chapter 1: chemistry and measurement
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Vanessa Prasad-Permaul Valencia College CHM 1045. Chapter 1: Chemistry and Measurement. Properties of Matter. Chemistry: The study of composition, properties, and transformations of matter Matter: Anything that has both mass & volume Hypothesis: Interpretation of results - PowerPoint PPT PresentationTRANSCRIPT
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Vanessa Prasad-PermaulValencia College
CHM 1045
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Properties of Matter
Chemistry: The study of composition, properties, and transformations of matter
Matter: Anything that has both mass & volume
Hypothesis: Interpretation of results
Theory: Consistent explanation of observations
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Conservation of Mass
Law of Mass Conservation: Mass is neither created nor destroyed in chemical reactions.
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Example 1: Conservation of Mass
C(s) + O2(g) CO2(g)
a) 12.3g C reacts with 32.8g O2, ?g CO2
12.3g + 32.8g = 45.1g
a) 0.238g C reacts with ?g O2 to make .873g CO2 0.238g + x = 0.873g = 0.873g-0.238g = 0.635g of O2
a) ?g C reacts with 1.63g O2 to make 2.24g CO2
x + 1.63g = 2.24g = 2.24g - 1.63g = 0.61g C
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Example 1: Conservation of Mass
Exercise 1.1 1.85g of wood is placed with 9.45g of air in a sealed vessel. It
isheated and the wood burns to produce ash and gases. The ashis weighed to yield 0.28g. What is the mass of the gases in thevessel?
1.85g Wood + 9.45g Air heat 0.28g Ash + ? g gases
1.85 + 9.45 - 0.28 = 11.02g of gases
What is the mass of wood that is converted to gas by the end of
the experiment?1.85g of Wood – 0.28g of ash = 1.57g
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Matter
Matter is any substance that has mass and occupies volume.
Matter exists in one of three physical states: solid liquid gas
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Solid
In a solid, the particles of matter are tightly packed together.
Solids have a definite, fixed shape.
Solids cannot be compressed and have a definite volume.
Solids have the least energy of the three states of matter.
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Liquid
In a liquid, the particles of matter are loosely packed and are free to move past one another.
Liquids have an indefinite shape and assume the shape of their container.
Liquids cannot be compressed and have a definite volume.
Liquids have less energy than gases but more energy than solids.
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Gases
In a gas, the particles of matter are far apart and uniformly distributed throughout the container.
Gases have an indefinite shape and assume the shape of their container.
Gases can be compressed and have an indefinite volume.
Gases have the most energy of the three states of matter.
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Phases
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Properties of Matter
A physical change is a change in the form of matter but not in its chemical identity
A chemical change or a chemical reaction is a change in which one of more kinds of matter are transformed into a new kind of matter or several new kinds of matter
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Properties of Matter
Physical Properties can be determined without changing the chemical makeup of the sample.
Some typical physical properties are: Melting Point, Boiling Point, Density, Mass, Touch,
Taste, Temperature, Size, Color, Hardness, Conductivity.
Some typical physical changes are: Melting, Freezing, Boiling, Condensation,
Evaporation, Dissolving, Stretching, Bending, Breaking.
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Properties of Matter
Chemical Properties are those that do change the chemical makeup of the sample.
Some typical chemical properties are: Burning, Cooking, Rusting, Color change,
Souring of milk, Ripening of fruit, Browning of apples, Taking a photograph, Digesting food.
Note: Chemical properties are actually chemical changes
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Properties of Matter
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Classifications of Matter
Matter can be divided into two classes: mixtures pure substances
Mixtures are composed of more than one substance and can be physically separated into its component substances.
Pure substances are composed of only one substance and cannot be physically separated.
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Pure Substances
There are two types of pure substances: Compounds Elements
A compound is a substance composed of two or more elements chemically combined Compounds can be chemically separated into
individual elements. Water is a compound that can be separated
into hydrogen and oxygen. An element cannot be broken down
further by chemical reactions.
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Dalton’s Atomic Theory
Law of Definite Proportions: Different samples of a pure chemical substance always contain the same proportion of elements by mass.
Any sample of H2O contains 2 hydrogen atoms for every oxygen atom
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Mixtures
There are two types of mixtures: homogeneous mixtures heterogeneous mixtures
Homogeneous mixtures have uniform properties throughout. Salt water is a homogeneous mixture.
Heterogeneous mixtures do not have uniform properties throughout. Sand and water is a heterogeneous mixture.
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Example 2: Matter
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Which of the following represents a mixture?
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Accuracy, Precision, and Significant Figures in Measurement
Accuracy is how close to the true value a given measurement is.
Precision is how well a number of independent measurements agree with one another.
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Example 8: Accuracy & Precision
Which of the following is precise but not accurate?
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Accuracy, Precision, and Significant Figures in Measurement
Significant Figures are the total number of digits in the measurement.
The results of calculations are only as reliable as the least precise measurement!!
Rules exist to govern the use of significant figures after the measurements have been made.
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Accuracy, Precision, and Significant Figures in Measurement
Rules for Significant Figures:
Zeros in the middle of a number are significant
Zeros at the beginning of a number are not significant
Zeros at the end of a number and following a period are significant
Zeros at the end of a number and before a period may or may not be significant.
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Example 4: Significant Figures
How many Significant Figures ?a) 0.000459 = 3
b) 12.36 = 4
c) 36,450 = 4
d) 8.005 = 4
e) 28.050 = 5
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Accuracy, Precision, and Significant Figures in Measurement
Rules for Calculating Numbers: During multiplication or division, the
answer can’t have more significant figures than any of the original numbers.
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Example 5: Significant Figures
a) 218.2 x 79 = 17237.8 = 1.7 x 104
a) 12.5 / 0.1272 = 94.33962264150943 = 94.3
b) 0.2895 x 0.29 = 0.083955 = 0.084
c) 32.567 / 22.98 = 1.417188859878155 = 1.417
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-During addition or subtraction, the answer can’t have more digits to the right of the decimal point than any of the original numbers.
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Accuracy, Precision, and Significant Figures in Measurement
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Example 6: Significant Figures
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a) 218.2 + 79 = 297.2 = 297
b) 12.5 - 0.1272 = 12.3728 = 12.4
c) 0.2895 + 0.29 = 0.5795 = 0.58
d) 32.567 - 22.98 = 55.547 = 55.55
e) 185.5+2.224 = 187.724 = 187.7
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Accuracy, Precision, and Significant Figures in Measurement
Rules for Rounding Numbers:
If the first digit removed is less than 5 round down (leave # same)
If the first digit removed is 5 or greater round up
Only final answers are rounded off, do not round intermediate calculations
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Example 7: Rounding and Significant Figures
Round off each of the following measurements
a) 3.774499 L to 4 sig. figs. = 3.774L
b) 255.0974 K to 3 sig. figs. = 255K
c) 55.265 kg to 4 sig. figs. = 55.27kg
d) 1.2151ml to 3 sig. figs. = 1.22ml
e) 1.2143g to 3 sig. figs. = 1.21g
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Exercise 1.3Give answers to the following arithmetic setups. Round to the correct number of significant figures:
a) 5.61 x 7.891 = 4.864671 = 4.9 9.1
b) 8.91 - 6.435 = 2.475 = 2.48
c) 6.81 – 6.730 = 0.08 = 0.08
d) 38.91 x (6.81-6.730) = 38.91 x 0.08 = 3.1128 = 3
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Scientific Notation
Changing numbers into scientific notation Large # to small # Moving decimal place to left, positive
exponent123,987 = 1.23987 x 105
Small # to large # Moving decimal place to right, negative
exponent0.000239 = 2.39 x 10-4
How to put into calculator
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Example 3: Scientific Notation
Put into or take out of scientific notation
a) 1973 = 1.973 x 103
b) 5.5423 x 10-4 = 0.00055423
c) 0.775 = 7.75 x 10-1
d) 3.55 x 107 = 35,500,000
e) 8500 = 8.5 x 103
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Measurement and Units
Physical Quantity Name of Unit Abbreviation Mass kilogram kg
Length meter m Temperature kelvin K
Amount of substance mole mol Time second s
Electric current ampere A Luminous intensity candela cd
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SI Units
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Measurement and Units
Factor Prefix Symbol 1,000,000,000 = 109 giga G
1,000,000 = 106 mega M 1,000 = 103 kilo k 100 = 102 hecto h 10 = 101 deka da 0.1 = 10-1 deci d
0.01 = 10-2 centi c 0.001 = 10-3 milli m
0.000,001 = 10-6 micro µ 0.000,000,001 = 10-9 nano n
0.000,000,000,001 = 10-12 pico p
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*
*
**
**
*
* Important
Some prefixes for multiples of SI units
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Exercise 1.4Express the following quantities using an SI prefix
and abase unit. For instance, 1.6 x 10-6m = 1.6m. A
quantity suchas0.000168g could be written 0.168mg or 168g.
a) 1.84 x 10-9 m = 1.84 nm (nanometer)b) 5.67 x 10-12 s = 5.67 ps (picosecond)c) 7.85 x 10-3 g = 7.85 mg (milligram)
d) 9.7 x 103 m = 9.7 km (kilometer)e) 0.000732 s = 0.732 ms (millisecond)
= 732us (microsecond)f) 0.000000000154 m = 0.154nm (nanometer)
= 154pm (picometer)
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Changes in Physical State
Most substances can exist as either a solid, liquid, or gas.
Water exists as a solid below 0 °C; as a liquid between 0 °C and 100 °C; and as a gas above 100°C.
A substance can change physical states as the temperature changes.
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Solid Liquid
When a solid changes to a liquid, the phase change is called melting.
A substance melts as the temperature increases.
When a liquid changes to a solid, the phase change is called freezing.
A substance freezes as the temperature decreases.
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Liquid Gas
When a liquid changes to a gas, the phase change is called vaporization.
A substance vaporizes as the temperature increases.
When a gas changes to a liquid, the phase change is called condensation.
A substance condenses as the temperature decreases.
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Solid Gas
When a solid changes directly to a gas, the phase change is called sublimation.
A substance sublimes as the temperatureincreases.
When a gas changes directly to a solid, the phasechange is called deposition.
A substance undergoes deposition as the temperature decreases.
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Temperature
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Diagram of the various phases of temperature change
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Temperature
Temperature Conversions:The Kelvin and Celsius scales have equal size units (a change of 1oC is equivalent to a change of 1K)
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100 K100 oC180oF
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Temperature Conversions: Celsius (°C) — Kelvin (K) temperature
conversion:Kelvin (K) = t°C x 1K + 273.15K
1oC Fahrenheit (°F) — Celsius (C) temperature
conversions: there are exactly 9oF for every 5oC. Knowing that 0oC = 32oF
tF = tC x 9oF + 32 5oC
tC = 5oC x (toF – 32) 9oF
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Example 9: Temp. Conversions
Carry out the indicated temperature conversions:
a) –78°C = ? K = (-78oC x 1K/1oC) +273.15K = 195.15 = 195K
b) 158°C = ? °F = (158oC x 9oF/5oC)+32oF = 316.4 = 316oF
c) 373.15 K = ? °C = (373.15K x 1oC/1K)– 273.15K = 100K
d) 98.6°F = ? °C = 5oC/9oF x (98.6oF – 32oF) = 37oC
e) 98.6°F = ? K = (37oC x 1K/1oC) +273.15K = 310.15 = 310K
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Exercise 1.5A person with a fever has temperature of
102.5oF.What is this temperature in oC? A cooling
mixtureof dry ice and isopropyl alcohol has a
temperatureof -78 oC. What is the temperature in kelvins?
a) oC = 5oC x (oF – 32 ) = 0.555 x (102.5 – 32) = 39.2oC
9oF
b) K = oC + 273.15 = -78 + 273.15 = 195 K
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Volume
Volume: how much three-dimensional space a substance (solid, liquid, gas) or shape occupies or contains often quantified numerically using the SI derived unit (m3) the cubic meter.
The volume of a container is generally understood to be the capacity of the container, i. e. the amount of fluid (gas or liquid) that the container could hold, rather than the amount of space the container itself displaces.
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Volume
units of Volume:
m3 or cm3 (cc)
Traditionally chemists use liter (L)
1cm3 = 1cc = 1mL48
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Measurement and Units
Density: relates the mass of an object to its volume.
Density = mass / Volume D = m / VV = m / Dm = V D
Density decreases as a substance is heated because the substance’s volume increases.
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Density
What is the density of glass (in mL) if a sample weighing 26.43 g has a volume of 12.40 cm3?
d = ?m = 26.43 gV = 12.40 cm3 = 12.40 mLd = m = 26.43 g = 2.13145 = 2.131
g/mL V 12.40 mL
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Density
What is the volume of an unknown solution if the mass is 12.567 g and the density is 14.621 g/mL ?
d = m/V V x d = m V = m/d
V = 12.567 g / 14.621 g/mL = 0.85952 mL
12.567g x 1mL = 0.85952 mL 14.621g
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Density
What is the mass of an unknown solution if the
volume is 20.2 mL and the density is 2.613 g/mL?
d = m/V m = d x V
m = 2.613g x 20.2 mL = 52.7826 = 52.8 g
mL
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Exercise 1.6A piece of metal wire has a volume of 20.2 cm3
and amass of 159 g. What is the density of the metal?
D = m = 159 g = 7.87128712 = 7.87 g /cm3
V 20.2cm3
We know that the following metals have the following
densities. Which metal is the wire made of?
Mn = 7.21 g/cm3
Fe = 7.87 g/cm3
Ni = 8.90 g/cm3
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Exercise 1.7
Ethanol (grain alcohol) has a density of 0.789 g/cm3.
What volume (mL) of ethanol must be poured into
a graduated cylinder to equal 30.3 g?
d = m/V V x d = m V = m / d
V = 30.3 g x 1 cm3 = 38.4cm3
0.789 g
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Dimensional Analysis & Units
Dimensional-Analysis method uses a conversion factor to express the relationship between units.
Original quantity x conversion factor = equivalentquantity
Example: express 2.50 kg lb.Conversion factor: 1.00 kg = 2.205 lb
2.50 kg x 2.205 lb = 6.00 lb 1.00 kg
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Exercise 1.8The oxygen molecule (O2) consists of two
oxygenatoms a distance of 121 pm apart. How manymillimeters (mm) is this distance?
121 pm x 10-12 m x 1mm = 1.21 x 10-7 mm
1 pm 10-3
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Exercise 1.9A large crystal is constructed by stacking small
identicalpieces of crystal. A unit cell is the smallest piece
fromwhich a crystal can be made. A unit cell of a
crystal ofgold metal has a volume of 67.6 A3. What is the
volumein dm3?
67.6 A3 x 10-10 m x 10 dm = 6.76 x 10 -26 dm3
1 A3 1 m
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Exercise 1.10Using the following definitions, obtain theconversion factor for yards to meters. How
manymeters are there in 3.54 yd?1 in = 2.54cm (exactly) 1 yd = 36in (exactly)
1 yd x 1 in x 1 cm = 1.093613298 = 1.094 yd/m
36 in 2.54 cm 10-2 m
3.54 yd x 1 m = 3.24 m 1.094 yd
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Conversions
a) 1.267 km m cm1.267km x 1000m x 100cm = 126700cm = 1.267
x 105
1km 1m
b) 0.784 L mL0.784L x 1000mL = 784L
1L
c) 3.67 x 105 cm in 3.67 x 105cm x 1in = 144488.1889in = 1.44 x
105in 2.54cm
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Conversions
d) 79 oz g79oz x 28.35g = 2239.65g = 2.2 x 103g
1oz
e) 9.63 x 10-3 yd ft9.63 x 10-3yd x 1m x 1km x 0.62137mile x
5280ft 1.0936yd 1000m 1km
1mile= 0.0289ft
f) 23.5 cm2 m2
23.5cm2 x 1m2 = 0.235m2
100cm2
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Conversions
g) 1.34 x 1012 pm m1.34 x 1012pm x 1m = 1.34 x 1024m
10-12pm
h) 4.67 x 10-7 nm pm4.67 x10-7nm x 1m x 10-12pm = 4.67 x 10-12pm
10-9nm 1m
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