Chapter 1 1.6 to 1.9 Chemical Foundations. Chapter 1 Table of Contents Return to TOC Copyright ©...
Transcript of Chapter 1 1.6 to 1.9 Chemical Foundations. Chapter 1 Table of Contents Return to TOC Copyright ©...
Chapter 11.6 to 1.9
Chemical Foundations
Chapter 1
Table of Contents
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1.6 Dimensional Analysis
1.7Temperature
1.8Density
1.9Classification of Matter
Section 1.6
Dimensional Analysis
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• Use when converting a given result from one system of units to another. To convert from one unit to another, use the
equivalence statement that relates the two units. Derive the appropriate unit factor by looking at the
direction of the required change (to cancel the unwanted units).
Multiply the quantity to be converted by the unit factor to give the quantity with the desired units.
Section 1.6
Dimensional Analysis
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Example #1
A golfer putted a golf ball 6.8 ft across a green. How many
inches does this represent?
To convert from one unit to another, use the equivalence
statement that relates the two units.
1 ft = 12 in
The two unit factors are:
Section 1.6
Dimensional Analysis
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• Derive the appropriate unit factor by looking at the direction of the required change (to cancel the unwanted units).
Example #1
A golfer putted a golf ball 6.8 ft across a green. How many inches does this represent?
Section 1.6
Dimensional Analysis
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• Multiply the quantity to be converted by the unit factor to give the quantity with the desired units.
Example #1
A golfer putted a golf ball 6.8 ft across a green. How many inches does this represent?
Section 1.6
Dimensional Analysis
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Example #2
An iron sample has a mass of 4.50 lb. What is the mass of this sample in grams?
(1 kg = 2.2046 lbs; 1 kg = 1000 g)
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• What data would you need to estimate the money you would spend on gasoline to drive your car from New York to Chicago? Provide estimates of values and a sample calculation.
React 3
Section 1.6
Dimensional Analysis
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Concept Check
What data would you need to estimate the money you would spend on gasoline to drive your car from New York to Los Angeles? Provide estimates of values and a sample calculation.
l Because you never learned dimensional analysis, you have been working at a fast food restaurant for the past 35 years wrapping hamburgers. Each hour you wrap 184 hamburgers. You work 8 hours per day. You work 5 days a week. you get paid every 2 weeks with a salary of $840.34. How many hamburgers will you have to wrap to make your first one million dollars?
Examples
l A senior was applying to college and wondered how many applications she needed to send. Her counselor explained that with the excellent grade she received in chemistry she would probably be accepted to one school out of every three to which she applied. She immediately realized that for each application she would have to write 3 essays, and each essay would require 2 hours work. Of course writing essays is no simple matter. For each hour of serious essay writing, she would need to expend 500 calories which she could derive from her mother's apple pies. Every three times she cleaned her bedroom, her mother would make her an apple pie. How many times would she have to clean her room in order to gain acceptance to 10 colleges?
Units to a Powerl How many m3 is 1500 cm3?
1500 cm33 1 m100 cm
1 m100 cm
1 m100 cm
1500 cm33 1 m
100 cm
33
Units to a Powerl How many cm2 is 15 m2?l 36 cm3 is how many mm3?
Section 1.7
Temperature
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• Fahrenheit• Celsius• Kelvin
Three Systems for Measuring Temperature
Section 1.7
Temperature
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The Three Major Temperature Scales
Section 1.7
Temperature
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Converting Between Scales
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• At what temperature does C = F? • Prove your answer.
React 4
ºC
ºF 9
5
0ºC is not 0ºF
ºC
ºF
(0,32)= (C1,F1)
ºC
ºF
(0,32) = (C1,F1)
(100,212) = (C2,F2)
Section 1.7
Temperature
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Exercise
At what temperature does C = F?
Section 1.7
Temperature
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• Since °C equals °F, they both should be the same value (designated as variable x).
• Use one of the conversion equations such as:
• Substitute in the value of x for both TC and TF. Solve for x.
Solution
Section 1.7
Temperature
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Solution
Section 1.8
Density
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• Mass of substance per unit volume of the substance.
• Common units are g/cm3 or g/mL.
Section 1.8
Density
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Example #1
A certain mineral has a mass of 17.8 g and a volume of 2.35 cm3. What is the density of this mineral?
Section 1.8
Density
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Example #2
What is the mass of a 49.6-mL sample of a liquid, which has a density of 0.85 g/mL?
Section 1.9
Classification of Matter
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• Anything occupying space and having mass.• Matter exists in three states.
Solid Liquid Gas
Matter
Section 1.9
Classification of Matter
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The Three States of Water
Section 1.9
Classification of Matter
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• Rigid• Has fixed volume and shape.
Solid
Section 1.9
Classification of Matter
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Structure of a Solid
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Section 1.9
Classification of Matter
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• Has definite volume but no specific shape.• Assumes shape of container.
Liquid
Section 1.9
Classification of Matter
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Structure of a Liquid
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Section 1.9
Classification of Matter
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• Has no fixed volume or shape.• Takes on the shape and volume of its
container.
Gas
Section 1.9
Classification of Matter
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Structure of a Gas
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Section 1.9
Classification of Matter
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• Have variable composition.
Mixtures
Homogeneous Mixture
Heterogeneous Mixture
Having visibly indistinguishable parts; solution.
Having visibly distinguishable parts.
Section 1.9
Classification of Matter
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Homogeneous Mixtures
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Section 1.9
Classification of Matter
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Homogeneous vs. Heterogeneous Mixtures
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Section 1.9
Classification of Matter
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Compound vs. Mixture
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Section 1.9
Classification of Matter
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Concept Check
Which of the following is a homogeneous mixture?
Pure water Gasoline Jar of jelly beans Soil Copper metal
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Chapter OnePrentice Hall © 2005 Hall © 2005General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Key Terms: Properties• A physical property is displayed
by a sample of matter without undergoing any change in the composition of the matter.٭ Physical properties include mass,
color, volume, temperature, density, melting point, etc.
• Chemical property – displayed by a sample of matter as it undergoes a change in composition.٭ Flammability, toxicity, reactivity,
acidity are all chemical properties.
Copper is red-brown,
opaque, solid: physical
properties.
Ethanol is flammable: a
chemical property.
Section 1.9
Classification of Matter
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• Change in the form of a substance, not in its chemical composition. Example: boiling or freezing water
• Can be used to separate a mixture into pure compounds, but it will not break compounds into elements. Distillation Filtration Chromatography
Physical Change
Section 1.9
Classification of Matter
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• A given substance becomes a new substance or substances with different properties and different composition. Example: Bunsen burner (methane reacts with
oxygen to form carbon dioxide and water)
Chemical Change
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Chapter OnePrentice Hall © 2005 Hall © 2005General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Key Terms: Properties• In a physical change, there is no
change in composition.• No new substances are formed.
٭ Examples include: evaporation; melting; cutting a piece of wood; dissolving sugar in water.
• In a chemical change or chemical reaction, the matter undergoes a change in composition.
• New substances are formed.٭ Examples include: burning
gasoline; dissolving metal in acid; spoilage of food.
The liquid fuel evaporates: a
physical change.
The vapor burns,
combining with oxygen: a chemical change.
Section 1.9
Classification of Matter
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Concept Check
How many of the following are examples of a chemical change?
Pulverizing (crushing) rock salt Burning of wood Dissolving of sugar in water Melting a popsicle on a warm summer day
Section 1.9
Classification of Matter
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The Organization of Matter
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Chapter OnePrentice Hall © 2005 Hall © 2005General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Classifying Matter
Figure 1.9
Atoms Molecules
make up
ALL MATTER
which exists as
Substances Mixtures
which may be
Elements Compounds Heterogeneous
which may be
Homogeneous
Pure
The Organization of MatterThe Organization of Matter
MatterMatter
Mixtures:a) Homogeneous (Solutions)b) Heterogeneous
Pure SubstancesPure Substances
Compounds ElementsElements
AtomsAtoms
NucleusNucleus ElectronsElectrons
Protons NeutronsNeutrons
QuarksQuarks QuarksQuarks
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Chapter OnePrentice Hall © 2005 Hall © 2005General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Classifying Matter
• A (pure) substance has a definite or fixed composition that does not vary from one sample to another.
• All substances are either elements or compounds.
• An element cannot be broken down into other simpler substances by chemical reactions.٭ About 100 elements known at this time
٭ Each element has a chemical symbol: O, H, Ag, Fe, Cl, S, Hg, Au, U, etc.
• A compound is made up of two or more elements in fixed proportions, and can be broken down into simpler substances.٭ Carbon dioxide, sodium chloride, sucrose (sugar), etc.
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Chapter OnePrentice Hall © 2005 Hall © 2005General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Classifying Matter• A mixture does not have a fixed composition.
• A homogeneous mixture has the same composition throughout, though the composition of different homogeneous mixtures may vary.٭ Soda pop, salt water, 14K gold, and many plastics are
homogeneous mixtures.
٭ 10K gold and 14K gold have different compositions but both are homogeneous.
• A heterogeneous mixture varies in composition and/or properties from one part of the mixture to another.٭ Adhesive tape, CD, pen, battery, chair, and people are
examples of heterogeneous mixtures.
• Most everyday “stuff” consists of mixtures.
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The Organization of Matter
Phase Phase DifferencesDifferences
SolidSolid – definite volume and shape; particles packed in fixed positions.LiquidLiquid – definite volume but indefinite shape; particles close together but not in fixed positions
GasGas – neither definite volume nor definite shape; particles are at great distances from one another
PlasmaPlasma – high temperature, ionized phase of matter as found on the sun.
Properties of MatterProperties of Matter
Volume
MassEnergy Content (think Calories!)
Extensive propertiesExtensive properties depend on the amount of matter that is present.
Intensive propertiesIntensive properties do not depend on the amount of matter present.
Melting point
Boiling point
Density
Separation of a MixtureSeparation of a Mixture
The constituents of the mixture retain The constituents of the mixture retain their identity and may be separated their identity and may be separated by physical means.by physical means.
Separation of a MixtureSeparation of a Mixture
The components of The components of dyes such as ink dyes such as ink may be separated may be separated by paper by paper chromatography.chromatography.
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Figure 1.15a A Line of the Mixture to be Separated is Placed at One End of a Sheet
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Figure 1.15b The Paper Acts as a Wick to Draw up the Liquid
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Figure 1.15c Component with the Weakest Attraction for the Paper Travels Faster
Separation of a Mixture Separation of a Mixture by Distillationby Distillation
Separation of a CompoundSeparation of a CompoundThe Electrolysis of water
Water Hydrogen + Oxygen
2 H2O 2 H2 + O2
Reactant Products
Compounds must be separated by chemicalchemical means.
With the application of electricity, water can be separated into its elements
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• Sketch a magnified view (showing atoms/molecules) of each of the following:– a heterogeneous mixture of two different
compounds.– a homogeneous mixture of an
element and a compound.
React 5