf hek Exerise 4.6 a. KI b. Mg 3 N Al hapter 5 elf hek Exerise 5.1 a. rubidium oxide b. strontium iodide. potassium sulfide elf hek Exerise 5. a. he ompound PbBr must ontain Pb 1 named lead(ii) to balane the harges of the two Br ions. hus the name is lead(ii) bromide. he ompound PbBr 4 must ontain Pb 41 named lead(iv) to balane the harges of the four Br ions. he name is therefore lead(iv) bromide. b. he ompound e ontains the ion (sulfide) and thus the iron ation present must be e 1, iron(ii). he name is iron(ii) sulfide. he ompound e 3 ontains three ions and two iron ations of unknown harge. We an determine the iron harge from the following: 1? Iron harge harge In this ase,? must represent 3 beause hus e 3 ontains e 31 and, and its name is iron(iii) sulfide.. he ompound AlBr 3 ontains Al 31 and Br. Beause aluminum forms only one ion (Al 31 ), no Roman numeral is required. he name is aluminum bromide. d. he ompound Na ontains Na 1 and ions. he name is sodium sulfide. (Beause sodium forms only Na 1, no Roman numeral is needed.) e. he ompound ol 3 ontains three l ions. hus the obalt ation must be o 31, whih is named obalt(iii) beause obalt is a transition metal and an form more than one type of ation. hus the name of ol 3 is obalt(iii) hloride. elf hek Exerise 5.3 Individual ompound Names Prefixes Name a. l 4 arbon none arbon hloride tetra- tetrahloride b. N nitrogen none nitrogen oxide di- dioxide. I 5 iodine none iodine fluoride penta- pentafluoride elf hek Exerise 5.4 a. silion dioxide. xenon hexafluoride b. dioxygen difluoride elf hek Exerise 5.5 a. hlorine trifluoride d. manganese(iv) oxide b. vanadium(v) fluoride e. magnesium oxide. opper(i) hloride f. water elf hek Exerise 5.6 a. alium hydroxide. potassium permanganate b. sodium phosphate d. ammonium dihromate e. obalt(ii) perhlorate (Perhlorate has a 1 harge, so the ation must be o 1 to balane the two l 4 ions.) f. potassium hlorate g. opper(ii) nitrite (his ompound ontains two N (nitrite) ions and thus must ontain a u 1 ation.) elf hek Exerise 5.7 ompound Name a. Na 3 sodium hydrogen arbonate ontains Na 1 and 3 ; often alled sodium biarbonate (ommon name). b. Ba 4 barium sulfate ontains Ba 1 and 4.. sl 4 esium perhlorate ontains s 1 and l 4. d. Br 5 bromine pentafluoride Both nonmetals (ype III binary). e. NaBr sodium bromide ontains Na 1 and Br (ype I binary). f. Kl potassium hypohlorite ontains K 1 and l _3_ans-se_A0-A6.indd 9

3 A10 g. Zn 3 (P 4 ) zin(ii) phosphate ontains Zn 1 and P 4 3 ; Zn is a transition metal and offiially requires a Roman numeral. owever, beause Zn forms only the Zn 1 ation, the II is usually left out. hus the name of the ompound is usually given as zin phosphate. elf hek Exerise 5. Name hemial ormula a. ammonium sulfate (N 4 ) 4 wo ammonium ions (N 41 ) are required for eah sulfate ion ( 4 ) to ahieve harge balane. b. vanadium(v) fluoride V 5 he ompound ontains V 51 ions and requires five ions for harge balane.. disulfur dihloride l he prefix di- indiates two of eah atom. d. rubidium peroxide Rb Beause rubidium is in Group 1, it forms only 11 ions. hus two Rb 1 ions are needed to balane the harge on the peroxide ion ( ). e. aluminum oxide Al 3 Aluminum forms only 31 ions. wo Al 31 ions are required to balane the harge on three ions. hapter 6 elf hek Exerise 6.1 a. Mg(s) 1 (l) Mg() (s) 1 (g) Note that magnesium (whih is in Group ) always forms the Mg 1 ation and thus requires two anions for a zero net harge. b. Ammonium dihromate ontains the polyatomi ions N 4 1 and r 7 (you should have these memorized). Beause N 4 1 has a 11 harge, two N 4 1 ations are required for eah r 7, with it harge, to give the formula (N 4 ) r 7. hromium(iii) oxide ontains r 31 ions signified by hromium(iii) and (the oxide ion). o ahieve a net harge of zero, the solid must ontain two r 31 ions for every three ions, so the formula is r 3. Nitrogen gas ontains diatomi moleules and is written N (g), and gaseous water is written (g). hus the unbalaned equation for the deomposition of ammonium dihromate is 1N 4 r 7 1s r 3 1s 1 N 1g 1 1g. Gaseous ammonia, N 3 (g), and gaseous oxygen, (g), reat to form nitrogen monoxide gas, N(g), plus gaseous water, (g). he unbalaned equation is N 3 1g 1 1g N1g 1 1g elf hek Exerise 6. tep 1 he reatants are propane, (g), and oxygen, (g); the produts are arbon dioxide, (g), and water, (g). All are in the gaseous state. tep he unbalaned equation for the reation is 1g 1 1g 1g 1 1g tep 3 We start with beause it is the most ompliated moleule. ontains three arbon atoms per moleule, so a oeffiient of 3 is needed for. 1g 1 1g 3 1g 1 1g Also, eah moleule ontains eight hydrogen atoms, so a oeffiient of 4 is required for. 1g 1 1g 3 1g 1 4 1g he final element to be balaned is oxygen. Note that the left side of the equation now has two oxygen atoms, and the right side has ten. We an balane the oxygen by using a oeffiient of 5 for. tep 4 hek: 1g 1 5 1g 3 1g 1 4 1g 3,, 10 3,, 10 Reatant atoms Produt atoms We annot divide all oeffiients by a given integer to give smaller integer oeffiients. elf hek Exerise 6.3 a. N 4 N (s) N (g) 1 (g) (unbalaned) N 4 N (s) N (g) 1 (g) (balaned) b. N(g) N (g) 1 N (g) (unbalaned) 3N(g) N (g) 1 N (g) (balaned). N 3 (l) N (g) 1 (l) 1 (g) (unbalaned) 4N 3 (l) 4N (g) 1 (l) 1 (g) (balaned) hapter 7 elf hek Exerise 7.1 a. he ions present are Ba 1 1aq 1 N 3 1aq 1 Na 1 1aq 1 l 1aq Ions in Ba(N 3 ) (aq) Ions in Nal(aq) Exhanging the anions gives the possible solid produts Bal and NaN 3. Using able 7.1, we see that both substanes are very soluble (rules 1,, and 3). hus no solid forms. b. he ions present in the mixed solution before any reation ours are Na 1 1aq 1 1aq 1 u 1 1aq 1 N 3 1aq Ions in Na (aq) Ions in u(n 3 ) (aq) Exhanging the anions gives the possible solid produts u and NaN 3. Aording to rules 1 and in able 7.1, NaN 3 is soluble, and by rule 6, u should be insoluble. hus u will preipitate. he balaned equation is Na 1aq 1 u 1N 3 1aq u1s 1 NaN 3 1aq 99030_3_ans-se_A0-A6.indd 10

4 A11. he ions present are N 41 1aq 1 l 1aq 1 Pb 1 1aq 1 N 3 1aq Ions in N 4 l(aq) Ions in Pb(N 3 ) (aq) Exhanging the anions gives the possible solid produts N 4 N 3 and Pbl. N 4 N 3 is soluble (rules 1 and ) and Pbl is insoluble (rule 3). hus Pbl will preipitate. he balaned equation is N 4 l 1aq 1 Pb 1N 3 1aq Pbl 1s 1 N 4 N 3 1aq elf hek Exerise 7. a. Moleular equation: Na 1aq 1 u 1N 3 1aq u 1s 1 NaN 3 1aq omplete ioni equation: Na 1 1aq 1 1aq 1 u 1 1aq 1 N 3 1aq u1s 1 Na 1 1aq 1 N 3 1aq Net ioni equation: 1aq 1 u 1 1aq u 1s b. Moleular equation: N 4 l 1aq 1 Pb 1N 3 1aq Pbl 1s 1 N 4 N 3 1aq omplete ioni equation: N 41 1aq 1 l 1aq 1 Pb 1 1aq 1 N 3 1aq Pbl 1s 1 N 41 1aq 1 N 3 1aq Net ioni equation: l 1aq 1 Pb 1 1aq Pbl 1s elf hek Exerise 7.3 a. he ompound NaBr ontains the ions Na 1 and Br. hus eah sodium atom loses one eletron (Na Na 1 1 e ), and eah bromine atom gains one eletron (Br 1 e Br ). Na + Na + Br Br (Na + Br ) + (Na + Br ) e e b. he ompound a ontains the a 1 and ions. hus eah alium atom loses two eletrons (a a 1 1 e ), and eah oxygen atom gains two eletrons ( 1 e ). a + a + (a + ) + (a + ) e e elf hek Exerise 7.4 a. oxidation redution reation; ombustion reation b. synthesis reation; oxidation redution reation; ombustion reation. synthesis reation; oxidation redution reation d. deomposition reation; oxidation redution reation e. preipitation reation (and double displaement) f. synthesis reation; oxidation redution reation g. aid base reation (and double displaement) h. ombustion reation; oxidation redution reation hapter elf hek Exerise.1 he average mass of nitrogen is amu. he appropriate equivalene statement is 1 N atom amu, whih yields the onversion fator we need: 3 N atoms 3 (exat) elf hek Exerise amu N atom 5 3. amu he average mass of oxygen is amu, whih gives the equivalene statement 1 atom amu. he number of oxygen atoms present is amu 3 1 atom atoms amu elf hek Exerise.3 Note that the sample of atoms of hromium is less than 1 mole ( atoms) of hromium. What fration of a mole it represents an be determined as follows: atoms r 3 1 mol r atoms r mol r Beause the mass of 1 mole of hromium atoms is 5.00 g, the mass of atoms an be determined as follows: mol r 3 elf hek Exerise g r 1 mol r g r Eah moleule of 3 l ontains two arbon atoms, three hydrogen atoms, and one hlorine atom, so 1 mole of 3 l moleules ontains moles of atoms, 3 moles of atoms, and 1 mole of l atoms. Mass of mol atoms: g Mass of 3 mol atoms: g Mass of 1 mol l atoms: g g he molar mass of 3 l is 6.49 g (rounding to the orret number of signifiant figures). elf hek Exerise.5 he formula for sodium sulfate is Na 4. ne mole of Na 4 ontains moles of sodium ions and 1 mole of sulfate ions. 1 mole of Na 4 1 mole of Na 1 4 Na 1 mol Na 1 1 mol 4 Mass of mol Na g Mass of 1 mol (16.00) g Mass of 1 mol Na g he molar mass for sodium sulfate is g _3_ans-se_A0-A6.indd 11

5 A1 A sample of sodium sulfate with a mass of g represents more than 1 mol. (ompare g to the molar mass of Na 4.) We alulate the number of moles of Na 4 present in g as follows: g Na mol Na g Na mol Na 4 elf hek Exerise.6 irst we must ompute the mass of 1 mole of 4 moleules (the molar mass). Beause 1 mole of 4 ontains moles of atoms and 4 moles of atoms, we have mol g mol 4 mol g mol g g Mass of 1 mole of 4 : g 5 molar mass Using the equivalene statement g mole 4, we alulate the moles of 4 units in 135 g of eflon. 135 g 4 units 3 1 mol g mol 4 units Next, using the equivalene statement 1 mol units, we alulate the number of 4 units in 135 mol of eflon. 135 mol units 1 mol elf hek Exerise units he molar mass of peniillin is omputed as follows: : 14 mol : 0 mol N: mol : 1 mol : 4 mol g mol g g mol g g mol 5.0 g g mol g g mol g Mass of 1 mole of 14 0 N g g Mass perent of 5 Mass perent of 5 Mass perent of N 5 Mass perent of g 31.4 g 14 0 N % % 0.16 g 31.4 g 14 0 N % %.0 g N 31.4 g 14 0 N % 5.969% 3.07 g 31.4 g 14 0 N % % Mass perent of g 31.4 g 14 0 N % % hek: he perentages add up to 99.99%. elf hek Exerise. tep g lead and g hlorine 1 mol Pb tep 0.64 g Pb mol Pb 07. g Pb 1 mol l g l mol l g l mol Pb tep mol Pb mol l mol l hese numbers are very lose to integers, so step 4 is unneessary. he empirial formula is Pbl. elf hek Exerise.9 tep g, g N, g, and.133 g tep g 3 1 mol mol 1.01 g tep g N 3 1 mol N mol N g N g 3 1 mol mol 1.00 g.133 g 3 1 mol mol g mol mol N mol mol mol mol N mol mol he empirial formula is N 3. elf hek Exerise.10 tep 1 In g of Nylon-6 the masses of elements present are 63.6 g, 1.3 g N, 9.0 g, and g. tep 63.6 g 3 1 mol mol 1.01 g 1.3 g N 3 1 mol N mol N g N 9.0 g 3 1 mol mol 1.00 g g 3 1 mol mol g 99030_3_ans-se_A0-A6.indd 1

6 A13 tep mol mol N mol mol mol mol N mol mol he empirial formula for Nylon-6 is 6 N 11. elf hek Exerise.11 tep 1 irst we onvert the mass perents to mass in grams. In g of the ompound, there are g of hlorine, 4.7 g of arbon, and 4.07 g of hydrogen. tep We use these masses to ompute the moles of atoms present g l g 3 1 mol 1.01 g 4.07 g 3 1 mol 1.00 g 1 mol l 5.01 mol l g l 5.01 mol mol tep 3 Dividing eah mole value by.01 (the smallest number of moles present), we obtain the empirial formula l. o determine the moleular formula, we must ompare the empirial formula mass to the molar mass. he empirial formula mass is l: : 1.01 : l : empirial formula mass he molar mass is known to be We know that Molar mass 5 n 3 1empirial formula mass o we an obtain the value of n as follows: Molar mass Empirial formula mass Moleular formula 5 1l 5 l 4 his substane is omposed of moleules with the formula l 4. we derive the equivalene statement 1 mol 5 3 mol he appropriate onversion fator (moles of must anel) is 3 mol y1 mol, and the alulation is hus we an say 4.30 mol 3 3 mol 1 mol mol elf hek Exerise mol yields 1.9 mol he problem an be skethed as follows: 1g 1 5 1g 3 1g 1 4 1g 96.1 g Grams of Use molar mass Use molar mass of of Moles of Use mole Moles of ratio between 3 and We have already done the first step in Example g 1 mol g.1 mol o find out how many moles of an be produed from.1 moles of, we see from the balaned equation that 3 moles of is produed for eah mole of reated. he mole ratio we need is 3 mol y1 mol. he onversion is therefore.1 mol 3 3 mol 1 mol mol Next, using the molar mass of, whih is g, we alulate the mass of produed mol g 1 mol 5 g he sequene of steps we took to find the mass of arbon dioxide produed from 96.1 g of propane is summarized in the following diagram. hapter g 1 mol g.1 mol elf hek Exerise 9.1 he problem an be stated as follows:.1 mol 3 mol 1 mol 6.54 mol 4.30 mol? mol yields 6.54 mol g 1 mol g rom the balaned equation Mass Moles 1g 1 5 1g 3 1g 1 4 1g 99030_3_ans-se_A0-A6.indd 13

7 A14 elf hek Exerise 9.3 We sketh the problem as follows: 1g 1 5 1g 3 1g 1 4 1g 96.1 g 1 mol g Moles of hen we do the alulations g.1 mol.7 mol 4 mol 1 mol 1 mol g 4 mol 1 mol 1.0 g mol Grams of 1.0 g mol Moles of.1 mol.7 mol 157 g herefore, 157 g of is produed from 96.1 g. elf hek Exerise 9.4 a. We first write the balaned equation. i 1s 1 41aq i 4 1g 1 1l he map of the steps required is i 1s 1 41aq i 4 1g 1 1l 5.6 g i Use molar mass of i Moles of Use mole i ratio between and i We onvert 5.6 g of i to moles as follows: Grams of Use molar mass of Moles of 5.6 g i 3 1 mol i g i mol i Using the balaned equation, we obtain the appropriate mole ratio and onvert to moles of mol i 3 4 mol 1 mol i mol inally, we alulate the mass of by using its molar mass mol g mol g b. he map for this problem is i 1s 1 4 1aq i 4 1g 1 1l 5.6 g i Use molar mass of i Moles of Use mole i ratio between and i Grams of Use molar mass of Moles of We have already aomplished the first onversion in part a. Using the balaned equation, we obtain moles of as follows: mol i 3 mol 1 mol i mol he mass of water formed is mol g mol elf hek Exerise g In this problem, we know the mass of the produt to be formed by the reation 1g 1 1g 3 1l and we want to find the masses of reatants needed. he proedure is the same one we have been following. We must first onvert the mass of 3 to moles, then use the balaned equation to obtain moles of and needed, and then onvert these moles to masses. Using the molar mass of 3 (3.04 g/mol), we onvert to moles of 3. irst we onvert kilograms to grams. 6.0 kg g kg g 3 Next we onvert g 3 to moles of 3, using the onversion fator 1 mol 3 y3.04 g g mol g mol 3 hen we have two questions to answer:? mol of? mol of required to produe required to produe mol mol 3 o answer these questions, we use the balaned equation 1g 1 1g 3 1l to obtain mole ratios between the reatants and the produts. In the balaned equation the oeffiients for both and 3 are 1, so we an write the equivalene statement 1 mol 5 1 mol _3_ans-se_A0-A6.indd 14

8 A15 Using the mole ratio 1 mol /1 mol 3, we an now onvert from moles of 3 to moles of mol mol 1 mol mol o alulate the moles of required, we onstrut the equivalene statement between 3 and, using the oeffiients in the balaned equation. mol 5 1 mol 3 Using the mole ratio mol /1 mol 3, we an onvert moles of 3 to moles of mol 3 3 mol 1 mol mol We now have the moles of reatants required to produe 6.0 kg of 3. ine we need the masses of reatants, we must use the molar masses to onvert from moles to mass mol 3.01 g 1 mol g mol g 1 mol herefore, we need g to reat with g to form g (6.0 kg) of 3. his whole proess is mapped in the following diagram g 3 1 mol g mol 3-1 mol 1 mol 3 mol 1 mol mol mol.016 g.01 g 1 mol 1 mol g g elf hek Exerise 9.6 tep 1 he balaned equation for the reation is 6i 1s 1 N 1g i 3 N 1s tep o determine the limiting reatant, we must onvert the masses of lithium (atomi mass g) and nitrogen (molar mass 5.0 g) to moles g i 3 1 mol i 5.07 mol i g i 56.0 g N 3 1 mol N.0 g N 5.00 mol N tep 3 Using the mole ratio from the balaned equation, we an alulate the moles of lithium required to reat with.00 moles of nitrogen..00 mol N 3 6 mol i 1 mol N mol i herefore, 1.0 moles of i is required to reat with.00 moles of N. owever, we have only.07 mol of i, so lithium is limiting. It will be onsumed before the nitrogen runs out. tep 4 Beause lithium is the limiting reatant, we must use the.07 moles of i to determine how many moles of i 3 N an be formed..07 mol i 3 mol i 3N 6 mol i 5.69 mol i 3 N tep 5 We an now use the molar mass of i 3 N (34.3 g) to alulate the mass of i 3 N formed. elf hek Exerise mol i 3 N g i 3N 1 mol i 3 N g i 3N a. tep 1 he balaned equation is il 4 1g 1 1g i 1s 1 l 1g tep he numbers of moles of reatants are g il mol il g il mol il g 3 1 mol 3.00 g mol tep 3 In the balaned equation both il 4 and have oeffiients of 1, so and 1 mol il mol mol il mol 1 mol il mol required We have moles of, so the is in exess and the il 4 is limiting. his makes sense. il 4 and reat in a 1:1 mole ratio, so the il 4 is limiting beause fewer moles of il 4 are present than moles of. tep 4 We will now use the moles of il 4 (the limiting reatant) to determine the moles of i that would form if the reation produed 100% of the expeted yield (the theoretial yield) mol il mol i 1 mol il mol i he mass of i expeted for 100% yield is mol i g i 1 mol i g i his amount represents the theoretial yield _3_ans-se_A0-A6.indd 15

9 A16 b. Beause the reation is said to give only a 75.0% yield of i, we use the definition of perent yield, to write the equation Atual yield 3 100% 5 % yield heoretial yield Atual yield g i 3 100% % yield We now want to solve for the atual yield. irst we divide both sides by 100%. Atual yield g i 3 100% 100% hen we multiply both sides by g i g i 3 Atual yield g i g i Atual yield g i g i hus g of i (s) is atually obtained in this reation. hapter 10 elf hek Exerise 10.1 he onversion fator needed is elf hek Exerise al, and the onversion is 4.14 J.4 J 3 1 al al 4.14 J We know that it takes 4.14 J of energy to hange the temperature of eah gram of water by 1, so we must multiply 4.14 by the mass of water (454 g) and the temperature hange ( ) elf hek Exerise 10.3 J g g J rom able 10.1, the speifi heat apaity for solid gold is 0.13 J/g. Beause it takes 0.13 J to hange the temperature of one gram of gold by one elsius degree, we must multiply 0.13 by the sample size (5.63 g) and the hange in temperature ( ) J g J g We an hange this energy to units of alories as follows:.1 J 3 1 al al 4.14 J elf hek Exerise 10.4 able 10.1 lists the speifi heat apaities of several metals. We want to alulate the speifi heat apaity (s) for this metal and then use able 10.1 to identify the metal. Using the equation Q 5 s 3 m 3 D we an solve for s by dividing both sides by m (the mass of the sample) and by D: In this ase, so Q m 3 D 5 s Q 5 energy (heat) required J m 5. g D 5 temperature hange s 5 Q m 3 D J 1. g J /g able 10.1 shows that silver has a speifi heat apaity of 0.4 J/g. he metal is silver. elf hek Exerise 10.5 We are told that 165 kj of energy is released when 4 moles of e reats. We first need to determine what number of moles 1.00 g e represents g e 3 1 mol 55.5 g mol e mol e kj kj 4 mol e hus 7.39 kj of energy (as heat) is released when 1.00 g of iron reats. elf hek Exerise 10.6 Noting the reatants and produts in the desired reation 1s 1 1g 1g We need to reverse the seond equation and multiply it by 1. his reverses the sign and uts the amount of energy by a fator of. or g 1g kj 1g 4 D 5 3 1g 1g 1 1 1g Now we add this reation to the first reation. (s) 1 3 (g) 3 (g) D kj D kj 3 (g) (g) 1 1 (g) D kj (s) 1 (g) (g) D kj 99030_3_ans-se_A0-A6.indd 16

10 A17 hapter 11 elf hek Exerise 11.1 a. irular pathways for eletrons in the Bohr model. b. hree-dimensional probability maps that represent the likelihood that the eletron will oupy a given point in spae.. he surfae that ontains 90% of the total eletron probability. d. A set of orbitals of a given type of orbital within a prinipal energy level. or example, there are three sublevels in prinipal energy level 3 (s, p, d). elf hek Exerise 11. Eletron Element onfiguration rbital Diagram 1s s p 3s 3p Al 1s s p 6 3s 3p 1 [Ne]3s 3p 1 i [Ne]3s 3p P [Ne]3s 3p 3 [Ne]3s 3p 4 l [Ne]3s 3p 5 Ar [Ne]3s 3p 6 elf hek Exerise 11.3 : 1s s p 5 or [e]s p 5 i: 1s s p 6 3s 3p or [Ne]3s 3p s: 1s s p 6 3s 3p 6 4s 3d 10 4p 6 5s 4d 10 5p 6 6s 1 or [Xe]6s 1 Pb: 1s s p 6 3s 3p 6 4s 3d 10 4p 6 5s 4d 10 5p 6 6s 4f 14 5d 10 6p or [Xe]6s 4f 14 5d 10 6p I: 1s s p 6 3s 3p 6 4s 3d 10 4p 6 5s 4d 10 5p 5 or [Kr]5s 4d 10 5p 5 ilion (i): In Group 4 and Period 3, it is the seond of the 3p elements. he onfiguration is 1s s p 6 3s 3p, or [Ne]3s 3p. esium (s): In Group 1 and Period 6, it is the first of the 6s elements. he onfiguration is 1s s p 6 3s 3p 6 4s 3d 10 4p 6 5s 4d 10 5p 6 6s 1, or [Xe]6s 1. ead (Pb): In Group 4 and Period 6, it is the seond of the 6p elements. he onfiguration is [Xe]6s 4f 14 5d 10 6p. Iodine (I): In Group 7 and Period 5, it is the fifth of the 5p elements. he onfiguration is [Kr]5s 4d 10 5p 5. hapter 1 elf hek Exerise 1.1 Using the eletronegativity values given in ig. 1.3, we hoose the bond in whih the atoms exhibit the largest differene in eletronegativity. (Eletronegativity values are shown in parentheses.) a.. P.. N (.1)(.5) (.1)(.1) (.5)(3.5) (3.0)(3.5) b. I. d. N. i (3.5)(.5) (3.5)(4.0) (3.0)(.1) (1.)(.1) elf hek Exerise 1. has one eletron, and l has seven valene eletrons. his gives a total of eight valene eletrons. We first draw in the bonding pair: il, whih ould be drawn as : l We have six eletrons yet to plae. he already has two eletrons, so we plae three lone pairs around the hlorine to satisfy the otet rule. elf hek Exerise 1.3 l or l tep 1 3 : 3(6) 5 1 valene eletrons tep tep 3 and his moleule shows resonane (it has two valid ewis strutures). elf hek Exerise 1.4 ee able A.1 for the answer to elf hek Exerise 1.4. elf hek Exerise 1.5 a. N 4 1 he ewis struture is N (ee elf hek Exerise 1.4.) here are four pairs of eletrons around the nitrogen. his requires a tetrahedral 1 arrangement of eletron pairs. he N 4 ion has a tetrahedral moleular struture (row 3 in able 1.4), beause all eletron pairs are shared. b. 4 he ewis struture is (ee elf hek Exerise 1.4.) he four eletron pairs around the sulfur require a tetrahedral arrangement. he 4 has a tetrahedral moleular struture (row 3 in able 1.4).. N 3 he ewis struture is N (ee elf hek Exerise 1.4.) he four pairs of eletrons on the nitrogen require a tetrahedral arrangement. In this ase, only three of the pairs are shared with the fluorine atoms, leaving one lone pair. hus the moleular struture is a trigonal pyramid (row 4 in able 1.4). d. he ewis struture is (ee elf hek Exerise 1.4.) he four pairs of eletrons around the sulfur require a tetrahedral arrangement. In this ase, two pairs are shared with hydrogen atoms, leaving two lone pairs. hus the moleular struture is bent or V-shaped (row 5 in able 1.4). e. l 3 he ewis struture is l (ee elf hek Exerise 1.4.) he four pairs of eletrons require a tetrahedral arrangement. In this ase, three pairs are shared with oxygen atoms, leaving one lone pair. hus the moleular struture is a trigonal pyramid (row 4 in able 1.4) _3_ans-se_A0-A6.indd 17

11 A1 Moleule or Ion otal Valene Eletrons Draw ingle Bonds alulate Number of Eletrons Remaining Use Remaining Eletrons to Ahieve Noble Gas onfigurations Atoms hek Eletrons a. N (7) = 6 N 6 6 = 0 N N b. (6) = 1 1 = = = d. P (1) = P 6 = P P e. (1) + 6 = 4 = 4 f (6) + = 3 3 = 4 + g. N (1) 1 = N = 0 N N h. l (6) + 1 = 6 l 6 6 = 0 l l i. 6 + (6) = = 14 and able A.1 Answer to elf hek Exerise 1.4 f. Be he ewis struture is Be he two eletron pairs on beryllium require a linear arrangement. Beause both pairs are shared by fluorine atoms, the moleular struture is also linear (row 1 in able 1.4). hapter 13 elf hek Exerise 13.1 We know that atm mm g. o 55 mm g 3 elf hek Exerise atm atm mm g Initial onditions inal onditions P torr P 5 75 torr V V 5? olving Boyle s law (P 1 V 1 5 P V ) for V gives V 5 V 1 3 P 1 P torr 75 torr 5 1. Note that the volume dereased, as the inrease in pressure led us to expet. elf hek Exerise 13.3 Beause the temperature of the gas inside the bubble dereases (at onstant pressure), the bubble gets smaller. he onditions are Initial onditions K V m 3 inal onditions K V 5? 99030_3_ans-se_A0-A6.indd 1

12 A19 olving harles s law, for V gives V V V 5 V m K 5 m K elf hek Exerise 13.4 Beause the temperature and pressure of the two samples are the same, we an use Avogadro s law in the form V 1 n 1 5 V n he following information is given: ample 1 ample V V n mol n 5? We an now solve Avogadro s law for the value of n (the moles of N in sample ): n 5 n 1 3 V mol mol V ere n is smaller than n 1, whih makes sense in view of the fat that V is smaller than V 1. Note: We isolate n from Avogadro s law as given above by multiplying both sides of the equation by n and then by n 1 /V 1, to give n 5 n 1 3 V /V 1. elf hek Exerise 13.5 an 3 n 1 V 1 b 5 an V 1 n 3 n 1 V b 1 V 1 n We are given the following information: P atm V n mol We solve for by dividing both sides of the ideal gas law by nr: to give PV nr 5 nr nr 5 PV nr atm mol a0.006 atm K mol b 5 99 K he temperature of the helium is 99 K, or elf hek Exerise 13.6 We are given the following information about the radon sample: n mol V K P 5? We solve the ideal gas law (PV 5 nr) for P by dividing both sides of the equation by V: P 5 nr V mol a0.006 atm K mol b 1306 K 5 1. atm elf hek Exerise o solve this problem, we take the ideal gas law and separate those quantities that hange from those that remain onstant (on opposite sides of the equation). In this ase, volume and temperature hange, and number of moles and pressure (and, of ourse, R) remain onstant. o PV 5 nr beomes V/ 5 nr/p, whih leads to ombining these gives We are given hus V nr P V nr P Initial onditions K V inal onditions K V 5? V 5 V and V 5 nr P 5 V or V V 1359 K K hek: Is the answer sensible? In this ase, the temperature was inreased (at onstant pressure), so the volume should inrease. he answer makes sense. Note that this problem ould be desribed as a harles s law problem. he real advantage of using the ideal gas law is that you need to remember only one equation to do virtually any problem involving gases. elf hek Exerise 13. We are given the following information: Initial onditions P atm K V inal onditions P atm K V 5? In this ase, the number of moles remains onstant. hus we an say P 1 V 1 5 nr and P V 5 nr _3_ans-se_A0-A6.indd 19

13 A0 or olving for V gives V 5 V P 1 1 P 5.01 elf hek Exerise 13.9 P 1 V P V a 39 K atm b a 6 K 1.1 atm b As usual when dealing with gases, we an use the ideal gas equation PV 5 nr. irst onsider the information given: P atm 5 P total V K Given this information, we an alulate the number of moles of gas in the mixture: n total 5 n N 1 n. olving for n in the ideal gas equation gives n total 5 P totalv R atm 1.0 a0.006 atm b 19 K K mol We also know that mole of N is present. Beause n total 5 n N 1 n mol (0.050 mol) we an alulate the moles of present mol 1 n mol mol n mol mol mol Now that we know the moles of oxygen present, we an alulate the partial pressure of oxygen from the ideal gas equation. P 5 n R V atm mol a0.006 atm b 19 K K mol.0 Although it is not requested, note that the partial pressure of the N must be 0.6 atm, beause elf hek Exerise atm atm atm P N P P total he volume is 0.500, the temperature is 5 (or K), and the total pressure is given as atm. f this total pressure, 4 torr is due to the water vapor. We an alulate the partial pressure of the beause we know that P total 5 P 1 P atm 4 torr Before we arry out the alulation, however, we must onvert the pressures to the same units. onverting P to atmospheres gives hus and 4 torr atm atm torr P total 5 P 1 P atm 5 P atm P atm 0.03 atm atm Now that we know the partial pressure of the hydrogen gas, we an use the ideal gas equation to alulate the moles of. n 5 P V R atm a0.006 atm b 19 K K mol mol mol he sample of gas ontains mole of, whih exerts a partial pressure of 0.91 atm. elf hek Exerise We will solve this problem by taking the following steps: Grams of zin Moles of zin Moles of Volume of tep 1 Using the atomi mass of zin (65.3), we alulate the moles of zin in 6.5 g. 6.5 g Zn 3 1 mol Zn mol Zn 65.3 g Zn tep Using the balaned equation, we next alulate the moles of produed mol Zn 3 1 mol 1 mol Zn mol tep 3 Now that we know the moles of, we an ompute the volume of by using the ideal gas law, where P atm V 5? n mol R atm/k mol K V 5 nr P of elf hek Exerise mol a0.006 atm b 19 K K mol 1.50 atm Although there are several possible ways to do this problem, the most onvenient method involves using the molar volume at P. irst we use the ideal gas equation to alulate the moles of N 3 present: n 5 PV R 99030_3_ans-se_A0-A6.indd 0 9/19/13 6:0 PM

14 A1 where P atm, V , and K. n atm a0.006 atm b 19 K K mol mol We know that at P eah mole of gas oupies.4. herefore, 3.07 mol has the volume 3.07 mol mol 5 6. he volume of the ammonia at P is 6.. hapter 14 elf hek Exerise 14.1 Energy to melt the ie: 15 g 3 1 mol 1 g mol 0.3 mol kj kj mol Energy to heat the water from 0 to 100 : 4.1 J 3 15 g J g 6300 J 3 1 kj kj 1000 J Energy to vaporize the water at 100 : otal energy required: 0.3 mol elf hek Exerise 14. kj 5 34 kj mol 5.0 kj kj 1 34 kj 5 45 kj a. ontains 3 moleules a moleular solid. b. ontains Ba 1 and ions an ioni solid.. ontains Au atoms an atomi solid. hapter 15 elf hek Exerise 15.1 Mass perent 5 mass of solute mass of solution 3 100% or this sample, the mass of solution is 135 g and the mass of the solute is 4.73 g, so Mass perent g solute 135 g solution 3 100% % elf hek Exerise 15. Using the definition of mass perent, we have Mass of solute Mass of solution 5 grams of solute 3 100% % grams of solute 1 grams of solvent here are 45 grams of solute (formaldehyde). ubstituting, we have 45 g 3 100% % 45 g 1 grams of solvent We must now solve for grams of solvent (water). his will take some patiene, but we an do it if we proeed step by step. irst we divide both sides by 100%. 45 g 45 g 1 grams of solvent 3 100% 100% % 100% Now we have 45 g 45 g 1 grams of solvent Next we multiply both sides by (45 g 1 grams of solvent). 145 g 1 grams of solvent 3 his gives 45 g 45 g 1 grams of solvent g 1 grams of solvent 45 g g 1 grams of solvent arrying out the multipliation gives 45 g g grams of solvent Now we subtrat 170. g from both sides, 45 g 170. g g 170. g grams of solvent 55 g grams of solvent and divide both sides by We finally have the answer: 55 g grams of solvent g 5 63 g 5 grams of solvent elf hek Exerise mass of water needed he moles of ethanol an be obtained from its molar mass (46.1) g mol g mol 5 Volume in liters m m Molarity of 5 5 moles of 5 liters of solution mol M 99030_3_ans-se_A0-A6.indd 1 9/19/13 6:0 PM

15 A elf hek Exerise 15.4 When Na 3 and Al ( 4 ) 3 dissolve in water, they produe ions as follows: ( l ) Na 3 1s n Na 1 1aq 1 3 1aq ( l ) Al s n Al 31 1aq aq herefore, in a 0.10 M Na 3 solution, the onentration of Na 1 ions is M M and the onentration of 3 ions is 0.10 M. In a M Al ( 4 ) 3 solution, the onentration of Al 31 ions is M M and the onentration of 4 ions is M M. elf hek Exerise 15.5 When solid All 3 dissolves, it produes ions as follows: ( l ) All 3 1s n Al 31 1aq 1 3l 1aq so a M All 3 solution ontains M Al 31 ions and M l ions. o alulate the moles of l ions in 1.75 of the M All 3 solution, we must multiply the volume by the molarity solution M l elf hek Exerise solution mol l solution mol l mol l We must first determine the number of moles of formaldehyde in.5 of 1.3 M formalin. Remember that volume of solution (in liters) times molarity gives moles of solute. In this ase, the volume of solution is.5 and the molarity is 1.3 moles of per liter of solution..5 solution mol solution 5 31 mol Next, using the molar mass of (30.0 g), we onvert 31 moles of to grams. 31 mol g 1 mol g herefore,.5 of 1.3 M formalin ontains g of formaldehyde. We must weigh out 930 g of formaldehyde and dissolve it in enough water to make.5 of solution. elf hek Exerise 15.7 We are given the following information: M mol M mol V 1 5? 1what we need to find V Using the fat that the moles of solute do not hange upon dilution, we know that M 1 3 V 1 5 M 3 V olving for V 1 by dividing both sides by M 1 gives and elf hek Exerise 15. V 1 5 M 0.5 mol 3 V M 1 1 mol V m tep 1 When the aqueous solutions of Na 4 (ontaining Na 1 and 4 ions) and Pb(N 3 ) (ontaining Pb 1 and N 3 ions) are mixed, solid Pb 4 is formed. Pb 1 (aq) 1 4 (aq) n Pb 4 (s) tep We must first determine whether Pb 1 or 4 is the limiting reatant by alulating the moles of Pb 1 and 4 ions present. Beause M Pb(N 3 ) ontains M Pb 1 ions, we an alulate the moles of Pb 1 ions in 1.5 of this solution as follows: mol Pb mol Pb 1 he M Na 4 solution ontains M 4 ions, and the number of moles of 4 ions in.00 of this solution is mol mol 4 tep 3 Pb 1 and 4 reat in a 1:1 ratio, so the amount of 4 ions is limiting beause 4 is present in the smaller number of moles. tep 4 he Pb 1 ions are present in exess, and only mole of solid Pb 4 will be formed. tep 5 We alulate the mass of Pb 4 by using the molar mass of Pb 4 (303.3 g) mol Pb g Pb 4 1 mol Pb g Pb 4 elf hek Exerise 15.9 tep 1 Beause nitri aid is a strong aid, the nitri aid solution ontains 1 and N 3 ions. he K solution ontains K 1 and ions. When these solutions are mixed, the 1 and reat to form water. 1 1aq 1 1aq 1l tep he number of moles of present in 15 m of M K is 15 m mol mol 1000 m tep 3 1 and reat in a 1:1 ratio, so we need mole of 1 from the M N 3. tep mole of requires mole of 1 to form mole of. herefore, V mol mol _3_ans-se_A0-A6.indd 9/19/13 6:0 PM

16 A3 where V represents the volume in liters of M N 3 required. olving for V, we have V mol mol elf hek Exerise m 5 63 m rom the definition of normality, N 5 equiv/, we need to alulate (1) the equivalents of K and () the volume of the solution in liters. o find the number of equivalents, we use the equivalent weight of K, whih is 56.1 g (see able 15.). 3.6 g K 3 1 equiv K 56.1 g K Next we onvert the volume to liters equiv K 755 m m inally, we substitute these values into the equation that defines normality. Normality 5 equiv elf hek Exerise equiv N o solve this problem, we use the relationship where We solve the equation N aid 3 V aid 5 N base 3 V base N aid equiv V aid 5? N base equiv V base N aid 3 V aid 5 N base 3 V base for V aid by dividing both sides by N aid. N aid 3 V aid N aid V aid 5 N base 3 V base N aid 5 V aid N base 3 V base N aid a0.0 equiv b equiv herefore, 0.40 of 0.50 N 4 is required to neutralize 0.50 of 0.0 N K. hapter 16 elf hek Exerise 16.1 he onjugate aid base pairs are and, 3 1 Base onjugate aid 3, 3 Aid onjugate base he members of both pairs differ by one 1. elf hek Exerise 16. Beause [ 1 ][ ] , we an solve for [ 1 ] M his solution is basi: [ ] M is greater than [ 1 ] M. elf hek Exerise 16.3 a. Beause [ 1 ] M, we get p beause p 5 log[ 1 ] 5 log[ ] b. Beause [ ] M, we an find [ 1 ] from the K w expression K w M p 5 log[ 1 ] 5 log[ ] elf hek Exerise 16.4 p 1 p p p p elf hek Exerise 16.5 tep 1 p tep p tep 3 inv log [ 1 ] M elf hek Exerise 16.6 tep 1 p tep p tep 3 inv log [ ] M elf hek Exerise 16.7 Beause l is a strong aid, it is ompletely dissoiated: M l M 1 and M l so M. p 5 log _3_ans-se_A0-A6.indd 3 9/19/13 6:0 PM

17 A4 hapter 17 elf hek Exerise 17.1 Applying the law of hemial equilibrium gives elf hek Exerise 17. oeffiient oeffiient of N of K 5 3N N d oeffiient of oeffiient of N 3 a. K 5 [ ] 3 he solids are not inluded. b. K 5 [N ][ ] he solid is not inluded. Water is gaseous in this reation, so it is inluded.. K he solids are not inluded. d. K elf hek Exerise 17.3 Water and 4 are pure liquids and so are not inluded. When rain is imminent, the onentration of water vapor in the air inreases. his shifts the equilibrium to the right, forming ol? 6 (s), whih is pink. elf hek Exerise 17.4 a. No hange. Both sides of the equation ontain the same number of gaseous omponents. he system annot hange its pressure by shifting its equilibrium position. b. hifts to the left. he system an inrease the number of gaseous omponents present, and so inrease the pressure, by shifting to the left.. hifts to the right to inrease the number of gaseous omponents and thus its pressure. elf hek Exerise 17.5 a. hifts to the right away from added. b. hifts to the right to replae removed 3.. hifts to the right to derease its pressure. d. hifts to the right. Energy is a produt in this ase, so a derease in temperature favors the forward reation (whih produes energy). elf hek Exerise 17.6 a. Ba 4 (s) Ba 1 (aq) 1 4 (aq); K sp 5 [Ba 1 ][ 4 ] b. e() 3 (s) e 31 (aq) 1 3 (aq); K sp 5 [e 31 ][ ] 3. Ag 3 P 4 (s) 3Ag 1 (aq) 1 P 4 3 (aq); K sp 5 [Ag 1 ] 3 [P 4 3 ] elf hek Exerise 17.7 ( ) K sp elf hek Exerise 17. Pbr 4 (s) Pb 1 (aq) 1 r 4 (aq) K sp 5 [Pb 1 ][r 4 ] [Pb 1 ] 5 x [r 4 ] 5 x K sp x x 5 [Pb 1 ] 5 [r 4 ] hapter 1 elf hek Exerise 1.1 a. u ontains u 1 and ions, so opper is oxidized (u u 1 1 e ) and oxygen is redued ( 1 e ). b. s ontains s 1 and ions. hus esium is oxidized (s s 1 1 e ) and fluorine is redued ( 1 e ). elf hek Exerise 1. a. 3 We assign oxygen first. Eah is assigned an oxidation state of, giving a total of 6 (3 3 ) for the three oxygen atoms. Beause the moleule has zero harge overall, the sulfur must have an oxidation state of 16. hek: () 5 0 b. 4 As in part a, eah oxygen is assigned an oxidation state of, giving a total of (4 3 ) on the four oxygen atoms. he anion has a net harge of, so the sulfur must have an oxidation state of 16. hek: () 5 4 has a harge of, so this is orret.. N 5 We assign oxygen before nitrogen beause oxygen is more eletronegative. hus eah is assigned an oxidation state of, giving a total of 10 (5 3 ) on the five oxygen atoms. herefore, the oxidation states of the two nitrogen atoms must total 110 beause N 5 has no overall harge. Eah N is assigned an oxidation state of 15. hek: (15) 1 5() 5 0 d. P 3 irst we assign the fluorine an oxidation state of 1, giving a total of 3 (3 3 1) on the three fluorine atoms. hus P must have an oxidation state of 13. hek: (1) 5 0 e. 6 In this ase, it is best to reognize that hydrogen is always 11 in ompounds with nonmetals. hus eah is assigned an oxidation state of 11, whih means that the six atoms aount for a total of 16 (6 3 11). herefore, the two arbon atoms must aount for 6, and eah arbon is assigned an oxidation state of 3. hek: (3) 1 6(11) 5 0 elf hek Exerise 1.3 We an tell whether this is an oxidation redution reation by omparing the oxidation states of the elements in the reatants and produts: N 1 3 N 3 a xidation states: (eah ) Nitrogen goes from 0 to 3. hus it gains three eletrons and is redued. Eah hydrogen atom goes from 0 to 11 and is thus oxidized, so this is an oxidation redution reation. he oxidizing agent is N (it takes eletrons from ). he reduing agent is (it gives eletrons to N ). N 1 3 N 3 6e 99030_3_ans-se_A0-A6.indd 4 9/19/13 6:0 PM

18 A5 elf hek Exerise 1.4 he unbalaned equation for this reation is u1s 1 N 3 1aq u1n 3 1aq 1 1l 1 N1g opper Nitri aid Aqueous Water Nitrogen metal opper(ii) monoxide nitrate (ontains u 1 ) tep 1 he oxidation half-reation is u 1 N 3 u1n 3 xidation states: (eah 0) (eah 0) he opper goes from 0 to 1 and thus is oxidized. his redution reation is N 3 N xidation states: (eah 0) In this ase, nitrogen goes from 15 in N 3 to 1 in N and so is redued. Notie two things about these reations: 1. he N 3 must be inluded in the oxidation halfreation to supply N 3 in the produt u(n 3 ).. Although water is a produt in the overall reation, it does not need to be inluded in either half-reation at the beginning. It will appear later as we balane the equation. tep Balane the oxidation half-reation. u 1 N 3 u1n 3 a. Balane nitrogen first. u 1 N 3 u1n 3 b. Balaning nitrogen also aused oxygen to balane.. Balane hydrogen using 1. u 1 N 3 u1n d. Balane the harge using e. u 1 N 3 u1n e his is the balaned oxidation half-reation. Balane the redution half-reation. N 3 N a. All elements are balaned exept hydrogen and oxygen. b. Balane oxygen using. N 3 N 1. Balane hydrogen using N 3 N 1 d. Balane the harge using e. 3e N 3 N 1 his is the balaned redution half-reation. tep 3 We equalize eletrons by multiplying the oxidation halfreation by 3: 3 3 3u 1 N 3 u1n e 4 gives 3u 1 6N 3 3u1N e Multiplying the redution half-reation by : gives 3 33e N 3 N 1 4 6e N 3 N 1 4 tep 4 We an now add the balaned half-reations, whih both involve a six-eletron hange. 3u 1 6N 3 3u1N e 6e N 3 N 1 4 6e u 1 N 3 3u1N 3 1 N e aneling speies ommon to both sides gives the balaned overall equation: 3u1s 1 N 3 1aq 3u1N 3 1aq 1 N1g 1 4 1l tep 5 hek the elements and harges. Elements 3u,, N, 4 3u,, N, 4 harges 0 0 hapter 19 elf hek Exerise 19.1 a. An alpha partile is a helium nuleus, 4 e. We an initially represent the prodution of an a partile by 6 Ra as follows: 6 Ra 4 e 1 A ZX Beause we know that both A and Z are onserved, we an write A and Z 1 5 olving for A gives and for Z gives 6, so A ZX is 6X. Beause Rn has Z 5 6, A ZX is 6Rn. he overall balaned equation is 6 Ra 4 e 1 6Rn hek: Z 5 Z A 5 6 A b. Using a similar strategy, we have 14 Pb 0 1e 1 A ZX Beause Z 1 5, Z 5 3, and beause A , A herefore, A ZX Bi. he balaned equation is 14 Pb 0 1e Bi hek: Z 5 Z A 5 14 A _3_ans-se_A0-A6.indd 5 9/19/13 6:0 PM

19 A6 elf hek Exerise 19. a. he missing partile must be 4 (an a partile), beause 6Rn 1 4Po 1 4 e is a balaned equation. hek: Z 5 6 Z A 5 A b. he missing speies must be 15 7X or 15 7N, beause the balaned equation is N 1 0 1e hek: Z 5 Z A 5 15 A elf hek Exerise 19.3 et s do this problem by thinking about the number of half-lives required to go from mole to mole of Ra mol h mol h irst eond half-life half-life mol h mol hird half-life It takes three half-lives, then, for the sample to go from mole of Ra to mole of Ra. rom able 19.3, we know that the half-life of Ra is 6.7 years. herefore, the elapsed time is 3(6.7 years) years, or years when we use the orret number of signifiant figures. hapter 0 elf hek Exerise 0.1 he alkane with ten arbon atoms an be represented as 3 ( ) 3 and its formula is 10. he alkane with fifteen arbons, has the formula elf hek Exerise 0. 3 ( ) a Methyl 3 Ethyl his moleule is 5-ethyl-3-methylotane b. 3 3 Methyl 3 3 Ethyl his moleule is 5-ethyl-3-methylheptane. Note that this hain ould be numbered from the opposite diretion to give the name 3-ethyl-5-methylheptane. hese two names are equally orret. elf hek Exerise 0.3 he root name deate indiates a ten-arbon hain. here is a methyl group at the number 4 position and an isopropyl group at the number 5 position. he strutural formula is Methyl 3 3 elf hek Exerise 0.4 Isopropyl 3 a. he longest hain has eight arbon atoms with a double bond, so the root name is otene. he double bond exists between arbons 3 and 4, so the name is 3-otene. here is a methyl group on the number- arbon. he name is -methyl-3-otene. b. he arbon hain has five arbons with a triple bond between arbons 1 and. he name is 1-pentyne. elf hek Exerise 0.5 a. -hloronitrobenzene or o-hloronitrobenzene b. 4-phenyl--hexene elf hek Exerise 0.6 a. 1-pentanol; primary alohol b. -methyl--propanol (but this alohol is usually alled tertiary butyl alohol); tertiary alohol. 5-bromo--hexanol; seondary alohol elf hek Exerise 0.7 a. 4-ethyl-3-hexanone Beause the ompound is named as a hexanone, the arbonyl group is assigned the lowest possible number. b. 7-isopropyldeanal _3_ans-se_A0-A6.indd 6 9/19/13 6:0 PM