Chemistry And Life In The Laboratory: Experiments, 6th Edition Solution Manual

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Experiment 1 - “Thinking Metric”Reagents -NoneSpecial Equipment -meter sticks8½"x 11" pieces of paperstring and scissorsrectangular objects such as boxes or blockspieces of chalknickel coins100 tablets (aspirin, vitamin) or other small objectspebbles, rubber stoppers, or metal shotSuggestions and Precautions -The key is to have the students “think metric” rather than to stress conversions.Prelaboratory Questions -1.For: allow for easier importation and exportation of trade products.Against: expense of conversion; resistance to re-education.2.mile, kilometer, meter, foot, inch, centimeter, micron, Angstrom.3.barrel, cubic foot, gallon, liter, milliliter, cubic milliliter.4.The density is less than that of water.5.Overall density, including air cavity, is less; otherwise greater. Bodies can be made tofloat or sink depending on chest expansion.Observations and Results -Part 3- Source of discrepancy - balance is off; personal error. A gram is about a fifth of a nickel’smass. Imagine a nickel that is cut into five pieces. Average weight.Part 4- Wood samples of same kind? - yes; samples of different wood? - no. Relative densities ofwood and water - density of wood is less than that of water. Wood would float onmercury because wood’s density is less than 13.6 g/ml.

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Questions and Problems -1.3.785124 gal •90.8411 gal=2.110 km1 mi•68.4 mi/hrhr1.609 km=3.2.5 cm • 12 cm • 20 cm = 600 cu cm; 600 g water4.4 years • 186,300 mi/sec • 60 sec/min • 24 hr/da • 365 da/yr = 23,500,000,000,000 miles5.wt (lb)÷2.2 lb/kg6.60 min/hr x 24 hr/da x da/10 cda = 144 min/cda365 day/yr x 1 kda/1000 da = 0.365 yr/kda7.(a) Probably safe(b) 0.5 L/sec x 60 sec/min x 60 min/hr x 8 hr = 14,400 L(c) 14,400 L x 1.2 g/L = 17,280 g; 12,200 g = 12.2 kg(d) 17,280 g x 1/106 = 0.017 g of X; 17 mg of X(e) Move information needed

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Experiment 2 - “Molecular Motion - The Particulate Nature of Matter”Reagents -Each Student100 Studentsmethyl salicylate, C8H8O3need 0.5 ml per demonstrationhydrochloric acid, HCl, 12 M500 mlammonium hydroxide, NH4OH, 15 M500 mlacetone, C3H6O1 ml100 mlsodium chloride, NaCl4 g400 gp-dichlorobenzene, C6H4Cl20.25 g25 giodine, I20.3 g30 gpotassium permanganate, KMnO40.2 g20 gSpecial Equipment -watch glass (for group experiment)hot plate (for group experiment)time piece with second hand (for group experiment)chalkstring and meter stick60 cm x 2 cm glass tube (for demonstration)cotton plugs (for demonstration)thermometer, °Clined, white paperSuggestions and Precautions -To avoid pre-release, bring the correct amount of oil of wintergreen in a closed vial and drop itdirectly onto the watch glass on a heated hotplate. Have no drafts in the room (open windows,hood drafts, etc.). Use only small crystals of iodine or the vapor will cause problems in the room.Define the size of a small pea.Prelaboratory Questions -1.Molecules of perfume vaporize and travel across the room to sensory cells in the nose.2.The intermolecular distance in a gas is much greater.3.Solids: touching each other in a regular three-dimensional patter. Liquids: randomarrangement with molecules in close contact with each other.4.Faster. More energy is being added to the system.5.Slower.6.Although their speed is great, because of collisions they travel by a circuitous route.Observations and Results -Part 1- Explain how the odor came: Vaporized molecules traveled by means of a zig-zag coursedue to molecular collisions. Difficult if you assume matter is continuous; could saymatter travels in a continuous ray. You would expect ammonia to travel faster because its

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molecular mass is less, thus its velocity will be faster. Suggest several explanations -different sensitivities of noses; drafts or different air circulation patterns. How does thisexperiment prove: The product (NH4Cl) was formed in the air above the liquid levels ofthe reactants - therefore, the reactant molecules had to travel through the air.Part 2- How do you know - because acetone evaporates faster (disappears faster). What dorelative rates of evaporation - the attractive forces between water molecules is greaterthan the forces between acetone molecules. Explain this observation - as those moleculeswith higher energies vaporize first, molecules with lower energy (the cooler ones) are leftbehind. Suggest a reason why molecules - in a concentrated solution there are manyparticles which get in the way of the water molecules escaping, thus increasing theamount of energy (and increasing the boiling point) needed to vaporize the solvent.Part 3- Solid mothballs activated your nose because molecules vaporized from the solid and thendiffused through the air to the nose.Part 4- Typical pathway for a particle:Questions and Problems -1.colorless2.bluish purple3.yes - sound is the displacement of air molecules and it travels in waves from the pointwhere the sound originated to the eardrum - the displaced molecules displace othermolecules, which eventually lead to collision with the ear drum. This sets off nerveimpulses which the brain interprets as a particular sound.4.(a) yes; no change in matter(b) no; odor is transfer of molecules – matter lost5.Molecules in one cup, 7.5 x 1024. Volume of oceans, 2.8 x 1020L. Fraction of moleculesfrom original cup times molecules in cup:0.25L2.8x1020Lx 7.5 x 1024molecules = about 7,000 molecules original6.Each received, 7 x 1023atoms / 7 x 109persons = 1 x 1014atoms/person$.01/1 x 104atoms = $1 x 10-6/atom1 x 1014atoms/person x 1 x 10-6$/atom = $108= 100 million dollars

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Experiment 3 - “What Is a Chemical Reaction?”Reagents -Each Student100 Studentsiron dust, Fe (reagent, free of sulfur)1 g100 gsulfur, S1 g100 gcarbon disulfide, CS2(demonstration)5 mlhydrochloric acid, HCl, 6 M0.5 ml50 mlcamphor, C10H16O0.2 g20 gcopper sulfate, CuSO4, solution, 0.1 M1 ml100 mlpotassium chromate, K2CrO4, solution, 0.1 M0.5 ml50 mllead acetate, Pb(C2H3O2)2, solution, 0.1 M0.5 ml50 mlcalcium carbonate, CaCO30.1 g10 gacetic acid, HC2H3O2, 1 M1 ml100 mlsaturated calcium hydroxide, Ca(OH)2, solution2 ml200 mliron nail1/student100 nailsSpecial Equipment -magnetmortar and pestleplastic strawSuggestions and Precautions -Use only reagent iron dust which is essentially free of sulfur. Inform the students that a faint odorof H2S from the reaction of iron with HCl is due to contamination with sulfide. The camphorshould be heated sufficiently to melt it but not enough to make it burn or physical and chemicalchanges will be confused.Prelaboratory Questions -1.Change in appearance, change in odor.2.(a) fermenting sugar; (c) burning wood; (e) cooking meat; (h) rusting iron.3.They get used up (consumed) and changed into different substances.4.Convenience, clarity and brevity; a chemical reaction is what actually takes place - anequation is a written statement of that occurrence.5.endothermic: absorbs heat (energy); exothermic: evolves heat.6.Attractive (electrostatic) forces which hold the atoms in a molecule together. Bonds arebroken and new bonds formed.Observations and Results -Part 1- This proves that the change which occurred with CS2was physical because the sulfur wasrecoverable in its original form. Two simple tests are magnet for iron and solubility inCS2for sulfur. Fe + S→FeS. Formula for this gas: H2S. The fact that the gas formedwhen treating the heated iron-sulfur mixture with acid had a different odor than the gas

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which was formed from iron and acid (sulfur doesn’t react in acid). FeS + 2HCl→H2S +FeCl2.Part 2- Burning of methane- bonds broken: C-H, O═O; bonds formed: C═O, H-O. CuSO4+ Fe→FeSO4+ Cu.Pb(C2H3O2)2+ K2CrO4→2KC2H3O2+ PbCrO4.Calcium carbonate is insoluble in water. A physical property of acetic acid is its odor.The evidence for the disappearance of the reactants is that CaCO3dissolves. Theevidence for the new products is the evolution of CO2.CaCO3+ 2HC2H3O2→Ca(C2H3O2)2+ CO2+ H2OQuestions and Problems -1.internal combustion engine; gas stove; food metabolism.2.bonds broken: H-H, O═O; bonds formed: O-H. Formed bonds are stronger.3.add acid to the substance - if effervescence occurs, it is CaCO3.4.use vinegar (or other household acid) to dissolve.5.H2, hydrogen gas, 2 atoms/molecules.6.internal combustion engine (a spark); 2H2+ O2→2H2O; cooking of food.7.6 bonds broken, 6 bonds formed.

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Experiment 4 - “The Calorie: A Unit of Energy”Reagents -Each Student100 Studentsammonium chloride, NH4Cl10 g1000 gcalcium chloride, CaCl210 g1000 gchipped ice100 g10 kgisopropyl alcohol1 ml100 mlSpecial Equipment -thermometer, °Cstyrofoam cup, 6 oz. cups work nicely; 4 oz. size is o.k.Suggestions and Precautions -In Part 3, the volume of water must be determined immediately when the temperature no longerdrops (approximately 2°C). All of Part 3 must be done rapidly to avoid heat loss to thesurroundings.Prelaboratory Questions -1.calorie, erg, joule, BTU, foot-pound.2.Calorie: The amount of heat (energy) required to raise the temperature of one gram ofwater one degree centigrade (from 14.5 to 15.5°C).Specific heat: The amount of energy required to raise one gram of a substance one degreecentigrade.Heat of fusion: The energy required for converting a solid to a liquid (per gram or permole).3.It is used to change the liquid to a vapor (steam) by supplying the energy needed toovercome the attractive forces between the molecules. Heat of vaporization is the amountof energy required to change a liquid to a gas (per gram or per mole).4.(40 g) (50°C - 25°C) (1 cal/g - °C) = 1000 calories.5.(100 g) (x - 20°C) (1 cal/g - °C) = 1000 calories; x = 30°C.Observations and Results -NoneQuestions and Problems -1.100 ml1 g•(70º - 30º ) (1 cal/g-ºC)500111ml20, 000 kilocalories1000 cal/kilocalorie=i2.(5 g) (80 cal/g) + (5 g) (100° - 0°C) (1 cal/g - °C) + (5 g) (540 cal/g) = 3600 calories3.(80 g) (80 cal/g) = 6400 calories used up in melting ice. (20 g) (540 cal/g) = 10,800calories liberated in condensing steam. (10,800 - 6400 calories) = (80 g) (x° - 0°C)(1 cal/g-°C); x = 55°C; mixing 80 g water at 55°C with 20 g water at 100°C = 64°C.

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Experiment 5 - “Reliability of Data: The Composition of a Compound”Reagents -Each Student100 Studentscopper sulfate pentahydrate, CuSO4• 5H2O1 g100 gSpecial Equipment –NoneSuggestions and Precautions -Heat the CuSO4• 5H2O with a low flame only as long as a blue color is present.Prelaboratory Questions -1.compounds - a definite percentage composition by weight, whereas, in mixtures, it isvariable. Examples: compounds - salt (NaCl), sugar (C6H12O6), ammonia (NH3);mixtures - alloys, sand and salt, air.2.15.12 g - 14.72 g • 1002.72%14.72 g=3.10.23110.27810.25610.36310.282 g4+++=10.282 - 10.231 = 0.05110.278 - 10.282 = 0.004average deviation = 0.02710.256 - 10.282 = 0.02610.363 - 10.282 = 0.08110.363 is not reliable because its deviation from the mean is at least three times theaverage of the other three values. Therefore, take the mean of the first three values,10.255.4.a.Na-2; S-1; O-14; H-20; total 37 atomsb.14 atoms • 10037.8%37 atoms=c.10 water molecules5.add together(0.2464 • 2.53) = 0.579 g of sand(0.0423 • 2.35) = 0.099 g of clay(0.7113 • 2.35) = 1.672 g of ash; Grind together with mortar and pestle.Observation and Results -Sources of error:balance poorly adjusted - either wayexperimenter’s technique - either waynot heating CuSO4enough - lowpicking up moisture - lowlosses due to spattering – high

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CuSO4+ 5H2O→CuSO4• 5H2OThe data obtained support the law of constant composition because they give fairly constantpercentages of water in this hydrate, thus showing that the components are present in thecompound in a definite ratio.Questions and Problems -1.2 (18 g/mole)10020.9%172 g/mole=i2.Range = 24.45% - 24.27% = 0.18%24.4024.32(24.45) 224.27Mean =24.38%5+++=Average deviation =24.40 - 24.38 = 0.0224.38 - 24.32 = 0.0624.45 - 24.38 = 0.0724.45 - 24.38 = 0.0724.38 - 24.27 = 0.110.33 = 0.066%53.63.524.45 - 24.3810025.45%1004.2% error249.525.45==ii4.324.010048.62%666.43=i48.7048.65 48.62 - 48.711000.18%48.6249.5047.99194.8448.71% experimental4==i48.7048.70 - 48.450.2548.6448.65 - 48.450.2047.9948.45 - 47.990.46145.340.9148.450.3033=====5.Cyclamates, saccharine, various pesticides, tris; opinions vary; thalidomide.

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Experiment 6 - “Getting Acquainted with Metals”Reagents -Each Student100 Studentsiron, Fe, strips or pieces0.5 g50 gtin, Sn, strips or pieces0.5 g50 glead, Pb, strips or pieces0.5 g50 gcopper, Cu, strips or pieces1 g100 gmagnesium, Mg, strips or pieces0.5 g50 galuminum, Al, strips or pieces0.5 g50 gzinc, Zn, strips or pieces0.5 g50 gsodium, Na, strips or pieces0.5 g50 giron nail2/student200 nailssteel wool1 g100 ghydrochloric acid, HCl, 1 M10 ml1000 mlhydrochloric acid, HCl, 6 M10 ml1000 mlhydrochloric acid, HCl, 12 M5 ml500 mlcopper sulfate, CuSO4, solution, 0.1 M10 ml1000 mlferrous sulfate, FeSO4, solution, 0.1 M10 ml1000 mlsilver nitrate, AgNO3, solution, 0.1 M10 ml1000 mlSpecial Equipment-knife (for demonstration)Suggestions and Precautions-It is very worthwhile to let students observe the instructor cutting sodium metal so they can seehow lustrous it is at first and how rapidly it tarnishes. Let them understand why it is stored undera liquid hydrocarbon. Note how the exothermic reaction causes sodium pieces to heat up andmelt as they float on a cushion of hydrogen over the water. At the end, there is usually a smallignition of the hydrogen.

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Prelaboratory Questions -1.activemoderately activeinactivemagnesiumleadcopperpotassiumirongoldsodiumtinsilvercalciumaluminumplatinumnickel2.Cr - chrome plating, stainless steelTa - synthetic bone parts in humansNi - coinage, stainless steelHg - thermometers, barometersZn - galvanizing coatingSn - coating for steel cansRa - radiation for cancer treatmentAg - jewelry, coinage, instrumentation3.Cu, Cu2+; Cu metal is red, lustrous; Cu2+solutions are blue.H2, H3O+; H2is a gas; H3O+is an ion in solution having acidic properties.4.a. Beneficial Mg+2, Ca+2, Fe+2b. Harmful Hg+2, Pb+2, Cd+2Observations and ResultsPart 1-3- 4Na + O2→2Na2O. Also, 2Na + 2H2O→2NaOH + H2.Sodium is stored under a hydrocarbon to protect it from oxidation by the air. Kerosene,toluene, or mineral oil are selected because they have low vapor pressures andconvenient liquid ranges.2Zn + O2→2ZnO2Al + 3/2O2→Al2O3Magnesium heated in air produces MgO.Make a sketch:Mg lost electrons; other atoms accepted electrons from Mg, such as oxygen. Can youexplain your observation - increasing temperature increases the energy of the reactants,allowing more frequent and more effective collisions. Magnesium would make a goodnight flare. Would any metals react differently? - All those which reacted with air shouldreact faster and more vigorously since the concentration of oxygen is increasedsignificantly. Steel wool has a greater surface area.

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2Fe + 3/2O2→Fe2O3Al + 3H+→Al3++ 3/2 H2(g)(Cu - no reaction)Silver is used in jewelry; sodium would be least suitable. All are true, but just the firstfour are based on the statement only.Part 4- a. Fe + CuSO4→FeSO4+ Cub. n.r.c. Cu + 2AgNO3→Cu(NO3)2+ 2AgWhich is more active: Fe; Fe; CuArrange the 3 metals: Fe, Cu, AgArrange all metals: Na, Mg, Al, Zn, Fe, Sn, Pb, H, Cu, AgQuestions and Problems-1.K, Ba, Ca, Na, Mg, Al, Sn, Pb.2.The most active metals are located toward the bottom left portion of the chart; activenonmetals are top right; most active metal: Cs; most active non-metal: F.3.NaOH, Mg(OH)2, Al(OH)3, Zn(OH)2, Fe(OH)2, Sn(OH)4, Pb(OH)2, Cu(OH)2.4.Ag2S.5.8×107tons of gold; density, 1.2×103lb/ft3; so, 67,000 cubic ft; e.g., building 10×80×80.

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Experiment 11 - “Weights of Reactants and Products: Preparation of a Metal”Reagents -Each Student100 Studentslead carbonate, PbCO32 g200 gwood charcoal1 g100 gSpecial Equipment-NoneSuggestions and Precautions -The type of flame is critical to the success of the experiment. The flame must be cool and yellow- a hot flame produces extremely hard material which cannot be removed from the crucible. Theyellow flame is easily disturbed by drafts in the lab. Students must be instructed to keep theflame close to the bottom of the crucible. Stirring to obtain a large globule of lead is important;otherwise, lead “sand” will be obtained, which is difficult to separate from the carbon.Prelaboratory Questions1.4 + 6 = 10.2.123.9 g/mole P219.9 g/mole P4O6•100=56.3%P;96 g/mole O219.9 g/mole P4O6•100=43.7% O.3.7.5 g PbCO3267.9 g/mole PbCO3= 0.028 mole PbCO3; 0.028 mole PbO; 0.028 mole Pb;0.014 mole C; YesIn the experiment:4 g PbCO3267.9 g/mole PbCO3= 0.0149 mole PbCO3= 0.0149 molePbO - will require 0.01492mole C = 0.00747 mole C = 0.090 g C and0300 g - 0.090 g = 0.21 g excess.4.2ZnS + 3O2→2ZnO + 2SO22ZnO + C→2Zn + CO25.A hard, fused lead oxide would form which could not be removed from the crucible.PbCO3would not be heated sufficiently hot to give PbO.

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Observations and Results -If the weight of PbO is too high, the PbCO3may not have been heated either hot enough or longenough. A low value suggests mechanical loss.Questions and Problems -1.Let x = weight of ore required to yield one ton of lead. Then:x • 0.147(% lead in ore) • 0.95 (% recovery) = 1 ton.x = 7.16 tons of ore.2.M. Wt. PbCO3(Cerussite) = 267 g/mole. 207.2 g/mole267.9 g/mole • 100 = 77.3% PbM. Wt. PbS (Galena) = 239.2 g/mole. 207.2 g/mole267.9 g/mole • 100 = 86.6% Pb3.2PbS + 3O2→2PbO + 2SO24.grams of lead/yr = 3 gal/day • 365 days/yr • 0.4 g/day = 438 g/yr.5.fishing sinkers; storage batteries. Discontinued: lead in paint; tetraethyl lead in gasoline;lead shot in some shotgun shells.

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Experiment 12 - “The pH Scale and the Use of Indicators”Reagents -Each Pair100 Studentsdistilled water100 ml10,000 mlvinegar, 10% solution35 ml1750 mlgrapefruit juice35 ml1750 ml7-Up, 50% solution35 ml1750 mllow alkalinity shampoo, 10% solution35 ml1750 mlordinary shampoo, 10% solution35 ml1750 mlliquid laundry detergent, 5% solution25 ml1250 mlhousehold ammonia25 ml1250 mlbaking soda, NaHCO3, solution, 5%25 ml1250 mlaspirin tablet (in 20 ml water)1 tab/pair50 tabletsBufferin (in 20 ml water)1 tab/pair50 tabletsbuffers - pH 425 ml for demonstrationpH 925 ml for demonstrationhydrochloric acid, HCl, 0.01 M5 ml250 ml- be sure to use boiled distilled watersodium hydroxide, NaOH, 0.01 M5 ml250 ml- be sure to use boiled distilled watermethyl orange3 ml150 ml- 1 g methyl orange per liter waterphenolphthalein3 ml150 ml- 0.05 g phenolphthalein per 50 ml reagent grade ethanol and 50 ml waterbromthymol blue3 ml150 ml- 0.01 g bromthyrnol blue per 16 ml 0.01 N NaOH and 234 ml water(this can also be obtained commercially prepared)alizarin yellow3 ml150 ml- 0.1% in methanol or 1 g alizarin yellow per 1 liter methanolmethyl red3 ml150 ml- 1 g methyl red per 600 ml reagent grade ethanol, diluted with 400 ml waterSpecial Equipment -pH meter

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