The Math: Very Very Important

Converting Units:

Unit (dimensional) analysis is the technique we will be using. The name indicates that we will be watch / working with the units. And you will see this.

Common Conversion Factors:

Length	Volume	Mass
1 m = 1.0936 yd	1 L = 1.0567 qt	1 kg = 2.2046 lb
1 in = 2.54 cm (exact)	1 qt = 0.94635 L	1 lb = 453.59 g
1 km = 0.62137 mi	1ft³ = 28.317 L	1 (Avoirdupois) oz = 28.349 g
1 mi = 1609.3 m	1 tbsp = 14.787 mL

These are "KNOW KNOWS"

We will be using a lots of equivalences (conversion factors):

Example1: 1 ft = 12 inches

That could be written and used either as: $\frac{1 f t}{12 i n c h e s} o r \frac{12 i n c h e s}{1 f t}$ $\frac{1 f t}{12 i n c h e s} o r \frac{12 i n c h e s}{1 f t}$

Example 2: 100 cm = 1 m

That could be written and used either as: $\frac{1 m}{100 c m} o r \frac{100 c m}{1 m}$ $\frac{1 m}{100 c m} o r \frac{100 c m}{1 m}$

Example 3: 1 mile = 1.609 km

That could be written and used either as: $\frac{1 m i l e}{1.609 k m} o r \frac{1.609 k m}{1 m i l e}$ $\frac{1 m i l e}{1.609 k m} o r \frac{1.609 k m}{1 m i l e}$

How this works

We will take the known quantity x conversion factor (equivalence) = new unit

Steps:

unknown and known
conversion factors
set up

repeat above until you are in the right units.

Example 1: Convert 63.5 inches to feet

XX feet = 63.5 inches
12 inches = 1 foot

Start with known: 63.5 inches

$63.5 i n c h e s x (\frac{1 f t}{12 i n c h e s}) =$ $63.5 i n c h e s x (\frac{1 f t}{12 i n c h e s}) =$

ANS 5.29 ft (3 sig fig because of 63.5 inches)

Example 2: Convert 2.4 gallons to mL

XX mL = 2.4 gallon
1 gallon = 4 quarts
1 L = 1.06 qt
1 L = 1000 mL

Start with known: 2.4 gallons

$2.4 g a l l o n x (\frac{4 q u a r t s}{1 g a l l o n}) =$ $2.4 g a l l o n x (\frac{4 q u a r t s}{1 g a l l o n}) =$ 9.6 quarts

not in mL yet

$9.6 q u a r t s x (\frac{1 L}{1.06 q u a r t s}) =$ $9.6 q u a r t s x (\frac{1 L}{1.06 q u a r t s}) =$ 9.0566 L

not in mL yet

$9.0566 L x (\frac{1000 m L}{1 L}) =$ $9.0566 L x (\frac{1000 m L}{1 L}) =$ 9056.6 mL

EUREKA!!! mL

2.4 gallons has 2 s.f. our answer is then 9100 mL

Example 3: Convert 23.0 pounds to grams

XX grams = 23.0 lbs
1 kg = 2.205 lb
1000 g = 1 kg

Start with known: 23.0 lbs

$23.0 l b x \frac{1 k g}{2.205 l b} =$ $23.0 l b x \frac{1 k g}{2.205 l b} =$ 10.43 kg

Not there yet

$10.43 k g x (\frac{1000 g}{1 k g}) =$ $10.43 k g x (\frac{1000 g}{1 k g}) =$ 10,431 g 3 sig.fig ANS = 10,400 g or 1.04 x 10⁴ g

Example 4: Convert 35 miles/hour to meters/second

$\frac{X X m e t e r}{1 \sec} = \frac{35 m i l e s}{1 h r}$ $\frac{X X m e t e r}{1 \sec} = \frac{35 m i l e s}{1 h r}$
1 mile = 1.609 km
1 km = 1000 m
1 hr = 60 min
1 min = 60 sec

Start with known: 35 mil/hr

$\frac{35 m i l e s}{1 h r} x \frac{1.609 k m}{1 m i l e} =$ $\frac{35 m i l e s}{1 h r} x \frac{1.609 k m}{1 m i l e} =$ 56.32 km/hr

$\frac{56.32 k m}{1 h r} x (\frac{1000 m}{1 k m}) =$ $\frac{56.32 k m}{1 h r} x (\frac{1000 m}{1 k m}) =$ 56,320 m/hr

$\frac{56, 320 m}{1 h r} x (\frac{1 h r}{60 min}) x (\frac{1 min}{60 \sec}) =$ $\frac{56, 320 m}{1 h r} x (\frac{1 h r}{60 min}) x (\frac{1 min}{60 \sec}) =$ 16 m/s (2 s.f.)

Unit analysis 1 Video

Unit Analysis 2 Video
		Unit Conversions Worksheet.docx Dimensional Analysis 1 worksheet.doc

Density:

defined as mass per volume

$D = \frac{m a s s}{v o l u m e} = \frac{m}{V}$ $D = \frac{m a s s}{v o l u m e} = \frac{m}{V}$ (Intensive Property)

Solids and Liquids units are usually: $\frac{g}{m L} o r \frac{g}{c m^{3}} o r \frac{k g}{m^{3}}$ $\frac{g}{m L} o r \frac{g}{c m^{3}} o r \frac{k g}{m^{3}}$

Gases units: $\frac{g}{L}$ $\frac{g}{L}$

rearrange equation:

$m = D x V$

$y = m (x) + b$ $y = m (x) + b$

01 density.jpeg $s l o p e = \frac{r i s e}{r u n} = (\frac{y_{2} - y_{1}}{x_{2} - x_{1}})$ $s l o p e = \frac{r i s e}{r u n} = (\frac{y_{2} - y_{1}}{x_{2} - x_{1}})$

The slope is the density of the material used.

Density Table

substance	density	unit analysis
H₂O	1.00 g/mL	1.00 g = 1.00 mL
Aluminum	2.64 g/mL	2.64 g = 1.00 mL
Copper	8.91 g/mL	8.91 g = 1.00 mL
Iron	7.21 g/mL	7.21 g = 1.00 mL
lead	11.34 g/mL	11.34 g = 1.00 mL
zinc	7.14 g/mL	7.14 g = 1.00 mL

Example 1: Determine the volume that 125 g of copper will occupy. Density of copper is 8.91 g/mL

XX volume = 125 g
8.91 g = 1.00 mL

$125 g x (\frac{1.00 m L}{8.91 g}) = m L = V o l u m e$ $125 g x (\frac{1.00 m L}{8.91 g}) = m L = V o l u m e$ = 14.0 mL (3 s.f.)

Determine the volume by water displacement is a technique used for solid.

01 water displacement.jpeg (the volume of the hammer is: 69.9 mL - 65.0 mL = 4.9 mL)

Example 2: A 25.0 g piece of metal was placed carefully into a graduated cylinder that had an initial volume of 11.3 mL. The volume now read 19.2 mL after the metal was added. Determine the density.

$D = \frac{m a s s}{v o l u m e} = \frac{m}{V}$ $D = \frac{m a s s}{v o l u m e} = \frac{m}{V}$

The volume is just the volume of the metal: V = 19.2 mL - 11.3 mL = 7.9 mL

m = 25.0 g

V = 7.9 mL

$D = \frac{25.0 g}{7.9 m L}$ $D = \frac{25.0 g}{7.9 m L}$ = 3.2 g/mL

Density Video
		Density worksheet.doc

Temperature:

it is the hotness or coldness of an object
it is the measure of the average kinetic energy

Scales:

Fahrenheit
Celsius
Kelvin: absolute scale

Temperature scales.jpeg

Conversion of: $F ⟺ C$ $F ⟺ C$

$\frac{F - 32}{C} = \frac{9}{5}$ $\frac{F - 32}{C} = \frac{9}{5}$ ( the $\frac{9}{5}$ $\frac{9}{5}$ comes from $\frac{180}{100}$ $\frac{180}{100}$ the difference in boiling and freezing points of water)

Conversion of : $C ⟺ K$ $C ⟺ K$

$C + 273.15 = K$ $C + 273.15 = K$

Example 1: Convert 65^oF to ^oC

rearrange equation: $\frac{F - 32}{C} = \frac{9}{5}$ $\frac{F - 32}{C} = \frac{9}{5}$ to solve for C

$\frac{5}{9} (F - 32) = C$ $\frac{5}{9} (F - 32) = C$ (plug in 65^oF)

$\frac{5}{9} (65 - 32) = C$ $\frac{5}{9} (65 - 32) = C$ solution = 18^oC

Temperature Conversion Video
		Temperature Conversion worksheet.doc

Energy:

is the ability to do work
work = force x distance (math equation)

Types of Energy

Kinetic Energy

Energy in motion
$K E = \frac{1}{2} m v^{2}$ $K E = \frac{1}{2} m v^{2}$ (m = mass in kg, v = velocity in m/s)

Potential Energy

Stored energy
$P . E . = U = m g h$ $P . E . = U = m g h$ (m = mass in kg, g = 9.8 m/s², v = velocity in m/s)

Unit of Heat or Energy

Joules (derived unit) $\frac{k g m^{2}}{s^{2}} = j o u l e$ $\frac{k g m^{2}}{s^{2}} = j o u l e$
calories
BTU (British Thermal Unit)

Conversions:

1 calories = 4.184 Joules
1 BTU = 1055 Joules
Food 1000 calories = 1 kcal = 1 C (C is a food calorie which is really a kilocalorie)

*** ALL chemical processes occur with a change in energy ***

q or Q = energy

Q = (mass) x (specific heat) x ( change in temperature) or $Q = (m) x (c) x (Δ T)$ $Q = (m) x (c) x (Δ T)$

Q is the energy in Joules or Calories

m is the mass

c is the specific heat (an intensive property)

$Δ T$ $Δ T$ is the change in temperature = (final temperature - initial temperature)

material	specific heat in $\frac{J o u l e}{(g)^{o} C}$ $\frac{J o u l e}{(g)^{o} C}$
H₂O	4.184
Iron	0.45
aluminum	0.905
copper	0.384
silver	0.236

Two other terms of importance:

exothermic: energy is given off to the surroundings (means there is a negative Q, -Q)
endothermic: energy is gained by the surroundings (means there is a positive Q, +Q)

Example 1: Calculate the amount of heat (energy) required to heat 250. g of water from 20.0^oC to 80.0^oC.

$Q = (m) x (c) x (Δ T)$ $Q = (m) x (c) x (Δ T)$

The specific heat determines the units

c = 4.284 J / g ^oC

m = 250. g

$Δ T$ $Δ T$ = (80.0 - 20.0)^oC = 60.0^oC

$Q = 250 g x (\frac{4.184 J}{g^{o} C}) x {60.0}^{o} C =$ $Q = 250 g x (\frac{4.184 J}{g^{o} C}) x {60.0}^{o} C =$ 62800 joules or 62.8 kJ

Example 2: Determine the specific heat of a metal when it loses 5575 J of energy to the surroundings and the temperature of the metal drops from 100^oC to 27^oC. The mass of the metal is 80.0 g.

Q = -5575 J

m = 80.0 g

c = xxx

$Δ T$ $Δ T$ = (27^oC - 100^oC) = -73^oC

$Q = (m) x (c) x (Δ T)$ $Q = (m) x (c) x (Δ T)$

rearrange to solve to (c)

$\frac{Q}{(m) Δ T} = c$ $\frac{Q}{(m) Δ T} = c$ (plug in numbers)

$\frac{- 5575 J}{(80.0 g) (- 73^{o} C)} = c$ $\frac{- 5575 J}{(80.0 g) (- 73^{o} C)} = c$ = 0.905 J/g ^oC (from chart could be aluminum)

Conservation of Energy: Though cannot create nor destroy energy but it can change forms

for heat: If something loses heat something else must gain heat

Q_lost = Q_gained