Scientific NotationUsed for numbers that are really big or really small

A number in exponential form consists of a coefficient multiplied by a power of 10

10,000

1,000,000

546,000

0.00001

0.00751

0.00000029

12,450

15,230,000

0.0884

1 x 104

1 x 106

5.46 x 105

1 X 10-5

7.51 x 10-3

2.9 x 10-7

1.245 x 104

1.523 x 107

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

0.0001 x 1012

4,490 x 10-

7

0.0065 x 103

0.090 x 10-5

0.0112 x 108

150 X 1012

200 x 10-2

1 X 105

1 X 108

4.49 x 10-4

6.5 x 100 = 6.5

9 X 10 -7

1.12 x 106

1.5 x 1014

2 X 100

calculations involving scientific notation

Enter the coefficient, then EE or ExpDo not enter “x

10”

(2.4 x 106) (3.1 x 103 ) =

(9.5 x 10-7) (5 x 10-4 ) =

4.8 x 109 = 2.4 x 102

7.5 x 10-5 + 4.2 x 10-6 =

7.44 x 109

4.75 x 10-10

2 X 107

7.92 x 10-5

Two types of dataQualitati

vedescriptiveEx: the burner flame is hot

Quantitative numerical Ex: the flame is

1000 ° CMeasurements should be both accurate and precise

Accuracyhow close the experimental value is to the accepted or true value

calculating percent error

% E = |accepted –experimental|accepted

X 100

Precisionhow close the measurements are to each other when the experiment is repeated

Ex: a student does an experiment todetermine the density of lead; she repeatsthe experiment two more times and getsthese results:

10.1 g/cm3 9.4 g/cm3 and 8.5 g/cm3

comment on her precision

If the actual density of lead is 11.4 g/cm3, calculate her percent error using her average density as the experimental value

poor

|11.4 – 9.3|% E =

11.4100 = 18.4 %Avg.= 9.3

Mass (weight)

Quantities and their units

gram (g)kilogram (kg) milligram (mg)

Length (distanc

e)

meter (m)kilometer

(km)millimeter

(mm)

Example units

Volume liter (L)milliliter (ml)any unit of length that is cubed

1 mL = cm3

1m

1 m

1 m

Vol. = L x W x H

Vol. = 1 m3

Temperature

Celsius, ° C Kelvin, K

HeatJoules (J)Calorie (Cal)

number of

particles

mole

1 mole = 6.02 x 1023

Metric Conversionsuse dimensional analysis to convert between these units

1. 12 inches = ? cm 1 in. = 2.54 cm

in.

cm12 in.

1

2.5430.48 cm

2. 95 miles = ? km 1 km = 0.62 mi

3. 400 lbs. = ? kg 1 kg = 2.2 lbs

4. 250 grams = ? oz.1 oz = 28 g

95mi

mi

km

lbs

lbs

kg

g

g

oz

1

0.62

400

2.2

1

250

28

1

153 km

182 kg

8.9 oz

5. 500 cm = ? in.1 in. = 2.54 cm

kilo (k)

1000 times larger

10 times smaller (1/10)

centi (c)

deci (d)

100 times smaller (1/100)

500 cmcm

in

2.54

1197 in

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

m

cm

1

100500 cm

1.50 L = ? mL

1 X 103 dg = ? g

0.025 g

1500 mL

Ex: 25 mg = ? g

25

mg

g

1000

1mg

1.50

L

mL

1

1000L

1x 103 dg

g

10

1dg100 g

2.4 g = ? cg

3 X 1012 µg = ? g

0.9 dm = m

2.4 g

g

cg

1

100240 cg

3 x 1012µg

µg

g

1 X 106

13 X 106 g

0.9dm

dm

m

10

10.09 m

7.7 x 105 µm = ? m

200 g = ? kg

7.7 x 105 µm

µm

m1

1 X 1060.77 m

200 g

g0.2 kg

kg1

1000

0.25 L = ? mL

0.25 L

L

ml1000

1250 mL

3.5 x 10-4 g = ? mg

800 µL = L

200 m = ? cm

200 m

m1

cm10020,000 cm

3.5 x 10-4 g

g1

mg1000 0.35 mg

800 µL

1 x106

L10.0008 dL

µL

Significant Figuresall the numbers in a measurement that are knownwith certainty plus one that is estimated

6.3 6.4

6.35

uncertain

3 sig figs

6.3500

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 figureEx: 213.5 m has

412, 567 m has

52. Zeros between nonzero numbers are significant figures

Ex: 205 g has 3 10.0002 g has 63. 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 35.Zeros at the end of a number are not significant figures unless the decimal point is shownEx: 1200 has

21200. 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:

1.______ 250 9.______ 13,979

2.______ 35,029 10.______ 3.00 x 102

3.______ 0.0075 11.______ 0.6000

4. ______ 9000 12.______ 50.

5.______ 0.0080 13.______ 4500

6.______ 10.00 14.______ 0.002

7._______ 3.6 x 105 15.______ 3.040

8._______ 15,000

2

5

2

1

2

4

2

2

5

3

4

2

2

1

4

Rounding off numbersBegin counting from the first significant figure on the left; ifthe 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 figs0.0598

round 43,925 to 2 sig figs44,000 or 4.4 x 104

round 545,858 to 4 sig figs545,900 or 5.459 x 105

round 9.9992 x 10-4 to 2 sig figs1.0 x 10-3

Round each number to 3 sig figs, then to 2 then 1 sig fig

1. 6.77510

2. 0.04031

3. 18.298

4. 0.0011299

5. 892.153

6. 57,320

6.78 6.8 7

18.3 18 20

0.00113 0.0011 0.001

892 890 900

57,320 57,000 60,000

0.0403 0.040 0.04

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

(5.00 x 103) ( 7.2598 x 102) =

7.581 = 7.6

21.9125 =

3.6299 x 106 =

3.63 x 106

Perform each calculation and round to the correctnumber of significant figures

1. 520 x 367 =

2. 2.5 x 9.821 =

3. 0.02430 = 0.95880

4. 4 x 10-8 =

1.5 x 10-2

190,840 = 190,000 or 1.9 x 105

24.5525 = 25

0.02534418 = 0.02534

2.6666 x 10-6 = 3 X 10-6

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 =

515.3215 - 30.42 =

1425 - 820.95 =

21.35 =

484.9015 =

604.05 =

21.4

484.90

604

Perform each calculation and round off tothe correct number of significant figures

1. 20.5 + 8.263 =

2. 0.88 + 3.104 =

3. 0.005 + 0.0066 =

4. 2291.7 - 1512.015 =

28.763 =

3.984 =

0.0116 =

779.685 =

28.8

3.98

0.012

779.7

Density: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 g50 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 = L x W x HV = (2.55 cm)(2.55 cm)(4.80 cm)

V = 31.212 cm3

D = 234.61 g 7.516667.52 g/cm3

31.212 cm3

A marble weighing53.87 g is placed in agraduated cylindercontaining 40.0 mL ofwater. If the water risesto 64.9 mL, what is the density of the marble?

V= 64.9 – 40.0 =

24.9 mL

D = 53.87 g

24.9 mL2.16345 2.16 g/mL

1.

Calculate the mass of a piece of aluminum having avolume of 8.45 cm3. The density of aluminum is 2.7 g/cm3

(8.45 cm3)D = M V M= D x V

M = (2.7 g/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 =

M DV = 25.0 g

13.6 g/mLV = 1.838235 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 VM = (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 =

M DV = 2.79 g

0.25 g/cm3

11.16 = 11 cm3

3 scales

B.P. water

F.P. water

AndersCelsius

LordKelvin

Celsius Scale:

Water freezes at 0°C , boils at 100°C

Kelvin Scale:Water freezes at 273 K and boils at 373 K

Absolute Zero: Lowest temperature that can be

reached; all molecular motion stopsK = °C +

273

0 Kelvin

Ex: 25 ° C = ? K

298 K37 °C = ? K

310 K-50 °C = ? K

223 K

0 K = ?°C-273 °C

600 K = ? °C327°C

Heat

Energy that flows from a region of higher tempto lower temp

units: Joules (J) , Calories (Cal)

Specific Heat Capacity: C,

A measure of how wellsomething stores heat

specific heat of water: 1.00 Cal/g°C

1 Cal = 4.18 J

or 4.18 J/g°C

high compared to most substances; water heats upslowly and cools off slowly.

Heat Calculations: q = mCΔTheat

mass

tempchange

specific heat

ProblemsHow 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

m

ΔTC

q = (50.0 g) (75.0 – 24.5) (4.18 J/g°C)

50.5°Cq = 10,554.5 10, 600 J

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

mΔTmΔT

C = q m ΔT

C = 362 J

28 g125- 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

CΔTCΔT

m = q CΔT

m = 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 = (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

q= mCΔT