Download Scientific Notation

Scientific Notation
Scientific Notation is a method of writing numbers so that only significant digits are shown.
e.g. the distance from the earth to the moon is reported as:
in common notation: 382 000 km
in scientific notation: 3.82 105 km
In scientific notation a number has one nonzero digit to the left of the decimal point and is
multiplied by the appropriate power of 10 (either positive or negative).
Scientific notation is especially useful in concisely reporting very large or very small numbers.
e.g. the number of electrons in a coulomb of charge is:
in common notation: 6 242 000 000 000 000 000
in scientific notation: 6.242  1018
Web Users please note:
the mass of a proton is :
in common notation: 0.000 000 000 000 000 000 000 000 001 672 kg
in scientific notation: 1.672  10-27 kg
the number of particles in a mole of substance is:
in common notation: 602 200 000 000 000 000 000 000
in scientific notation: 6.022  1023
Answers to these questions are on page 23 & 24, i.e., at the end of this unit.
Practice:
1. Express the following in scientific notation.
a)
b)
c)
68090
0.000 023
5400
________________________
________________________
________________________
d)
0.80
________________________
e)
0.050 40
________________________
2. Express the following in common notation.
a)
7  101
_______________________
b)
4.098  10-3
_______________________
c)
8.3  107
_______________________
d)
10.1  10-5
____________________________
Units & Errors
1
SIGNIFICANT FIGURES (Sig Figs) = all digits known for sure plus 1 more digit
1.
During the manufacture of steel from iron ore, the ore is fed into a large kiln whose
temperature increases from less than 400 C at the top to about 1600 C at the bottom. The
thermometer shown below indicates various kiln temperatures A through E. For each of these,
record the temperature in common notation, the number of significant digits, and the
temperature in scientific notation.
1000
C
800
1200
B
D
600
1400
E
A
400
1600
ºC
KILN
TEMPERATURE
common
notation
number of
sig figs
scientific
notation
A
B
C
D
E
Using a thermometer which is more finely calibrated gives readings with more sig figs.
Repeat the previous exercise for temperatures A and C using the thermometer shown below.
1000
C
800
1200
B
D
600
1400
E
A
400
1600
ºC
KILN
TEMPERATURE
common
notation
number of
sig figs
scientific
notation
A
C
Units & Errors
2
SIG FIGS (continued)
2 a) An automobile tachometer measures engine speed in revolutions per minute (rpm). In
the tachometer shown, engine speeds to the nearest 100 rpm are displayed, i.e., two tailing
zeros are present in all readings- they never change. In case A, below, an automobile is idling
at rest and, in case B, the automobile is moving at a steady speed of 90 km/h. For both cases,
report the rpm reading in common notation, the number of significant digits, and the rpm
reading in scientific notation.
Digital instruments do not allow the observer to estimate a final digit.
Assume that the first digit to the left of the zeros is a number which is estimated by the
instrument.
700
rpm
2300
rpm
A
B
common
notation
number of
sig figs
scientific
notation
A
B
b) Repeat the previous exercise assuming that the tachometer reads to the nearest 10 rpm's.
common
notation
number of
sig figs
scientific
notation
A
B
c) Repeat the previous exercise assuming that the tachometer reads to the nearest 1 rpm.
common
notation
number of
sig figs
scientific
notation
A
B
Units & Errors
3
Sig Figs (continued)
3.
The length of a desk top is measured with 3 different meter sticks (A, B, and C) as
shown below. For each meter stick, record the length of the desk top in common notation, the
number of sig figs, and the length in scientific notation.
0
A
1
0
.1
.2
.3
.4
.5
.6
.7
.8
.9
1.0
.1
.2
0
.1
.2
.3
.4
.5
.6
.7
.8
.9
1.0
.1
.2
common
notation
number of
sig figs
B
C
scientific
notation
A
B
C
How does the sensitivity of a measuring instrument affect the number of significant figures
obtainable in a measurement?
Note that by using ruler A we are able to exactly measure the object's length to the
nearest meter and estimate the length to the nearest 0.1 meters (2 sig figs).
Note that by using ruler B we are able to exactly measure the object's length in meters
and in 0.1 meters and estimate the length to the nearest 0.01 meters (3 sig figs).
Note that by using ruler C we are able to exactly measure the object's length in meters,
0.1 meters, and 0.01 meters and estimate the length to the nearest 0.001 meters (4 sig figs).
In general, the smallest unit we can exactly measure is the smallest division of our measuring
instrument. We may then estimate 1 digit more. All of these (and only these) digits are
significant digits (sig figs).
Units & Errors
4
Sig Figs (continued)
4.
For each of the automobile gages shown below, record the gage reading in common
notation, the number of sig figs, and the length in scientific notation.
180
170
190
200
160
ENGINE
TEMPERATURE
150
210
ºF
70
60
50
80
90
40
100
30
110
20
10
120
SPEED
km / h
common
notation
130
number of
sig figs
scientific
notation
engine temp. (ºF)
speed (km/h)
Units & Errors
5
Rules for Significant Digits in Numbers
1. Nonzero digits: 1, 2, 3, 4, 5, 6, 7, 8, and 9 are always significant.
6.2
16.2
16.25
two significant digits
three significant digits
four significant digits
2. Leading zeros: zeros that appear at the start of a number, are never significant because
they act only to fix the position of the decimal point in a number less than 1.
`
0.564
0.0564
three significant digits
three significant digits
3. Confined zeros: that appear between nonzero numbers are always significant.
104
1004
three significant digits
four significant digits
4. Trailing zeros: zeros at the end of a number are significant only if:
a) the number contains a decimal point or
b) the number contains an overbar
15400
three significant digits
1540.0
five significant digits
15.4000
six significant digits
5,600
four significant digits or 5.600  103
5,600
three significant digits or 5.60  103
Practice: Determine the number of significant digits in the following numbers:
a)
345
_______
i)
b)
32000
c)
0.0078
_______
d)
9.0068
_______
e)
5.0200
_______
g)
h)
10700
122.0
10.04
Units & Errors
_______
_______
j)
f)
20
0.604
_______
_______
_______
_______
6
Uncertainty in Measurement
In scientific work, we recognize two kinds of numbers:
exact numbers
- values known exactly
- numbers which are defined or counted
- defined: 12 donuts in a dozen
1000 grams in a kilogram
-counted: 25 students in a classroom
inexact numbers
- values are uncertain
- numbers which are measured
eg. - speed of an automobile
- temperature of a cup of soup
- volume of water in a beaker
Exact numbers are assigned an infinite
number of significant figures for
calculations
Inexact numbers have a number of
significant figures equal to the number of
digits known for certain plus one more.
Significant Figures in Calculations
One additional digit (one more than the number of sig figs) is usually carried on all values
throughout a calculation and the answer is rounded to the correct number of sig figs.
In calculations involving multiplication, division, roots, and powers: round off the
answer so that it has only as many significant figures as the value in the calculation
with the fewest significant digits.
e.g.
35.63  0.5481  0.0530
 100%  88.5470578%  88.5%
1.1689
The answer only contains 3 significant figures since this is the least number of sig figs
amoung the values in the calculation. 100 % is an exact number having an infinite
number of significant digits.
In addition and subtraction: the answer is only as precise as the least precise
measurement. The answer has as many decimal places as the value with the least.
e.g. Suppose we have 3 measurements of length to be added, i.e.,
6.6 m
18.74 m
0.766 m
26.106 m  26.1 m
Note that the least precise measurement, 6.6 m, was only reported to the nearest 0.1m,
therefore the answer cannot be reported with greater precision.
Note that the correct answer has more significant digits than one of the measurements.
Remember that the answer of additions or subtractions cannot have more precision
than the least precise measurement. For example, if approx. 1 L of Pepsi is divided
equally between 3 people, each person will not get 0.333 L, but rather 0.3L.
Units & Errors
7
Accuracy and Precision
Accuracy refers to how closely a measured value agrees with the correct value.
Precision refers to how closely individual measurements agree with each other.
A group of measurements may be very precise (agree closely) but still be inaccurate. Ideally,
all measurements should be both precise and accurate.
precise but not accurate
accurate and precise
not accurate or precise
Rounding off Numbers
If the digit following the last significant digit is greater than 5, the number is rounded up to
the next higher digit:
9.47 to 2 significant digits = 9.5
If the digit following the last significant digit is less than 5, the number is rounded to the
present value of the last significant digit:
9.43 to 2 significant digits = 9.4
If the last digit is exactly 5, the number is rounded to the nearest even digit:
8.65 to 2 significant digits = 8.6
8.75 to 2 significant digits = 8.8
8.55 to 2 significant digits = 8.6
Practice: Round the following numbers to 2 sig figs.
a)
2249
_____
d)
b)
2250
_____
e)
0.07750
____
c)
2251
_____
f)
0.07749
____
Units & Errors
2250.001
____
8
SI Base Units
Quantity
Name of unit
Symbol for unit
length
metre
m
mass
kilogram
kg
time
second
s
amount of substance
mole
mol
temperature
kelvin
K
electric current
ampere
A
luminous intensity
candela
cd
Conversion Factors
Unit
Equivalent
1 inch (in)
2.54 cm
1 foot (ft)
12 in
1 yard (yd)
0.9 144 m
1 metre (m)
39.37 in = 3.28084 ft
1 mile
1 760 yd = 1.609 km
1 pound (lb)
453.6 g
1 kilogram (kg)
2.205 lb
1 grain (gr)
64.8 mg
1 ton (ton) - short
2 000 lb = 907 kg
1 ton (ton) - long
2 240 lb
1 tonne (t) - metric
2 205 lb = 1000 kg
1 imperial gallon (gal)
4.546 L
1 U.S. gallon (gal)
3.785 L
1 imperial fluid ounze (fl oz)
28.41 mL or cc
1 U.S. fluid ounze (fl oz)
29.57 mL or cc
1 imperial quart (qt)
1.136 L
1 U.S. quart (qt)
0.946 L
Memorize:
Units & Errors
1mL = 1cm3 = 1cc
1 L = 1000 mL = 1000 cm3 = 1000 cc = 1 dm3
1000 L = 1m3
9
Decimal Multipliers
Prefix
Symbol
Mult. Factor
e.g. meters
e.g. meters
exa
E
1018
1 Em = 1018 m
10-18 Em = 1 m
peta
P
1015
1 Pm =1015 m
10-15 Pm = 1 m
tera
T
1012
1 Tm = 1012 m
10-12 Tm = 1 m
giga
G
109
1 Gm =109 m
10-9 Gm = 1 m
mega
M
106
1 Mm = 106 m
10-6 Mm = 1 m
kilo
k
103
1 km = 103 m
10-3 km = 1 m
hecto
h
102
1 hm = 102 m
10-2 hm = 1 m
deca
da
10
1 dam = 10 m
10-1 dam = 1 m
1
1m=1m
1m=1m
base unit
deci
d
10-1
1 dm = 10-1 m
10 dm = 1 m
centi
c
10-2
1 cm = 10-2 m
102 cm =1 m
milli
m
10-3
1 mm = 10-3 m
103 mm = 1 m
micro

10-6
1 m = 10-6 m
106 m = 1 m
nano

10-9
1 m = 10-9 m
109 m = 1 m
pico

10-12
1 m = 10-12 m
1012 m = 1 m
femto
f
10-15
1 fm = 10-15 m
1015 fm = 1 m
atto
a
10-18
1 am = 10-18 m
1018 am = 1 m
Units & Errors
10
DECIMAL MULTIPLIERS APPLIED TO UNITS
1 Ms =
s
1s=
Ms
1 kg =
g
1g=
kg
1 dL =
L
1L=
dL
1 cm =
m
1m=
cm
1 mol =
mol
1 mol.=
1 m =
m
1m=
m
1 mg =
g
1s=
Gs
1 g =
g
1m=
mm
Units & Errors
mol.
11
Converting Units Using the ‘Unit Factor’ Method (‘Dimensional Analysis’)
Multiply the given unit by conversion factors. We will start with a simple example and then
progress to more difficult examples. Learn the procedure by working through the examples.
Example 1:
Convert 3 minutes to seconds.
CONVERSION FACTORS
Step 1: Write down the problem as shown below and decide what
conversion factors you will need.
? s = 3 min
60 s = 1 min
Step 2: Get the conversion factors. Include units.
Step 3: Set up the calculation using the units. Then cancel the units (as if
they were numbers) to ensure that the conversion factors are set up
correctly.

? s  3 min  

s 

min 
s
If the factor is upside down, the units of the answer make no
sense. Check this. Do not leave out the units as the units
will help you avoid errors. For example:
This unit makes no sense so
 min 
? s  3 min  

s 

min 2
.
s
the conversion factor must be
upside-down.
Step 4: Put the numbers in the conversion factors and calculate the
correct answer.
 60 s 
? s  3 min  

 1 min 
180 s
Steps 3 and 4 can be done simultaneously but do not leave
out the units. They are the key to success.
Plan your conversions
using arrows. There will
be one conversion factor
for each arrow.
Example 2:
Convert 4 days to seconds.
3
Step1:
?s
2
min
1
h
=
2
1
Step 4:
2
1 h = 60 min
3
 h   min   s 
? s  4 da  



 da   h   min 
Step 3:
Step 2:
1 min = 60 s
4 da
1
1 da = 24 h
s
3
 24 h   60 min   60 s 
? s  4da  


  345,600 s
 1 da   1 h   1 min 
Units & Errors
12
Example 3:
Convert 37 mi/h to m/s
2
Step 1:
km
? m/s
=
1
mi/h
Step 2:
3
Step 3 and 4:
2
1
1 mi = 1.609 km
1 km = 103 m
1 h = 3600 s
3
m
mi  1.609 km   10 m   1 h 
m

?  37  
  
  16.5

s
h  1 mi   1 km   3600 s 
s
3
Example 4:
Gold has a mass density of 19.30 g/cm3.
Convert this to units of kg/m3.
1
? kg/m3
Step 1:
Step 2:
19.30 g/cm3
=
1 kg = 103 g
(1 m = 102 cm)3
1 m3 = 106 cm3
2
Step 3 and 4:
1
2
kg
g  1 kg   10 6 cm3 
kg
  19300 3

? 3  19.30

m
cm3  103 g   1 m3 
m
The Metric Multipliers
Technology students are expected to memorize the metric multipliers shown below. Use them to solve the conversion
problems on the following page.
giga
mega
kilo
G
M
k
109
106
103
Units & Errors
base
unit deci centi milli
d
1
c
m
10-1 10-2 10-3
micro
nano
pico


p
10-6
10-9
10-12
13
Convert and express in correct scientific notation.
_______
1.
2.0  1010 L  dL
_______
2.
5.08  106 in  km
_______
3.
5.91  10-3 ks  cs
_______
4.
3.0  10-2 m3  mL
_______
5.
9.00  102 g  lb
_______
6.
4.56  10-4 Ms  ms
_______
7.
0.000 002 km  fm
_______
8.
101 g  g
_______
9.
25 dm  mm
_______
10.
8.42 mmol  kmol
_______
11.
5.0  105 mol  mol
_______
12.
7.10  103 kgm-3  g / L
Units & Errors
14
Unit Conversion Problems
1. Convert 142 Ms to leap years. ( 366 da = 1 lp. yr )
2. A window sized air conditioner rated at 8000 B.T.U. will remove
8000 B.T.U. of heat in one hour. How many calories of heat is this?
1 B.T.U. = 1.0541010 ergs
1 erg = 10-7 joules
4.184 joules = 1 cal.
3. The driver of a Honda Civic claims he gets 50 miles per gallon
(mpg). Convert this to litres per 100 km.
1 mile = 1.609 km.
4.546 L. = 1 Imp. gallon
4. The density of pure gold is 19.32 g / mL. Calculate its density in units of
kg / m3.
Units & Errors
15
Problem Solving By Logical Statements:
Relying on the method of Ratio and Proportion to solve technology problems will often prove uncessful.
No chemistry text book written within the last 20 years has taught ratio and proportion as a method for
converting units. With minimal investment of time (practice), you will be convinced that the unit factor
method (just presented) is much quicker and more effective than ratio and proportion for converting
units.
Also, for general problem solving in technology, ratio and proportion is inadequate and will often lead to
incorrect answers. An adaptation of ratio and proportion, called ‘solving by logical statements’ is easy to
use and will improve your exam scores because you will correctly solve more problems using it. Logical
statements are especially well suited for dilution and concentration calculations in chemistry.
We will apply the method of logical statements, first, to a simple problem. The same pattern is then
applied to all other problems.
The Method of Logical Statements:
Problem: 12 eggs cost 60¢. How much do 6 eggs cost?
Solution:

You will intuitively know that 6 eggs cost ½ as much as 12 eggs so you will divided 60¢ by 2 and get
30¢.

This is, of course, correct, but why did you divide by 2? Because, in your head you calculated:
6 eggs
 1/2
12 eggs
or
12 eggs
2
6 eggs
You determined a ratio or ‘factor’ of the two known values, then multiplied or divided by that factor
60¢  ½
i.e.,

or
60¢  2
Don’t you wish all chemistry problems were that simple? The truth is that most are if use this same
simple, logical method to solve your problems. Let’s write out this method in a standard format.
12 eggs cost 60¢
 6 eggs 
1
 = 60¢    = 30¢.
6 eggs cost x¢ = 60¢  
 2
 12 eggs 
1. Write 2 statement lines, one immediately below the other, inserting ‘x’ for the value to be
solved for. At this point it looks like the set up for ratio and proportion, but do not cross
multiply.
2. Bring down the known value (60¢) with its unit. See the arrow. Do not leave out the unit.
The known value that has the same unit as the answer will have.
3. Multiply the known value (60¢) by a ratio (or factor) of the two egg values. You only have
 6 eggs 
 12 eggs 
 or 
 . Choose the one that makes sense, i.e.,
two choices, either 
 12 eggs 
 6 eggs 
fewer eggs should cost less. Include the units but note that the units of eggs will cancel
leaving the correct unit (¢).
Units & Errors
16
There are two reasons why you will successfully solve more problems by this method than by ratio
and proportion:
a) The units are carried. Some will cancel leaving the correct unit for the answer. In ratio and
proportion, units are never carried because when they are cross multiplied they make no
sense. In the example above, the unit obtained by cross multiplying is ‘eggs¢’, which has
no meaning.
b) You make a logical decision in setting up the ratio, i.e., you multiply by ½ not by 2, because
fewer eggs must cost less money. Ratio and proportion is a mechanical, mindless operation,
i.e., simply cross multiplying. In many technology problems this will be incorrect as we will
illustrate in the following examples.
Example 2:
A Honda Civic with a 1.3 L engine gets 50 miles per gallon (mpg). Calculate the gas mileage (mpg) of a
5.0 L Ford Mustang assuming gas mileage depends only upon engine size.
Solution:
Honda:
1.3 L engine gets 50 mpg
 1.3 L 
Mustang:
5.0 L engine gets x mpg = 50 mpg  
 = 13 mpg
 5.0 L 
The ratio (1.3/5.0) is chosen because it makes sense that a larger engine gets lower gas mileage.
Crossmultiplying, as one might do using ratio and proportion, yields a ridiculous answer (192 mpg)!

This problem is an example of an inverse relationship. Inverse relationships are common in
technology, e.g., pressure-volume, mass-velocity, volume-concentration, age-beauty. For such
problems one parameter is increasing while the other is decreasing.

The egg problem was an example of a direct relationship. Direct relationships are those in which
both parameters increase or decrease simultaneously. Cross multiplying and the unit factor method
work for direct relationships but not for indirect relationships. When converting a distance of 1 km
to meters, the relationship can be thought of as direct even though the distance is not changing. But
note that the greater the value in km, the greater will be the value in meters.
Direct Relationship
eggs , cost 
M i l e a g e (mpg)
Cost
# eggs
Units & Errors
Inverse Relationship
engine size , mpg 
engine size (L)
17
Practice Problems:
1. In 10 s a well trained sprinter weighing 100 lbs. runs 100 m. After a year of overeating the
sprinter now weighs 250 lbs. Assuming that his weight is the only factor affecting his
performance, calculate his new distance for a 10 s run. Show your steps in your calculations.
2. Using a regulation hardball weighing 150 grams, American league pitcher, Nolan Ryan, has
delivered pitches clocked up to 180 km/h. Assuming ball weight to be the only factor affecting
the speed of his pitch, calculate the top speed which this pitcher could throw a 4.5 kg bowling
ball.
3. A liquid chemical storage tank can be filled in 15 minutes using a pump which transfers liquid
at a rate of 120 L/min. Calculate a) the volume of the storage tank in litres and b) the number of
minutes it would take to fill the tank using a pump which transfers 180 L/min.
4. Average adults have a lung capacity of 10 L at sea level (1.0 atmospheres pressure). Calculate
the volume of the lungs when a skin diver dives to a depth at which the pressure increases to 2.5
atmospheres.
Units & Errors
18
Solving Concentration Problems by Logical Statements:
Problem:
Calculate the grams of solute in 250 mL of 10%w/v KI soln.
Solution:
Start by writing out ‘what the given concentration unit means’. This is the first of
the two logical statements:
10% w/v means:
100 mL soln. contains 10 g solute
therefore:
 250 mL 
250 mL soln. contains x g solute = 10 g  
 = 25 g KI
 100 mL 
Problem:
Calculate the concentration in g/kg for a soln. of 10 g HCl + 240 g H2O.
Solution:
10 g HCl + 240 g H2O = 250 g soln.
st
1 statement:
250 g soln. contains 10 g HCl
2nd statement:
 1000 mL 
1000 g soln. (1 kg) contains x g HCl = 10 g  
 = 40 g/kg
 250 mL 
Problem:
Calculate the mass of KOH (MW = 56.1 g/ mol) to prepare 600 mL 0.45 M KOH
Solution:
 56.1 g 
0.45 M KOH means: 1000 mL (1 L) contains 0.45 moles KOH  
 = 25.2 g KOH
 1 mol 
(Note this first calculation uses the unit factor method to convert moles to grams. This is valid
because the quantity is unchanged; we are simply converting its units)
1st Statement:
1000 mL soln. contains 25.2 g KOH
2nd Statement:
 600 mL 
600 mL soln. contains x g KOH = 25.2 g  
 = 15.1 g KOH
 1000 mL 
Try these:
1. Calculate the %w/w of a soln. made with 20 g NaCl in 100 g H2O
2. Convert 50 g NaCl in 200 mL soln ( = 1.12 g/mL) to g/kg.
3. Calculate the mass of Na2CO3 (MW = 106 g/mol) in 150 mL of 2.0 M Na2CO3 soln.
4. Calculate the mass and vol. of 30.0% H2SO4 (=1.22 g/mL) that contains 50.0 g H2SO4
Answers:
1. 16.7 %w/w NaCl
2. 220 g/kg
3. 32 g Na2CO3
4. 167 g soln. and 137 mL soln.
Units & Errors
19
Solving Dilution Problems by Logical Statements:
Although dilution problems do not necessarily need two logical statements to solve, the
calculation using ‘dilution factors’ is similar and uses the same reasoning skills. Some examples
follow:
Problem:
A 5.0 mL aliquot from a 50 ppm Pb stock soln. is pipetted into a 25 mL
volumetric flask and the flask is made up to the mark with water and mixed.
Calculate the concentration of the final soln.
Solution:
The final solution is less concentrated, so we simply multiply the original
concentration by a ‘dilution factor’, i.e., a ratio of the two volumes.
 5 mL 
 25 mL 
Only two ratios are possible, either 
 or 
.
 25 mL 
 5 mL 
Of course we choose the first ratio (1/5) since makes the final concentration value lower
than the original concentration. We know the final concentration is lower because a
dilution was performed.
dilution factor
 5 mL 
50 ppm  
 = 10 ppm
 25 mL 
Problem:
A 20 g/mL Ni soln. was prepared by diluting 15 mL of a stock soln. to 50 mL.
Calculate the concentration of the stock soln.
Solution:
 50 mL 
20 g/mL Ni  
 = 67 g/mL Ni
 15 mL 
factor
The ratio is set up so that the stock concentration will be greater than the diluted soln.
concentration.
Problem:
50 mg. Cl- is contained in 250 mL solution. How many mg Cl- are contained in a 10 mL
aliquot of this solution.
Solution:
 10 mL 
50 mg Cl-  
 = 2 mg Cl250
mL


or
250 mL solution contain 50 mg Cl-
 10 mL 
 = 2 mg Cl250
mL


10 mL solution contain x mg Cl- = 50 mg Cl-  
Try these:
1. 4 mL of a stock soln. was diluted to 100 mL. 5 mL of the diluted soln. contains 0.002 g
F- Calculate the %w/v of F- in the stock soln. (answer = 1 % w/v)
2. 25 mL of a brine solution was diluted to 500 mL. From this 1st dilution, a 25 mL aliquot
was diluted to 500 mL. A 20 mL aliquot of the 2nd dilution was titrated with an AgNO3
solution. The mass of Cl- as NaCl was determined to be 0.0350 g in the 20 mL aliquot that
was titrated. Calculate the conc. (g/L) NaCl in the brine soln. (answer = 7.00  102 g/L)
Units & Errors
20
Rearranging Equations
1.
2.
3.
EH = m  c  T,
solve for T
T = T2 - T1, solve for T1
v2 = v1 + a  t,
solve for a
Often we must rearrange an equation before we can solve for a term. We rearrange an
equation so as to isolate the term to be solved for. By isolating a term we mean the term is alone,
positive (+), and in the numerator on either side of the equation. Move an unwanted term to the
opposite side of the equation by applying its reverse operation, then cancel to simplify.
Multiplication is the reverse operation to division.
Addition is the reverse operation to subtraction.
1.
EH = m  c  T
3.
v2 = v1 + a t
(v2) - v1 = (v1 + a t) - v1
2.
E H m  c  T

mc
mc
EH
 T
mc
v2 -v1 = a t

T = T2 - T1
(T) + T1 = (T2 -T1) + T1
T + T1 = T2
(T + T1) - T = (T2) -T
T1 = T2 - T
v2  v1 a  t

t
t

v2  v1
a
t
Note: In an equation, the whole left side = the whole right side.
Any operation can be carried out on an equation without changing its equality so long as the
same operation is applied to all terms on both sides.
In the following equations, isolate and solve for the unknown where:
L = loonie (100¢), Q = quarter (25 ¢), D = dime (10 ¢), and N = nickel ( 5 ¢).
Wrong way:
L
 10 = 60
2
L + 10 = 60  2
L = 120 - 10
L  110
Units & Errors
Right Way:
L
 10  60
2
( 2L  10)  2 = (60)  2
L + 20 = 120
L = 120 - 20 = 100
21
Practice:
Solve the following equations as indicated by rearranging the equations and isolating the
desired term. Check your answers by substituting the coin values in for each letter.
1. Solve for N:
3.
5.
2 L  10 N
Q
6
Solve for D: L = 3Q + 2D + N
Solve for Q:
Units & Errors
2 L 6Q

5
D
D
2. Solve for L:
L  5N
Q
3
4.
Solve for L: 1 
6.
Solve for Q:
5 N 3Q

L
L
L
2
D
8
 

QN N QN Q
22
Scientific Notation - Answers
Answers to Units Problems
Page 1
6.8091  104
1a)
2a)
70
b)
2.3  10-5
b)
0.004098
c)
5.4  103
c)
83,000,000
d)
8.0  10-1
d)
0.000101
e)
5.04  10-2
Page 2
A
B
C
D
E
common notation
440
710
1070
1105
1590
number of sig figs
2
2
3
4
3
scientific notation
4.4  102
7.1  102
1.07  103
1.105  103
1.59  103
A
C
common notation
442
1059
number of sig figs
3
4
scientific notation
4.42  102
1.059  103
Page 3
A
B
common notation
700
2300
number of sig figs
1
2
scientific notation
7  102
2.3  103
common notation
1.1
1.08
1.089
number of sig figs
2
3
4
scientific notation
1.1  100
1.08  100
1.089  100
Page 4
A
B
C
Page 5
engine temp. (ºF)
speed (km/h)
common notation
183.7
97
number of sig figs
4
2
scientific notation
1.837  102
9.7  101
Page 6
a)
3
e)
5
i)
1
b) 2
f)
3
j)
3
c)
2
g) 4
d) 5
h) 4
Units & Errors
23
Scientific Notation - Answers
Page 8
a)
2200
c)
b) 2200
2300
e)
0.078
d) 2300
f)
0.077
2.5  103 mm
Page 14
1.
2.0  105 dL
5.
1.98 lb
9.
2.
1.29  102 km
6.
4.56  105 ms
10. 8.42  10-6 kmol
3.
5.91  102 cs
7.
2  102 fm
11. 5.0  10-1 mol
4.
3.0  104 mL
8.
1.01  1011 g
12. 7.10  103 g/L
Page 15
1.
4.49 leap years
3.
5.7 L/100 km
2.
2.0  10 cal.
4.
1.932  104 kg/m3
6
Page 18
1.
40 m
3.
a) 1800 L, b) 10 min.
2.
6 km/h
4.
4L
Units & Errors
24