Chemistry Q3Amazing 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

– Changing altitudes affect pressure.

Atmospheric Pressure

Pressure

One standard atmosphere (abbreviated 1 atm) is equal to: 1.000 atm

760.0 mm Hg 1.000 atm

760.0 mm Hg 760.0 torr (Torricelli) 1.000 atm

760.0 torr 101325 Pa (Pascal) 1.000 atm

101325 Pa 14.69 psi (lbs/ in. square) 14.69 psi

1.000 atm

Units of Pressure

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 is independent of the type of atoms or molecules in container.

The pressure of the gas (8.4 atm) is affected by the number of particles in atoms or molecules.

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 T = 0 C + 273 K = 273 K

RT P = 1.5 atm

V = 8.56 L

1.5 atm (8.56 L) = 0.57 mol

0.08206 L atm (273 K)

K mol

Example 5:

Standard 4d: Know the values and meanings of standard temperature and pressure (STP)

Molar Volume

Gas Stoichiometry

• Molar volume of an ideal gas at STP is 22.4 L = 1.000 mol or 22.4 L

1.000 mol

• Standard Temperature and Pressure (abbreviated STP) is 0oC (or 273 K) at 1 atm with 1 mole of gas “X”

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 x 22.4 L CO2 = 34.1 L CO2

100.1 g 1 mol CaCO3 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 (energy) is a measure of the random motions of the components of a substance.

Temperature and Heat

Hot water Cold water

(90. oC) (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 thesolution 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

Parts per million (ppm) indicates the number of molecules of solute for every million molecules of solution.

millionperpartsxmasstotal

solutemass__000,000,1

_

_

Example: Parts per million (ppm)

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 = 58.8 g K2Cr2O7

1 L K2Cr2O7 1 mol 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

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.

The Arrhenius Model:

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

Acid – proton (H+) donor Base – proton (H+) acceptor

The general reaction for an acid dissolving in water is

The Bronsted-Lowry Model

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.