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

Name: Chemistry And Life In The Laboratory: Experiments, 6th Edition Solution Manual
Brand: Academic Documents Platform
Price: 10 USD
Availability: InStock
Author: Michael Davis

Chemistry And Life In The Laboratory: Experiments, 6th Edition Solution Manual is an essential tool for mastering your textbook, offering detailed solutions and helpful explanations.

Michael Davis

Contributor

4.5

3 days ago

Preview (16 of 88)

Experiment 1 - “Thinking Metric”
Reagents -
None
Special Equipment -
meter sticks
8½" x 11" pieces of paper
string and scissors
rectangular objects such as boxes or blocks
pieces of chalk
nickel coins
100 tablets (aspirin, vitamin) or other small objects
pebbles, rubber stoppers, or metal shot
Suggestions 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 to
float 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’s
mass. 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 of
wood and water - density of wood is less than that of water. Wood would float on
mercury because wood’s density is less than 13.6 g/ml.

Questions and Problems -
1. 3.7851
24 gal • 90.841
1 gal =
2. 110 km 1 mi
• 68.4 mi/hr
hr 1.609 km =
3. 2.5 cm • 12 cm • 20 cm = 600 cu cm; 600 g water
4. 4 years • 186,300 mi/sec • 60 sec/min • 24 hr/da • 365 da/yr = 23,500,000,000,000 miles
5. wt (lb) ÷ 2.2 lb/kg
6. 60 min/hr x 24 hr/da x da/10 cda = 144 min/cda
365 day/yr x 1 kda/1000 da = 0.365 yr/kda
7. (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

Experiment 2 - “Molecular Motion - The Particulate Nature of Matter”
Reagents - Each Student 100 Students
methyl salicylate, C8 H8O3 need 0.5 ml per demonstration
hydrochloric acid, HCl, 12 M 500 ml
ammonium hydroxide, NH4 OH, 15 M 500 ml
acetone, C3H6 O 1 ml 100 ml
sodium chloride, NaCl 4 g 400 g
p-dichlorobenzene, C6 H4Cl 2 0.25 g 25 g
iodine, I 2 0.3 g 30 g
potassium permanganate, KMnO4 0.2 g 20 g
Special Equipment -
watch glass (for group experiment)
hot plate (for group experiment)
time piece with second hand (for group experiment)
chalk
string and meter stick
60 cm x 2 cm glass tube (for demonstration)
cotton plugs (for demonstration)
thermometer, °C
lined, white paper
Suggestions and Precautions -
To avoid pre-release, bring the correct amount of oil of wintergreen in a closed vial and drop it
directly 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: random
arrangement 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 course
due to molecular collisions. Difficult if you assume matter is continuous; could say
matter travels in a continuous ray. You would expect ammonia to travel faster because its

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 this
experiment prove: The product (NH4Cl) was formed in the air above the liquid levels of
the reactants - therefore, the reactant molecules had to travel through the air.
Part 2- How do you know - because acetone evaporates faster (disappears faster). What do
relative rates of evaporation - the attractive forces between water molecules is greater
than the forces between acetone molecules. Explain this observation - as those molecules
with higher energies vaporize first, molecules with lower energy (the cooler ones) are left
behind. Suggest a reason why molecules - in a concentrated solution there are many
particles which get in the way of the water molecules escaping, thus increasing the
amount 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 then
diffused through the air to the nose.
Part 4- Typical pathway for a particle:
Questions and Problems -
1. colorless
2. bluish purple
3. yes - sound is the displacement of air molecules and it travels in waves from the point
where the sound originated to the eardrum - the displaced molecules displace other
molecules, which eventually lead to collision with the ear drum. This sets off nerve
impulses which the brain interprets as a particular sound.
4. (a) yes; no change in matter
(b) no; odor is transfer of molecules – matter lost
5. Molecules in one cup, 7.5 x 1024 . Volume of oceans, 2.8 x 10 20L. Fraction of molecules
from original cup times molecules in cup:
0.25L
2.8 x10 20 L x 7.5 x 1024 molecules = about 7,000 molecules original
6. Each received, 7 x 1023 atoms / 7 x 10 9 persons = 1 x 10 14 atoms/person
$.01/1 x 10 4 atoms = $1 x 10 -6 /atom
1 x 1014 atoms/person x 1 x 10 -6 $/atom = $108 = 100 million dollars

Experiment 3 - “What Is a Chemical Reaction?”
Reagents - Each Student 100 Students
iron dust, Fe (reagent, free of sulfur) 1 g 100 g
sulfur, S 1 g 100 g
carbon disulfide, CS2 (demonstration) 5 ml
hydrochloric acid, HCl, 6 M 0.5 ml 50 ml
camphor, C10 H16 O 0.2 g 20 g
copper sulfate, CuSO4 , solution, 0.1 M 1 ml 100 ml
potassium chromate, K2 CrO4 , solution, 0.1 M 0.5 ml 50 ml
lead acetate, Pb(C2 H3 O2 ) 2 , solution, 0.1 M 0.5 ml 50 ml
calcium carbonate, CaCO3 0.1 g 10 g
acetic acid, HC2 H3 O2 , 1 M 1 ml 100 ml
saturated calcium hydroxide, Ca(OH) 2 , solution 2 ml 200 ml
iron nail 1/student 100 nails
Special Equipment -
magnet
mortar and pestle
plastic straw
Suggestions and Precautions -
Use only reagent iron dust which is essentially free of sulfur. Inform the students that a faint odor
of H2 S from the reaction of iron with HCl is due to contamination with sulfide. The camphor
should be heated sufficiently to melt it but not enough to make it burn or physical and chemical
changes 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 - an
equation 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 are
broken and new bonds formed.
Observations and Results -
Part 1- This proves that the change which occurred with CS 2 was physical because the sulfur was
recoverable in its original form. Two simple tests are magnet for iron and solubility in
CS2 for sulfur. Fe + S → FeS. Formula for this gas: H2 S. The fact that the gas formed
when treating the heated iron-sulfur mixture with acid had a different odor than the gas

Loading page 6...

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(C2 H3 O2 ) 2 + K2 CrO4 → 2KC2 H3 O2 + PbCrO 4 .
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 CaCO3 dissolves. The
evidence for the new products is the evolution of CO 2 .
CaCO3 + 2HC2 H3 O2 → Ca(C2 H3 O2 ) 2 + CO2 + H2 O
Questions 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 → 2H2 O; cooking of food.
7. 6 bonds broken, 6 bonds formed.

Loading page 7...

Experiment 4 - “The Calorie: A Unit of Energy”
Reagents - Each Student 100 Students
ammonium chloride, NH4 Cl 10 g 1000 g
calcium chloride, CaCl 2 10 g 1000 g
chipped ice 100 g 10 kg
isopropyl alcohol 1 ml 100 ml
Special Equipment -
thermometer, °C
styrofoam 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 longer
drops (approximately 2°C). All of Part 3 must be done rapidly to avoid heat loss to the
surroundings.
Prelaboratory Questions -
1. calorie, erg, joule, BTU, foot-pound.
2. Calorie: The amount of heat (energy) required to raise the temperature of one gram of
water 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 degree
centigrade.
Heat of fusion: The energy required for converting a solid to a liquid (per gram or per
mole).
3. It is used to change the liquid to a vapor (steam) by supplying the energy needed to
overcome the attractive forces between the molecules. Heat of vaporization is the amount
of 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 -
None
Questions and Problems -
1.
100 ml 1 g
• (70º - 30º ) (1 cal/g-ºC)5001 11 ml 20, 000 kilocalories
1000 cal/kilocalorie =
i
2. (5 g) (80 cal/g) + (5 g) (100° - 0°C) (1 cal/g - °C) + (5 g) (540 cal/g) = 3600 calories
3. (80 g) (80 cal/g) = 6400 calories used up in melting ice. (20 g) (540 cal/g) = 10,800
calories 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.

Loading page 8...

Experiment 5 - “Reliability of Data: The Composition of a Compound”
Reagents - Each Student 100 Students
copper sulfate pentahydrate, CuSO4 • 5H2 O 1 g 100 g
Special Equipment –
None
Suggestions and Precautions -
Heat the CuSO 4 • 5H2 O 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 is
variable. Examples: compounds - salt (NaCl), sugar (C6 H12 O6), ammonia (NH3 );
mixtures - alloys, sand and salt, air.
2. 15.12 g - 14.72 g • 100 2.72%
14.72 g =
3. 10.231 10.278 10.256 10.363 10.282 g
4
+ + + =
10.282 - 10.231 = 0.051
10.278 - 10.282 = 0.004 average deviation = 0.027
10.256 - 10.282 = 0.026
10.363 - 10.282 = 0.081
10.363 is not reliable because its deviation from the mean is at least three times the
average 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 atoms
b. 14 atoms • 100 37.8%
37 atoms =
c. 10 water molecules
5. 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 way
experimenter’s technique - either way
not heating CuSO4 enough - low
picking up moisture - low
losses due to spattering – high

Loading page 9...

CuSO4 + 5H 2 O → CuSO 4 • 5H2 O
The data obtained support the law of constant composition because they give fairly constant
percentages of water in this hydrate, thus showing that the components are present in the
compound in a definite ratio.
Questions and Problems -
1. 2 (18 g/mole) 100 20.9%
172 g/mole =i
2. Range = 24.45% - 24.27% = 0.18%
24.40 24.32 (24.45) 2 24.27
Mean = 24.38%
5
+ + + =
Average deviation =
24.40 - 24.38 = 0.02
24.38 - 24.32 = 0.06
24.45 - 24.38 = 0.07
24.45 - 24.38 = 0.07
24.38 - 24.27 = 0.11
0.33 = 0.066%
5
3. 63.5 24.45 - 24.38
100 25.45% 100 4.2% error
249.5 25.45
= =i i
4. 324.0 100 48.62%
666.43 =i
48.70
48.65 48.62 - 48.71 100 0.18%
48.6249.50
47.99
194.84 48.71% experimental
4
=
=
i
48.70 48.70 - 48.45 0.25
48.64 48.65 - 48.45 0.20
47.99 48.45 - 47.99 0.46
145.34 0.91
48.45 0.30
3 3
=
=
=
= =
5. Cyclamates, saccharine, various pesticides, tris; opinions vary; thalidomide.

Loading page 10...

Experiment 6 - “Getting Acquainted with Metals”
Reagents - Each Student 100 Students
iron, Fe, strips or pieces 0.5 g 50 g
tin, Sn, strips or pieces 0.5 g 50 g
lead, Pb, strips or pieces 0.5 g 50 g
copper, Cu, strips or pieces 1 g 100 g
magnesium, Mg, strips or pieces 0.5 g 50 g
aluminum, Al, strips or pieces 0.5 g 50 g
zinc, Zn, strips or pieces 0.5 g 50 g
sodium, Na, strips or pieces 0.5 g 50 g
iron nail 2/student 200 nails
steel wool 1 g 100 g
hydrochloric acid, HCl, 1 M 10 ml 1000 ml
hydrochloric acid, HCl, 6 M 10 ml 1000 ml
hydrochloric acid, HCl, 12 M 5 ml 500 ml
copper sulfate, CuSO4 , solution, 0.1 M 10 ml 1000 ml
ferrous sulfate, FeSO4 , solution, 0.1 M 10 ml 1000 ml
silver nitrate, AgNO3 , solution, 0.1 M 10 ml 1000 ml
Special Equipment-
knife (for demonstration)
Suggestions and Precautions-
It is very worthwhile to let students observe the instructor cutting sodium metal so they can see
how lustrous it is at first and how rapidly it tarnishes. Let them understand why it is stored under
a liquid hydrocarbon. Note how the exothermic reaction causes sodium pieces to heat up and
melt as they float on a cushion of hydrogen over the water. At the end, there is usually a small
ignition of the hydrogen.

Loading page 11...

Prelaboratory Questions -
1. active moderately active inactive
magnesium lead copper
potassium iron gold
sodium tin silver
calcium aluminum platinum
nickel
2. Cr - chrome plating, stainless steel Ta - synthetic bone parts in humans
Ni - coinage, stainless steel Hg - thermometers, barometers
Zn - galvanizing coating Sn - coating for steel cans
Ra - radiation for cancer treatment Ag - jewelry, coinage, instrumentation
3. Cu, Cu 2+
; Cu metal is red, lustrous; Cu
2+ solutions are blue.
H2 , H3 O
+
; H 2 is a gas; H3O
+ is an ion in solution having acidic properties.
4. a. Beneficial Mg
+2 , Ca +2 , Fe +2
b. Harmful Hg
+2 , Pb +2 , Cd
+2
Observations and Results
Part 1-3- 4Na + O2 → 2Na 2 O. Also, 2Na + 2H2 O → 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 and
convenient liquid ranges.
2Zn + O2 → 2ZnO 2Al + 3/2O2 → Al 2 O3
Magnesium heated in air produces MgO.
Make a sketch:
Mg lost electrons; other atoms accepted electrons from Mg, such as oxygen. Can you
explain your observation - increasing temperature increases the energy of the reactants,
allowing more frequent and more effective collisions. Magnesium would make a good
night flare. Would any metals react differently? - All those which reacted with air should
react faster and more vigorously since the concentration of oxygen is increased
significantly. Steel wool has a greater surface area.

Loading page 12...

2Fe + 3/2O2 → Fe 2 O3
Al + 3H
+ → Al 3+ + 3/2 H2 (g) (Cu - no reaction)
Silver is used in jewelry; sodium would be least suitable. All are true, but just the first
four are based on the statement only.
Part 4- a. Fe + CuSO4 → FeSO 4 + Cu
b. n.r.
c. Cu + 2AgNO3 → Cu(NO3 ) 2 + 2Ag
Which is more active: Fe; Fe; Cu
Arrange the 3 metals: Fe, Cu, Ag
Arrange all metals: Na, Mg, Al, Zn, Fe, Sn, Pb, H, Cu, Ag
Questions 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; active
nonmetals 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. Ag 2 S.
5. 8 × 10 7 tons of gold; density, 1.2 × 103 lb/ft 3 ; so, 67,000 cubic ft; e.g., building 10 × 80 × 80.

Loading page 13...

Experiment 11 - “Weights of Reactants and Products: Preparation of a Metal”
Reagents - Each Student 100 Students
lead carbonate, PbCO3 2 g 200 g
wood charcoal 1 g 100 g
Special Equipment-
None
Suggestions 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. The
yellow flame is easily disturbed by drafts in the lab. Students must be instructed to keep the
flame 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 Questions
1. 4 + 6 = 10.
2. 123.9 g/mole P
219.9 g/mole P4 O 6
• 100 = 56.3%P; 96 g/mole O
219.9 g/mole P4 O 6
• 100 = 43.7% O.
3. 7.5 g PbCO 3
267.9 g/mole PbCO 3
= 0.028 mole PbCO3 ; 0.028 mole PbO; 0.028 mole Pb;
0.014 mole C; Yes
In the experiment: 4 g PbCO 3
267.9 g/mole PbCO 3
= 0.0149 mole PbCO3 = 0.0149 mole
PbO - will require 0.0149
2 mole C = 0.00747 mole C = 0.090 g C and
0300 g - 0.090 g = 0.21 g excess.
4. 2ZnS + 3O2 → 2ZnO + 2SO2
2ZnO + C → 2Zn + CO2
5. A hard, fused lead oxide would form which could not be removed from the crucible.
PbCO3 would not be heated sufficiently hot to give PbO.

Loading page 14...

Observations and Results -
If the weight of PbO is too high, the PbCO3 may not have been heated either hot enough or long
enough. 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/mole
267.9 g/mole • 100 = 77.3% Pb
M. Wt. PbS (Galena) = 239.2 g/mole. 207.2 g/mole
267.9 g/mole • 100 = 86.6% Pb
3. 2PbS + 3O2 → 2PbO + 2SO2
4. 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.

Loading page 15...

Experiment 12 - “The pH Scale and the Use of Indicators”
Reagents - Each Pair 100 Students
distilled water 100 ml 10,000 ml
vinegar, 10% solution 35 ml 1750 ml
grapefruit juice 35 ml 1750 ml
7-Up, 50% solution 35 ml 1750 ml
low alkalinity shampoo, 10% solution 35 ml 1750 ml
ordinary shampoo, 10% solution 35 ml 1750 ml
liquid laundry detergent, 5% solution 25 ml 1250 ml
household ammonia 25 ml 1250 ml
baking soda, NaHCO 3 , solution, 5% 25 ml 1250 ml
aspirin tablet (in 20 ml water) 1 tab/pair 50 tablets
Bufferin (in 20 ml water) 1 tab/pair 50 tablets
buffers - pH 4 25 ml for demonstration
pH 9 25 ml for demonstration
hydrochloric acid, HCl, 0.01 M 5 ml 250 ml
- be sure to use boiled distilled water
sodium hydroxide, NaOH, 0.01 M 5 ml 250 ml
- be sure to use boiled distilled water
methyl orange 3 ml 150 ml
- 1 g methyl orange per liter water
phenolphthalein 3 ml 150 ml
- 0.05 g phenolphthalein per 50 ml reagent grade ethanol and 50 ml water
bromthymol blue 3 ml 150 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 yellow 3 ml 150 ml
- 0.1% in methanol or 1 g alizarin yellow per 1 liter methanol
methyl red 3 ml 150 ml
- 1 g methyl red per 600 ml reagent grade ethanol, diluted with 400 ml water
Special Equipment -
pH meter

Loading page 16...

15 more pages available. Scroll down to load them.

Preview Mode

100%

Study Now!

XY-Copilot AI

Unlimited Access

Secure Payment

Instant Access

24/7 Support

Document Chat

Document Details

Subject

Chemistry

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

Study Now!

Document Details

Related Documents

Quiz 2: Chemical Reactions, Atmospheric Science, and Physical Properties

Analysis and Estimation of Molar Flow Rate in a Humidification Process: Degree-of-Freedom and Discrepancy Evaluation

Barron's Chemistry Practice Plus: 400+ Online Questions and Quick Study Review (2022)

Test Bank For Chemistry in Context, 8th Edition

Organic Chemistry 11th Edition Test Bank

Investigating The Effect Of Molecular Size On London Dispersion Forces Using Adhesive-Backed Shelf Paper

Introduction To Chemical Nomenclature And Bonding: A Comprehensive Assignment

Application Of Gas Laws In Various Thermodynamic Processes

Solution Manual for Forensic Chemistry, 2nd Edition

Balancing And Completing Chemical Reactions: A Comprehensive Analysis

Company

Explore

Study Tools