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UWW CHEM 104 - Review for Chapters 12-13

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Chem 104 1st Edition Lecture 21 Outline of Last Lecture I. Continuation of lecture on rates.a. Measuring reaction rateb. Continuous monitoringc. Sampling reaction mixture at specific timesd. Factors affecting reaction ratei. Nature of reactantsii. Temperatureiii. Catalystsiv. Reactant concentrationOutline of Current LectureI. Review of practice examCurrent LectureI. Review of practice exam:a. 1. Determine the partial pressure of oxygen necessary to form an aqueous solution that is 4.1 × 10-4 M O2 at 25°C. The Henry's law constant for oxygen in water at 25°C is 1.3 × 10-3 M/atm.Sgas=kHPgasPO2=SgaskH=4.1∗10−4M1.3∗1 0−3Ma tm=0.32 atmb. 3. Calculate the molality of a solution formed by dissolving 27.8 g of LiI in 500.0 mL of water. m=¿mol solutemass kg solventM=¿molesV(l)¿moles LiI=27.8 g∗(1 mol133.8 g)=.207 mol500.0 ml H2O∗(1 g1 ml)∗(1 kg1000 g)=.500 kgm=.207 mol.500 kg=.415 mc. 2. Parts per billion requires a multiplication factor of ________.109d. 4. What mass (in g) of NH3 must be dissolved in 475 g of methanol to make a 0.250 m solution?m=¿mol solutemass kg solvent.250=¿mol solute.475 kg=.11875= .119 mol.119mol∗17 MM=2.01875 g=2.02 ge. 5. Determine the molality of a solution prepared by dissolving 0.500 moles of CaF2 in 11.5 moles H2O.f. 6. Commercial grade HCl solutions are typically 39.0% (by mass) HCl in water. Determine the molarity of the HCl, if the solution has a density of 1.20 g/mL.M=¿molessoluteV(l)solution39 g∗(1 mol36.4 g)=1.07 molvolume=massdensity=100 g1.20 g /ml=83.3 ml83.3 ml∗(1 L1000 ml)=.0833 LM=1.07 mol.0833 L=12.8 Mg. 7. A solution is prepared by dissolving 98.6 g of NaCl in enough water to form 875mL of solution. Calculate the mass % of the solution if the density of the solutionis 1.06 g/mL.% mass=mass solute ∈solutionmass solution∗100mass=density∗volume=1.06gml∗875 ml=927.5 g% mass=98.6 g NaClmass solution∗100=98.6 g NaCl927.5 g∗100=10.6 %h. 8. Calculate the mole fraction of total ions in an aqueous solution prepared by dissolving 0.400 moles of MgCl2 in 850.0 g of water.mole fractionof A , XA=moles solutetotal moles∈solutionmole fractionof Xtotal ions=nMg+nClnMg+nCl+nH 2Omole fractionof Xtotal ions=.400 mol+(2).400 mol.400 mol +(2).400 mol+47.2 mol=0.0252


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UWW CHEM 104 - Review for Chapters 12-13

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