Used for numbers that are really big or really small
A number in exponential form consists of a coefficient
multiplied by a power of 10
1 x 104
10,000
0.00000029
2.9 x 10-7
1,000,000
546,000
1 x 106
5.46 x
105
0.00001
1 X 10-5
0.00751
7.51 x 10-3
12,450
1.245 x 104
15,230,000
1.523 x 107
0.0884
8.84 x 10-2
If the coefficient does not fall between 1 and 10 ,
it must be re-written correctly
If you move the decimal to the right, subtract
from the exponent
If you move the decimal to the left, add to the
exponent
100 x 103
1 X 105
0.090 x 10-5 9 X 10 -7
0.0001 x 1012
1 X 108
4,490 x 10-7
4.49 x 10-4
0.0065 x 103 6.5 x 100 = 6.5
0.0112 x 108 1.12 x 106
150 X 1012
1.5 x 1014
200 x 10-2
2 X 100
calculations involving
scientific notation
Enter the coefficient, then
EE
or
Do not enter “x 10”
(2.4 x 106) (3.1 x 103 ) =
(9.5 x 10-7) (5 x 10-4 ) =
7.44 x 109
4.75 x 10-10
4.8 x 109 = 2 X 107
2.4 x 102
7.5 x 10-5
+ 4.2 x 10-6 = 7.92 x 10-5
Exp
Two types of data
Qualitative descriptive Ex: the burner flame is hot
Quantitative
numerical
Ex: the flame is 1000 ° C
Measurements should be both accurate and precise
how close the experimental value is
to the accepted or true value
calculating percent error
% E = |accepted –experimental| X 100
accepted
how close the measurements are to
each other when the experiment is
repeated
Ex: a student does an experiment to
determine the density of lead; she repeats
the experiment two more times and gets
these results:
10.1 g/cm3
9.4 g/cm3
and 8.5 g/cm3
comment on her precision poor
If the actual density of lead is 11.4 g/cm3, calculate
her percent error using her average density as
the experimental value
|11.4 – 9.3|
100 = 18.4 %
Avg.= 9.3
%E=
11.4
Quantities and their units
Example units
Mass
(weight)
gram (g)
kilogram (kg)
milligram (mg)
Length
(distance)
meter (m)
kilometer (km)
millimeter (mm)
Volume
liter (L)
milliliter (ml)
any unit of length
that is cubed
1m
Vol. = L x W x H
1m
1m
Temperature
Heat
Vol. = 1 m3
1 mL = cm3
Celsius, ° C
Kelvin, K
Joules (J)
Calorie (Cal)
number of
particles
mole
1 mole = 6.02 x 1023
use dimensional analysis to convert between
these units
1.
12 inches = ? cm
12 in.
2.54 cm
1
in.
1 in. = 2.54 cm
30.48 cm
2.
95 miles = ? km
1
95 mi
km
0.62 mi
3.
400 lbs. = ? kg
1
400 lbs
153 km
1 kg = 2.2 lbs
kg
2.2 lbs
4. 250 grams = ? oz.
250 g
1 km = 0.62 mi
1
oz
28
g
182 kg
1 oz = 28 g
8.9 oz
5.
500 cm = ? in. 1 in. = 2.54 cm
500 cm
1
in
2.54 cm
kilo (k)
197 in
1000 times larger
deci (d)
10 times smaller (1/10)
centi (c)
100 times smaller (1/100)
milli(m)
1000 times smaller (1/1000)
micro (µ)
1 x 106 times smaller(1/106)
nano (n)
1 x 109 times smaller(1/109)
Ex: 5 m = ? cm
5 m
100 cm
1
m
500 cm
Ex:
25 mg = ? g
25 mg
1
g
1000 mg
0.025 g
1.50 L = ? mL
1.50 L
1000 mL
1
L
1500 mL
1 X 103 dg = ? g
1x 103 dg
1
g
10 dg
100 g
2.4 g = ? cg
2.4 g
100
cg
1
g
240 cg
3 X 1012 µg = ? g
3 x 1012 µg
1
g
1 X 106µg
0.9 dm = m
0.9 dm 1 m
10 dm
3 X 106 g
0.09 m
7.7 x 105 µm = ? m
1
7.7 x 105 µm
m
1 X 106 µm
0.77 m
200 g = ? kg
200 g
1
kg
1000 g
0.2 kg
0.25 L = ? mL
0.25 L
1000 ml
1 L
250 mL
200 m = ? cm
100 cm
200 m
1
m
20,000 cm
3.5 x 10-4 g = ? mg
3.5 x 10-4 g
1000 mg
1
g
0.35 mg
800 µL = L
800 µL
1
1
x106
L
µL
0.0008 dL
all the numbers in a measurement that are known
with certainty plus one that is estimated
6.35
uncertain
3 sig figs
6.3500
6.3
6.4
incorrect
Rules for determining which numbers in a
measurement are significant figures
1. Any number in a measurement that is not zero is
a significant figure
Ex: 213.5 m has
4
12, 567 m has
5
2. Zeros between nonzero numbers are significant
figures
Ex: 205 g has 3
10.0002 g has
6
3. Zeros to the left of a number are not significant
figures
Ex: 0.078 L has 2
0.00005 has
1
4. Zeros to the right of a number and to the right of
the decimal are significant figures
Ex: 2.00 has 3
0.00100 has
3
5.Zeros at the end of a number are not significant
figures unless the decimal point is shown
Ex: 1200 has
2
1200. has
4
For numbers in exponential form, look only at the
coefficient and not the exponent
4.50 x 108 has 3
Identify the number of significant figures in each measurement:
2 250
1.______
5 13,979
9.______
5 35,029
2.______
3 3.00 x 102
10.______
2 0.0075
3.______
4 0.6000
11.______
1
4. ______ 9000
2 50.
12.______
2 0.0080
5.______
2 4500
13.______
4
1
6.______ 10.00
14.______ 0.002
2 3.6 x 105
7._______
4 3.040
15.______
2 15,000
8._______
Rounding off numbers
Begin counting from the first significant figure on the left; if
the number being left off is 5 or higher, round up.
Ex: round 65.31890 to 3 sig figs:
65.3
round 0.05981 to 3 sig figs
0.0598
round 43,925 to 2 sig figs
44,000 or 4.4 x 104
round 545,858 to 4 sig figs 545,900 or 5.459 x 105
round 9.9992 x 10-4 to 2 sig figs 1.0 x 10-3
Round each number to 3 sig figs, then to 2 then 1 sig fig
1. 6.77510
6.78
6.8
7
2. 0.04031
0.0403
3. 18.298
18.3
4. 0.0011299
0.00113
5. 892.153
892
6.
57,320
57,320
0.040
18
0.04
20
0.0011
890
57,000
0.001
900
60,000
When rounding off the answer after a calculation,
the answer cannot be more accurate than your
least accurate measurement
Multiplication and Division
Rule: The number of sig figs in the answer
is determined by the number with the
fewest sig figs
Ex: 1.33 x 5.7 =
0.153
0.08
=
7.581 =
1.9125 =
(5.00 x 103) ( 7.2598 x 102) =
7.6
2
3.6299 x 106 =
3.63 x 106
Perform each calculation and round to the correct
number of significant figures
1. 520 x 367 =
190,840 =
190,000 or 1.9 x 105
2.
2.5 x 9.821 = 24.5525 =
25
3.
0.02430 =
0.95880
0.02534418 =
0.02534
2.6666 x 10-6 =
3 X 10-6
4. 4 x 10-8
1.5 x 10-2
=
Addition and
Subtraction
Rule: The number of decimal places in the
answer is determined by the number with
the fewest decimal places
Ex: 10.25 + 11.1 =
21.35 =
515.3215 - 30.42 = 484.9015 =
1425 - 820.95 =
604.05 =
21.4
484.90
604
Perform each calculation and round off to
the correct number of significant figures
1. 20.5 + 8.263 =
28.763 =
28.8
2. 0.88 + 3.104 =
3.984 =
3.98
3. 0.005 + 0.0066 = 0.0116 =
0.012
4. 2291.7 - 1512.015 = 779.685 =
779.7
Ratio of an object’s mass to
its volume
Density of water = 1 g/mL or 1 g/cm3
Which is more dense: the water in a
tub or in a small cup ?
both water
samples have the
same density
Sample problems
Calculate the density of a liquid if 50 mL of the liquid
weighs 46.25 grams.
D=M
V
D = 46.25 g
50 mL
0.925
0.9 g/mL
A block of metal has the dimensions: 2.55 cm x 2.55 cm x 4.80 cm.
If the mass of the block is 234.61 g, what is the density?
V=LxWxH
V = (2.55 cm)(2.55 cm)(4.80 cm)
234.61 g
V = 31.212 cm3 D =
31.212 cm3
7.51666
7.52 g/cm3
1. A marble weighing
53.87 g is placed in a
graduated cylinder
containing 40.0 mL of
water. If the water rises
to 64.9 mL, what is the
density of the marble?
V= 64.9 – 40.0 =
D=
53.87 g
24.9 mL
24.9 mL
2.16345
2.16 g/mL
Calculate the mass of a piece of aluminum having a
volume of 8.45 cm3. The density of aluminum is 2.7 g/cm3
D=M
V
M= D x V
M = (2.7 g/cm3) (8.45 cm3)
22.815
23 g
What volume of mercury weighs 25.0 grams?
Density of mercury = 13.6 g/mL
D=M
V
M= D x V
V = 25.0 g
13.6 g/mL
V = 1.838235
V=M
D
1.84 mL
1. What mass of gold (density= 19.3 g/cm3) has
a volume of 12.80 cm3 ?
D=M
V
M= D x V
M = (19.3 g/cm3)(12.80 cm3)
M= 247.04 = 247 g
2. Calculate the volume of a cork if its mass is 2.79 grams and
it has a density of 0.25 g/cm3
D=M
V
M= D x V
V = 2.79 g
0.25 g/cm3
V=M
D
11.16 = 11 cm3
3 scales
B.P. water
F.P. water
Anders
Celsius
Lord
Kelvin
Celsius Scale:
Kelvin Scale:
Water freezes at 0°C , boils at 100°C
Water freezes at 273 K and
boils at 373 K
Absolute Zero: Lowest temperature that can be
reached; all molecular motion stops
0 Kelvin
K = °C + 273
Ex: 25 ° C = ? K
298 K
0 K = ?°C
-273 °C
37 °C = ? K
310 K
600 K = ? °C
327°C
-50 °C = ? K
223 K
Energy that flows from a region of higher temp
to lower temp
units: Joules (J) , Calories (Cal)
1 Cal = 4.18 J
A measure of how well
Specific Heat Capacity: C,
something stores heat
specific heat of water: 1.00 Cal/g°C or 4.18 J/g°C
high compared to most substances; water heats up
slowly and cools off slowly.
Heat Calculations:
Problems
heat
q = mCΔT
mass
m
temp
change
specific heat
How many Joules of heat energy are needed to raise the
temperature of 50.0 grams of water from 24.5°C to 75.0°C ?
specific heat of water = 4.18 J/g °C
q= mCΔT
C
ΔT
q = (50.0 g) (4.18 J/g°C) (75.0 – 24.5)
q = 10,554.5
10, 600 J
50.5°C
A piece of gold weighing 28 grams cools from
125°C to 23°C. To do this it must lose 362 Joules of
heat energy. Calculate the specific heat of gold.
q= mCΔT
q= mCΔT
mΔT mΔT
C= q
m ΔT
362 J
C=
28 g 125- 23
102°C
C = 0.12675
0.13 J/g°C
What mass of graphite can be heated from 30°C to
80°C by the addition of 1500 Joules of heat?
Specific heat of graphite = 0.709 J/g°C
q= mCΔT
q= mCΔT
CΔT
m=
CΔT
m= q
CΔT
1500 J
(0.709 J/g°C) (80-30)
50°C
42.3131
40 g
1. How many Joules of heat are needed to increase the
temperature of 100.0 grams of iron metal by 80.0°C?
specific heat of iron = 0.45 J/g°C
q= mCΔT
q = (100.0 g)(0.45 J/g°C)(80.0°C)
q= 3600 J
2.
25 grams of water absorb 150 calories
of heat. What will be the temp change of the water?
T= q
mC
150 Cal______
(25 g)(1.00 Cal/g °C)
6°C