General Chemistry 11th Edition Ebbing Solution Manual – Full Solutions, Exercises & Answers

CHAPTERV1
ChemistryVandVMeasurement


■ SOLUTIONSVTOVEXERCISES
NoteVonVsignificantVfigures:VIfVtheVfinalVanswerVtoVaVsolutionVneedsVtoVbeVroundedVoff,VitVisVgive
nVfirstVwithVoneVnonsignificantVfigure,VandVtheVlastVsignificantVfigureVisVunderlined.VTheVfinalVans
werVisVthenVroundedVtoVtheVcorrectVnumberVofVsignificantVfigures.VInVmultistepVproblems,Vintermed
iateVanswersVareVgivenVwithVatVleastVoneVnonsignificantVfigure;Vhowever,VonlyVtheVfinalVanswerVha
sVbeenVroundedVoff.

1.1. FromVtheVlawVofVconservationVofVmass,
MassVofVwoodV+VmassVofVairV=VmassVofVashV+Vmass
VofVgasesVSubstituting,VyouVobtain

1.85VgramsV+V9.45VgramsV=V0.28VgramsV+VmassVofVgases
or,
MassVofVgasesV=V(1.85V+V9.45V−V0.28)VgramsV=V11.02Vgrams
Thus,VtheVmassVofVgasesVinVtheVvesselVatVtheVendVofVtheVexperimentVisV11.02Vgrams.

1.2. PhysicalVproperties:Vsoft,Vsilvery-coloredVmetal;VmeltsVatV64°C.
ChemicalVproperties:VreactsVvigorouslyVwithVwater,VwithVoxygen,VandVwithVchlorine.

1.3. a.
TheVfactorV9.1VhasVtheVfewestVsignificantVfigures,VsoVtheVanswerVshouldVbeV
reportedVtoVtwoVsignificantVfigures.
5.61V V 7.89
=V4.86V=V4.9
1
9.1
b. TheVnumberVwithVtheVleastVnumberVofVdecimalVplacesVisV8.91.VTherefore,VroundVtheV
answerVtoVtwoVdecimalVplaces.
8.91V−V6.435V=V2.475V=V2.48
c. TheVnumberVwithVtheVleastVnumberVofVdecimalVplacesVisV6.81.VTherefore,VroundVtheV
answerVtoVtwoVdecimalVplaces.
6.81V−V6.730V=V0.080V=V0.08
d. YouVfirstVdoVtheVsubtractionVwithinVparentheses.VInVthisVstep,VtheVnumberVwithVth
eVleastVnumberVofVdecimalVplacesVisV6.81,VsoVtheVresultVofVtheVsubtractionVhasVtw
oVdecimalVplaces.VTheVleastVsignificantVfigureVforVthisVstepVisVunderlined.
38.91VV(6.81V−V6.730)V=V38.91VV0.080
Next,VperformVtheVmultiplication.VInVthisVstep,VtheVfactorV0.080VhasVtheVfewestVsignif
icantVfigures,VsoVroundVtheVanswerVtoVoneVsignificantVfigure.

©V2017VCengageVLearning.VAllVRightsVReserved.VMayVnotVbeVscanned,VcopiedVorVduplicated,VorVpostedVtoVaVpubliclyVaccessibleVwebsite,Vi
nVwholeVorVinVpart.

, 38.91VV0.080V=V3.11V=V3

1




©V2017VCengageVLearning.VAllVRightsVReserved.VMayVnotVbeVscanned,VcopiedVorVduplicated,VorVpostedVtoVaVpubliclyVaccessibleVwebsite,Vi
nVwholeVorVinVpart.

,
, 2 ChapterV 1:V ChemistryV andV Measureme
nt

1.4. a. 1.84VxV10−9VmV=V1.84Vnm
b. 5.67VxV10−12VsV=V5.67Vps
c. 7.85VxV10−3VgV=V7.85Vmg
d. 9.7VxV103VmV=V9.7Vkm
e. 0.000732VsV=V0.732Vms,VorV732Vµs
f. 0.000000000154VmV=V0.154Vnm,VorV154Vpm

1.5. a. Substituting,VweVfindVthat
5CV 5CV
t C V =V V(tFV −V32°F)V=V V(102.5°FV−V32°F)V=V39.167°C
9FV 9FV
=V39.2°C
b. Substituting,VweVfindVth
at
V 1VKVV V 1VKVV
TKV =VV c tV V V  +V273.15 VKV=V −78CV V +V273.15VKV=V195.15VK
1CV 1CV
   
=V195VK

1.6. RecallVthatVdensityVequalsVmassVdividedVbyVvolume.VYouVsubstituteV159VgVforVtheVmass
VandV20.2Vg/cm3VforVtheVvolume.

dV = V 159Vg
= =V7.871Vg/cm3V=V7.87Vg/cm3
m 20.2Vcm
3
V
TheVdensityVofVtheVmetalVequalsVthatVofViron.

1.7. RearrangeVtheVformulaVdefiningVtheVdensityVtoVobtainVtheVvolume.
m
VV=V
d
SubstituteV30.3VgVforVtheVmassVandV0.789Vg/cm3VforVtheVdensity.
30.3Vg
VV= =V38.40Vcm3V=V38.4Vcm3
0.789Vg/cm
3




1.8. SinceVoneVpmV=V10−12Vm,VandVtheVprefixVmilli-VmeansV10−3,VyouVcanVwrite
−12V 1VmmV V
121VpmV 10 V =V1.21VV10−7Vmm
−3V
 mV1V 10 m
pm
3
V10−10V mV
67.6VÅ VV  V 1Vd V V =V6.76VV10−26Vdm3

3
1.9. m 3

  −

V10V mV 
1V
V V 1VÅ 

1.10. FromVtheVdefinitions,VyouVobtainVtheVfollowingVconversionVfactors:
©V2017VCengageVLearning.VAllVRightsVReserved.VMayVnotVbeVscanned,VcopiedVorVduplicated,VorVpostedVtoVaVpubliclyVaccessibleVwebsite,Vi
nVwholeVorVinVpart.