4. SCIENTIFIC NOTATION
LARGE NUMBERS
SCIENT. NOTATION
(to 2 sig. digs)
SMALL NUMBERS
7,000,000
0.000 000 40
84,600
0.000 321
344,100, 000
0.0009646
100,000,000, 000
0.010
1,678
0.10
100
0.000 000 000 000 026
10
0.00100
SCIENT. NOTATION
5. SIGNIFICANT DIGITS
*Quantities that are counted are exact; there is no uncertainty about
them…they have UNLIMITED significant digits. Two examples are found in
these statements:
i)
ii)
There are 22 people in the room.
Ten dimes have the same value as one dollar.
*Measured quantities are much different because they are never exact.
Suppose you are finding the mass of a friend in kilograms. You might try a
bathroom scale and get 56 kg, then a scale in a physician’s office and get 55.8
kg, and finally a sensitive electronic scale in a scientific laboratory and get
55.778 kg. Not one of the measurements is exact; even the most precise
measurement, the last one, has an uncertainty.
*(↑ decimal places = ↑ precision)
*In any measurement, the digits that are reliably known are called significant digits. These include the digits known for
certain and the single last digit that is ESTIMATED.
Rules for Determining the Number of Significant Digits
1. All NON-ZERO numbers are significant. (e.g., 1,292 mm has four significant digits.)]
2. “Captive” zeros(i.e. zeroes placed between other digits) ARE significant. (e.g. 400,006 cm has 6 sig. digs).
3. Leading zeros are never significant. (e.g. 0.0000089 kg has 2 significant digits.)
4. Trailing zeros:
a) ARE significant if they indicate a measurement(i.e. to the right of a decimal: e.g. 5.800 km, 900.0 mL
and 703.0 N all have four significant digits.
*b) Zeroes at the end of a number are significant only if they are indicated to be so using scientific notation. For example,
5 800 000 km may have anywhere form 2 to 7 significant digits. By using scientific notation, we can judge which digit is
the estimated one, so we can determine the number of significant digits.
* 5.8 x 106 km has 2 significant digits
* 5.800 x 106 km has 4 significant digits
* 5.800 000 x 106 km has 7 significant digits
5. Counted quantities are exact and infinite number of significant figures. (i.e. there is NO uncertainty about them.)
Activity 1:
*Determine the number of significant digits in the following numbers:
NUMBER
0.02
0.020
501
501.0
5.000 x 103
5 000.0
6 052.00
10 800 000
SIG. DIGS
NUMBER
0.0005
0.1020
10 001
3.50 x 104
2 x 109
8.4572 x1023
8040
3.01 x 1021
SIG. DIGS
NUMBER
80405
0.0300
699.5
2.000 x102
0.90100
90 100
4.7 x 10-8
0.000 410
SIG. DIGS.
Activity 2 : Round the following to the indicated number of significant digits.
*When performing calculations that involve a series of mathematical steps, don’t round off numbers until you are ready to
state the final answer…use all the decimal places reported by your calculator.
a) 2150 (2 s.d.) -___________
b) 0.0256 (2 s.d.) - _________
c) 0.0346 (2 s.d.) - __________
d) 0.0050129 (3 s.d.) - _________
e) 1.994 (3 s.d.) - _________
f) 2149.99 (3 s.d.) - ________
CALCULATIONS INVOLVING MEASUREMENTS
a) ADDING or SUBTRACTING:
RULE: Round off the final answer to the same number of _____________ places as the least precise number.
e.g. Add the following measurements: 123.0 cm + 12.40 cm + 5.380 cm
Solution:
123.0
cm
12.40 cm
5.380 cm
140.780 cm
*Thus, 140.780 cm should be rounded to _____________ cm since 123.0 cm has only one decimal place.
b) MULTIPLYING or DIVIDING:
RULE: Round off the final answer to the same number of ______________ digits as the least precise number.
e.g. A cyclist travels 4.00 x 103 m on a racetrack in 292.4 s. Calculate the average speed of the cyclist.
Solution: Average speed (v) is distance (d) divided by time (t):
v=
d
4.00 x 103 m
=
= 13.67989056 m/s
t
292.4 s
*Thus, the average speed should be reported to _____ significant digits – 13.7 m/s.
Activity 3:
*Perform the following operations expressing the answer in the correct number of significant digits.
a) 1.35  2.267 =
b) 1 035  42 =
c) 12.01 + 35.2 + 6 =
d) 55.46 – 28.9 =
e) 0.021  3.2  100.1 =
f) 1.278  103  1.4267  102 =
g) 0.2129 + 0.002 + 0.03 =
h) 101.4 + 25 + 201 =
i) 1.0 + 2.04 + 5.03 =
j) 2.5 x 1.1111 =
k) 8.314 x 2.5x10-2 =
l) 35.45 x 2.25 =
m) 150  4 =
n) 505 – 450.25 =
o) 1.252  0.115  0.012 =
p) 0.15 + 1.15 + 2.051 =
q) 41.11 + 20.5 + 18.333 =
*r) 3.76x105 – 276 =