Dimensional Analysis and Units
Physical quantities often (in fact, almost always) are expressed in terms of a certain number of characteristic units of measure representing the physical quantity itself. For example, a person's height may be expressed in the common American units of length, feet and inches, as 5 ft 9 in. This is equivalent to 5.75 ft, of course, and in the system of units used in the rest of the world (as well as in Chemistry 5 and 6), this same height is 1.7526 m, using the metric system's basic length unit, the meter. Thus, one physical quantity can often be expressed with different units, and it is completely correct to write 5.75 ft = 1.7526 m even though 5.75 ≠ 1.7526. The unit specifier makes all the difference, and without it, the pure number itself is meaningless.
This brings us to our first and most important rule:
Always write the units associated with any physical quantity.
The value of any physical quantity (such as 5.75 ft) is the product of a numerical value (5.75) and a unit (ft). Both are necessary.
The Most Common Units
The metric system (officially known as the SI system, for "Le Système International d'Unités") has seven fundamental or base quantities. Here they are, but only the first six (in bold) will interest us in Chem 5 and 6. The seventh, luminous intensity, is fairly obscure and won't interest us.
SI Base Units
Physical quantity 
Name 
Symbol for quantity 
length 
meter 
m 
mass 
kilogram 
kg 
time 
second 
s 
electric current 
ampere 
A 
absolute temperature 
kelvin 
K 
amount of substance 
mole 
mol 
luminous intensity 
candela 
cd 
In addition to these base units, we will often encounter units for important physical quantities such as pressure and energy that are given a single name but which are quantities that are composite combinations of the fundamental units.
For example, the common unit of energy we will use is the joule, symbolized J. (Note that the symbol is capitalized, but the name, joule, isn't, even though it is named after James Joule. Likewise, the temperature unit, kelvin, is not capitalized while the symbol, K, is.)
Energy can be decomposed into the following set of base units: kg m^{2} s^{–2} or "kilogram meter squared per second squared." If you recall the elementary equation that states kinetic energy is mass times speed squared (and speed is length per unit time), you can see the source of this decomposition.
The second composite quantity we will use often is pressure, and the fundamental composite SI unit for pressure is the pascal, or Pa (named after Blaise Pascal). Pressure is a force per unit area, and if you recall Newton's law that force is mass times acceleration, you can see that pressure is kg m^{–1} s^{–2}. There are several other units used to express pressure, such as the atmosphere (abbreviated atm), the bar, and the torr (abbreviated Torr after Evangelista Torricelli). But no matter which unit is used to express a pressure, they are all decomposable into kg m^{–1} s^{–2} base units.
Dimensional Algebra
The final two areas we cover here concern the common "power of ten prefixes" that are used to express quantities that may vary over wide magnitudes and the approach we take to convert from one unit to another.
For example, our common pressure unit of atmosphere is related to the fundamental SI pascal unit as follows: 1 atm = 101,325 Pa (exactly). Many common units have exact conversion factors, such as the length conversion 1 ft = 12 in, which is one you probably know. Conversion factors you have not memorized need to be looked up, and your text (and your exams) will contain all of them you'll need.
The conversion factor above connecting two pressure units is not only a fairly long number, six digits, it is also a fairly large number. We humans like numbers that are smaller in magnitude (usually those in the 0.1 to 10.0 range feel most comfortable), and the familiar exponential power of ten notation is used to achieve that. In this case, we write 101,325 Pa = 1.01325 x 10^{5} Pa. This notation is also convenient for the (very common) situation where we do not need the full precision of the conversion factor. We may often be completely satisfied with the approximate relation 1 atm ≅ 1.0 x 10^{5} Pa.
We can further simplify this factor if we use one of the many power of ten prefixes that are gathered in the table below. We will most frequently use only the seven shown in bold text, but you should be aware of them all. In our pressure example, we see that 1 atm is approximately 0.1 MPa ("megapascal") because 1 MPa = 10^{6} Pa. The prefixed units we will use most often include mL ("milliliter"  note that "liter" is symbolized "L"), kJ ('kilojoule"), and the base unit kg ("kilogram") as well as mg ("milligram").
Power of Ten Prefixes
Power

Prefix

Symbol

Origin

+24

yotta

Y

Variant of yocto

+21

zetta

Z

Variant of zepto

+18

exa

E

Variant of Greek hex, six (18 = 3 x 6)

+15

peta

P

Greek pente, five (15 = 3 x 5)

+12

tera

T

Greek teras, monster

+9

giga

G

Greek gigas, giant

+6

mega

M

Greek megas, great

+3

kilo

k

Greek khilioi, thousand

+2

hecto

h

Greek hekaton, hundred

+1

deca

da

Greek deka, ten

–1

deci

d

Latin decimus, tenth

–2

centi

c

Latin centum, hundred

–3

milli

m

Latin mille, thousand

–6

micro

μ

Greek mikros, small

–9

nano

n

Latin nanus, dwarf

–12

pico

p

Spanish pico, small quantity

–15

femto

f

Danish or Norwegian femten, fifteen

–18

atto

a

Danish or Norwegian atten, eighteen

–21

zepto

z

Blend of z and Latin septem, seven (21 = 3 x 7)

–24

yocto

y

Blend of y and Greek octo, eight (24 = 3 x 8)

Using Conversion Factors
Finally, we turn to the derivation of conversion factors and the general algebra of unit combination. Just as the cliché tells us "you can't add apples and oranges," neither can you add a length to a mass, for example. You can, however, multiply or divide any two quantities, no matter what the units of each, and when you do, the quantity that results carries the units of the product or ratio of the original quantities' units, as in the following simple example: (3 kg) x (2 m s^{–1})^{2} = 12 kg m^{2} s^{–2} = 12 J. This follows from our earlier observation that a physical quantity can be considered the product of a numerical value and a unit. Here, we have moreover simplified our otherwise perfectly good result, 12 kg m^{2} s^{–2}, by recognizing that kg m^{2} s^{–2} is equivalent to the joule unit, J.
For conversion from one unit to another of the same physical quantity, we can follow these steps. We use a common situation for our example. One can find tabulated values of the vapor pressure, p, of many liquids at various temperatures, and among these tables, one finds that the vapor pressure of water at 20 °C is p = 2.33 kPa. Since the Pa may not be a very familiar pressure unit, let's convert this value into atm units. We found earlier that 1 atm ≅ 1.0 x 10^{5} Pa, and if we don't need great precision, this approximate relation will suffice. Note that we can write our original value as p = 2.33 x 10^{3} Pa and our conversion equation as 1 Pa = 1/(1.0 x 10^{5})^{ }atm. We simply substitute this conversion equation into our original expression (where every step is shown for clarity) to find p = 2.33 x 10^{3} Pa = 2.33 x 10^{3} (1 Pa) = 2.33 x 10^{3} (1/(1.0 x 10^{5})^{ }atm) = (2.33 x 10^{3})/(1.0 x 10^{5})^{ }atm = 2.33 x 10^{–2} atm, a value we can perhaps relate to much better than 2.33 kPa. It's a small pressure!
