Test Bank for Financial Statement Analysis and Security Valuation, 5th Edition

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1REVIEWMEASUREMENT AND CALCULATIONS IN CHEMISTRYQuestions1.A random error has equal probability of being too high or too low. This type of error occurswhen estimating the value of the last digit of a measurement.A systematic error is one thatalways occurs in the same direction, either too high or too low.For example, this type oferror would occur if the balance you were using weighed all objects 0.20 g too high, that is, ifthe balance wasn’t calibrated correctly.A random error is an indeterminate error, whereas asystematic error is a determinate error.2.Precision: reproducibility; accuracy: the agreement of a measurement with the true value.a.Imprecise and inaccurate data: 12.32 cm, 9.63 cm, 11.98 cm, 13.34 cmb.Precise but inaccurate data: 8.76 cm, 8.79 cm, 8.72 cm, 8.75 cmc.Precise and accurate data: 10.60 cm, 10.65 cm, 10.63 cm, 10.64 cmData can be imprecise if the measuring device is imprecise as well as if the user of themeasuring device has poor skills. Data can be inaccurate due to a systematic error in themeasuring device or with the user. For example, a balance may read all masses as weighing0.2500 g too high or the user of a graduated cylinder may read all measurements 0.05 mL toolow.A set of measurements that are imprecise implies that all the numbers are not close to eachother. If the numbers aren’t reproducible, then all the numbers can’t be very close to the truevalue. Some say that if the average of imprecise data gives the true value, then the data areaccurate; a better description is that the data takers are extremely lucky.3.Accuracy refers to how close a measurement or series of measurements are to an accepted ortrue value.Precision refers to how close a series of measurements of the same item are toeach other (reproducible). The results, average = 14.91 ±0.03%, are precise (are close to eachother) but are not accurate (are not close to the true value).4.Volume readings are estimated to one place past the markings on the glassware.Theassumed uncertainty is ±1 in the estimated digit.For glassware a, the volume would beestimated to the tenths place since the markings are to the ones place.A sample readingwould be 4.2 with an uncertainty of ±0.1.This reading has two significant figures.Forglassware b, 10.52 ±0.01 would be a sample reading and the uncertainty; this reading has foursignificant figures.For glassware c, 18 ±1 would be a sample reading and the uncertainty,with the reading having two significant figures.5.Significant figures are the digits we associate with a number. They contain all of the certaindigits and the first uncertain digit (the first estimated digit). What follows is one thousand

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2REVIEWMEASUREMENT AND CALCULATIONS IN CHEMISTRYindicated to varying numbers of significant figures: 1000 or 1 × 103(1 S.F.); 1.0 × 103(2S.F.); 1.00 × 103(3 S.F.); 1000. or 1.000 × 103(4 S.F.).To perform the calculation, the addition/subtraction significant figure rule is applied to 1.51.0.Theresultofthisistheone-significant-figureanswerof0.5.Next,themulti-plication/division rule is applied to 0.5/0.50. A one-significant-figure number divided by atwo-significant-figure number yields an answer with one significant figure (answer = 1).6.In both sets of rules, the least precise number determines the number of significant figures inthe final result. For multiplication/division, the number of significant figures in the result isthe same as the number of significant figures in the least precise number used in thecalculation. For addition/subtraction, the result has the same number of decimal places as theleast precise number used in the calculation (not necessarily the number with the fewestsignificant figures).7.In a subtraction, the result gets smaller, but the uncertainties add. If the two numbers are veryclose together, the uncertainty may be larger than the result.For example, let’s assume wewant to take the difference of the following two measured quantities, 999,999 ±2 and 999,996±2.The difference is 3 ±4.Because of the uncertainty, subtracting two similar numbers ispoor practice.8.Consider gold with a density of 19.32 g/cm3. The two possible conversion factors are:3cm1g32.19org32.19cm13Use the first form when converting from the volume of gold in cm3to the mass of gold, anduse the second form when converting from mass of gold in grams to the volume of gold.When using conversion factors, concentrate on the units crossing off.9.Straight line equation:y=mx+b,wheremis the slope of the line andbis they-intercept. Forthe TFvs. TCplot:TF= (9/5)TC+ 32y=mx+bThe slope of the plot is 1.8 (= 9/5) and they-intercept is 32F.For the TCvs. TKplot:TC=TK273Y=mx+bThe slope of the plot is 1, and they-intercept is273C.10.To convert from Celsius to Kelvin, a constant number of 273 is added to the Celsiustemperature. Because of this,T(C) =T(K). When converting from Fahrenheit to Celsius,one conversion that must occur is to multiply the Fahrenheit temperature by a factor less thanone (5/9). Therefore, the Fahrenheit scale is more expansive than the Celsius scale, and 1Fwould correspond to a smaller temperature change than 1C or 1 K.

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REVIEWMEASUREMENT AND CALCULATIONS IN CHEMISTRY311.a.coffee; saltwater; the air we breathe (N2+ O2+ others); brass (Cu + Zn)b.book; human being; tree; deskc.sodium chloride (NaCl); water (H2O); glucose (C6H12O6); carbon dioxide (CO2)d.nitrogen (N2); oxygen (O2); copper (Cu); zinc (Zn)e.boiling water; freezing water; melting a popsicle; dry ice sublimingf.Electrolysis of molten sodium chloride to produce sodium and chlorine gas; the explosivereaction between oxygen and hydrogen to produce water; photosynthesis, which convertsH2O and CO2into C6H12O6and O2; the combustion of gasoline in our car to produce CO2and H2O12.a.b.13.The law of conservation of energy states that energy can be converted from one form toanother but can neither be destroyed nor created; the energy content of the universe isconstant. The two categories of energy are kinetic energy and potential energy. Kineticenergy is energy due to motion and potential energy is stored energy due to position. Forexample, the energy stored in the bonds between atoms in a molecule is potential energy.14.Counting by weighing utilizes the average mass of a particular unit of substance. Consider alarge sample size of marbles which will contain many different individual masses for thevarious marbles. However, the very large sample size will have an average mass so that thegas element (monoatomic)atoms/molecules far apart;random order; takes volumeof containeratoms/molecules closetogether; somewhatordered arrangement;takes volume of containerliquid elementsolid elementatoms/moleculesclose together;ordered arrangement;has its own volume2 compoundscompound and element (diatomic)

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4REVIEWMEASUREMENT AND CALCULATIONS IN CHEMISTRYmarbles behave as if each individual marble has that average mass. This assumption is validas long as the sample size is very large. When a large sample of marbles is weighed, onedivides the total mass of marbles by the average mass of a marble, and this will give a verygood estimate of the number of marbles present.Because we can’t count individual atoms, we “count” the atoms by weighing; convertingthesample mass in grams to the number of atoms in the sample by using the average mass of thesubstance. The mole scale of atoms is a huge number (6.022 × 1023atoms = 1 mole), so theassumption that a weighable sample size behaves as a bunch of atoms, each with the sameaverage mass, is valid and allows one to count atoms by weighing. The mole scale also giveseasier numbers to deal with. So instead of saying you have 12 × 1023atoms of helium, youcan instead say that you have 2.0 moles of helium.ExercisesSignificant Figures and Unit Conversions15.a.exactb.inexactc.exactd.inexact (π has an infinite number of decimal places.)16.a.one significant figure (S.F.). The implied uncertainty is ±1000 pages. More significantfigures should be added if a more precise number is known.b.two S.F.c.four S.F.d.two S.F.e.infinite number of S.F. (exact number)f.one S.F.17.a.6.07 ×1510; 3 S.F.b.0.003840; 4 S.F.c.17.00; 4 S.F.d.8 × 108; 1 S.F.e.463.8052; 7 S.F.f.300; 1 S.F.g.301; 3 S.F.h.300.; 3 S.F.18.a.100; 1 S.F.b.1.0 × 102; 2 S.F.c.1.00 × 103; 3 S.F.d.100.; 3 S.F.e.0.0048; 2 S.F.f.0.00480; 3 S.F.g.4.80 ×310; 3 S.F.h.4.800 ×310; 4 S.F.19.When rounding, the last significant figure stays the same if the number after this significantfigure is less than 5 and increases by one if the number is greater than or equal to 5.a.3.42 ×410b.1.034 × 104c.1.7992 × 101d.3.37 × 10520.a.4 × 105b.3.9 × 105c.3.86 × 105d.3.8550 × 10521.Volume measurements are estimated to one place past the markings on the glassware. Thefirst graduated cylinder is labeled to 0.2 mL volume increments, so we estimate volumes tothe hundredths place. Realistically, the uncertainty in this graduated cylinder is±0.05 mL.The second cylinder, with 0.02 mL volume increments, will have an uncertainty of±0.005

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REVIEWMEASUREMENT AND CALCULATIONS IN CHEMISTRY5mL.The approximate volume in the first graduated cylinder is 2.85 mL, and the volume inthe other graduated cylinder is approximately 0.280 mL. The total volume would be:2.85 mL+0.280 mL3.13 mLWe should report the total volume to the hundredths place because the volume from the firstgraduated cylinder is only read to the hundredths (read to two decimal places).The firstgraduated cylinder is the least precise volume measurement because the uncertainty of thisinstrument is in the hundredths place, while the uncertainty of the second graduated cylinderis to the thousandths place.It is always the least precise measurement that limits theprecision of a calculation.22.a.Volumes are always estimated to one position past the marked volume increments. Theestimated volume of the first beaker is 32.7 mL, the estimated volume of the middlebeaker is 33 mL, and the estimated volume in the last beaker is 32.73 mL.b.Yes, all volumes could be identical to each other because the more precise volumereadings can be rounded to the other volume readings.But because the volumes are inthree different measuring devices, each with its own unique uncertainty, we cannot saywith certainty that all three beakers contain the same amount of water.c.32.7 mL33 mL32.73 mL98.43 mL = 98 mLThe volume in the middle beaker can only be estimated to the ones place, which dictatesthat the sum of the volume should be reported to the ones place.As is always the case,the least precise measurement determines the precision of a calculation.23.For addition and/or subtraction, the result has the same number of decimal places as thenumber in the calculation with the fewest decimal places. When the result is rounded to thecorrect number of significant figures, the last significant figure stays the same if the numberafter this significant figure is less than 5 and increases by one if the number is greater than orequal to 5. The underline shows the last significant figure in the intermediate answers.a.212.2 + 26.7 + 402.09 = 640.99 = 641.0b.1.0028 + 0.221 + 0.10337 = 1.32717 = 1.327c.52.331 + 26.010.9981 = 77.3429 = 77.34d.2.01 × 102+ 3.014 × 103= 2.01 × 102+ 30.14 × 102= 32.15 × 102= 3215When the exponents are different, it is easiest to apply the addition/subtraction rule whenall numbers are based on the same power of 10.e.7.2556.8350 = 0.42 = 0.420 (first uncertain digit is in the third decimal place).24.For multiplication and/or division, the result has the same number of significant figures as thenumber in the calculation with the fewest significant figures.

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6REVIEWMEASUREMENT AND CALCULATIONS IN CHEMISTRYa.2.26352.261.012730.08210.102b.0.14 × 6.022 × 1023= 8.431 × 1022= 8.4× 1022;since 0.14 only has two significantfigures, the result should only have two significant figures.c.4.0 × 104× 5.021 ×310× 7.34993 × 102= 1.476 × 105= 1.5 × 105d.1212761067.6106766.61000.31000.225.a.Here, apply the multiplication/division rule first; then apply the addition/subtraction ruleto arrive at the one-decimal-place answer. We will generally round off at intermediatesteps in order to show the correct number of significant figures.However, you shouldround off at the end of all the mathematical operations in order to avoid round-off error.The best way to do calculations is to keep track of the correct number of significantfigures during intermediate steps, but round off at the end.For this problem, weunderlined the last significant figure in the intermediate steps.4326.0705.80623.0470.01.3526.2= 0.8148 + 0.7544 + 186.558 = 188.1b.Here, the mathematical operation requires that we apply the addition/subtraction rulefirst, then apply the multiplication/division rule.6.191.2404.61.177.1891.2404.6= 12c.6.071 ×5108.2 ×6100.521 ×410= 60.71 ×6108.2 ×61052.1 ×610= 0.41 ×610= 4 ×710d.2612131212131313121312103.6107610241063104100.41038103.6104100.4108.3e.4755.194175.38.21.45.9= 4.89 = 4.9Uncertainty appears in the first decimal place. The average of several numbers can onlybe as precise as the least precise number. Averages can be exceptions to the significantfigure rules.f.925.8002.0100925.8905.8925.8× 100 = 0.2226.a.6.022 × 1023× 1.05 × 102= 6.32 × 1025b.1798341082.71054.210998.2106262.6

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REVIEWMEASUREMENT AND CALCULATIONS IN CHEMISTRY7c.1.285 ×210+ 1.24 ×310+ 1.879 ×110= 0.1285 ×110+ 0.0124 ×110+ 1.879 ×110= 2.020 ×110When the exponents are different, it is easiest to apply the addition/subtraction rule whenall numbers are based on the same power of 10.d.2723231029.21002205.600138.01002205.6)00728.100866.1(e.122222101.810010875.910080.010010875.910795.910875.9f.310625.110824.010942.0310625.110234.81042.9333322= 1.130 × 10327.a.8.43 cm ×mm3.84mmm1000cm100m1b.2.41× 102cm ×cm100m1= 2.41 mc.294.5 nmcm10945.2mcm100nm101m159d.1.445 × 104m ×mkm10001= 14.45 kme. 235.3 m ×mmm1000= 2.353 × 105mmf.903.3 nmmμ9033.0mmμ101nm101m16928.a.1 Tgkg101g1000kg1Tgg101912b.6.50 × 102Tmnm1050.6mnm101Tmm10123912c.25 fgkg102.5kg1025g1000kg1fg101g1171815d.8.0 dm3×31dmL= 8.0 L(1 L = 1 dm3= 1000 cm3= 1000 mL)e.1 mLLμ101LLμ101mL0001L136f.1 μgpg101gpg101gμ101g16126

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8REVIEWMEASUREMENT AND CALCULATIONS IN CHEMISTRY29.a.Appropriate conversion factors are found in Appendix 6. In general, the number ofsignificant figures we use in the conversion factors will be one more than the number ofsignificant figures from the numbers given in the problem. This is usually sufficient toavoid round-off error.3.91 kg ×kg4536.0lb1= 8.62 lb; 0.62 lb ×lboz16= 9.9 ozBaby’s weight = 8 lb and 9.9 oz or, to the nearest ounce, 8 lb and 10. oz.51.4 cm ×cm54.2in1= 20.2 in ≈ 20 1/4 in = baby’s heightb.25,000 mi ×mikm61.1= 4.0 × 104km; 4.0 × 104km ×kmm1000= 4.0 × 107mc.V = 1 × w × h = 1.0 m ×dm10m1dm1.2cm100m1cm6.5= 1.2 ×210m31.2 ×210m3×33dmL1mdm01= 12 L12 L ×33cm54.2in1Lcm1000= 730 in3; 730 in3×3in12ft1= 0.42 ft330.a.908 oz ×lbkg4536.0oz16lb1= 25.7 kgb.12.8 L ×qt4gal1L9463.0qt1= 3.38 galc.125 mL ×L9463.0qt1mL1000L1= 0.132 qtd.2.89 gal ×L1mL1000qt057.1L1gal1qt4= 1.09 × 104mLe.4.48 lb ×lb1g6.453= 2.03 × 103gf.550 mL ×Lqt06.1mL1000L1= 0.58 qt31.a.1.25 mi ×mifurlongs8= 10.0 furlongs; 10.0 furlongs ×furlongrods40= 4.00 × 102rods4.00 × 102rods ×cm100m1incm54.2ydin36rodyd5.5= 2.01 × 103m

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REVIEWMEASUREMENT AND CALCULATIONS IN CHEMISTRY92.01 × 103m ×m1000km1= 2.01 kmb.Let's assume we know this distance to ±1 yard. First, convert 26 miles to yards.26 mi ×ft3yd1mift5280= 45,760. yd26 mi + 385 yd = 45,760. yd + 385 yd = 46,145 yards46,145 yard ×yd5.5rod1= 8390.0 rods; 8390.0 rods ×rods40furlong1= 209.75 furlongs46,145 yard ×cm100m1incm54.2ydin36= 42,195 m; 42,195 m ×m1000km1= 42.195 km32.a.1 ha ×22m1000km1ham000,10= 122km10b.5.5 acre ×22cm100m1incm54.2ydin36rodyd5.5acrerod160= 2.2 × 104m22.2 × 104m2×24m101ha1= 2.2 ha; 2.2 × 104m2×2m1000km1= 0.022 km2c.Area of lot = 120 ft × 75 ft = 9.0 × 103ft29.0 × 103ft2×22rod160acre1yd5.5rod1ft3yd1= 0.21 acre;acre000,31$acre21.0500,6$We can use our result from (b) to get the conversion factor between acres and hectares(5.5 acre = 2.2 ha.). Thus 1 ha = 2.5 acre.0.21 acre ×acre5.2ha1= 0.084 ha; the price is:ha000,77$ha084.0500,6$33.a.1 troy lb ×g1000kg1graing0648.0pwgrains24oztroypw20lbtroyoztroy12= 0.373 kg1 troy lb = 0.373 kg ×kglb205.2= 0.822 lbb.1 troy oz ×graing0648.0pwgrains24oztroypw20= 31.1 g1 troy oz = 31.1 g ×g200.0carat1= 156 carats

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10REVIEWMEASUREMENT AND CALCULATIONS IN CHEMISTRYc.1 troy lb = 0.373 kg; 0.373 kg ×g3.19cm1kgg10003= 19.3 cm334.a.1 grain ap ×apdramg888.3scruples3apdram1apgrain20scruple1= 0.06480 gFrom the previous question, we are given that 1 grain troy = 0.0648 g = 1 grain ap. So thetwo are the same.b.1 oz ap ×g1.31*troyoz1apdramg888.3apozapdram8= 1.00 oz troy;*see Exercise 33b.c.5.00 × 102mg ×apdramscruples3g888.3apdram1mg1000g1= 0.386 scruple0.386 scruple ×scrupleapgrains20= 7.72 grains apd.1 scruple ×apdramg888.3scruples3apdram1= 1.296 g35.15.6 g ×g65.0capsule1= 24 capsules36.1.5 teaspoons ×teaspoon50.0acetmg.80= 240 mg acetaminophenkg454.0lb1lb24acetmg240= 22 mg acetaminophen/kgkg454.0lb1lb35acetmg240= 15 mg acetaminophen/kgThe range is from 15 to 22 mg acetaminophen per kg of body weight.37.warp 1.71 =h/yd2000knot1hmin60mins60myd094.1sm1000.300.58= 2.95 × 109knotshmin60mins60km609.1mi1m1000km1sm1000.300.58= 3.36 × 109mi/h38.s58.9m.100= 10.4 m/s;hmin60mins60m1000km1s58.9m.100= 37.6 km/hydft3myd0936.1s58.9m.100= 34.2 ft/s;hmin60mins60ft5280mi1sft2.34= 23.3 mi/h

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REVIEWMEASUREMENT AND CALCULATIONS IN CHEMISTRY111.00 × 102yd ×m.100s58.9yd0936.1m1= 8.76 s39.kmmi6214.0hkm65= 40.4 = 40. mi/hTo the correct number of significant figures (2), 65 km/h does not violate a 40 mi/h speedlimit.40.112 km ×mi65h1kmmi6214.0= 1.1 h = 1 h and 6 min112 km ×galL785.3mi28gal1kmmi6214.0= 9.4 L of gasoline41.euro32.1$lb2046.2kg1kgeuros45.2= $1.47/lbOne pound of peaches costs $1.47.42.For the gasoline car:500. migal50.3$mi0.28gal1= $62.5For the E85 car:500. migal85.2$mi5.22gal1= $63.3The E85 vehicle would cost slightly more to drive 500. miles as compared to the gasolinevehicle ($63.3 versus $62.5).43.Volume of lake = 100 mi2×2mift5280× 20 ft = 6 × 1010ft36 × 1010ft3×mLμg0.4cmmL1incm2.54ftin1233= 7 × 1014μg mercury7 × 1014μgg101kg1gμ101g136= 7 × 105kg of mercury44.Volume of room = 18 ft × 12 ft × 8 ft = 1700 ft3(carrying one extra significant figure)1700 ft33333m48cm100m1incm54.2ftin1248 m3COgμ101COg1mCOgμ000,40063= 19 g = 20 g CO (to 1 sig. fig.)

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12REVIEWMEASUREMENT AND CALCULATIONS IN CHEMISTRYTemperature45.a.TC=95(TF32) =95(459F32) =273C; TK= TC+ 273 =273C + 273 = 0 Kb.TC=95(40.F32) =40.C; TK=40.C + 273 = 233 Kc.TC=95(68F32) = 20.C; TK= 20.C + 273 = 293 Kd.TC=95(7 × 107F32) = 4 × 107C; TK= 4 × 107C + 273 = 4 × 107K46.96.1°F ±0.2°F; first, convert 96.1°F to °C. TC=95(TF32) =95(96.132) = 35.6°CA change in temperature of 9°F is equal to a change in temperature of 5°C.So theuncertainty is:±0.2°F ×F9C5= ±0.1°C. Thus 96.1 ±0.2°F = 35.6 ±0.1°C.47.a.TF=59× TC+ 32 =59× 39.2°C + 32 = 102.6°F(Note: 32 is exact.)TK= TC+ 273.2 = 39.2 + 273.2 = 312.4 Kb.TF=59× (25) + 32 =13°F; TK=25 + 273 = 248 Kc.TF=59× (273) + 32 = -459°F; TK=273 + 273 = 0 Kd.TF=59×801 + 32 = 1470°F;TK= 801 + 273 = 1074 K48.a.TC= TK273 = 233273 =40.°CTF=59× TC+ 32 =59× (40.) + 32 =40.°Fb.TC= 4273 =269°C; TF=59× (269) + 32 =452°Fc.TC= 298273 = 25°C; TF=59× 25 + 32 = 77°Fd.TC= 3680273 = 3410°C; TF=59× 3410 + 32 = 6170°F49.TF=59× TC+ 32; from the problem, we want the temperature where TF= 2TC.Substituting:

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REVIEWMEASUREMENT AND CALCULATIONS IN CHEMISTRY132TC=59× TC+ 32, (0.2)TC= 32, TC=2.032= 160CTF= 2TCwhen the temperature in Fahrenheit is 2(160) = 320F. Because all numbers whensolving the equation are exact numbers, the calculated temperatures are also exact numbers.50.TC=95(TF–32) =95(72–32) = 22CTC= TK–273 = 313–273 = 40.CThe difference in temperature between Jupiter at 313 K and Earth at 72F is 40.C–22C =18C.51.a.A change in temperature of 160°C equals achange in temperature of 100°A.SoA100C160is our unit conversion for adegree change in temperature.At the freezing point: 0°A =45°CCombining these two pieces of information:TA= (TC+ 45°C) ×C160A100= (TC+ 45°C) ×C8A5or TC= TA×A5C845°Cb.TC= (TF32) ×95; TC= TA×5845 = (TF32) ×95TF32 =594558TA=2572TA81, TF= TA×A25F7249Fc.TC= TA×5845 and TC= TA;so TC= TC×5845,53TC= 45, TC= 75°C = 75°Ad.TC= 86°A ×A5C845°C = 93°C; TF= 86°A ×A25F7249°F = 199°F = 2.0 × 102°Fe.TA= (45°C + 45°C) ×C8A5= 56°A52.a.A change in temperature of 140C is equal to 50X. Therefore,X50C140oois the unit con-version between a degree on the X scale to a degree on the Celsius scale. To account forthe different zero points,10must be subtracted from the temperature on the X scale toget to the Celsius scale. The conversion betweenX toC is:TC= TXX50C140oo10C, TC= TXX5C14oo10C100oA115oC160oC-45oC100oA0oA

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14REVIEWMEASUREMENT AND CALCULATIONS IN CHEMISTRYThe conversion betweenC toX would be:TX= (TC+ 10C)C14X5oob.Assuming 10C andC14X5ooare exact numbers:TX= (22.0C + 10C)C14X5oo= 11.4Xc.Assuming exact numbers in the temperature conversion formulas:TC= 58.0XX5C14oo10C = 152CTK= 152C + 273 = 425 KTF=C5F9oo152C + 32F = 306FDensity53.Mass=350 lblbg6.453= 1.6 × 105g; V = 1.2 × 104in33incm54.2= 2.0 × 105cm3Density =3355g/cm80.0cm100.2g101volumemassBecause the material has a density less than water, it will float in water.54.3333cm52.0g0.2d;cm52.0)cm50.0(14.334r34V= 3.8 g/cm3The ball will sink.55.333353cm104.1mcm100kmm1000km100.714.334r34VDensity =33336cm104.1kgg1000kg102volumemass= 1.4 × 106g/cm3= 1 × 106g/cm356.V = l × w × h = 2.9 cm × 3.5 cm × 10.0 cm = 1.0 × 102cm3d = density =332cmg2.6cm100.1g0.615

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REVIEWMEASUREMENT AND CALCULATIONS IN CHEMISTRY1557.a.5.0 caratg51.3cm1caratg200.03= 0.28 cm3b.2.8 mLg200.0carat1cmg51.3mLcm133= 49 carats58.For ethanol: 100. mL ×mLg789.0= 78.9 gFor benzene: 1.00 LmLg880.0LmL1000= 880. gTotal mass = 78.9 g + 880. g = 959 g59.V = 21.6 mL12.7 mL = 8.9 mL; density =mL9.8g42.33= 3.8 g/mL = 3.8 g/cm360.5.25 gg5.10cm13= 0.500 cm3= 0.500 mLThe volume in the cylinder will rise to 11.7 mL (11.2 mL + 0.500 mL = 11.7 mL).61.a.Both have the same mass of 1.0 kg.b.1.0 mL of mercury; mercury is more dense than water.Note: 1 mL = 1 cm3.1.0 mLmLg6.13= 14 g of mercury; 1.0 mLmLg998.0= 1.0 g of waterc.Same; both represent 19.3 g of substance.19.3 mLmLg9982.0= 19.3 g of water; 1.00 mLmLg32.19= 19.3 g of goldd.1.0 L of benzene (880 g versus 670 g)75 mLmLg96.8= 670 g of copper; 1.0 LmLg880.0LmL1000= 880 g of benzene62.a.1.50 qtmLg789.0LmL1000qt0567.1L1= 1120 g ethanolb.3.5 in333cmg6.13incm54.2= 780 g mercury63.a. 1.0 kg feather; feathers are less dense than lead.b. 100 g water; water is less dense than gold.c. Same; both volumes are 1.0 L.64.a.H2(g): V = 25.0 g ×g000084.0cm13= 3.0 × 105cm3[H2(g) = hydrogen gas.]b.H2O(l): V = 25.0 g ×g9982.0cm13= 25.0 cm3[H2O(l) = water.]

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16REVIEWMEASUREMENT AND CALCULATIONS IN CHEMISTRYc.Fe(s): V = 25.0 g ×g87.7cm13= 3.18 cm3[Fe(s) = iron.]Notice the huge volume of the gaseous H2sample as compared to the liquid and solidsamples. The same mass of gas occupies a volume that is over 10,000 times larger than theliquid sample. Gases are indeed mostly empty space.65.V = 1.00 × 103g ×g57.22cm13= 44.3 cm344.3 cm3= 1 × w × h = 4.00 cm × 4.00 cm × h, h = 2.77 cm66.V = 22 g ×g96.8cm13= 2.5 cm3; V = πr2×l, wherel= length of the wire2.5 cm3= π ×22mm25.0×2mm10cm1×l,l= 5.1 × 103cm = 170 ftClassification and Separation of Matter67.A gas has molecules that are very far apart from each other, whereas a solid or liquid hasmolecules that are very close together.An element has the same type of atom, whereas acompound contains two or more different elements.Picture i represents an element thatexists as two atoms bonded together (like H2or O2or N2). Picture iv represents a compound(like CO, NO, or HF).Pictures iii and iv contain representations of elements that exist asindividual atoms (like Ar, Ne, or He).a.Picture iv represents a gaseous compound.Note that pictures ii and iii also contain agaseous compound, but they also both have a gaseous element present.b.Picture vi represents a mixture of two gaseous elements.c.Picture v represents a solid element.d.Pictures ii and iii both represent a mixture of a gaseous element and a gaseous compound.68.Solid: rigid; has a fixed volume and shape; slightly compressibleLiquid: definite volume but no specific shape; assumes shape of the container; slightlycompressibleGas: no fixed volume or shape; easily compressiblePure substance: has constant composition; can be composed of either compounds or ele-mentsElement: substances that cannot be decomposed into simpler substances by chemical orphysical means.Compound: a substance that can be broken down into simpler substances (elements) bychemical processes.Homogeneous mixture: a mixture of pure substances that has visibly indistinguishable parts.Heterogeneous mixture: a mixture of pure substances that has visibly distinguishable parts.

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REVIEWMEASUREMENT AND CALCULATIONS IN CHEMISTRY17Solution: a homogeneous mixture; can be a solid, liquid or gasChemical change: a given substance becomes a new substance or substances with differentproperties and different composition.Physical change: changes the form (g, l, or s) of a substance but does no change the chemicalcomposition of the substance.69.Homogeneous: Having visibly indistinguishable parts (the same throughout).Heterogeneous: Having visibly distinguishable parts (not uniform throughout).a.heterogeneous (due to hinges, handles, locks, etc.)b.homogeneous (hopefully; if you live in a heavily polluted area, air may beheterogeneous.)c.homogeneousd.homogeneous (hopefully, if not polluted)e.heterogeneousf.heterogeneous70.a.heterogeneousb.homogeneousc.heterogeneousd.homogeneous (assuming no imperfections in the glass)e.heterogeneous (has visibly distinguishable parts)71.a.pureb.mixturec.mixtured.puree.mixture (copper and zinc)f.pureg.mixtureh.mixturei.mixtureIron and uranium are elements. Water (H2O) is a compound because it is made up of two ormore different elements.Table salt is usually a homogeneous mixture composed mostly ofsodium chloride (NaCl), but will usually contain other substances that help absorb watervapor (an anticaking agent).72.Initially, a mixture is present.The magnesium and sulfur have only been placed together inthe same container at this point, but no reaction has occurred. When heated, a reaction occurs.Assuming the magnesium and sulfur had been measured out in exactly the correct ratio forcomplete reaction, the remains after heating would be a pure compound composed ofmagnesium and sulfur.However, if there were an excess of either magnesium or sulfur, theremains after reaction would be a mixture of the compound produced and the excess reactant.73.Chalk is a compound because it loses mass when heated and appears to change into anothersubstance with different physical properties (the hard chalk turns into a crumbly substance).74.Because vaporized water is still thesame substanceas solid water (H2O), no chemicalreaction has occurred. Sublimation is a physical change.75.A physical change is a change in the state of a substance (solid, liquid, and gas are the threestates of matter); a physical change does not change the chemical composition of thesubstance.A chemical change is a change in which a given substance is converted intoanother substance having a different formula (composition).

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18REVIEWMEASUREMENT AND CALCULATIONS IN CHEMISTRYa.Vaporization refers to a liquid converting to a gas, so this is a physical change.Theformula (composition) of the moth ball does not change.b.This is a chemical change since hydrofluoric acid (HF) is reacting with glass (SiO2) toform new compounds that wash away.c.This is a physical change since all that is happening during the boiling process is theconversion of liquid alcohol to gaseous alcohol. The alcohol formula (C2H5OH) does notchange.d.This is a chemical change since the acid is reacting with cotton to form new compounds.76.a.Distillation separates components of a mixture, so the orange liquid is a mixture (has anaverage color of the yellow liquid and the red solid).Distillation utilizes boiling pointdifferences to separate out the components of a mixture. Distillation is a physical changebecause the components of the mixture do not become different compounds or elements.b.Decomposition is a type of chemical reaction.The crystalline solid is a compound, anddecomposition is a chemical change where new substances are formed.c.Tea is a mixture of tea compounds dissolved in water. The process of mixing sugar intotea is a physical change.Sugar doesn’t react with the tea compounds, it just makes thesolution sweeter.Additional Exercises77.Avogadro’s number of dollars = 6.022 × 1023dollars/mol dollarspeople107dollarsmoldollars106.022dollarsmol1923= 8.6 × 1013= 9 × 1013dollars/person1 trillion = 1,000,000,000,000 = 1 × 1012; each person would have 90 trillion dollars.78.Density3cm32.0g384.0volumemass1.2 g/cm3; from the table, the other ingredient is caffeine.79.Because each pill is 4.0% Lipitor by mass, for every 100.0 g of pills, there are 4.0 g of Lipitorpresent. Note that 100 pills is assumed to be an exact number.100 pillsg1000kg1pillsg0.100Lipitorg0.4pillg5.2= 0.010 kg Lipitor80.A chemical change involves the change of one or more substances into other substancesthrough a reorganization of the atoms. A physical change involves the change in the form ofa substance, but not its chemical composition.a.physical change (Just smaller pieces of the same substance.)

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REVIEWMEASUREMENT AND CALCULATIONS IN CHEMISTRY19b.chemical change (Chemical reactions occur.)c.chemical change (Bonds are broken.)d.chemical change (Bonds are broken.)e.physical change (Water is changed from a liquid to a gas.)f.physical change (Chemical composition does not change.)81.Total volume =mcm100m.300mcm100m.200× 4.0 cm = 2.4 × 109cm3Volume of topsoil covered by 1 bag =incm54.2in0.1incm54.2ftin12ft.10222= 2.4 × 104cm32.4 × 109cm334cm104.2bag1= 1.0 × 105bags topsoil82.a.No; if the volumes were the same, then the gold idol would have a much greater massbecause gold is much more dense than sand.b.Mass = 1.0 Lg1000kg1cmg32.91Lcm000133= 19.32 kg (= 42.59 lb)It wouldn't be easy to play catch with the idol because it would have a mass of over 40pounds.83.1 light year = 1 yrsmi000,186mins60hmin60dayh24yrday365= 5.87 × 1012miles9.6 parsecskmm1000mikm609.1yrlightmi1087.5parsecyrlight26.312= 3.0 × 1017m84.mift5280hmi65min60h1s60min1s1= 95.3 ft = 100 ftIf you take your eyes off the road for one second traveling at 65 mph, your car travelsapproximately 100 feet.85.18.5 cm ×cm25.5F0.10o= 35.2F increase; Tfinal= 98.6 + 35.2 = 133.8FTC= 5/9 (133.8–32) = 56.56C

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20REVIEWMEASUREMENT AND CALCULATIONS IN CHEMISTRY86.Massbenzene= 58.80 g25.00 g = 33.80 g; Vbenzene= 33.80 gg880.0cm13= 38.4 cm3Vsolid= 50.0 cm338.4 cm3= 11.6 cm3;density =3cm6.11g00.25= 2.16 g/cm387.a.Volume × density = mass;the orange block is more dense.Because mass (orange) >mass (blue) and because volume (orange) < volume (blue), the density of the orangeblock must be greater to account for the larger mass of the orange block.b.Which block is more dense cannot be determined.Because mass (orange) > mass (blue)and because volume (orange) > volume (blue), the density of the orange block may ormay not be larger than the blue block. If the blue block is more dense, its density cannotbe so large that its mass is larger than the orange block’s mass.c.The blue block is more dense. Because mass (blue) = mass (orange) and because volume(blue) < volume (orange), the density of the blue block must be larger in order to equatethe masses.d.The blue block is more dense.Because mass (blue) > mass (orange) and because thevolumes are equal, the density of the blue block must be larger in order to give the blueblock the larger mass.88.Circumference = c= 2πr; V=23336c2c343r4Largest density =23π6)in00.9(oz25.5=3in3.12oz25.5=3inoz427.0Smallest density =23π6)in25.9(oz00.5=3in4.13oz00.5=3inoz73.0Maximum range is:3inoz)427.0373.0(or 0.40 ±0.03 oz/in3Uncertainty is in 2nd decimal place.89.V = VfinalVinitial; d =333cm4.3g90.28cm4.6cm8.9g90.28= 8.5 g/cm3dmax=minmaxVmass;we get Vminfrom 9.7 cm36.5 cm3= 3.2 cm3.dmax=33cmg0.9cm2.3g93.28; dmin=maxminVmass=333cmg0.8cm3.6cm9.9g87.28The density is 8.5 ±0.5 g/cm3.

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REVIEWMEASUREMENT AND CALCULATIONS IN CHEMISTRY2190.We need to calculate the maximum and minimum values of the density, given the uncertaintyin each measurement. The maximum value is:dmax=33cm03.0cm00.25g002.0g625.19=3cm97.24g627.19= 0.7860 g/cm3The minimum value of the density is:dmin=33cm03.0cm00.25g002.0g625.19=3cm03.25g623.19= 0.7840 g/cm3The density of the liquid is between 0.7840 and 0.7860 g/cm3.These measurements aresufficiently precise to distinguish between ethanol (d = 0.789 g/cm3) and isopropyl alcohol(d = 0.785 g/cm3).91.Letx= mass of copper andy= mass of silver.105.0 g =x+yand 10.12 mL =5.1096.8yx; solving:5.100.10596.812.10xx× 8.96 × 10.5, 952.1 = (10.5)x+ 940.8(8.96)x(carrying 1 extra sig. fig.)11.3 = (1.54)x,x= 7.3 g; mass % Cu =g0.105g3.7× 100 = 7.0% Cu92.In general, glassware is estimated to one place past the markings.a.128.7 mL glasswareb.18 mL glasswarec.23.45 mL glasswareread to tenth’s placeread to one’s placeread to two decimal placesTotal volume = 128.7 + 18 + 23.45 = 170.15 = 170. (Due to 18, the sum would beknown only to the ones place.)1301291281273020102324

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22REVIEWMEASUREMENT AND CALCULATIONS IN CHEMISTRY93.Ball A: PE = mgz = 2.00 kg ×2sm81.9× 10.0 m =22smkg196= 196 JAt point I:All this energy is transferred to ball B. All of B's energy is kinetic energy at thispoint.Etotal= KE = 196 J.At point II, the sum of the total energy will equal196 J.At point II: PE = mgz = 4.00 kg ×2sm81.9× 3.00 m = 118 JKE = EtotalPE = 196 J − 118 J = 78 J94.The bubbles of gas is air in the sand that is escaping; methanol and sand are not reacting. Wewill assume that the mass of trapped air is insignificant.Mass of dry sand = 37.3488 g22.8317 g = 14.5171 gMass of methanol = 45.2613 g37.3488 g = 7.9125 gVolume of sand particles (air absent) = volume of sand and methanolvolume of methanolVolume of sand particles (air absent) = 17.6 mL10.00 mL = 7.6 mLDensity of dry sand (air present) =mL0.10g5171.14= 1.45 g/mLDensity of methanol =mL00.10g9125.7= 0.7913 g/mLDensity of sand particles (air absent) =mL6.7g5171.14= 1.9 g/mLChemWork Problems95.4,145 mi ×fathom100lengthcable1ft6fathom1mift5280= 3.648 × 104cable lengths4,145 mi ×kmm1000mi0.62137km1= 6.671 × 106m3.648 × 104cable lengths ×lengthscable10milenautical1= 3,648 nautical miles96.km1m1000mi0.6214km1s119.2mi1.25= 16.9 m/s97.TC=95(TF32) =95(134°F32) = 56.7°C; phosphorus would be a liquid.

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REVIEWMEASUREMENT AND CALCULATIONS IN CHEMISTRY2398.3243123cm101.4mcm100pmm101pm6914.334rπ34VDensity =32423cm101.4g103.35volumemass= 24 g/cm399.a.False; sugar is generally considered to be the pure compound sucrose, C12H22O11.b.False; elements and compounds are pure substances.c.True; air is a mixture of mostly nitrogen and oxygen gases.d.False; gasoline has many additives, so it is a mixture.e.True; compounds are broken down to elements by chemical change.100.The rusting of iron is the only change listed where chemical formulas change, so it is the onlychemical change. The others are physical properties. Note that the red glow of a platinumwire assumes no reaction between platinum and oxygen; the red glow is just hot Pt.

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24CHAPTER 1CHEMICAL FOUNDATIONSQuestions11.A law summarizes what happens, e.g., law of conservation of mass in a chemical reaction orthe ideal gas law, PV = nRT. A theory (model) is an attempt to explain why somethinghappens. Dalton’s atomic theory explains why mass is conserved in a chemical reaction. Thekinetic molecular theory explains why pressure and volume are inversely related at constanttemperatureandmolesofgaspresent,aswellasexplainingtheothermathematicalrelationships summarized in PV = nRT.12.a.At 8 a.m., approximately 57 cars pass through the intersection per hour.b.At 12 a.m. (midnight), only 1 or 2 cars pass through the intersection per hour.c.Traffic at the intersection is limited to less than 10 cars per hour from 8 p.m. to 5 a.m.Starting at 6 a.m., there is a steady increase in traffic through the intersection, peaking at8 a.m. when approximately 57 cars pass per hour.Past 8 a.m. traffic moderates to about40 cars through the intersection per hour until noon, and then decreases to 21 cars perhour by 3 p.m.Past 3 p.m. traffic steadily increases to a peak of 52 cars per hour at 5p.m., and then steadily decreases to the overnight level of less than 10 cars through theintersection per hour.d.The traffic pattern through the intersection is directly related to the work schedules of thegeneral population as well as to the store hours of the businesses in downtown.e.Run the same experiment on a Sunday, when most of the general population doesn’twork and when a significant number of downtown stores are closed in the morning.13.The fundamental steps are(1) making observations;(2) formulating hypotheses;(3) performing experiments to test the hypotheses.The key to the scientific method is performing experiments to test hypotheses. If after the testof time the hypotheses seem to account satisfactorily for some aspect of natural behavior,then the set of tested hypotheses turns into a theory (model). However, scientists continue toperform experiments to refine or replace existing theories.Hence, science is a dynamic oractive process, not a static one.14.A compound will always contain the same numbers (and types) of atoms. A given amount ofhydrogen will react only with a specific amount of oxygen. Any excess oxygen will remainunreacted.15.Law of conservation of mass: Mass is neither created nor destroyed. The total mass before achemical reaction always equals the total mass after a chemical reaction.

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CHAPTER 1CHEMICAL FOUNDATIONS25Law of definite proportion: A given compound always contains exactly the same proportionof elements by mass. For example, water is always 1 g H for every 8 g oxygen.Law of multiple proportions: When two elements form a series of compounds, the ratios ofthe mass of the second element that combine with 1 g of the first element always can bereduced to small whole numbers.For CO2and CO discussed in Section 1.4, the mass ratiosof oxygen that react with 1 g carbon in each compound are in a 2 : 1 ratio.16.Yes, 1.0 g H would react with 37.0 g37Cl, and 1.0 g H would react with 35.0 g35Cl.No, the mass ratio of H/Cl would always be 1 g H/37 g Cl for37Cl and 1 g H/35 g Cl for35Cl.As long as we had pure37Cl or pure35Cl, the ratios will always hold. If we have a mixture(such as the natural abundance of chlorine), the ratio will also be constant as long as thecomposition of the mixture of the two isotopes does not change.17.Natural niacin and commercially produced niacin have the exact same formula of C6H5NO2.Therefore, both sources produce niacin having an identical nutritional value.There may beother compounds present in natural niacin that would increase the nutritional value, but thenutritional value due to just niacin is identical to the commercially produced niacin.18.a.The smaller parts are electrons and the nucleus. The nucleus is broken down into protonsand neutrons, which can be broken down into quarks. For our purpose, electrons,neutrons, and protons are the key smaller parts of an atom.b.All atoms of hydrogen have 1 proton in the nucleus. Different isotopes of hydrogen have0, 1, or 2 neutrons in the nucleus. Because we are talking about atoms, this implies aneutral charge, which dictates 1 electron present for all hydrogen atoms. If charged ionswere included, then different ions/atoms of H could have different numbers of electrons.c.Hydrogen atoms always have 1 proton in the nucleus, and helium atoms always have 2protons in the nucleus. The number of neutrons can be the same for a hydrogen atom anda helium atom. Tritium (3H) and4He both have 2 neutrons. Assuming neutral atoms, thenthe number of electrons will be 1 for hydrogen and 2 for helium.d.Water (H2O) is always 1 g hydrogen for every 8 g of O present, whereas H2O2is always 1g hydrogen for every 16 g of O present. These are distinctly different compounds, eachwith its own unique relative number and types of atoms present.e.A chemical equation involves a reorganization of the atoms. Bonds are broken betweenatoms in the reactants, and new bonds are formed in the products. The number and typesof atoms between reactants and products do not change. Because atoms are conserved ina chemical reaction, mass is also conserved.19.J. J. Thomson’s study of cathode-ray tubes led him to postulate the existence of negativelycharged particles that we now call electrons. Ernest Rutherford and his alpha bombardment ofmetal foil experiments led him to postulate the nuclear atoman atom with a tiny dense centerof positive charge (the nucleus) with electrons moving about the nucleus at relatively largedistances away; the distance is so large that an atom is mostly empty space.

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CHAPTER 1CHEMICAL FOUNDATIONS2620.The atom is composed of a tiny dense nucleus containing most of the mass of the atom. Thenucleus itself is composed of neutrons and protons. Neutrons have a mass slightly larger thanthat of a proton and have no charge. Protons, on the other hand, have a 1+ relative charge ascompared to the 1charged electrons; the electrons move about the nucleus at relatively largedistances. The volume of space that the electrons move about is so large, as compared to thenucleus, that we say an atom is mostly empty space.21.The number and arrangement of electrons in an atom determine how the atom will react withother atoms. The electrons determine the chemical properties of an atom. The number ofneutrons present determines the isotope identity.22.Density = mass/volume; if the volumes are assumed equal, then the much more massiveproton would have a much larger density than the relatively light electron.23.For lighter, stable isotopes, the number of protons in the nucleus is about equal to the numberof neutrons.When the number of protons and neutrons is equal to each other, the massnumber (protons + neutrons) will be twice the atomic number (protons).Therefore, forlighter isotopes, the ratio of the mass number to the atomic number is close to 2.Forexample, consider28Si, which has 14 protons and (28–14 =) 14 neutrons.Here, the massnumber to atomic number ratio is 28/14 = 2.0. For heavier isotopes, there are more neutronsthan protons in the nucleus.Therefore, the ratio of the mass number to the atomic numberincreases steadily upward from 2 as the isotopes get heavier and heavier.For example,238Uhas 92 protons and (238–92 =) 146 neutrons.The ratio of the mass number to the atomicnumber for238U is 238/92 = 2.6.24.Some elements exist as molecular substances. That is, hydrogen normally exists as H2molecules, not single hydrogen atoms. The same is true for N2, O2, F2, Cl2, Br2, and I2.ExercisesDevelopment of the Atomic Theory25.a.The composition of a substance depends on the numbers of atoms of each elementmaking up the compound (depends on the formula of the compound) and not on thecomposition of the mixture from which it was formed.b.Avogadro’s hypothesis(law) implies that volume ratios are equal to molecule ratios atconstant temperature and pressure.H2(g) + Cl2(g) → 2 HCl(g).From the balancedequation, the volume of HCl produced will be twice the volume of H2(or Cl2) reacted.26.Avogadro’s hypothesis (law) implies that volume ratios are equal to molecule ratios atconstant temperature and pressure.Here, 1 volume of N2reacts with 3 volumes of H2toproduce 2 volumes of the gaseous product or in terms of molecule ratios:1 N2+ 3 H22 productIn order for the equation to be balanced, the product must be NH3.

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CHAPTER 1CHEMICAL FOUNDATIONS2727.From the law of definite proportions, a given compound always contains exactly the sameproportion of elements by mass. The first sample of chloroform has a total mass of 12.0 g C+ 106.4 g Cl + 1.01 g H = 119.41 g (carrying extra significant figures). The mass percent ofcarbon in this sample of chloroform is:totalg41.119Cg0.12× 100 = 10.05% C by massFrom the law of definite proportions, the second sample of chloroform must also contain10.05% C by mass. Letx= mass of chloroform in the second sample:xCg0.30× 100 = 10.05,x= 299 g chloroform28.A compound will always have a constant composition by mass. From the initial data given,the mass ratio of H : S : O in sulfuric acid (H2SO4) is:02.200.64:02.207.32:02.202.2= 1 : 15.9 : 31.7If we have 7.27 g H, then we will have 7.27 × 15.9 = 116 g S and 7.27 × 31.7 = 230. g O inthe second sample of H2SO4.29.Hydrazine: 1.44 ×110g H/g N;ammonia: 2.16 ×110g H/g N;hydrogen azide:2.40 ×210g H/g N.Let's try all of the ratios:0240.0144.0= 6.00;0240.0216.0= 9.00;0240.00240.0= 1.00;144.0216.0= 1.50 =23All the masses of hydrogen in these three compounds can be expressed as simple whole-number ratios. The g H/g N in hydrazine, ammonia, and hydrogen azide are in the ratios6 : 9 : 1.30.The law of multiple proportions does not involve looking at the ratio of the mass of oneelement with the total mass of the compounds. To illustrate the law of multiple proportions,we compare the mass of carbon that combines with 1.0 g of oxygen in each compound:compound 1:27.2 g C and 72.8 g O(100.027.2 = mass O)compound 2:42.9 g C and 57.1 g O(100.042.9 = mass O)The mass of carbon that combines with 1.0 g of oxygen is:compound 1:Og8.72Cg2.27= 0.374 g C/g Ocompound 2:Og1.57Cg9.42= 0.751 g C/g O12374.0751.0; this supports the law of multiple proportions because this carbon ratio is a wholenumber.

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CHAPTER 1CHEMICAL FOUNDATIONS2831.For CO and CO2, it is easiest to concentrate on the mass of oxygen that combines with 1 g ofcarbon.From the formulas (two oxygen atoms per carbon atom in CO2versus one oxygenatom per carbon atom in CO), CO2will have twice the mass of oxygen that combines pergram of carbon as compared to CO.For CO2and C3O2, it is easiest to concentrate on themass of carbon that combines with 1 g of oxygen. From the formulas (three carbon atoms pertwo oxygen atoms in C3O2versus one carbon atom per two oxygen atoms in CO2), C3O2willhave three times the mass of carbon that combines per gram of oxygen as compared to CO2.As expected, the mass ratios are whole numbers as predicted by the law of multipleproportions.32.Compound I:Qg00.1Rg67.4Qg00.3Rg0.14;compound II:Qg00.1Rg56.1Qg50.4Rg00.7The ratio of the masses of R that combine with 1.00 g Q is:56.167.4= 2.99 ≈ 3As expected from the law of multiple proportions, this ratio is a small whole number.Because compound I contains three times the mass of R per gram of Q as compared withcompound II (RQ), the formula of compound I should be R3Q.33.Mass is conserved in a chemical reaction because atoms are conserved.Chemical reactionsinvolve the reorganization of atoms, so formulas change in a chemical reaction, but thenumber and types of atoms do not change.Because the atoms do not change in a chemicalreaction, mass must not change.In this equation we have two oxygen atoms and fourhydrogen atoms both before and after the reaction occurs.34.Mass is conserved in a chemical reaction.ethanol + oxygenwater + carbon dioxideMass:46.0 g96.0 g54.0 g?Mass of reactants = 46.0 + 96.0 = 142.0 g = mass of products142.0 g = 54.0 g + mass of CO2, mass of CO2= 142.0–54.0 = 88.0 g35.To get the atomic mass of H to be 1.00, we divide the mass of hydrogen that reacts with 1.00g of oxygen by 0.126; that is,126.0126.0= 1.00. To get Na, Mg, and O on the same scale, we dothe same division.Na:126.0875.2= 22.8; Mg:126.0500.1= 11.9;O:126.000.1= 7.94HONaMgRelative value1.007.9422.811.9Accepted value1.00816.0022.9924.31For your information, the atomic masses of O and Mg are incorrect. The atomic masses of Hand Na are close to the values given in the periodic table.Something must be wrong aboutthe assumed formulas of the compounds.It turns out the correct formulas are H2O, Na2O,and MgO. The smaller discrepancies result from the error in the assumed atomic mass of H.

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CHAPTER 1CHEMICAL FOUNDATIONS2936.If the formula is InO, then one atomic mass of In would combine with one atomic mass of O,or:Og000.1Ing784.400.16A, A = atomic mass of In = 76.54If the formula is In2O3, then two times the atomic mass of In will combine with three timesthe atomic mass of O, or:Og000.1Ing784.400.16)3(A2, A = atomic mass of In = 114.8The latter number is the atomic mass of In used in the modern periodic table.The Nature of the Atom37.From section 1-7, the nucleus has “a diameter of about 1013cm” and the electrons “moveabout the nucleus at an average distance of about 108cm from it.”We will use thesestatements to help determine the densities. Density of hydrogen nucleus (contains one protononly):Vnucleus=3403143cm105)cm105()14.3(34r34d = density =31534024g/cm103cm105g1067.1Density of H atom (contains one proton and one electron):Vatom=32438cm104)cm101()14.3(34d =33242824g/cm0.4cm104g109g1067.138.Because electrons move about the nucleus at an average distance of about 1 ×810cm, thediameter of an atom will be about 2 ×810cm. Let's set up a ratio:cm102cm101modelofdiametermm1atomofdiameternucleusofdiameter813; solving:diameter of model = 2 × 105mm = 200 m39.C101.602chargeelectron1C1093.51918= 37 negative (electron) charges on the oil drop40.First, divide all charges by the smallest quantity, 6.40 ×1310:13121040.61056.2=4.00;640.068.7= 12.0;640.084.3= 6.00

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CHAPTER 1CHEMICAL FOUNDATIONS30Because all charges are whole-number multiples of 6.40 ×1310zirkombs, the charge on oneelectron could be 6.40 ×1310zirkombs.However, 6.40 ×1310zirkombs could be thecharge of two electrons (or three electrons, etc.).All one can conclude is that the charge ofan electron is 6.40 ×1310zirkombs or an integer fraction of 6.40 ×1310zirkombs.41.Z is the atomic number and is equal to the number of protons in the nucleus.A is the massnumber and is equal to the number of protons plus neutrons in the nucleus. X is the symbolof the element.See the front cover of the text which has a listing of the symbols for thevarious elements and corresponding atomic number or see the periodic table on the cover todetermine the identity of the various atoms. Because all of the atoms have equal numbers ofprotons and electrons, each atom is neutral in charge.a.Na2311b.F199c.O16842.The atomic number for carbon is 6.14C has 6 protons, 146 = 8 neutrons, and 6 electrons inthe neutral atom.12C has 6 protons, 12–6 = 6 neutrons, and 6 electrons in the neutral atom.The only difference between an atom of14C and an atom of12C is that14C has two additionalneutrons.43.a.7935Br: 35 protons, 7935 = 44 neutrons. Because the charge of the atom is neutral,the number of protons = the number of electrons = 35.b.8135Br: 35 protons, 46 neutrons, 35 electronsc.23994Pu: 94 protons, 145 neutrons, 94 electronsd.13355Cs: 55 protons, 78 neutrons, 55 electronse.31H: 1 proton, 2 neutrons, 1 electronf.5626Fe: 26 protons, 30 neutrons, 26 electrons44.a.23592U: 92 p, 143 n, 92 eb.2713Al: 13 p, 14 n, 13 ec.5726Fe: 26 p, 31 n, 26 ed.20882Pb: 82 p, 126 n, 82 ee.8637Rb: 37 p, 49 n, 37 ef.4120Ca: 20 p, 21 n, 20 e45.a.Element 8 is oxygen. A = mass number = 9 + 8 = 17;178Ob.Chlorine is element 17.3717Clc. Cobalt is element 27.6027Cod.Z = 26; A = 26 + 31 = 57;5726Fee. Iodine is element 53.13153If.Lithium is element 3.73Li

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