Hóa học đại cương của Julia Burdge (2009)

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INTRODUCTION OF THE BOOK “CHEMISTRY” 2009 BY JULIA BURDGE
Represented and redistributed by Mr. Truong Minh Chien
[email protected]
Introduction
My name is Truong Minh Chien (Trương Minh Chiến), CEO of NKMB co., ltd. Sometime, the people knew me as CEO of Ninh Kieu Tera Net co., ltd.
I also be a scientist and an economist. My favourist field is telecom science. Therefore, the chemistry is not one of my hoppies, a relative major only.

My blogsite : http://mba-programming.blogspot.com
My email : [email protected]
My facebook : losedtales
My company website : http://chodientu.vn/nko
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Preface
I found out this book “Chemistry, Julia Burdge, Second Edition. McGraw Hill.” through internet and Rapidshare network.

I think that the Chemistry book will be very helpful and useful for who is the teachers, instructors or students in high school level in Vietnam, either in other degrees as a perfect hardcover reference book.

So that I attempt to gather more lecture notes and other possible information resources to represent this chemistry book of Dr. Julia Burdge to other Vietnamese.
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Preface 2
To completed entire this chemistry resources (included ebook, hardcover book, lecture notes and software simulation…), I accessed some authority information resources as worse as pass through some private information or hidden resources. Hence, if you use this lecture notes or other information as redistriputed or reproduced, could you change some word or themes or font size, please!. Because the copyright term and illegal internet security policy.

This e-learning resources were reproduced from two hardcover book which are Chemistry, Mrs. Julia Burdge, 2nd Edition. McGraw Hill and Chemistry: A Molecular Approach, Mr. Nivaldo Tro, 1st Ed., Prentice Hall
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How to use the book resources
The hardcover book in English is available at website : http://tailieu.vn/losedtales . The entired book was divided and separated into 40 parts for uploading and management purposes.

The lecture notes and other resources are available at website : http://www.violet.vn (user : losedtales)
or http://www.violet.vn/losedtales
or http://losedtales.violet.vn
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How to use the book resources 2
The text book website :
http://www.mharis.com/
( http://highered.mcgraw-hill.com/classware/infoCenter.do?isbn=0073025542 )

E-learning resource :
https://elearning.utdallas.edu/webct/entryPageIns.dowebct

Other useful resources :
http://www.imperial.edu/index.php?pid=1232
http://faculty-staff.ou.edu/K/Phillip.E.Klebba-1/
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How to use the book resources 3
The software simulation are available on :
http://hydrachem.sourceforge.net/
http://genchemlab.sourceforge.net/
http://sourceforge.net/projects/organicbuilder/
http://organicchemi.sourceforge.net/
http://jbf.sourceforge.net/
http://basiclife.sourceforge.net/

Or you can review the copyright application at :
http://www.sci.ccny.cuny.edu/~mceruso/index.html

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Significant Figures
When using our calculators we must determine the correct answer; our calculators are mindless drones and don’t know the correct answer.
There are 2 different types of numbers
Exact
Measured
Exact numbers are infinitely important
Measured number = they are measured with a measuring device (name all 4) so these numbers have ERROR.
When you use your calculator your answer can only be as accurate as your worst measurement…Doohoo 
Chapter Two
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9
Exact Numbers
An exact number is obtained when you count objects or use a defined relationship.

Counting objects are always exact
2 soccer balls
4 pizzas
Exact relationships, predefined values, not measured
1 foot = 12 inches
1 meter = 100 cm
For instance is 1 foot = 12.000000000001 inches? No
1 ft is EXACTLY 12 inches.
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Learning Check
A. Exact numbers are obtained by
1. using a measuring tool
2. counting
3. definition

B. Measured numbers are obtained by
1. using a measuring tool
2. counting
3. definition
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Solution
A. Exact numbers are obtained by
2. counting
3. definition

B. Measured numbers are obtained by
1. using a measuring tool
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Learning Check
Classify each of the following as an exact or a
measured number.
1 yard = 3 feet
The diameter of a red blood cell is 6 x 10-4 cm.
There are 6 hats on the shelf.
Gold melts at 1064°C.
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Classify each of the following as an exact (1) or a
measured(2) number.
This is a defined relationship.
A measuring tool is used to determine length.
The number of hats is obtained by counting.
A measuring tool is required.

Solution
2.4 Measurement and Significant Figures
Every experimental measurement has a degree of uncertainty.
The volume, V, at right is certain in the 10’s place, 10mLThe 1’s digit is also certain, 17mLA best guess is needed for the tenths place.
Chapter Two
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What is the Length?
We can see the markings between 1.6-1.7cm
We can’t see the markings between the .6-.7
We must guess between .6 & .7
We record 1.67 cm as our measurement
The last digit an 7 was our guess...stop there
Learning Check
What is the length of the wooden stick?
1) 4.5 cm
2) 4.54 cm
3) 4.547 cm
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8.00 cm or 3 (2.2/8)
?
Measured Numbers
Do you see why Measured Numbers have error…you have to make that Guess!
All but one of the significant figures are known with certainty. The last significant figure is only the best possible estimate.
To indicate the precision of a measurement, the value recorded should use all the digits known with certainty.
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Chapter Two
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Below are two measurements of the mass of the same object. The same quantity is being described at two different levels of precision or certainty.
Note the 4 rules
When reading a measured value, all nonzero digits should be counted as significant. There is a set of rules for determining if a zero in a measurement is significant or not.
RULE 1. Zeros in the middle of a number are like any other digit; they are always significant. Thus, 94.072 g has five significant figures.
RULE 2. Zeros at the beginning of a number are not significant; they act only to locate the decimal point. Thus, 0.0834 cm has three significant figures, and 0.029 07 mL has four.

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Chapter Two
RULE 3. Zeros at the end of a number and after the decimal point are significant. It is assumed that these zeros would not be shown unless they were significant. 138.200 m has six significant figures. If the value were known to only four significant figures, we would write 138.2 m.
RULE 4. Zeros at the end of a number and before an implied decimal point may or may not be significant. We cannot tell whether they are part of the measurement or whether they act only to locate the unwritten but implied decimal point.
Chapter Two
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Practice Rule #1 Zeros
45.8736
.000239
.00023900
48000.
48000
3.982106
1.00040
6
3
5
5
2
4
6
All digits count
Leading 0’s don’t
Trailing 0’s do
0’s count in decimal form
0’s don’t count w/o decimal
All digits count
0’s between digits count as well as trailing in decimal form
2.5 Scientific Notation
Scientific notation is a convenient way to write a very small or a very large number.
Numbers are written as a product of a number between 1 and 10, times the number 10 raised to power.
215 is written in scientific notation as:
215 = 2.15 x 100 = 2.15 x (10 x 10) = 2.15 x 102
Chapter Two
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Chapter Two
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Two examples of converting standard notation to scientific notation are shown below.
Chapter Two
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Two examples of converting scientific notation back to standard notation are shown below.
Scientific notation is helpful for indicating how many significant figures are present in a number that has zeros at the end but to the left of a decimal point.
The distance from the Earth to the Sun is 150,000,000 km. Written in standard notation this number could have anywhere from 2 to 9 significant figures.
Scientific notation can indicate how many digits are significant. Writing 150,000,000 as 1.5 x 108 indicates 2 and writing it as 1.500 x 108 indicates 4.
Scientific notation can make doing arithmetic easier. Rules for doing arithmetic with numbers written in scientific notation are reviewed in Appendix A.
Chapter Two
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2.6 Rounding Off Numbers
Often when doing arithmetic on a pocket calculator, the answer is displayed with more significant figures than are really justified.
How do you decide how many digits to keep?
Simple rules exist to tell you how.
Chapter Two
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Once you decide how many digits to retain, the rules for rounding off numbers are straightforward:
RULE 1. If the first digit you remove is 4 or less, drop it and all following digits. 2.4271 becomes 2.4 when rounded off to two significant figures because the first dropped digit (a 2) is 4 or less.
RULE 2. If the first digit removed is 5 or greater, round up by adding 1 to the last digit kept. 4.5832 is 4.6 when rounded off to 2 significant figures since the first dropped digit (an 8) is 5 or greater.
If a calculation has several steps, it is best to round off at the end.
Chapter Two
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Practice Rule #2 Rounding
Make the following into a 3 Sig Fig number
1.5587
.0037421
1367
128,522
1.6683 106
1.56
.00374
1370
129,000
1.67 106
Your Final number must be of the same value as the number you started with,
129,000 and not 129
Examples of Rounding
For example you want a 4 Sig Fig number
4965.03
 
780,582
 
1999.5
0 is dropped, it is <5

8 is dropped, it is >5; Note you must include the 0’s
5 is dropped it is = 5; note you need a 4 Sig Fig
4965

780,600

2000.
RULE 1. In carrying out a multiplication or division, the answer cannot have more significant figures than either of the original numbers.
Chapter Two
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RULE 2. In carrying out an addition or subtraction, the answer cannot have more digits after the decimal point than either of the original numbers.
Chapter Two
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Multiplication and division
32.27  1.54 = 49.6958
3.68  .07925 = 46.4353312
1.750  .0342000 = 0.05985
3.2650106  4.858 = 1.586137  107
6.0221023  1.66110-24 = 1.000000
49.7
46.4
.05985
1.586 107
1.000
Addition/Subtraction
25.5 32.72 320
+34.270 ‑ 0.0049 + 12.5
59.770 32.7151 332.5
59.8 32.72 330
__ ___ __
Addition and Subtraction
.56 + .153 = .713
82000 + 5.32 = 82005.32
10.0 - 9.8742 = .12580
10 – 9.8742 = .12580
.71
82000
.1
0
Look for the last important digit
Mixed Order of Operation
8.52 + 4.1586  18.73 + 153.2 =




(8.52 + 4.1586)  (18.73 + 153.2) =
239.6



2180.
= 8.52 + 77.89 + 153.2 = 239.61 =
= 12.68  171.9 = 2179.692 =
Try
Find the standard deviation for the following numbers: 7.691 g, 7.23 g, 7.892 g
Try
7.691 g
7.23 g
7.892 g
22.813 g
22.81 g = 7.603 g
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7.691 g, 7.23 g, 7.892 g
Try
 = 7.603 g
7.691 g – 7.603 g = .088 g  .0077
7.23 g – 7.603 g = -.37 g  .14
7.892 g – 7.603 g = .289 g  .0835
.23
Try
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