SURVIVAL SKILLS
FOR
GENERAL CHEMISTRY
L.E. Chirlian
These units will help you teach yourself the skills you need to succeed in General
Chemistry. Some of them will be very easy, some you may never have seen before.
Every person's preparation for this course is different and these units are designed to bring
everyone to the same level. Each unit has a series of associated problems. Complete
these problems to test your mastery of the material.
TOPICS FOR STUDY
•
Algebra and arithmetic
•
Scientific notation
•
Logarithms
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UNIT I
ALGEBRA AND ARITHMETIC
Using positive and negative numbers
•
Adding a negative number = Subtracting a positive number
4 + (–3) = 4 – 3 = 1
•
Subtracting a negative number = Adding a positive number
4 – (–3) = 4 + 3 = 7
•
Multiplying a negative number and a positive number gives a negative result
4 x (–3) = –12
•
Multiplying a negative number and a negative number gives a positive result
(–4) x (–3) = 12
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PROBLEMS: Using positive and negative numbers
1) Without actually doing the calculation, circle the sign (+/–) of the answer.
=
+
–
b) (–81.7)(–17.3)(47.9) =
38.28166.189.5
=
20.2 11.8
+
–
d)
a) (15.7)(–38.92)(2.8)
c)
3.70 2.18.5
=
0.21.8
+
–
+
–
2. Perform the following calculations:
a) 33.82 + (–75.1) + 88.382 =
b) –3828.28 – (–4838.9) + (83.21) =
c) (–81.7)(–17.3)(47.9) =
d)
38.28166.189.5
=
20.2 11.8
e)
3.70 2.18.5
=
0.21.8
3) The temperature in Bryn Mawr reached 102oF in July. In January, the temperature fell to
–12oF. What was the change in temperature between July and January?
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4) An electron has a charge of –1. What is the total charge of 3 electrons?
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Manipulating Mathematical Expressions
RULE
•
Any operation may be performed on an equation,
provided the same operation is applied to both sides of
the equals sign.
Example:
Given the equation:
2x  4y  12
Through mathematical manipulations, this expression can be transformed into one which
defines x in terms of y.
•
Step 1: Start with the original formula:
2x  4y  12
•
Step 2: Examine the formula. What changes need to be made to obtain an expression
for x in terms of y? In this case,
a) Subtract 4y from both sides:
2x  4y  4y  12  4y
2x  12  4y
b) Divide both sides by 2:
2x 12  4y

2
2
•
Step 3: Determine the final formula:
x  6  2y
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If the value of y is known, the value of x may be determined using this formula. For
example if y is 2, the value of x is 2.
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PROBLEMS: Manipulating mathematical expressions
1) Solve each equation for x:
a) 3x  2  2x 1
b) 7x  2y  23
c) 8x 2  48
d)
4x  6y
 32
2
e)
144
9
4x 2
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Using formulas
Chemistry problems are often solved using formulas. Generally, the formula is
manipulated to give the appropriate expression for the problem in question and then the
given values are substituted into the formula. The area of a rectangle is found by
multiplying the lengths of the base and height together. The formula for the area of a
rectangle is:
A  bh
If the base of a rectangle is 10 cm and the height is 4 cm the area is given by:
A  10cm  4cm
A  40cm 2
(Note: the units multiply along with the numbers. cm  cm  cm 2 )
Manipulating formulas—I
If the area and the base of the rectangle are given, the height can be found by
manipulating the original formula.
Finding the height of a rectangle if the Area and base are known
•
Step 1: Start with the original formula:
A  bh
•
Step 2: Examine the formula. What changes need to be made to obtain an expression
for the length of the height (h) in terms of the other variables (Area (A) and length of
base (b))?
Divide both sides by the length of the base b:
A bh

b
b
•
Step 3: Determine the final formula:
A
h
b
If the area of a rectangle is 36 cm2 and the base is 6 cm, the height can be found using this
new formula:
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2
A 36cm
h 
 6cm
b
6cm
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Manipulating formulas—II
Some formulas require more manipulation, but the basic procedure is the same. Consider
the formula for the area of circle:
Area    (radius) 2
rad ius
Finding the radius of a circle if the area is known
•
Step 1: Start with the original formula:
A  r
•
2
Step 2: Examine the formula. What changes need to be made to obtain an expression
for the radius(r) in terms of the other variable (Area (A))? In this case, a constant  is
also involved.
a) Divide both sides by :
A

A


r2

 r2
b) Take the square root of both sides:
A

•
 r2
Step 3: Determine the final formula:
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r
A

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PROBLEMS: Manipulating formulas
1) Rearrange EACH equation to solve for the underlined variable.
a) y  mx  b
b) G  H  TS
c) PV  nRT
d)
1
nMv2  nRT (solve for v; not v2)
3
e) Q  mc T
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f) u  3RT Mm

n2 a 

g) P  2 V  nb  nRT

V 

n2 a 
h) P  2 V  nb   nRT

V 
i)
o
F
9o
C  32
5

1
1 
2
j) E  RH  2  2  (solve for n f not n f )
n f ni 
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UNIT II
SCIENTIFIC NOTATION
Very large/very small numbers
Very large numbers can be awkward to write. For example, the approximate
distance from the earth to the sun is ninety three million miles. This is commonly written
as the number "93" followed by six zeros signifying that the "93" is actually 93 million
miles and not 93 thousand miles or 93 miles.
}
placeholders
provide no additional information
93,000,000 miles
Scientific notation (also called exponential notation) provides a more compact
method for writing very large (or very small) numbers. In scientific notation, the distance
from the earth to the sun is 9.3 x 107miles.
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Very small numbers can be as awkward to write as large numbers. A paper clip
weighs a bit more than one thousandth of a pound (0.0011 LB). This would be expressed
in scientific notation as 1.1 x 10-3 lb. The negative sign indicates that the decimal point
is moved to the left.
Numbers are customarily written with one digit to the left of the decimal point. Numbers
may be correctly represented in other ways.
Representing a number using scientific notation
•
The number 2,398,730,000,000 can be written in scientific notation as:
2.39873 x 1012
23.9873 x 1011
239.873
x 1010
•
most common
The number 0.003,483 can be written in scientific notation as:
0.3483
3.483
34.83
348.3
3483.
x 10-2
x 10-3
x 10-4
x 10-5
x 10-6
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most common
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Scientific notation and multiplication/division
Multiplication and division of large or small numbers is simplified using scientific
notation. The decimal parts of the two numbers are multiplied or divided as appropriate
to give the decimal part of the answer. The exponents are added together (in the case of
multiplication) or subtracted (for division) and provide the exponent for the answer.
The answer is adjusted so that only one digit is to the left of the decimal point in the
decimal part.
Multiplying numbers written in scientific notation
•
To multiply 4 10 4 and 6 105 :
•
Step 1:
Multiply the decimal parts together.
4  6  24
•
Step 2: Add the two exponents.
4 5  9
•
Step 3
Examine the result.
9
24 10
•
Step 4:
Adjust the result so only one digit is to the left of the decimal point (if
necessary).
9
10
24 10  2.4 10
General Rule
(a 10 x )(b 10 y )  ab 10 xy
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Dividing numbers written in scientific notation
•
To divide 6 105 by 4 10 4 :
•
Step 1:
Perform the appropriate division of the decimal parts.
6
 1.5
4
•
Step 2: Subtract the exponents
5 4 1
•
Step 3
Examine the result.
1
1.5 10
•
Step 4: Adjust the result so only one digit is to the left of the decimal point
(if necessary).
1
1.5 10
(no adjustment necessary)
General Rule
a  10 x a
 10 x y
y 
b  10
b
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Scientific notation and addition/subtraction
Adding and subtracting numbers written in scientific notation is more complicated
than multiplication/division. Consider adding 0.0034 and 0.021:
0.0034
+ 0.021 in scientific notation ——>
0.0244
3.4 x 10-3
+ 2.1 x 10-2
2.44 x 10-2
Now, neither the decimal part or the exponential part combine together in any obvious
manner (as they did with multiplication and division). When adding or subtracting
numbers written in exponential notation, the numbers must first be rewritten so the
exponents are identical. Then, the numbers can be added or subtracted normally
Adding/subtracting numbers written in scientific notation
•
To add 3.4 10 3 to 2.1 102 :
•
Step 1:
Adjust one of the numbers so that its exponent is equivalent to the
other number. In this case change 2.1 102 into a number which
has 10-3 as its exponential part.
2
2.1 10
•
3
 21 10
Step 2: Add the decimal parts together.
21 + 3.4 = 24.4
• Step 3
parts
The exponential part of the result is the same as the exponential
of the two numbers, in this case, 10-3.
3
24.4 10
•
Step 4: Adjust the result so only one digit is to the left of the decimal point
(if necessary).
2
2.44 10
•
Subtraction follows the same procedure except one number is subtracted from the
other in Step 2.
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PROBLEMS: Scientific notation
1) Write the following numbers using scientific notation.
a) 382,000,000,000
b) 9,882
c) 76.124
d) 0.132
e) 0.000,000,000,009
2) Express the following as ordinary numbers.
a) 7.995 105
b) 4.21 104
c) 1.2  108
d) 8.02 102
e) 3.481763 10 6
3) Perform the following calculations. Express the answers scientific notation. Do NOT
use a calculator.
a) 400  2, 000 
b) 7,000, 000  0.003 
c)
25, 000

0.05
d)
0.000, 008

0.002
e) 4.3 10 3  5.8 10 4 
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f) 7.9 10 4  9.8 10 3 
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UNIT III
LOGARITHMS
Common Logarithms
The common logarithm (written: log x) of a number is the power to which 10
must be raised to equal that number. The logarithm of 1000 (written log 1000) is 3
because 10 cubed (103) is equal to 1000.
1000  10  10  10  103
log1000  3
Logarithms for numbers are generally found using a calculator since the
relationship between a number and its logarithm is not generally obvious. Logarithms of
numbers larger than one are positive while logarithms of numbers smaller than 1 are
negative. (Logarithms are not defined for numbers less than zero.) Writing numbers in
scientific notation clarifies this point as the logarithm of a number will be close in value
to the value of its exponent.
log 3021  3.480
3021  3.021 10
3
3
log 3.021  10  3.480
For numbers larger than one, the first digit of the logarithm is equal to the exponent
(when the number is written with one digit to the left of the decimal point). For numbers
smaller than one, the first digit may be one less than the value of the exponent.
log 0.0000378  4.132
5
0.0000378  7.38  10
log 7.38  10 5  4.132
Antilogarithms
If the logarithm is known, the corresponding number is found by raising 10 to the
logarithm (anti–logarithm). Suppose the logarithm of an unknown number is 3.63. The
unknown number is found as follows:
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log x  3.63
10 log x  103.63
x  4.27  10 3
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Natural Logarithms
Natural logarithms (written: ln x) are identical to common logarithms, except the
natural log of a number is the power to which e (2.718...) must be raised to get that
number. While this number appears arbitrary it appears in many mathematical formulas.
Natural logarithms and common logarithms are related as follows:
ln x  ln10  log x  2.303log x
The natural antilogarithm of x is found by raising e to the power of x. Raising e to a
power is called taking the exponential.
Logarithmic and Exponential Relationships
•
Some useful properties of logarithms:
log ab  log a  log b
•
ln ab  ln a  ln b
log
a
 log a  log b
b
ln
a
 ln a  ln b
b
log
1
  log x
x
ln
1
  ln x
x
log y x  x log y
ln y x  x ln y
log10 x  x
ln e x  x
10 logx  x
e logx  x
Some useful properties of exponents:
a
b
a b
a
10  10  10
b
e e  e
a
a
10
ab
b  10
10
e
a b
b e
e
25
a b
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\PROBLEMS: Logarithms
1) Without using a calculator, find the common logarithms of:
a) 10,000
b) 0.000,0001
c) 1.0 107
d) 1.0 1012
2) Find the common logarithms of: (calculators may be used)
a) 28,383,233
b) 0.766
c) 0.000,000,822
d) 9,595 105
3) Solve for y:
a) 0.852  log y
b) 2.72  log y
c) 3.21  log y
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d) 5.91  ln y
e) 6.01  ln y
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Manipulating Formulas Containing Logarithms
The general rules for manipulating formulas apply to formulas containing
logarithms. At times, these formulas may seem more confusing. The appropriate
application of logarithms and antilogarithms are used to isolate the variable of interest.
Example:
Rearrange the following equation to find E when k, A, T and R are known.
ln k  ln A 
•
Step 1: Start with the original formula:
E
RT
E
RT
Step 2: Examine the formula. What changes need to be made to obtain an expression
for E in terms of the other variables?
ln k  ln A 
•
Subtract ln A from both sides:
ln k  ln A  ln A 
ln k  ln A  
E
 ln A
RT
E
RT
Multiply both sides by –1
 E 
1 ln k  ln A  1  
RT 
E
ln A  ln k 
RT
NOTE: the left–hand side of the equation can be rewritten using one of the logarithmic
a
relationships given above: ln a  ln b  ln .
b
A
E
ln 
k RT
Multiply both sides by RT .
RT ln
•
A
 E
 RT  
k
RT 
Step 3: Determine the final formula
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General Chemistry Survival Skills
E  RT ln
A
k
Example:
Rearrange the following equation to find A when k, E, T and R are known.
ln k  ln A 
•
Step 1: Start with the original formula:
ln k  ln A 
•
E
RT
E
RT
Step 2: Examine the formula. What changes need to be made to obtain an expression
for A in terms of the other variables?
Add
E
to both sides:
RT
E
E
E
ln k 
 ln A 

RT
RT RT
E
ln k 
 ln A
RT
Now the expression is for ln A. To remove the natural log take the
exponential of each side:
e
•
ln k 
E
RT
e
ln A
Step 3: Determine the final formula
ln k
Ae
E
RT
The formula may be further simplified using the properties of exponents
ab
 e aeb and e ln x  x .
given above: e
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ln k 
Ae
E
RT
A
E
ln k RT
e e
A
ke RT
E
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PROBLEMS: Manipulating formulas containing logarithms
1) Rearrange EACH equation to solve for the underlined variable.
 
a) pH   log H
A 
b) ln  0  kt
 A 
c) E  E o 
RT
ln Q
nF
k'
E 1 1 
d) ln  a   ' 
k
R T T
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o
e) G  RT ln K
33