Welcome to CHEMISTRY !!!
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An Observational Science
An Experimental Science
A Laboratory Science
An Interesting Science
An Important Science
A “Hard” Science
What Happened To The Balloon?
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It was Whimpy and Broke!
It was fearful of all of the people!
Campbell scared it!
It got zapped by Klingons!
Hydrogen burns!
2H2 (g) + O2 (g) = 2 H2O (g) +
Energy
• Hydrogen and oxygen
are diatomic gases!
• Water can be a gas!
• ENERGY was given
off!-- This is
characteristic of an
Exothermic Reaction!
• This is a balanced
chemical reaction!
CHEMISTRY
The Study of
Matter and its Properties,
the Changes that
Matter Undergoes,
and the Energy
Associated with
those Changes
Chemistry as the Central Science
Oceanography
Atmospheric
Sciences
Physics
Economics
Medicine
Governments
Chemistry
People
Geology
Biology
Anthropology
Politics
Astronomy
Chemistry 142B: Campbell / Callis
Text: Chemistry - The Molecular Nature of Matter
and Change - By Martin Silberberg
Chapter #1 : Keys to the Study of Chemistry
Chapter #2 : The Components of Matter
Chapter #3 : Stoichiometry: Mole - Mass Relationships in
Chemical Systems
Chapter #4 : The Major Classes of Chemical Reactions
Chapter #5 : Gases and the Kinetic - Molecular Theory
Chapter #6 : Thermochemistry : Energy Flow and Chemical
Change
Chapter #7 : Quantum Theory and Atomic Structure
Chapter #8 : Electron Configuration and Chemical Periodicity
Chemistry Homework !!!
“ Chemistry is not a spectator sport,
you must become involved, and that
means that you must do homework!”
Linus Pauling - 1967
CHEMISTRY 142B
SYLLABUS
INSTRUCTOR (1): Professor C. T. Campbell (campbell@chem.washington.edu).
Telephone:616-6085, Office Hours: Mon. 10:30-11:30, Weds. 2:30-3:30 in 227 Bagley.
INSTRUCTOR (2): Professor J. B. Callis (callis@cpac.washington.edu) Telephone:
543-1208, Office Hours: Wed. 2:30 - 4:20, 204 Bagley Hall
TEXTS: Silberberg, Chemistry, The Molecular Nature of Matter and Change, Second
Edition, 2000, McGraw-Hill (required for lecture)
Chemistry 142 Laboratory Manual, available at the Copy Center, Odegaard Library
(required for lab).
OPTIONAL: Weberg, Student Study Guide to Accompany Chemistry, The Molecular
Nature of Matter and Change, 1996, Moseby. On reserve: Chemistry Library.
CHEMISTRY STUDY CENTER - BAG 330: M - Th 9 a.m. - 6 p.m, F 9 am - 2 pm.
Work with / learn from fellow Chem. 142 students.
Receive help from the TA's.
27 computers with helpful general chemistry programs.
Photocopy machine is available at 10 cents per copy.
LAB GROUP NUMBERS: You are in:
Gp. 1 if you are in subsections D, G, H, I, J, K or L;
Gp. 2 if you are in subsections A, B, C, E or F.
COURSE SCHEDULE:
Week
Silberberg
Chapter
Appen. A
1
9/25-9/29
Ch1,
start Ch2
2
Ch2,
10/2-10/6 start Ch3
Lecture Topic, Lecturer
3
10/9-13
Stoichiometry: Mass-Mole Relationships in Chemical
Systems.
Ch3
Read Appendix A in back of book: math used in C142.
Keys to the Study of Chemistry. Start: Components of
Matter.
Components of Matter
Start: Stoichiometry.
4
10/1610/20
Ch3,
Stoichiometry, cont. The Major Classes of Chemical
Start Ch4 Reactions. Midterm #1, (Ch. 1-3 and part 4) 10/20:
bring calculator, follow seating chart to all exams).
5
10/2310/27
6
10/3011/3
7
11/6-10
Ch4,
The Major Classes of Chemical Reactions. Start: Gases
start Ch5 and the Kinetic-Molecular Theory.
Laboratory Topic
(your Lab Gp. #)
No labs this week.
Lab #1: Safety, Stats.,
Composition of Two
Compounds (Gp.1)
" (Gp.2)
Lab #2: Stoichiometry of
Rxn of Hydrogen
Peroxide with Bleach
(formal write-up) (Gp.1)
" (Gp.2)
Ch5,
Gases, cont. Start: Thermochemistry: Energy Flow and Lab #3: Oxidation –
Redn. Titration (Gp.1)
Start Ch6 Chemical Change.
Ch6
Thermochemistry: Energy Flow and Chemical Change.
Holiday: 11/10, No Lecture.
" (Gp.2)
Quantum Theory and Atomic Structure.
Lab #4: Molar Mass of a
Low Boiling Cpd. (Gp.1)
Quantum …, cont.. Midterm #2, covering Chs. 4 (last No labs this week.
part), 5-6, 7 on 11/22. Holidays: 11/23-4
8
11/13-17
Ch7
9
11/2011/24
10
11/27-1
Ch7
Ch8
Electron Configuration and Chemical Periodicity.
11
12/4-12/6
Ch8
12
12/11-15
Ch1 - 8
Periodicity, cont.. Review for Final: Last Day of Class
12/6.
Final Examination: Tuesday, Dec. 12, 8:30-10:20
am, Bagley Rm. 133
Lab #4: Molar Mass of a
Low Boiling Cpd. (Gp.2)
No labs this week.
POLICIES AND PROCEDURES
LECTURES: Read the material to be covered prior to the lecture.
QUIZ SECTIONS – Help + quiz.
EXPERIMENTAL LABS – Pick up your C142 Lab Manual well in advance and read it up
through 1st expt. carefully.
LAB PRELIMS: before you enter the laboratory for each new experiment:
(1) complete that lab’s Preliminary Exercise,
(2) Fill out “Purpose” and “Procedure”.
LAB REPORTS – For all 4 of the labs, complete all calculations and tabulate your results
in lab manual in lab.
For one of the labs: write a formal lab report (due the next week in lab).
EXAMS - Most of the questions on the exams will be similar to homework problems.
FINAL EXAM - 1st half = material presented since Midterm Exam 2.
2nd half = cumulative review of all lecture material in Ch. 1-8.
REGRADING OF HOMEWORK, LAB REPORTS AND EXAMSLATE POLICY and MAKE-UP WORK –
COURSE GRADING 2 midterm exams (1 hr. each, 100 pts. each)
Quizzes / homework (lowest HW and quiz score dropped)
Informal laboratory reports (4)
Formal laboratory report
Final exam (2 hr.)
TOTAL
34 %
17 %
12 %
4%
33 %
100 %
Safety: There is an element of hazard in any laboratory
course. You are required to follow the safety rules as outlined
in your laboratory manual. In particular you are required to
wear approved safety goggles during all experiments. Proper
clothing must be worn at all times. Unnecessarily-exposed
skin is at risk from accidental spills, therefore shorts, short
skirts, or open-toed shoes are not allowed in the laboratory. In
order to comply with this policy, I suggest that you keep a
pair of jogging pants and sneakers in one of the hall lockers,
along with safety goggles. The lab is not a good place to wear
your favorite clothes. Long hair should be tied back. You may
buy goggles and notebooks at the University Book Store.
Chapter# 1 : Keys to the Study of Chemistry
First: Read Appendix A at back of book:
Math review for math needed in Chem 142.
1.1 Some Fundamental Definitions
1.2 Chemical Arts and the Origins of
Modern Chemistry
1.3 The Scientific Approach: Developing a Model
1.4 Chemical Problem Solving
1.5 Measurement in Scientific Study
1.6 Uncertainty in Measurement: Significant Figures
Definitions-I
Matter - The “stuff” of the universe: books, planets,
trees, professors - anything that has mass
and volume.
Composition - The types and amounts of simpler
substances that make up a sample of matter.
Properties - The characteristics that give each
substance a unique identity.
Physical Properties - are those the substance shows by
itself, without interacting with another substance
( color, melting point, boiling point,density, etc.)
Chemical Properties - are those that the substance shows
as it interacts with, or transforms into, other
substances (flammability, corrosiveness, etc.)
Fig 1.1 (P. 3) The distinction between physical and
chemical change.
Solid and Liquid water
have the same chemical
composition.
STATES OF MATTER and The World around US
• SOLID - The Earth
• LIQUID - Water
• GAS - The Atmosphere
Fig1.2
demos
Energy Involved in Phase Changes
Liberates
Energy
Gas
Boiling
Condensation
Liquid
Melting
Requires
Energy
Freezing
Solid
Definitions - II
Energy - The capacity to do work!
Potential Energy - The energy due to the position
of the object.Or Energy from a
chemical reaction.
Kinetic Energy - The energy due to the motion of
the object (= 1/2 mass x velocity2)
Fig. 1.3
Fig.1.6
Fig.1.7
Aspects of The Scientific Approach
1. Quantitative, reproducible measurements.
2. Theories that explain these, and thus hope to
reveal the mysteries of how nature works.
3. Testing of these theories (esp. convincing when
making accurate predictions of new behavior).
Curiosity driven: desire to understand how nature
works (basic or fundamental research)
or
Application driven: desire to improve a product or
process or to cure a problem (applied research)
Units Used in Calculations
Length : A car is 12 feet long, not “12 long” !
A person is 6 feet tall, not “6 tall” !
Area : A carpet measuring 3 feet(ft) by 4 ft has an area of:
(3 ft)x(4 ft) = ( 3 x 4 )( ft x ft ) = 12 ft2
Speed and Distance : A car traveling 350 miles(mi) in
7 hours(hr) has a speed of:
In 3 hours the car travels:
Units Used in Calculations
Length : A car is 12 feet long, not “12 long” !
A person is 6 feet tall, not “6 tall” !
Area : A carpet measuring 3 feet(ft) by 4 ft has an area of:
(3 ft)x(4 ft) = ( 3 x 4 )( ft x ft ) = 12 ft2
Speed and Distance : A car traveling 350 miles(mi) in
7 hours(hr) has a speed of:
350 mi / 7 hr = (350/7) x (mi/hr) = 50 mi/hr
In 3 hours the car travels:
(3 hr) x (50 mi/hr) = 150 mi
Derived SI Units
Quantity
Definition of Quantity
SI unit
Area
Length squared
m2
Volume
Length cubed
m3
Density
Mass per unit volume
kg/m3
Speed
Distance traveled per unit time
m/s
Acceleration
Speed changed per unit time
m/s2
Force
Mass times acceleration of object
Pressure
Force per unit area
Energy
Force times distance traveled
kg * m/s2
( =newton, N)
kg/(ms2)
( = pascal, Pa)
kg * m2/s2
( = joule, J)
How to Solve Chemistry Problems
1) Problem: States all of the information needed to solve
the Problem.
2) Plan: Clarify the known and unknown.
Suggest the steps needed to find the solution.
Develop a “roadmap” solution.
3)Solution: Calculations appear in the same order as outlined.
4) Check: Is the result what you expect or at least in the same
order of magnitude!
5) Comment: Additional information as needed.
Conversion Factors : Unity Factors - I
Equivalent factors can be turned into conversion factors by
dividing one side into the other!
1 mile = 5280 ft
1 in = 2.54 cm
or 1 = 1 mile / 5280 ft = 5280 ft / 1 mi
or
1=
=
In converting one set of units for another, the one desired
is on top in the conversion factor, and the “old” one is
canceled out!
convert 29,141 ft into miles!
Conversion Factors : Unity Factors - I
Equivalent factors can be turned into conversion factors by
dividing one side into the other!
1 mile = 5280 ft
1 in = 2.54 cm
or 1 = 1 mile / 5280 ft = 5280 ft / 1 mi
or
1 = 1 in / 2.54 cm = 2.54 cm / 1 in
In converting one set of units for another, the one desired
is on top in the conversion factor, and the “old” one is
canceled out!
convert 29,141 ft into miles!
29,141 ft x 1 mi / 5280 ft = 5. 519 mi
Conversion Factors - II
1.61 km = 1 mi or
1=
Convert 5.519 miles in to kilometers:
Conversions in the metric system are easy, as 1 km = 1000 m
and 1 meter (m) = 100 centimeters (cm)
and 1 cm = 10 millimeters (mm).
Ex.: Convert 8.89 km into cm and mm:
Conversion Factors - II
1.61 km = 1 mi or
1 = 1.61 km / 1 mi
Convert 5.519 miles in to kilometers:
5.519 mi x 1.61 km / mi = 8.89 km
Conversions in the metric system are easy, as 1 km = 1000 m
and 1 meter (m) = 100 centimeters(cm)
and 1 cm = 10 millimeters(mm).
Ex.: Convert 8.89 km into cm and mm:
8.89 km x 1000m / 1 km = 8890 m
8890 m x 100 cm / m = 889000 cm
Conversion Factors - III
• Multiple conversion factors!
• Convert 3.56 lbs/hr into units of
milligrams/sec!
Conversion Factors - III
• Multiple conversion factors!
• Convert 3.56 lbs/hr into units of
milligrams/sec!
• 3.56 lbs/hr x (1kg/2.205 lbs) x(1000g/1kg) x
(1000mg/1g) x (1hr/60 min) x (1min/60 sec)
= 448 mg/sec
Fig. 1.9
m3
Liter = L
= dm3
cm3 = mL
Fig. 1.10
Conversion Factors - IV
metric volume to liters
•
•
•
•
1.35 x 109 km3 = volume of world’s oceans.
1 Liter = 1 L = 1 dm3
How many liters of water are in the oceans?
conversion factors: 1 km = 1000 m
1 L = 1 dm3 = 10-3 m3 or 1000 L = 1 m3
Conversion Factors - IV
metric volume to liters
•
•
•
•
1.35 x 109 km3 = volume of world’s oceans.
1 Liter = 1 L = 1 dm3
How many liters of water are in the oceans?
conversion factors: 1 km = 1000 m
1 L = 1 dm3 = 10-3 m3 or 1000 L = 1 m3
• 1.35 x 109 km3 x (103 m/1 km )3 x ( 103 L/m3)
= 1.35 x 1021 liters
Conversion Factors - V
Calculate the mass of 1.00 ft3 of Lead
(density=11.3 g/ cm3)?
• 1 cm3 of Pb = 11.3 g of Pb so
1 = (11.3 g of Pb)/(1 cm3 of Pb) = 11.3 g/cm3
Conversion Factors - V
Calculate the mass of 1.00 ft3 of Lead
(density=11.3 g/ cm3)?
• 1 cm3 of Pb = 11.3 g of Pb so
1 = (11.3 g of Pb)/(1 cm3 of Pb) = 11.3 g/cm3
• 1.00 ft3 x (12 in/ft)3 x (2.54 cm/in)3
= 28,316.84659 cm3
• 2.83 x 104 cm3 x 11.3 g/cm3 = 319,790.0000 g
• Ans. = 3.20 x 105 g = 3.20 x 102 kg
Fig1.11
Like Sample Problem 1.3 (p20)
The Volume of an irregularly shaped solid can be determined
from the volume of water it displaces. A graduated cylinder
contains 245.0 mL water. When a small piece of Pyite, an ore
of Iron, is submerged in the water, the volume increases to
315.8 mL . What is the volume of the piece of Pyrite in cm3
and in liters.
Vol (mL ) =
Vol (cm3) =
Vol (liters) =
Like Sample Problem 1.3 (p20)
The Volume of an irregularly shaped solid can be determined
from the volume of water it displaces. A graduated cylinder
contains 245.0 mL water. When a small piece of Pyite, an ore
of Iron, is submerged in the water, the volume increases to
315.8 mL . What is the volume of the piece of Pyrite in cm3
and in liters.
Vol (mL ) = 315.8 mL - 245.0 mL = 70.80 mL
Vol (cm3) = 70.80 mL x 1 cm3/ 1 mL = 70.80 cm3
Vol (liters) = 70.80 mL x 10 -3liters / mL = 7.08 x 10 -2 liters
Sample Problem 1.5 (p 23) - I
Lithium (Li) is a soft, gray solid that has the lowest density
of any metal. If a slab of Li weighs 1.49 x 103 mg and has
sides that measure 20.9 mm by 11.1 mm by 12.0 mm, what
is the density of Li in g/ cm3 ? Lengths (mm)
of sides
Mass (mg)
of Li
Lengths (cm)
of sides
Mass (g)
of Li
Volume (cm3)
Density (g/cm3) of Li
Sample Problem 1.5 - II
Mass (g) of Li = 1.49 x 103 mg
Length (cm) of one side = 20.9 mm
Similarly, the other side lengths are:
Volume (cm3) =
Density = mass/volume
Density of Li =
Sample Problem 1.5 - II
Mass (g) of Li = 1.49 x
103
mg x
1g
103 mg
= 1.49 g
Length (cm) of one side = 20.9 mm x 1cm / 10 mm = 2.09 cm
Similarly, the other side lengths are 1.11 cm and 1.20 cm
Volume (cm3) = 2.09 cm x 1.11 cm x 1.20 cm = 2.78 cm3
Density = mass/volume
Density of Li =
1.49 g
2.78 cm3
= 0.536 g/cm3
Like Sample Problem 1. 5 - Density of a Metal
Problem: Cesium is the most reactive metal in the periodic
table. What is its density if a 3.4969 kg cube of Cs has sides
of 125.00 mm each?
Plan: Calculate the volume from the dimensions of the cube,
and calculate the density from the mass and volume.
Solution:
length = 125.00 mm
mass = 3.4969 kg
Volume =
density =
mass
volume
Like Sample Problem 1. 5 - Density of a Metal
Problem: Cesium is the most reactive metal in the periodic
table. What is its density if a 3.4969 kg cube of Cs has sides
of 125.00 mm each?
Plan: Calculate the volume from the dimensions of the cube,
and calculate the density from the mass and volume.
Solution:
length = 125.00 mm = 12.500 cm
mass = 3.4969 kg x 1000g/kg = 3,496.9 g
Volume = (length)3 = (12.500 cm)3 = 1,953.125 cm3
density =
mass
volume
=
3496.9 g = 1.7904 g/cm3
1,953.125 cm3
Archemedies Principle Problem
Problem: Calculate the density of an irregulaly shaped metal
object that has a mass of 567.85 g if, when it is placed into a
2.00 liter graduated cylinder containing 900.00 mL of water,
the final volume of the water in the cylinder is 1,277.56 mL ?
Plan: Calculate the volume from the different volume
readings, and calculate the density using the mass that
was given.
Solution:
Volume =
Density =
mass
volume
Archemedies Principle Problem
Problem: Calculate the density of an irregulaly shaped metal
object that has a mass of 567.85 g if, when it is placed into a
2.00 liter graduated cylinder containing 900.00 mL of water,
the final volume of the water in the cylinder is 1,277.56 mL ?
Plan: Calculate the volume from the different volume
readings, and calculate the density using the mass that
was given.
Solution:
Volume = 1,277.56 mL - 900.00 mL = 377.56 mL
Density =
mass
volume
=
567.85 g
377.56 mL
= 1. 50 g / mL
Definitions - Mass & Weight
Mass - The quantity of matter an object contains
kilogram - ( kg ) - the SI base unit of mass, is a
platinum - iridium cylinder kept in
Paris as a standard!
Weight - depends upon an object’s mass and the strength
of the gravitational field pulling on it.
Sample Problem: Computer Chips
Future computers might use memory bytes which require an
area of a square with 0.25 mm sides. How many bytes could
be put on a 1 in x 1 in computer chip? If each byte required
that 25 % of its area to be coated with a gold film 10 nm thick,
what mass of gold would be needed to make one chip?
Mass (g) of gold
Area of byte
No. of bytes
Gold area on chip
Volume of gold
Sample Problem: Computer Chips
Future computers might use memory bytes which require an
area of a square with 0.25 mm sides. How many bytes could
be put on a 1 in x 1 in computer chip? If each byte required
that 25 % of its area to be coated with a gold film 10 nm thick,
what mass of gold would be needed to make one chip?
Mass (g) of gold
Area of byte
Total area / area of byte
No. of bytes
Gold area per
byte x No. bytes
Gold area on chip
Density Tab.1.5:
gold 19.5 g/cm3
Area x thickness
Volume of gold
Sample Problem: Chips, soln.
Area of byte =
No. of bytes = 1 in.2/chip
Gold area on a byte =
Gold area on chip =
Sample Problem: Chips, soln.
Area of byte = 0.25 x 10-6 m x 0.25 x 10-6 m
= 6.25 x 10-14 m2 x (100 cm/m)2 = 6.25 x 10-10 cm2 /byte
No. of bytes =
1 in.2/chip x (2.54 cm/in)2 / (6.25 x 10-10cm2/byte)
= 1.03 x 1010 bytes/chip
Gold area on a byte =
(6.25 x 10-10 cm2 /byte)(0.25 cm2 gold/cm2 of chip)
=1.56 x 10-10 cm2 gold / byte
Gold area on chip =
(1.56 x 10-10 cm2 gold / byte)x(1.03 x 1010 bytes/chip)
= 1.61 cm2 gold / chip
Sample Problem: Chips, soln. cont.
Gold volume on chip =
Density of gold: 19.3 g/cm3 from Table 1.5
Mass of gold =
Sample Problem: Chips, soln. cont.
Gold volume on chip = (1.61 cm2)x(10 nm) x (10-7 cm/nm)
= 1.61x10-6 cm3 gold
Density of gold: 19.3 g/cm3 from Table 1.5
Mass of gold = (1.61x10-6 cm3 gold)x(19.3 g gold/cm3 gold)
= 3.1x10-5 g gold = 31 mg gold
Fig.1.12
Fig.1.13
Temperature Scales and Interconversions
Kelvin ( K ) - The “Absolute temperature scale” begins at
absolute zero and only has positive values.
Celsius ( oC ) - The temperature scale used by science,
formally called centigrade and most
commonly used scale around the world,
water freezes at 0oC, and boils at 100oC.
Fahrenheit ( oF ) - Commonly used scale in America for
our weather reports, water freezes at 32oF,
and boils at 212oF.
T (in K) = T (in oC) + 273.15 T (in oF) = 9/5 T (in oC) + 32
T (in oC) = T (in K) - 273.15 T (in oC) = [ T (in oF) - 32 ] 5/9
Temperature Conversions
The boiling point of Liquid Nitrogen is -195.8 oC, what is
the temperature in Kelvin and degrees Fahrenheit?
T (in K) = T (in oC) + 273.15
T (in K) =
T (in oF) = 9/5 T (in oC) + 32
T (in oF) =
The normal body temperature is 98.6oF, what is it in Kelvin
and degrees Celsius?
T (in oC) = [ T (in oF) - 32] 5/9
T (in oC) =
T (in K) = T (in oC) + 273.15
T (in K) =
Temperature Conversions
The boiling point of Liquid Nitrogen is - 195.8 oC, what is
the temperature in Kelvin and degrees Fahrenheit?
T (in K) = T (in oC) + 273.15
T (in K) = -195.8 + 273.15 = 77.35 K = 77.4 K
T (in oF) = 9/5 T (in oC) + 32
T (in oF) = 9/5 ( -195.8oC) +32 = -320.4 oF
The normal body temperature is 98.6oF, what is it in Kelvin
and degrees Celsius?
T (in oC) = [ T (in oF) - 32] 5/9
T (in oC) = [ 98.6oF - 32] 5/9 = 37.0 oC
T (in K) = T (in oC) + 273.15
T (in K) = 37.0 oC + 273.15 = 310.2
Fig1.15
Fig. 1.16
Precision and Accuracy
Errors in Scientific Measurements
Precision - Refers to reproducibility or How close the
measurements are to each other!
Accuracy - Refers to how close a measurement is to the
real value!
Systematic error - produces values that are either all higher
or all lower than the actual value.
Random Error - in the absence of systematic error, produces
some values that are higher and some that
are lower than the actual value.
Fig. 1.17
Rules for Determining Which Digits Are
Significant
All digits are significant, except zeros that are used only to
position the decimal point.
1. Make sure that the measured quantity has a decimal point.
2. Start at the left of the number and move right until you
reach the first nonzero digit.
3. Count that digit and every digit to its right as significant.
Zeros that end a number and lie either after or before the
decimal point are significant; thus 1.030 mL has four
significant figures, and 5300. L has four significant figures
also. Numbers such as 5300 L have 2 sig. figs., but
5.30x103 L has 3. A terminal decimal point is often used to
clearify the situation, but scientific notation is clearer (best)!
Examples of Significant Digits in Numbers
Number
- Sig digits
0.0050 L
two
18.00 g
four
0.00012 kg
two
83.0001 L
five
0.006002 g
four
875,000 oz
three
30,000 kg
one
5.0000 m3
five
23,001.00 lbs seven
0.000108 g
three
1,470,000 L
three
Number
-
1.34000 x 107 nm
5600 ng
87,000 L
78,002.3 ng
0.000007800 g
1.089 x 10 -6L
0.0000010048 oz
6.67000 kg
2.70879000 mL
1.0008000 kg
1,000,000,000 g
Sig digits
six
two
two
six
four
four
five
six
nine
eight
one
Rules for Significant Figures in answers
1. For multiplication and division. The number with the
least certainty limits the certainty of the result. therefore, the
answer contains the same number of significant figures as
there are in the measurement with the fewest significant
figures. Multiply the following numbers:
9.2 cm x 6.8 cm x 0.3744 cm = 23.4225 cm3 = 23 cm3
2. For addition and subtraction. The answer has the same
number of decimal places as there are in the measurement
with the fewest decimal places. Example, adding two volumes
83.5 mL + 23.28 mL = 106.78 mL = 106.8 mL
Example subtracting two volumes:
865.9 mL - 2.8121393 mL = 863.0878607 mL = 863.1 mL
Rules for Rounding off numbers:
1. If the digit removed is more than 5, the preceding number increases
by 1 : 5.379 rounds to 5.38 if three significant figures are retained and to
5.4 if two significant figures are retained.
2. If the digit removed is less than 5, the preceding number is
unchanged : 0.2413 rounds to 0.241 if three significant figures are
retained and to 0.24 if two significant figures are retained.
3.If the digit removed is 5, the preceding number increases by 1 if it is
odd and remains unchanged if it is even: 17.75 rounds to 17.8, but
17.65 rounds to 17.6. If the 5 is followed only by zeros, rule 3 is
followed; if the 5 is followed by nonzeros, rule 1 is followed:
17.6500 rounds to 17.6, but 17.6513 rounds to 17.7
4. Be sure to carry two or more additional significant figures through a
multistep calculation and round off only the final answer. (In sample
problems and follow-up problems, we round off intermediate steps of
a calculation to show the correct number of significant figures.