Solution Manual for Precalculus , 10th Edition

Solution Manual for Precalculus, 10th Edition offers the best solutions to textbook problems, helping you prepare for exams and assignments.

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CHAPTER 21.a)qp=ΔH= 1 L×0.9982 kg L–1×2447 kJ kg–1= 2443 kJb)ΔT= 2443 kJ/(60 kg×4.184 kJ K–1kg–1) = 9.7 Kc)C12H22O11(s) + 12O2(g)→12 CO2(g) + 11H2O (l)ΔH= 11(–285.83 kJ mol–1) + 12(–393.51 kJ mol–1) – (–2222.10 kJ mol–1)= –5644 kJ mol–1Msucrose= 342.31 g mol–1msucrose= 2443 kJ×342.31 g mol–1/5644 kJ mol–1= 148.2 g2.a)1 kg of carbon is (1.0 kg×1000 g kg–1)/(12.011 g mol–1) = 83.3 mol83.3 mol CO2×8.3145 J mol–1K–1×298 K/(105Pa) = 2.06 m3CO22.06 m3CO2/(0.000390 m3CO2/m3air ) = 5292 m3airb)p =F/A=mg/A =105Pa.m=Ap/g= 1 m2×105Pa/(9.80662 m s–2) = 10197 kgMair= 0.8×0.0280 kg mol–1+ 0.2×0.0320 kg mol–1= 0.0288 kg/(mol air)nair= 10197 kg/0.0288 kg/(mol air) = 354069 mol airnCO2= 354069 mol air×0.000390 mol CO2/(mol air) = 138.09 mol CO2mC= 0.012011 kg/mol C×138.09 mol CO2= 1.659 kg Cc)1.659 kg C/(1 kg y–1) = 1.659 y.3.a)w=mgh= 10 kg×9.81 m s–2×10 m = 981 Jb)k=F/(x –x0) = 5.00 N/(0.105 m – 0.100 m) = 1.00×103N m–1w= (k/2)(x–x0)2= 1.00×103N m–1×(0.01 m)2/2 = 0.050 JΔd/dsurface×100% = 1.2%c)w= –p(V2–V1) = –105Pa×0.002 m3= –200 J.d)w= –2×10–4Je)w= –nRTln(V2/V1) = –p1V1ln(V2/V1) = –105Pa×10–3m3×ln 3 = –110 J.4.a)w=f dx=010∫(20x+x2)dx=010∫10x2+13x3010=(1000+333)×10−21J = 1.333×10−18Jb)w=1.333×10−18J×6.022×1023mol–1=8.03×105J mol–15.a)p =ρgh= 1025 kg m–3×9.80662 m s–2×2500 m = 2.513×107Pa = 251.3 barb)ΔV/V= –κΔp= –(2.513×107Pa)×4.9458×10−10Pa−1= –0.0124So the densityincreasesby a factor of 0.0124, or 1.24 %c)p1V1/T1=p2V2/T2, soV2=p1V1T2/(T1p2) = 1 bar×10 L×277 K/(293 K×252.3 bar)

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Precalculus by Michael Sullivan

Solution Manual for Precalculus , 10th Edition

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