General Chemistry, 10th Edition Solution Manual

Name: General Chemistry, 10th Edition Solution Manual
Brand: Academic Documents Platform
Price: 20 USD
Availability: InStock
Author: Benjamin Fisher

General Chemistry, 10th Edition Solution Manual provides expertly crafted summaries and insights to boost retention.

Benjamin Fisher

Contributor

4.6

2 days ago

Preview (16 of 1264)

1
CHAPTER 1
Chemistry and Measurement
■ SOLUTIONS TO EXERCISES
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with
one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to
the correct number of significant figures. In multistep problems, intermediate answers are given with at
least one nonsignificant figure; however, only the final answer has been rounded off.
1.1. From the law of conservation of mass,
Mass of wood + mass of air = mass of ash + mass of gases
Substituting, you obtain
1.85 grams + 9.45 grams = 0.28 grams + mass of gases
or,
Mass of gases = (1.85 + 9.45 − 0.28) grams = 11.02 grams
Thus, the mass of gases in the vessel at the end of the experiment is 11.02 grams.
1.2. Physical properties: soft, silvery-colored metal; melts at 64°C.
Chemical properties: reacts vigorously with water, with oxygen, and with chlorine.
1.3. a. The factor 9.1 has the fewest significant figures, so the answer should be reported to two
significant figures.
5.61 7.891
9.1
 = 4.86 = 4.9
b. The number with the least number of decimal places is 8.91. Therefore, round the answer to
two decimal places.
8.91 − 6.435 = 2.475 = 2.48
c. The number with the least number of decimal places is 6.81. Therefore, round the answer to
two decimal places.
6.81 − 6.730 = 0.080 = 0.08
d. You first do the subtraction within parentheses. In this step, the number with the least
number of decimal places is 6.81, so the result of the subtraction has two decimal places.
The least significant figure for this step is underlined.
38.91  (6.81 − 6.730) = 38.91  0.080
Next, perform the multiplication. In this step, the factor 0.080 has the fewest significant
figures, so round the answer to one significant figure.
38.91  0.080 = 3.11 = 3

2 Chapter 1: Chemistry and Measurement
1.4. a. 1.84 x 10−9 m = 1.84 nm
b. 5.67 x 10−12 s = 5.67 ps
c. 7.85 x 10−3 g = 7.85 mg
d. 9.7 x 103 m = 9.7 km
e. 0.000732 s = 0.732 ms, or 732 μs
f. 0.000000000154 m = 0.154 nm, or 154 pm
1.5. a. Substituting, we find that
tC = 5 C
9 F

  (tF − 32°F) = 5 C
9 F

  (102.5°F − 32°F) = 39.167°C
= 39.2°C
b. Substituting, we find that
TK = c
1 K
1 C
t
 
 
  + 273.15 K = 1 K
78 C 1 C
 
   
  + 273.15 K = 195.15 K
= 195 K
1.6. Recall that density equals mass divided by volume. You substitute 159 g for the mass and 20.2
g/cm3 for the volume.
d = m
V = 3
159 g
20.2 cm = 7.871 g/cm3 = 7.87 g/cm3
The density of the metal equals that of iron.
1.7. Rearrange the formula defining the density to obtain the volume.
V = m
d
Substitute 30.3 g for the mass and 0.789 g/cm3 for the density.
V = 3
30.3 g
0.789 g/cm = 38.40 cm3 = 38.4 cm3
1.8. Since one pm = 10−12 m, and the prefix milli- means 10−3, you can write
121 pm 
12
10 m
1 pm

 3
1 mm
10 m = 1.21  10−7 mm
1.9. 67.6 Å3 
310
10 m
1 Å

 
 
 

3
1
1 dm
10 m
 
 
  = 6.76  10−26 dm3
1.10. From the definitions, you obtain the following conversion factors:
1 = 36 in
1 yd 1 = 2.54 cm
1 in 1 =
-2
10 m
1 cm
The conversion factor for yards to meters is as follows:

Chapter 1: Chemistry and Measurement 3
1.000 yd x 36 in
1 yd x 2.54 cm
1 in x
-2
10 m
1 cm = 0.9144 m (exact)
Finally,
3.54 yd x 0.9144 m
1 yd = 3.237 m = 3.24 m
■ ANSWERS TO CONCEPT CHECKS
1.1. Box A contains a collection of identical units; therefore, it must represent an element. Box B
contains a compound because a compound is the chemical combination of two or more elements
(two elements in this case). Box C contains a mixture because it is made up of two different
substances.
1.2. a. For a person who weighs less than 100 pounds, two significant figures are typically used,
although one significant figure is possible (for example, 60 pounds). For a person who
weighs 100 pounds or more, three significant figures are typically used to report the weight
(given to the whole pound), although people often round to the nearest unit of 10, which
may result in reporting the weight with two significant figures (for example, 170 pounds).
b. Assuming a weight of 165 pounds, rounded to two significant figures this would be
reported as 1.7 x 102 pounds.
c. For example, 165 lb weighed on a scale that can measure in 100-lb increments would be
200 lb. Using the conversion factor 1 lb = 0.4536 kg, 165 lb is equivalent to 74.8 kg. Thus,
on a scale that can measure in 50-kg increments, 165 lb would be 50 kg.
1.3. a. If your leg is approximately 32 inches long, this would be equivalent to 0.81 m, 8.1 dm, or
81 cm.
b. One story is approximately 10 feet, so three stories is 30 feet. This would be equivalent to
approximately 9 m.
c. Normal body temperature is 98.6°F, or 37.0°C. Thus, if your body temperature were 39°C
(102°F), you would feel as if you had a moderate fever.
d. Room temperature is approximately 72°F, or 22°C. Thus, if you were sitting in a room at
23°C (73°F), you would be comfortable in a short-sleeve shirt.
1.4. Gold is a very unreactive substance, so comparing physical properties is probably your best
option. However, color is a physical property you cannot rely on in this case to get your answer.
One experiment you could perform is to determine the densities of the metal and the chunk of
gold. You could measure the mass of the nugget on a balance and the volume of the nugget by
water displacement. Using this information, you could calculate the density of the nugget. Repeat
the experiment and calculations for the sample of gold. If the nugget is gold, the two densities
should be equal and be 19.3 g/cm3.
Also, you could determine the melting points of the metal and the chunk of pure gold. The two
melting points should be the same (1338 K) if the metal is gold.

4 Chapter 1: Chemistry and Measurement
■ ANSWERS TO SELF-ASSESSMENT AND REVIEW QUESTIONS
1.1. One area of technology that chemistry has changed is the characteristics of materials. The liquid-
crystal displays (LCDs) in devices such as watches, cell phones, computer monitors, and
televisions are materials made of molecules designed by chemists. Electronics and
communications have been transformed by the development of optical fibers to replace copper
wires. In biology, chemistry has changed the way scientists view life. Biochemists have found
that all forms of life share many of the same molecules and molecular processes.
1.2. An experiment is an observation of natural phenomena carried out in a controlled manner so that
the results can be duplicated and rational conclusions obtained. A theory is a tested explanation of
basic natural phenomena. They are related in that a theory is based on the results of many
experiments and is fruitful in suggesting other, new experiments. Also, an experiment can
disprove a theory but can never prove it absolutely. A hypothesis is a tentative explanation of
some regularity of nature.
1.3. Rosenberg conducted controlled experiments and noted a basic relationship that could be stated
as a hypothesis—that is, that certain platinum compounds inhibit cell division. This led him to do
new experiments on the anticancer activity of these compounds.
1.4. Matter is the general term for the material things around us. It is whatever occupies space and can
be perceived by our senses. Mass is the quantity of matter in a material. The difference between
mass and weight is that mass remains the same wherever it is measured, but weight is
proportional to the mass of the object divided by the square of the distance between the center of
mass of the object and that of the earth.
1.5. The law of conservation of mass states that the total mass remains constant during a chemical
change (chemical reaction). To demonstrate this law, place a sample of wood in a sealed vessel
with air, and weigh it. Heat the vessel to burn the wood, and weigh the vessel after the
experiment. The weight before the experiment and that after it should be the same.
1.6. Mercury metal, which is a liquid, reacts with oxygen gas to form solid mercury(II) oxide. The
color changes from that of metallic mercury (silvery) to a color that varies from red to yellow
depending on the particle size of the oxide.
1.7. Gases are easily compressible and fluid. Liquids are relatively incompressible and fluid. Solids
are relatively incompressible and rigid.
1.8. An example of a substance is the element sodium. Among its physical properties: It is a solid, and
it melts at 98°C. Among its chemical properties: It reacts vigorously with water, and it burns in
chlorine gas to form sodium chloride.
1.9. An example of an element: sodium; of a compound: sodium chloride, or table salt; of a
heterogeneous mixture: salt and sugar; of a homogeneous mixture: sodium chloride dissolved in
water to form a solution.
1.10. A glass of bubbling carbonated beverage with ice cubes contains three phases: gas, liquid, and
solid.
1.11. A compound may be decomposed by chemical reactions into elements. An element cannot be
decomposed by any chemical reaction. Thus, a compound cannot also be an element in any case.

Chapter 1: Chemistry and Measurement 5
1.12. The precision refers to the closeness of the set of values obtained from identical measurements of
a quantity. The number of digits reported for the value of a measured or calculated quantity
(significant figures) indicates the precision of the value.
1.13. Multiplication and division rule: In performing the calculation 100.0 x 0.0634 ÷ 25.31, the
calculator display shows 0.2504938. We would report the answer as 0.250 because the factor
0.0634 has the least number of significant figures (three).
Addition and subtraction rule: In performing the calculation 184.2 + 2.324, the calculator display
shows 186.524. Because the quantity 184.2 has the least number of decimal places (one), the
answer is reported as 186.5.
1.14. An exact number is a number that arises when you count items or sometimes when you define a
unit. For example, a foot is defined to be 12 inches. A measured number is the result of a
comparison of a physical quantity with a fixed standard of measurement. For example, a steel rod
measures 9.12 centimeters, or 9.12 times the standard centimeter unit of measurement.
1.15. For a given unit, the SI system uses prefixes to obtain units of different sizes. Units for all other
possible quantities are obtained by deriving them from any of the seven base units. You do this by
using the base units in equations that define other physical quantities.
1.16. An absolute temperature scale is a scale in which the lowest temperature that can be attained
theoretically is zero. Degrees Celsius and kelvins have units of equal size and are related by the
formula
tC = (TK − 273.15 K)  1°C
1 K
1.17. The density of an object is its mass per unit volume. Because the density is characteristic of a
substance, it can be helpful in identifying it. Density can also be useful in determining whether a
substance is pure. It also provides a useful relationship between mass and volume.
1.18. Units should be carried along because (1) the units for the answers will come out in the
calculations, and (2), if you make an error in arranging factors in the calculation, this will become
apparent because the final units will be nonsense.
1.19. The answer is c, three significant figures.
1.20. The answer is a, 4.43 x 102 mm.
1.21. The answer is e, 75 mL.
1.22. The answer is c, 0.23 mg.
■ ANSWERS TO CONCEPT EXPLORATIONS
1.23. a. First, check the physical appearance of each sample. Check the particles that make up each
sample for consistency and hardness. Also, note any odor. Then perform on each sample
some experiments to measure physical properties such as melting point, density, and
solubility in water. Compare all of these results and see if they match.

Loading page 6...

6 Chapter 1: Chemistry and Measurement
b. It is easier to prove that the compounds were different by finding one physical property that
is different, say different melting points. To prove the two compounds were the same
would require showing that every physical property was the same.
c. Of the properties listed in part a, the melting point would be most convincing. It is not
difficult to measure, and it is relatively accurate. The density of a powder is not as easy to
determine as the melting point, and solubility is not reliable enough on its own.
d. No. Since neither solution reached a saturation point, there is not enough information to tell
if there was a difference in behavior. Many white powders dissolve in water. Their
chemical compositions are not the same.
1.24. Part 1
a. 3 g + 1.4 g + 3.3 g = 7.7 g = 8 g
b. First, 3 g + 1.4 g = 4.4 g = 4 g. Then, 4 g + 3.3 g = 7.3 g = 7 g.
c. Yes, the answer in part a is more accurate. When you round off intermediate steps, you
accumulate small errors and your answer is not as accurate.
d. The answer 29 g is correct.
e. This answer is incorrect. It should be 3 x 101 with only one significant figure in the answer.
The student probably applied the rule for addition (instead of for multiplication) after the
first step.
f. The answer 28.5 g is correct.
g. Don’t round off intermediate answers. Indicate the round-off position after each step by
underlining the least significant digit.
Part 2
a. The calculated answer is incorrect. It should be 11 cm3. The answer given has too many
significant figures. There is also a small round off error due to using a rounded-off value
for the density.
b. This is a better answer. It is reported with the correct number of significant figures (three).
It can be improved by using all of the digits given for the density.
c. V = 10 ball bearings
1  1.234 g
1 ball bearing 
3
1 cm
3.1569 g = 3.90889 = 3.909 cm3
d. There was no rounding off of intermediate steps; all the factors are as accurate as possible.
■ ANSWERS TO CONCEPTUAL PROBLEMS
1.25. a. Two phases: liquid and solid.
b. Three phases: liquid water, solid quartz, and solid seashells.
1.26. If the material is a pure compound, all samples should have the same melting point, the same
color, and the same elemental composition. If it is a mixture, these properties should differ
depending on the composition.

Loading page 7...

Chapter 1: Chemistry and Measurement 7
1.27. a. You need to establish two points on the thermometer with known (defined) temperatures—
for example, the freezing point (0C) and boiling point (100C) of water. You could first
immerse the thermometer in an ice-water bath and mark the level at this point as 0C. Then,
immerse the thermometer in boiling water, and mark the level at this point as 100C. As
long as the two points are far enough apart to obtain readings of the desired accuracy, the
thermometer can be used in experiments.
b. You could make 19 evenly spaced marks on the thermometer between the two original
points, each representing a difference of 5°C. You may divide the space between the two
original points into fewer spaces as long as you can read the thermometer to obtain the
desired accuracy.
1.28. a. b. c.
1.29. a. To answer this question, you need to develop an equation that converts between F and
YS. To do so, you need to recognize that one degree on the Your Scale does not
correspond to one degree on the Fahrenheit scale and that −100F corresponds to 0 on
Your Scale (different “zero” points). As stated in the problem, in the desired range of 100
Your Scale degrees, there are 120 Fahrenheit degrees. Therefore, the relationship can be
expressed as 120F = 100YS, since it covers the same temperature range. Now you need
to “scale” the two systems so that they correctly convert from one scale to the other. You
could set up an equation with the known data points and then employ the information from
the relationship above.
For example, to construct the conversion between YS and F, you could perform the
following steps:
Step 1: F = YS
Not a true statement, but one you would like to make true.
Step 2: °F = °YS  120°F
100°YS
This equation takes into account the difference in the size between the temperature unit on
the two scales but will not give you the correct answer because it doesn’t take into account
the different zero points.
Step 3: By subtracting 100F from your equation from Step 2, you now have the complete
equation that converts between F and YS.
F = (YS  120°F
100°YS ) − 100F
b. Using the relationship from part a, 66°YS is equivalent to
(66YS  120°F
100°YS ) − 100F = −20.8F = −21F

Loading page 8...

8 Chapter 1: Chemistry and Measurement
1.30. Some physical properties you could measure are density, hardness, color, and conductivity.
Chemical properties of sodium would include reaction with air, reaction with water, reaction with
chlorine, reaction with acids, bases, etc.
1.31. The empty boxes are identical, so they do not contribute to any mass or density difference. Since
the edge of the cube and the diameter of the sphere are identical, they will occupy the same
volume in each of the boxes; therefore, each box will contain the same number of cubes or
spheres. If you view the spheres as cubes that have been rounded by removing wood, you can
conclude that the box containing the cubes must have a greater mass of wood; hence, it must have
a greater density.
1.32. a. Since the bead is less dense than any of the liquids in the container, the bead will float on
top of all the liquids.
b. First, determine the density of the plastic bead. Since density is mass divided by volume,
you get
d = m
V =
2
3.92 10 g
0.043 mL

 = 0.911 g/mL = 0.91 g/mL
Thus, the glass bead will pass through the top three layers and float on the ethylene glycol
layer, which is more dense.
c. Since the bead sinks all the way to the bottom, it must be more dense than 1.114 g/mL.
1.33. a. A paper clip has a mass of about 1 g.
b. Answers will vary depending on your particular sample. Keeping in mind that the SI unit
for mass is kg, the approximate weights for the items presented in the problem are as
follows: a grain of sand, 1  10−5 kg; a paper clip, 1 x 10−3 kg; a nickel, 5  10−3 kg; a 5.0-
gallon bucket of water, 2.0  101 kg; a brick, 3 kg; a car, 1  103 kg.
1.34. When taking measurements, never throw away meaningful information even if there is some
uncertainty in the final digit. In this case, you are certain that the nail is between 5 and 6 cm. The
uncertain, yet still important, digit is between the 5 and 6 cm measurements. You can estimate
with reasonable precision that it is about 0.7 cm from the 5 cm mark, so an acceptable answer
would be 5.7 cm. Another person might argue that the length of the nail is closer to 5.8 cm, which
is also acceptable given the precision of the ruler. In any case, an answer of 5.7 or 5.8 should
provide useful information about the length of the nail. If you were to report the length of the nail
as 6 cm, you would be discarding potentially useful length information provided by the
measuring instrument. If a higher degree of measurement precision were needed (more significant
figures), you would need to switch to a more precise ruler—for example, one that had mm
markings.
1.35. a. The number of significant figures in this answer follows the rules for multiplication and
division. Here, the measurement with the fewest significant figures is the reported volume
0.310 m3, which has three. Therefore, the answer will have three significant figures. Since
Volume = L x W x H, you can rearrange and solve for one of the measurements, say the
length.
L = 
V
W H =
3
0.310 m
(0.7120 m) (0.52145 m) = 0.83496 m = 0.835 m

Loading page 9...

Chapter 1: Chemistry and Measurement 9
b. The number of significant figures in this answer follows the rules for addition and
subtraction. The measurement with the least number of decimal places is the result 1.509
m, which has three. Therefore, the answer will have three decimal places. Since the result is
the sum of the three measurements, the third length is obtained by subtracting the other two
measurements from the total.
Length = 1.509 m − 0.7120 m − 0.52145 m = 0.27555 m = 0.276 m
1.36. The mass of something (how heavy it is) depends on how much of the item, material, substance,
or collection of things you have. The density of something is the mass of a specific amount
(volume) of an item, material, substance, or collection of things. You could use 1 kg of feathers
and 1 kg of water to illustrate that they have the same mass yet have very different volumes;
therefore, they have different densities.
■ SOLUTIONS TO PRACTICE PROBLEMS
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with
one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to
the correct number of significant figures. In multistep problems, intermediate answers are given with at
least one nonsignificant figure; however, only the final answer has been rounded off.
1.37. By the law of conservation of mass:
Mass of sodium carbonate + mass of acetic acid solution = mass of contents of reaction vessel +
mass of carbon dioxide
Plugging in gives
15.9 g + 20.0 g = 29.3 g + mass of carbon dioxide
Mass of carbon dioxide = 15.9 g + 20.0 g − 29.3 g = 6.6 g
1.38. By the law of conservation of mass:
Mass of iron + mass of acid = mass of contents of beaker + mass of hydrogen
Plugging in gives
5.6 g + 15.0 = 20.4 g + mass of hydrogen
Mass of hydrogen = 5.6 g + 15.0 g − 20.4 g = 0.2 g
1.39. By the law of conservation of mass:
Mass of zinc + mass of sulfur = mass of zinc sulfide
Rearranging and plugging in give
Mass of zinc sulfide = 65.4 g + 32.1 g = 97.5 g
For the second part, let x = mass of zinc sulfide that could be produced. By the law of
conservation of mass:
20.0 g + mass of sulfur = x

Loading page 10...

10 Chapter 1: Chemistry and Measurement
Write a proportion that relates the mass of zinc reacted to the mass of zinc sulfide formed, which
should be the same for both cases.
mass zinc
mass zinc sulfide = 65.4 g
97.5 g = 20.0 g
x
Solving gives x = 29.81 g = 29.8 g
1.40. By the law of conservation of mass:
Mass of aluminum + mass of bromine = mass of aluminum bromide
Plugging in and solving give
27.0 g + Mass of bromine = 266.7 g
Mass of bromine = 266.7 g − 27.0 g = 239.7 g
For the second part, let x = mass of bromine that reacts. By the law of conservation of mass:
18.1 g + x = mass of aluminum bromide
Write a proportion that relates the mass of aluminum reacted to the mass of bromine reacted,
which should be the same for both cases.
mass aluminum
mass bromine = 27.0 g
239.7 g = 18.1 g
x
Solving gives x = 160.7 g = 161 g
1.41. a. Solid b. Liquid c. Gas d. Solid
1.42. a. Solid b. Solid c. Solid d. Liquid
1.43. a. Physical change
b. Physical change
c. Chemical change
d. Physical change
1.44. a. Physical change
b. Chemical change
c. Chemical change
d. Physical change
1.45. Physical change: Liquid mercury is cooled to solid mercury.
Chemical changes: (1) Solid mercury oxide forms liquid mercury metal and gaseous oxygen; (2)
glowing wood and oxygen form burning wood (form ash and gaseous products).
1.46. Physical changes: (1) Solid iodine is heated to gaseous iodine; (2) gaseous iodine is cooled to
form solid iodine.
Chemical change: Solid iodine and zinc metal are ignited to form a white powder.

Loading page 11...

Chapter 1: Chemistry and Measurement 11
1.47. a. Physical property
b. Chemical property
c. Physical property
d. Physical property
e. Chemical property
1.48. a. Physical property
b. Chemical property
c. Physical property
d. Chemical property
e. Physical property
1.49. Physical properties: (1) Iodine is solid; (2) the solid has lustrous blue-black crystals;
(3) the crystals vaporize readily to a violet-colored gas.
Chemical properties: (1) Iodine combines with many metals, such as with aluminum to give
aluminum iodide.
1.50. Physical properties: (1) is a solid; (2) has an orange-red color; (3) has a density of
11.1 g/cm3; (4) is insoluble in water.
Chemical property: Mercury(II) oxide decomposes when heated to give mercury and oxygen.
1.51. a. Physical process
b. Chemical reaction
c. Physical process
d. Chemical reaction
e. Physical process
1.52. a. Chemical reaction
b. Physical process
c. Physical process
d. Physical process
e. Chemical reaction
1.53. a. Solution
b. Substance
c. Substance
d. Heterogeneous mixture

Loading page 12...

12 Chapter 1: Chemistry and Measurement
1.54. a. Homogeneous mixture, if fresh; heterogeneous mixture, if spoiled
b. Substance
c. Solution
d. Substance
1.55. a. A pure substance with two phases present, liquid and gas.
b. A mixture with two phases present, solid and liquid.
c. A pure substance with two phases present, solid and liquid.
d. A mixture with two phases present, solid and solid.
1.56. a. A mixture with two phases present, solid and liquid.
b. A mixture with two phases present, solid and liquid.
c. A mixture with two phases present, solid and solid.
d. A pure substance with two phases present, liquid and gas.
1.57. a. four
b. three
c. four
d. five
e. three
f. four
1.58. a. five
b. four
c. two
d. four
e. three
f. four
1.59. 40,000 km = 4.0 x 104 km
1.60. 150,000,000 km = 1.50  108 km
1.61. a. 8.71 0.0301
0.031
 = 8.457 = 8.5
b. 0.71 + 89.3 = 90.01 = 90.0
c. 934  0.00435 + 107 = 4.0629 + 107 = 111.06 = 111
d. (847.89 − 847.73)  14673 = 0.16  14673 = 2347 = 2.3  103

Loading page 13...

Loading page 14...

Loading page 15...

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

General Chemistry, 10th Edition Solution Manual

Study Now!

Document Details

Related Documents

General Chemistry, 11th Edition Test Bank

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

Introduction To Chemical Nomenclature And Bonding: A Comprehensive Assignment

Exploring The Effects Of Salt On Freezing Point Depression: A Comparative Study Of Ice Water And Salt Water

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

Solution Manual for Chemistry For Changing Times, 14th Edition

Company

Explore

Study Tools