Chemistry Q3
Amazing Benchmark
Review
Example 1:
Standard 4a: Know random motion of molecules
and their collisions with a surface create the
observable pressure on that surface.
Pressure
Measuring Pressure
• Atmospheric
Pressure: results
from the mass of air
being pulled toward
the center of the
Earth by gravity.
Pressure
Atmospheric Pressure
–
Changing altitudes affect pressure.
Pressure
Units of Pressure
One standard atmosphere (abbreviated 1 atm) is equal to:
 1.000 atm


760.0 mm Hg

760.0 torr (Torricelli)

101325 Pa (Pascal)

14.69 psi (lbs/ in. square)

1.000 atm
760.0 mm Hg

1.000 atm
101325 Pa

1.000 atm
760.0 torr
14.69 psi
1.000 atm
Example 2:
Standard 4b: Know random motion of
molecules explains the diffusion of gases and
the rate of diffusion is affected by increase in
temperature.
Example: Gas
Example: Gas diffuses from high
concentration to less concentrated areas.

Diffusion is the movement of particles from a high
concentration to a lower concentration, in order to
reach equilibrium.

Molar Mass/Weight of a gas will affect the rate of
diffusion. The smaller you are the faster you move.
Example: Think of a mouse squeezing under a door
as compared to a rock, of the same size. A mouse
will more easily be able to squeeze through the gap.
Example 3:
Standard 4c: Apply the gas laws to relations
between the pressure, temperature, and volume
of any amount of an ideal gas or any mixture of
ideal gases.
Gas Laws:

Boyle’s Law: volume and pressure are
inversely proportional. P1V1 = P2V2

Charles’s Law: volume and temperature are
directly proportional.
V1 = V2
T1

T2
Avogadro’s Law: volume and moles are
directly proportional.
V1 = V2
n1
n2
The Ideal Gas Law
Rearranging the equation gives the Ideal Gas Law
PV = nRT
R = 0.08206 L atm
mol K (Universal Gas Constant)
P = pressure
V = volume
n = number of moles
T = temperature
Gases that obey this equation is said to behave ideally.
Dalton’s Law of Partial Pressures
Dalton’s Law of partial pressures
• For a mixture of gases in a container, the total pressure
exerted is the sum of the partial pressures of the gases
present:
P = nRT
V
P total = P1 + P2 + P3 ….
subscripts referring to individual gases (gas1, gas 2, etc.)
P total = O2 + CO2 + CH4 ….
Dalton’s Law of Partial Pressures

The pressure of the gas (8.4 atm) is affected by the
number of particles in atoms or molecules.
• The pressure is independent of the type of atoms or
molecules in container.
Example: Ideal Gas Law
Calculation

A sample of hydrogen gas has a volume of 8.56 L at a temperature of
0 C and a pressure of 1.5 atm. Calculate the number of moles of H2
present in this gas sample. Assume that the gas behave ideally.
PV = nRT

PV = n
RT
1.5 atm (8.56 L) = 0.57 mol
0.08206 L atm (273 K)
K mol
T = 0 C + 273 K = 273 K
P = 1.5 atm
V = 8.56 L
Example 5:
Standard 4d: Know the values and meanings of
standard temperature and pressure (STP)
Gas Stoichiometry
Molar Volume
• Standard Temperature and Pressure (abbreviated STP)
 is 0oC (or 273 K)
 at 1 atm
 with 1 mole of gas “X”
• Molar volume of an ideal gas at STP is
22.4 L = 1.000 mol
or
22.4 L
1.000 mol
Example: Gas Stoichiometry at
STP

Quicklime, CaO, is produced by heating calcium
carbonate, CaCO3. Calculate the volume of CO2
produced at STP from the decomposition of 152 g of
CaCO3 according to the reaction.
CaCO3 (s)  CaO(s) + CO2 (g)
152 g CaCO3 x 1 mol CaCO3 x 1 mol CO2
100.1 g
1 mol CaCO3
x 22.4 L CO2 = 34.1 L CO2
1 mol
Example 6:
Standard 7c: know energy is released when a
material condenses or freezes and is
absorbed when a material evaporates or
melts.
Temperature and Heat

Temperature (energy) is a measure of the
random motions of the components of a
substance.
Hot water
(90. oC)
Cold water
(10. oC)
System vs. Surrounding
• As a match burns, it loses thermal heat or energy (PE) to
the surrounding.
Energy (heat) is released when water
freezes. Exothermic Reaction
Liquid  Solid
Freezing
Energy (heat) is released when water
vapor condenses. Exothermic
Reaction
Gas  Liquid
Condensation
Energy (heat) is absorbed when ice
melts. Endothermic Reaction
Solid  Liquid
Melting
Energy (heat) is absorbed when water
evaporates. Endothermic Reaction
Liquid  Gas
Evaporation
Example 7:
Standard 6a: know the definitions of solute and
solvent.
What are the solvent and solutes?
What are the solvent and solutes?
What are the solvent and solutes?


10 kt, 14 kt, or 18 kt
are just gold alloy.
18kt is 75% gold and
25% metals (Cu, Ag).
24 kt is pure gold.
Example 8
Standard 6c: know temperature, pressure, and
surface area affect the dissolving process
Solubility is defined as the ability to
dissolve.

Ionic substances (NaCl) breakup into individual
cations (Na+) and anions (Cl-).
Factors Affecting the Rate of
Dissolving
When a solid is being dissolved in a liquid to form a solution, the
dissolving process may occur rapidly or slowly:
1.
Increase Surface area: allows more surface area to be exposed
to solvent.
2. Stirring: causes solutes and solvent to continuously mix.
3. Increase Temperature: causes solvent and solutes to move faster.
Factors Affecting the Rate of
Dissolving
When a gas is being dissolved in a liquid, the dissolving process
may occur rapidly or slowly:
1. Increase pressure: forces the gas molecules into the
solution since this will best relieve the pressure that has been applied.
Example 9:
Standard 6d: know how to calculate the
concentration of a solute in terms of g/L, Molarity,
parts per million, and percent composition.
Solution Composition: Mass Percent
• To express the mass of solute present in a given mass
of solution, we use mass percent.
 For example: If you dissolve 1.0 g of NaCl in 48 g of
water, the solution would have a mass of 49 g. Then,
the mass percent of NaCl would be 2.0%.
1.0 g solute x 100% = 2.0% NaCl Solution
49 g solution
Example: Mass Percent
Calculation

A solution is prepared by mixing 1.00 g of ethanol, C2H5OH, with
100.0 g of water. Calculate the mass percent of ethanol in this
solution.
Mass percent (Methanol) = Mass of solute x 100%
Mass of solution
1.00g C2H5OH
x 100% = 0.990% C2H5OH
100.0g H2O + 1.00g C2H5OH
Example: Parts per million (ppm)
Parts per million (ppm) indicates the number of
molecules of solute for every million molecules of
solution.
mass _ solute
x1,000,000  parts _ per _ million
total _ mass
1.00g NaOH
x 10^6 = 9900 ppm NaOH
100.0g H2O + 1.00g NaOH
Example : Calculating Molarity

Calculate the molarity (M) of a solution prepared by
dissolving 11.5 g of solid NaOH in enough water to make
1.50 L of solution.
Molarity =
moles of solute
liters of solution
11.5g NaOH x 1 mol NaOH = 0.288 mol  0.288 mol = 0.192 M NaOH
40.0 g
1.50 L solution
Example: Calculating mass from
Molarity (M)

To analyze the alcohol content of a certain substance, a
chemist needs 1.00 L of an aqueous 0.200 M K2Cr2O7
(potassium dichromate) solution. How much mass of
K2Cr2O7 must be weighted out to make this solution?
0.200 M = 0.200 mol K2Cr2O7
1 L K2Cr2O7
1.00 L K2Cr2O7 x 0.200 mol K2Cr2O7 x 294.2 g K2Cr2O7
1 L K2Cr2O7
1 mol K2Cr2O7
= 58.8 g K2Cr2O7
Example : Calculating g/L
You want to make lemonade.
You have 25 g of sugar and 200
mL of water. What is the
concentration of your lemonade
in g/L?
200 mL  0.2 L H2O
25 g
0.2 L
= 125 g/L
Example 10
Standard 5a: know the observable properties of
acids, bases, and salt solutions
Acids and Bases 1
The Arrhenius Model:

Acid – produces hydrogen ions (H+) in aqueous solution.
Acids are recognized as substances that taste sour such
as vinegar (acetic acid). It reacts with metals to produce
gas.

Base – produces hydroxide ions (OH-) in aqueous
solution.
Bases (aka: alkalis) are recognized by their bitter taste
and slimy/slippery feel such as hand soaps and salt.
Acids and Bases 2

Acid Property:
Acids make a blue vegetable dye called litmus
turn red.

Base Property:
Bases are substances which will restore the
original blue color of litmus after having been
reddened by an acid.
Acids and Bases 3


Acid Property: Acids conduct an electric current.
Base Property: Bases conduct an electric current.

This is a common property shared with salts. Acids,
bases, and salts are grouped together into a category
called electrolytes, meaning that an aqueous solution
of the given substance will conduct an electric current.

Non-electrolyte solutions cannot conduct a current.
The most common example of this is sugar dissolved in
water. Pure (bottled) water is a non-electrolyte.
Acids and Bases 4
We have learned that when a strong acid and a
strong base are mixed, the H+ and OH- react to
form H2O:
H+ (aq) + OH- (aq)  H2O (l) + Salt
Example: HCl + NaOH
--> HOH + NaCl
We call this reaction, Neutralization reaction,
because a neutral solution (pH = 7) will result.
Example 11
Standard 5b: know acids are H+ donating and
bases are H+ accepting substances
Acids and Bases
The Bronsted-Lowry Model



Acid – proton (H+) donor
Base – proton (H+) acceptor
The general reaction for an acid dissolving in
water is
Example 12
Standard 5c: know strong acids and bases fully
dissociate and weak acids and bases partially
dissociate.
Acid Strength

For example with strong acid (HA) – completely
ionized or completely dissociated
Acid Strength
• For example with a weak acid (HA) – most of the acid
molecules remain intact.