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8. Calculate the heat required to convert 5,5 g of ice at -25°C to steam at 118°C. 9=mcST onoH q=(5.5g)(2.0951g.oun) (o-(-25)°C) -287.37S J bebivo Hfus mol HzO +118°C 100 5.5g H20)=1,836KJ=\836J 18.0gimal =(6.0115)mol) q=mcAT 25°C q =C5.5g) (4.1831g.n)(lo0-0°c) = 2299 J q=Huap, mmol 4: (40,67nimai) tizO 18,0gima J=12427 J )= 12.427 h q=MCAT 4=Cs.Sg)(1.84 31g.6)Cl18-100°C)-182.16 J 1743 287.3755+18365+2299st124275182.16J=170313= phasiom erioslea nfog grilad arigit.er asri Jbloeo

8. Calculate the heat required to convert 5,5 g of ice at -25°C to steam at 118°C. 9=mcST onoH q=(5.5g)(2.0951g.oun) (o-(-25)°C) -287.37S J bebivo Hfus mol HzO +118°C 100 5.5g H20)=1,836KJ=\836J 18.0gimal =(6.0115)mol) q=mcAT 25°C q =C5.5g) (4.1831g.n)(lo0-0°c) = 2299 J q=Huap, mmol 4: (40,67nimai) tizO 18,0gima J=12427 J )= 12.427 h q=MCAT 4=Cs.Sg)(1.84 31g.6)Cl18-100°C)-182.16 J 1743 287.3755+18365+2299st124275182.16J=170313= phasiom erioslea nfog grilad arigit.er asri Jbloeo

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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8. Calculate the heat required to convert 5,5 g of ice at -25°C to steam at 118°C. 9=mcST onoH q=(5.5g)(2.0951g.oun) (o-(-25)°C) -287.37S J bebivo Hfus mol HzO +118°C 100 5.5g H20)=1,836KJ=\836J 18.0gimal =(6.0115)mol) q=mcAT 25°C q =C5.5g) (4.1831g.n)(lo0-0°c) = 2299 J q=Huap, mmol 4: (40,67nimai) tizO 18,0gima J=12427 J )= 12.427 h q=MCAT 4=Cs.Sg)(1.84 31g.6)Cl18-100°C)-182.16 J 1743 287.3755+18365+2299st124275182.16J=170313= phasiom erioslea nfog grilad arigit.er asri Jbloeo
Transcribed Image Text:8. Calculate the heat required to convert 5,5 g of ice at -25°C to steam at 118°C. 9=mcST onoH q=(5.5g)(2.0951g.oun) (o-(-25)°C) -287.37S J bebivo Hfus mol HzO +118°C 100 5.5g H20)=1,836KJ=\836J 18.0gimal =(6.0115)mol) q=mcAT 25°C q =C5.5g) (4.1831g.n)(lo0-0°c) = 2299 J q=Huap, mmol 4: (40,67nimai) tizO 18,0gima J=12427 J )= 12.427 h q=MCAT 4=Cs.Sg)(1.84 31g.6)Cl18-100°C)-182.16 J 1743 287.3755+18365+2299st124275182.16J=170313= phasiom erioslea nfog grilad arigit.er asri Jbloeo
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