Solution Manual For General Chemistry: Principles and Modern Applications, 11th Edition

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PetrucciHerringMaduraBissonnetteP r i n c iPl e sa n dM o d e r na P Pl i c a t i o n sGeneral Chemistrye l e v e n t h e d i t i o nAliciAPAterno PArsi•ArAsh PArsi•tomislAv PintAuer•lucio Gelmini•roberts W.hiltscoM Pl e t es o l u t i o n sMa n u a l05/02/1611:01 am

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COMPLETESOLUTIONSMANUALAlicia Paterno ParsiArash ParsiTomislav PintauerDuquesne UniversityDuquesne UniversityDuquesne UniversityLucio GelminiRobert W. HiltsGrant MacEwan CollegeGrant MacEwan CollegeGeneral ChemistryPrinciples and Modern ApplicationsEleventh EditionRalph H. PetrucciCalifornia State University, San BernardinoF. Geoffrey HerringUniversity of British ColumbiaJeffry D. MaduraDuquesne UniversityCarey BissonnetteUniversity of WaterlooA01_PETR5044_11_CSM_FM.pdf11/29/1610:02:33 PM

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ContentsPreface.............................................................................................................................................. ivChapter 1Matter—Its Properties and Measurement ..................................................................1Chapter 2Atoms and the Atomic Theory.................................................................................27Chapter 3Chemical Compounds..............................................................................................60Chapter 4Chemical Reactions ...............................................................................................110Chapter 5Introduction to Reactions in Aqueous Solutions ...................................................160Chapter 6Gases ......................................................................................................................204Chapter 7Thermochemistry ...................................................................................................256Chapter 8Electrons in Atoms.................................................................................................300Chapter 9The Periodic Table and Some Atomic Properties..................................................346Chapter 10Chemical Bonding I: Basic Concepts ....................................................................376Chapter 11Chemical Bonding II: Valence Bond and Molecular Orbital Theories .................444Chapter 12Intermolecular Forces: Liquids and Solids ............................................................501Chapter 13Spontaneous Change: Entropy and Gibbs Energy .................................................556Chapter 14Solutions and Their Physical Properties ................................................................610Chapter 15Principles of Chemical Equilibrium.......................................................................661Chapter 16Acids and Bases .....................................................................................................717Chapter 17Additional Aspects of Acid–Base Equilibria.........................................................786Chapter 18Solubility and Complex-Ion Equilibria..................................................................887Chapter 19Electrochemistry ....................................................................................................943Chapter 20Chemical Kinetics................................................................................................1011Chapter 21Chemistry of the Main-Group Elements I: Groups 1, 2, 13, and 14....................1065Chapter 22Chemistry of the Main-Group Elements II: Groups 18, 17, 16, 15,and Hydrogen.......................................................................................................1097Chapter 23The Transition Elements ......................................................................................1138Chapter 24Complex Ions and Coordination Compounds......................................................1166Chapter 25Nuclear Chemistry ...............................................................................................1201Chapter 26Structures of Organic Compounds.......................................................................1226Chapter 27Reactions of Organic Compounds .......................................................................1269Chapter 28Chemistry of the Living State ..............................................................................1306A01_PETR5044_11_CSM_FM.pdf31/29/1610:02:35 PM

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1CHAPTER 1MATTER—ITS PROPERTIES AND MEASUREMENTPRACTICE EXAMPLES1A(E)Convert the Fahrenheit temperature to Celsius and compare.5 °C5 °C9 °F9 °F°C°F32 °F350 °F32 °F177 °C.1B(E)We convert the Fahrenheit temperature toCelsius.5 °C5 °C9 °F9 °F°C°F32 °F15 °F32 °F26 °C. The antifreeze only protects to22 Cand thus it will not offer protection to temperatures as low as15 F =26.1 C.2A(E)The mass is the difference between the mass of the full and empty flask.291.4 g108.6 gdensity == 1.46 g/mL125 mL2B(E)First determine the volume required.V= (1.000 × 103g)(8.96 g cm–3) = 111.6 cm3.Next determine the radius using the relationship between volume of a sphere and radius.V= 43r3= 111.6 cm3= 43 (3.1416)r3r=3111.634(3.1416)= 2.987 cm3A(E)The volume of the stone is the difference between the level in the graduated cylinderwith the stone present and with it absent.mass28.4 g rockdensity === 2.76 g/mLvolume44.1 mL rock & water33.8 mL water= 2.76 g/cm33B(E)The water level will remain unchanged. The mass of the ice cube displaces the samemass of liquid water. A 10.0 g ice cube will displace 10.0 g of water. When the ice cubemelts, it simply replaces the displaced water, leaving the liquid level unchanged.4A(E)The mass of ethanol can be found using dimensional analysis.1000 mL0.71 g gasohol10 g ethanol1 kg ethanolethanol mass = 25 L gasohol1 L1 mL gasohol100 g gasohol1000 g ethanol= 1.8 kg ethanol4B(E)We use the mass percent to determine the mass of the 25.0 mL sample.100.0 g rubbing alcoholrubbing alcohol mass15.0 g (isopropyl alcohol)70.0 g (isopropyl alcohol)21.43 g rubbing alcohol21.4 grubbing alcohol density0.857 g/mL25.0 mLM01_PETR5044_11_CSM_C01.pdf11/29/1610:03:14 PM

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Chapter 1: Matter– Its Properties and Measurement25A(M)For this calculation, the value 0.000456 has the least precision (three significantfigures), thus the final answer must also be quoted to three significant figures.62.3560.0004566.42210= 21.335B(M)For this calculation, the value 1.3103has the least precision (two significant figures),thus the final answer must also be quoted to two significant figures.8.21101.3100.002364.07110= 1.11043266A(M)The number in the calculation that has the least precision is 102.1 (+0.1), thus the finalanswer must be quoted to just one decimal place. 0.236 +128.55102.1 = 26.76B(M)This is easier to visualize if the numbers are not in exponential notation.321.30210+ 952.71302 + 952.72255=== 15.615712.221451.571012.22INTEGRATIVE EXAMPLEA(D)Stepwise Approach:First, determine the density of the alloy by the oil displacement.Mass of oil displaced = mass of alloy in air – mass of alloy in oil= 211.5 g – 135.3 g = 76.2 gVOil=m/d= 76.2 g / 0.926 g/mL = 82.3 mL =VMg-AldMg-Al= 211.5 g / 82.3 mL = 2.57 g/cm3Now, since the density is a linear function of the composition,dMg-Al=mx+b, wherexis the mass fraction of Mg, andbis they-intercept.Substituting 0 for x (no Al in the alloy), everything is Mg and the equation becomes:1.74 =m0 +b. Therefore,b= 1.74Assuming 1 forx(100% by weight Al):2.70 = (m× 1) + 1.74, therefore,m= 0.96Therefore, for an alloy:2.57 = 0.96x+ 1.74x= 0.86 = mass % of AlMass % of Mg = 1 – 0.86 = 0.14, 14%M01_PETR5044_11_CSM_C01.pdf21/29/1610:03:14 PM

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Chapter 1: Matter– Its Properties and Measurement3B(M)Stepwise approach:Mass of seawater =dV= 1.027 g/mL × 1500 mL = 1540.5 g2.67 g NaCl39.34 g Na1540.5 g seawater16.18 g Na100 g seawater100 g NaClThen, convert mass of Na to atoms of Na23261 kg Na1 Na atom16.18 g Na4.23910Na atoms1000 g Na3.81710kg NaConversion Pathway:262.67 g NaCl39.34 g Na1 kg Na1 Na atom1540.5 g seawater100 g seawater100 g NaCl1000 g Na3.817510kg NaEXERCISESThe Scientific Method1.(E)One theory is preferred over another if it can correctly predict a wider range ofphenomena and if it has fewer assumptions.2.(E)No. The greater the number of experiments that conform to the predictions of the law,the more confidence we have in the law. There is no point at which the law is ever verifiedwith absolute certainty.3.(E)For a given set of conditions, a cause, is expected to produce a certain result or effect.Although these cause-and-effect relationships may be difficult to unravel at times (“God issubtle”), they nevertheless do exist (“He is not malicious”).4.(E)As opposed to scientific laws, legislative laws are voted on by people and thus aresubject to the whims and desires of the electorate. Legislative laws can be revoked by agrassroots majority, whereas scientific laws can only be modified if they do not accountfor experimental observations. As well, legislative laws are imposed on people, who areexpected to modify their behaviors, whereas, scientific laws cannot be imposed on nature,nor will nature change to suit a particular scientific law that is proposed.5.(E)The experiment should be carefully set up so as to create a controlled situation in whichone can make careful observations after altering the experimental parameters, preferably oneat a time. The results must be reproducible (to within experimental error) and, as more andmore experiments are conducted, a pattern should begin to emerge, from which acomparison to the current theory can be made.M01_PETR5044_11_CSM_C01.pdf31/29/1610:03:14 PM

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Chapter 1: Matter– Its Properties and Measurement46.(E)For a theory to be considered plausible, it must, first and foremost, agree with and/orpredict the results from controlled experiments. It should also involve the fewest number ofassumptions. The best theories predict new phenomena that are subsequently observed afterthe appropriate experiments have been performed.Properties and Classification of Matter7.(E)When an object displays a physical property it retains its basic chemical identity. Bycontrast, the display of a chemical property is accompanied by a change in composition.(a)Physical: The iron nail is not changed in any significant way when it is attracted to amagnet. Its basic chemical identity is unchanged.(b)Chemical: The paper is converted to ash, CO2(g), and H2O(g) along with the evolutionof considerable energy.(c)Chemical: The green patina is the result of the combination of water, oxygen, andcarbon dioxide with the copper in the bronze to produce basic copper carbonate.(d)Physical: Neither the block of wood nor the water has changed its identity.8.(E)When an object displays a physical property it retains its basic chemical identity. Bycontrast, the display of a chemical property is accompanied by a change in composition.(a)Chemical: The change in the color of the apple indicates that a new substance(oxidized apple) has formed by reaction with air.(b)Physical: The marble slab is not changed into another substance by feeling it.(c)Physical: The sapphire retains its identity as it displays its color.(d)Chemical: After firing, the properties of the clay have changed from soft and pliableto rigid and brittle. New substances have formed. (Many of the changes involvedriving off water and slightly melting the silicates that remain. These moltensubstances cool and harden when removed from the kiln.)9.(E)(a)Homogeneous mixture: Air is a mixture of nitrogen, oxygen, argon, and traces ofother gases. By “fresh,” we mean no particles of smoke, pollen, etc., are present. Suchspecies would produce a heterogeneous mixture.(b)Heterogeneous mixture: A silver-plated spoon has a surface coating of the elementsilver and an underlying baser metal (typically iron). This would make the coatedspoon a heterogeneous mixture.(c)Heterogeneous mixture: Garlic salt is simply garlic powder mixed with table salt.Pieces of garlic can be distinguished from those of salt by careful examination.(d)Substance: Ice is simply solid water (assuming no air bubbles).M01_PETR5044_11_CSM_C01.pdf41/29/1610:03:14 PM

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Chapter 1: Matter– Its Properties and Measurement510.(E)(a)Heterogeneous mixture: We can clearly see air pockets within the solid matrix.On close examination, we can distinguish different kinds of solids by their colors.(b)Homogeneous mixture: Modern inks are solutions of dyes in water. Older inks oftenwere heterogeneous mixtures: suspensions of particles of carbon black (soot) in water.(c)Substance: This is assuming that no gases or organic chemicals are dissolved in thewater.(d)Heterogeneous mixture: The pieces of orange pulp can be seen through a microscope.Most “cloudy” liquids are heterogeneous mixtures; the small particles impede thetransmission of light.11.(E)(a)If a magnet is drawn through the mixture, the iron filings will be attracted to themagnet and the wood will be left behind.(b)When the glass-sucrose mixture is mixed with water, the sucrose will dissolve,whereas the glass will not. The water can then be boiled off to produce pure sucrose.(c)Olive oil will float to the top of a container and can be separated from water, which ismore dense. It would be best to use something with a narrow opening that has theability to drain off the water layer at the bottom (i.e., buret).(d)The gold flakes will settle to the bottom if the mixture is left undisturbed. The waterthen can be decanted (i.e., carefully poured off).12.(E)(a)Physical: This is simply a mixture of sand and sugar (i.e., not chemically bonded).(b)Chemical: Oxygen needs to be removed from the iron oxide.(c)Physical: Seawater is a solution of various substances dissolved in water.(d)Physical: The water-sand slurry is simply a heterogeneous mixture.Exponential Arithmetic13.(E) (a)38950. = 8.95010 (4 sig. fig.)(b)410, 700. = 1.070010 (5 sig. fig.)(c)0.0240 = 2.40102(d)0.0047 = 4.7103(e)938.3 = 9.383102(f)275,482 = 2.7548210514.(E)(a)335.1210= 0.00512(b)558.0510= 0.0000805(c)44291.110= 0.02911(d)2272.110= 0.72115.(E)(a)34,000 centimeters / second = 3.4104cm/s(b)36378 km6.37810=´km(c)(trillionth = 110–12) hence, 7410–12m or 7.410–11m(d)323533(2.210 )(4.710 )2.7104.6105.8105.810M01_PETR5044_11_CSM_C01.pdf51/29/1610:03:14 PM

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Chapter 1: Matter– Its Properties and Measurement616.(E)(a)173 thousand trillionwattsW= 173,000,000,000,000,000= 1.731017W(b)oneten-millionth of a meter =7110, 000, 000 m = 1 10m(c)(trillionth = 110–12) hence, 14210–12m or 1.4210–10m(d)2432(5.0710 )1.8100.16== 1.60.0980.065 + 3.310Significant Figures17.(E)(a)An exact number—500 sheets in a ream of paper.(b)Pouring the milk into the bottle is a process that is subject to error; there can beslightly more or slightly less than one liter of milk in the bottle. This is a measuredquantity.(c)Measured quantity: The distance between any pair of planetary bodies can only bedetermined through certain astronomical measurements, which are subject to error.(d)Measured quantity: The internuclear separation quoted for O2is an estimated valuederived from experimental data, which contains some inherent error.18.(E)(a)The number of pages in the text is determined by counting; the result is an exactnumber.(b)An exact number: Although the number of days can vary from one month to another(say, from January to February), the month of January always has 31 days.(c)Measured quantity: The area is determined by calculations based on measurements.These measurements are subject to error.(d)Measured quantity: Average internuclear distance for adjacent atoms in a gold medalis an estimated value derived from X-ray diffraction data, which contain someinherent error.19.(E)Each of the following is expressed to four significant figures.(a)3984.63985(b)422.04422.0(c)186,000 = 1.860105(d)33,9003.390104(e)6.321104is correct(f)5.0472105.047104420.(E)(a)450 has two or three significant figures; trailing zeros left of the decimal areindeterminate, if no decimal point is present.(b)98.6 has three significant figures; non-zero digits are significant.(c)0.0033 has two significant digits; leading zeros are not significant.(d)902.10 has five significant digits; trailing zeros to the right of the decimal point aresignificant, as are zeros flanked by non-zero digits.(e)0.02173 has four significant digits; leading zeros are not significant.M01_PETR5044_11_CSM_C01.pdf61/29/1610:03:14 PM

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Chapter 1: Matter– Its Properties and Measurement7(f)7000 can have anywhere from one to four significant figures; trailing zeros left of thedecimal are indeterminate, if no decimal point is shown.(g)7.02 has three significant figures; zeros flanked by non-zero digits are significant.(h)67,000,000 can have anywhere from two to eight significant figures; there is no wayto determine which, if any, of the zeros are significant, without the presence of adecimal point.21.(E)(a)40.4060.0023 = 9.310(b)0.135716.800.096 = 2.2101(c)0.458 + 0.120.037 = 5.4101(d)32.18 + 0.0551.652 = 3.05810122.(M)(a)32024.90.080= 3.2102.49108.010= 1.0102125(b)432.76.50.002300620.103= 4.327106.52.300106.2101.0310= 1.02311(c)111132.44 + 4.90.3043.24410 + 4.93.0410== 4.471082.948.29410(d)8.002 + 0.304013.40.066 +1.02 =8.002 + 3.040101.34106.610+1.02 = 5.7910112123.(M)(a)42.4410(b)31.510(c)40.0(d)32.131 10(e)34.81024.(M)(a)17.510(b)126.310(c)34.610(d)11.05810(e)34.21025.(M)(a)The average speed is obtained by dividing the distance traveled (in miles) by theelapsed time (in hours). First, we need to obtain the elapsed time, in hours.24 h1 h1 h9 days= 216.000 h3min= 0.050 h44 s= 0.012 h1 d60 min3600 stotal time = 216.000 h + 0.050 h + 0.012 h = 216.062 h25, 012 mi1.609344 kmaverage speed == 186.30 km/h216.062 h1 miM01_PETR5044_11_CSM_C01.pdf71/29/1610:03:14 PM

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Chapter 1: Matter– Its Properties and Measurement8(b)First compute the mass of fuel remainingmassgalqtgalLqtmLLgmLlbglb= 14410.94641100010.7011453.6= 82Next determine the mass of fuel used, and then finally, the fuel consumption.Notice that the initial quantity of fuel is not known precisely, perhaps at best to thenearest 10 lb, certainly (“nearly 9000 lb”) is not to the nearest pound.0.4536 kgmass of fuel used = (9000 lb82 lb)4045 kg1 lb25, 012 mi1.609344 kmfuel consumption == 9.95 km/kg or ~10 km/kg4045 kg1 mi26.(M)If the proved reserve truly was an estimate, rather than an actual measurement, it wouldhave been difficult to estimate it to the nearest trillion cubic feet. A statement such as2,911,000 trillion cubic feet (or even183310ft) would have more realistically reflected theprecision with which the proved reserve was known.Units of Measurement27.(E)(a)0.12710001= 127LmLLmL(b)15.811000= 0.0158mLLmLL(c)98111000= 0.98133cmLcmL(d)3363100 cm2.65 m= 2.6510cm1 m28.(E)(a)31000 g2.35 kg= 2.3510g1 kg(b)1 kg792 g= 0.792 kg1000 g(c)1 cm3869 mm= 386.9 cm10 mm(d)10 mm0.043 cm= 0.43 mm1 cm29.(E)(a)2.54 cm68.4 in.= 174 cm1 in.(b)9412.12.541.1100= 29ftinftcminmcmm(c)1.42453.61= 644lbglbg(d)2480.45361= 112lbkglbkg(e)334 qt0.9464 dm1.85 gal7.00 dm1 gal1 qt(f)30.9464 L1000 mL3.72 qt= 3.5210mL1 qt1 LM01_PETR5044_11_CSM_C01.pdf81/29/1610:03:14 PM

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Chapter 1: Matter– Its Properties and Measurement930.(M)(a)22621000 m1.00 km= 1.0010m1 km(b)3363100 cm1.00 m= 1.0010cm1 m(c)22625280 ft12 in.2.54 cm1 m1.00 mi= 2.5910m1 mi1 ft1 in.100 cm31.(E)Express both masses in the same units for comparison.31 g10mg3245 g3.245 mg,61 g10gmmæöæö÷ç÷÷çç÷´´=÷çç÷÷ç÷çç÷èø÷çèøwhich is larger than 0.00515 mg.32.(E)Express both masses in the same units for comparison.1000 g0.000475 kg= 0.4751 kgg,which is smaller than 3257110= 3.2573mggmgg.33.(E)Conversion pathway approach:heighthandsinhandcminmcm= 154.12.541.1100= 1.5mStepwise approach:4 in.15 hands60 in.1 hand2.54 cm60 in.152.4 cm1 in.1 m152.4 cm= 1.524 m = 1.5 m100 cm34.(M)A mile is defined as being 5280 ft in length. We must use this conversion factor to findthe length of a link in inches.1 chain1 furlong1 mi5280 ft12 in.2.54 cm1.00 link= 20.1 cm100 links10 chains8 furlongs1 mi1 ft1 in.35.(M)(a)We use the speed as a conversion factor, but need to convert yards into meters.9.3 s1 yd39.37 in.time = 100.0 m= 10. s100 yd36 in.1 mThe final answer can only be quoted to a maximum of two significant figures.M01_PETR5044_11_CSM_C01.pdf91/29/1610:03:14 PM

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Chapter 1: Matter– Its Properties and Measurement10(b)We need to convert yards to meters.100 yd36 in.2.54 cm1 mspeed == 9.83 m/s9.3 s1 yd1 in.100 cm(c)The speed is used as a conversion factor.1min1000 m1 stime = 1.45 km= 2.5 min1 km9.83 m60 s36.(M)(a)mass25.0 gr1.0 g1000 mgmg= 2 tablets= 6.7101 tablet15 gr1 gmg(b)dosage26.710mg1 lb1000 grate == 9.2161 lb453.6 g1 kgmg aspirin/kg body weight(c)31000 g2 tablets1 daytime = 1.0 kg= 1.510days1 kg0.67 g2 tablets37.(D)222100 m100 cm1 in.1 ft1 mi640 acres1 hectare = 1 hm1 hm1 m2.54 cm12 in.5280 ft1 mi1 hectare = 2.47 acres38.(D)Here we must convert pounds per cubic inch into grams per cubic centimeter:density for metallic iron =30.284 lb1 in.454 g1 lb33(1 in.)(2.54 cm)= 7.873gcm39.(D)232232 lb453.6 g1 in.pressure == 2.210g/cm1 lb2.54 cm1 in.234222.210g1 kg100 cmpressure == 2.210kg/m1000 g1 m1 cm40.(D)First we will calculate the radius for a typical red blood cell using the equation for thevolume of a sphere.V= 4/3r3= 90.010–12cm3r3= 2.1510–11cm3andr= 2.7810–4cmThus, the diameter is 2r= 22.7810–4cm10 mm1 cm= 5.5610–3mmM01_PETR5044_11_CSM_C01.pdf101/29/1610:03:14 PM

Solution Manual For General Chemistry: Principles and Modern Applications, 11th Edition - Page 15 preview image

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Chapter 1: Matter– Its Properties and Measurement11Temperature Scales41.(E)low:9F9 °F5 °C5 °C(°F)(°C) + 32 °F10 °C + 32 °F14 °Ftthigh:9 °F9 °F5 °C5 °C(°F)(°C)32 °F50 °C32 °F122 °Ftt42.(E)high:5 °C5 °C9 °F9 °F(°C) =(°F)32 °F= 118 °F32 °F= 47.8 °C48 °Cttlow:5 °C5 °C9 °F9 °F(°C)(°F)32 °F17 °F32 °F8.3 °Ctt 43.(M)Let us determine the Fahrenheit equivalent of absolute zero.9 °F9 °F5 °C5 °C(°F)(°C)32 °F273.15 °C + 32 °F459.7 °Ftt A temperature of465 °Fcannot be achieved because it is below absolute zero.44.(M)Determine the Celsius temperature that corresponds to the highest Fahrenheittemperature,240 °F.5 °C5 °C9 °F9 °F(°C)(°F)32 °F240 °F32 °F116 °CttBecause116 °Cis above the range of the thermometer, this thermometer cannot beused in this candy-making assignment.45.(D)(a)From the data provided we can write down the following relationship:–38.9C = 0M and 356.9C = 100M. To find the mathematical relationship betweenthese two scales, we can treat each relationship as a point on a two-dimensional Cartesiangraph:Therefore, the equation for the line isy= 3.96x– 38.9. The algebraic relationshipbetween the two temperature scales ist(C) = 3.96(M) – 38.9 or rearranging,t(M) =( C)38.93.96tAlternatively, note that the change in temperature in °C corresponding to a change of100 °M is [356.9 – (–38.9)] = 395.8 °C, hence, (100 °M/395.8 °C) = 1 °M/3.96 °C.This factor must be multiplied by the number of degrees Celsius above zero on the Mscale. This number of degrees ist(°C) + 38.9, which leads to the general equationt(°M) = [t(°C) + 38.9]/3.96.The boiling point of water is 100C, corresponding tot(M) = 10038.93.96= 35.1M(b)t(°M) =273.1538.93.96= –59.2M would be the absolute zero on this scale.M01_PETR5044_11_CSM_C01.pdf111/29/1610:03:14 PM

Solution Manual For General Chemistry: Principles and Modern Applications, 11th Edition - Page 16 preview image

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Chapter 1: Matter– Its Properties and Measurement1246.(D)(a)From the data provided we can write down the following relationship:–77.75 = 0A and –33.35C = 100A. To find the mathematical relationshipbetween these two scales, we can treat each relationship as a point on a two-dimensional Cartesian graph.Therefore, the equation for the line isy= 0.444x– 77.75The algebraic relationship between the two temperature scales ist(C) = 0.444(A) – 77.75 or rearrangingt(A) =( C)77.750.444tThe boiling point of water (100C) corresponds tot(A) = 10077.750.444= 400A(b)t(A) =273.1577.750.444= –440ADensity47.(E)butyric acidmass2088 g1 Ldensity0.958 g/mLvolume2.18 L1000 mL48.(E)chloroformmass22.54 kg1 L1000 gdensity1.48 g/mLvolume15.2 L1000 mL1 kg49.(M)The mass of acetone is the difference in masses between empty and filled masses.Conversion pathway approach:437.5 lb75.0 lb453.6 g1 gal1 Ldensity0.790 g/mL55.0 gal1 lb3.785 L1000 mLStepwise approach:5437.5 lb75.0 lb362.5 lb453.6 g362.5 lb1.64410 g1 lbM01_PETR5044_11_CSM_C01.pdf121/29/1610:03:14 PM

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