Kilo   k   1000
hecto   h   100
deca   da     10
unit              1
deci   d         0.1
centi   c        0.01
milli   m        0.001

mass in grams (g)     A nickel weighs about 5g.

distance in meters (m)     A nickel = 2 mm thick.

volume in liters (L)     1.057quarts = 1.000 liters
 

Mass is a measure of the quantity of matter.  Weight is a measure of the force of gravity on an object.
 

English Units

Length
12 inches (in.) = 1 foot (ft)
3 feet (ft) = 1 yard (yd)
5280 feet (ft) = 1 mile (mi)

Mass
16 ounces (oz) = 1 pound (lb)
2000 lb = 1 ton

Volume
2 pints (pt) = 1 quart (qt)
4 (qt) = 1 gallon (gal)

English/Metric   Conversions +
1 inch = 2.54 cm
1 pound = 453.6 g
1 quart = 0.946 L
1cm³ = 1 mL

Dimensional Analysis

Convert 27.2 ft. to meters (m).

27.2 ft _x 12 in _x2.54 cm   _x 1 m = 8.29 m
1            1ft         1 in     100 cm
 

Convert 99 oz to kg.

99 oz _x 1 lb _x453.6 g  _x 1 kg    = 2.8 kg
1         16 oz     1 lb      1000 g
 

Convert 1.0 ft³ to L

1.0 ft³ _x12 in³ _x 2.54 cm³_x1 mL _x 1 L     = 28 L
1            1 ft³        1 in³  1 cm³  1000 ml
 

Density = mass/volume

solids;     g/cm³
liquids;    g/mL
gases;      g/L

Sample densities

Au 19.3 g/cm³
H₂O @ 4^o C 1.00 g/mL
Air 1.29 g/L

Calculate the mass of 2500 mL of methane (density 0.714 g/L)

_{d =} m ;     m = dv
      v
m = (2500mL) (0.714g/L) (1L/1000mL) = 1.785 g = 1.8 g
 

What is the mass in lbs. of 1.00 gal. H₂O.
 

1.00 gal _x 4 qt _x 1 L      _x 1000 mL _x 1.00 g _x1 lb      = 8.34 lb
                1 gal  1.057qt      1L         1 mL   453.6 g
 
 

Percentage = % = parts per 100

_{% =} part _x 100
      whole    1

Unless specified otherwise, we will use % by mass.

14-karat gold (Au) is 41.6 % copper (Cu). Calculate the mass of gold
in a 12.2 g 14-K gold ring. Let "R" stand for ring.
 

12.2 g R _x 58.4 g Au = 7.12 g Au
    1            100. g R

_{% =} part _x 100
      whole    1

_{58.4 =}  x       _x  100
        12.2 g         1

_{x =} (58.4) (12.2 g) = 7.12 g Au
             100

Temperature Scales
 

	oF	oC	K
boiling point of H2O	212	100	373
freezing point of  H2O	32	0	273
absolute zero	-460	-273	0

 

Temperature Conversions

K = ^oC + 273

^oF = 1.8^oC + 32

^oC = (^oF - 32 )
              1.8

Convert - 22^oF to ^oC & K

^oC = ^oF - 32
            1.8

^oC = (-22 - 32) = - 30.^oC
             1.8

K = 273 + ^oC

K = 273 + (-30) = 243 K

Energy = the capacity to perform work.
Common types of energy: mechanical, electrical, chemical, light, heat

Law of Conservation of Energy: In a non nuclear process energy can neither be created or destroyed.

Potential energy  = stored energy

Kinetic energy = energy due to the motion of matter

Units of Energy

J = Joule = kg m²
                  sec²

cal = calorie = heat needed to warm 1 g H₂O from 14.5^oC to 15.5^oC

BTU = heat to warm 1 lb of H₂O by 1^oF

Specific heat (Sp. Ht.) = heat energy needed to warm 1 g of a substance by 1^oC.

Examples
H₂O(s) 2.1 J/g^oC
H₂O() 4.184 J/g^oC
H₂O() 1.00 cal/g^oC
Fe(s) 0.444 J/g^oC
O₂(g) 0.92 J/g^oC

Heat is a measure of the quantity of energy while temperature is a measure of intensity.

Heat = (Sp.Ht.)(mass)( delta T)

 delta T = T_final - T_initial

Calculate the joules of energy needed to warm 55g Al from 25^oC to 79^o C

Heat = (Sp.Ht.)(mass)( delta T)

= 0.908 J _x 55g _x(79^oC - 25^oC)   = 2696.76 = 2700 J
      g^oC       1              1
 

Heat always flows from a warm body to a cold body. The heat lost by the warm body equals that gained by the colder body. Heat lost is negative by convention because T = T_f - T_i < O.

A 60.0g Fe bar at 99.0^oC is placed in

55.0 g H₂O at 22.0^oC. What is the final temperature.

- Heat Fe = + Heat H₂O

- 0.444 J _x (T_f - 99.0^oC) _x 60.0g  =  4.184 J _x(T_f -22^oC)_x 55.0₂
         g^oC            1             1            g^oC            1             1

T_f = 30.0^oC