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. 3

Pressure Atmospheric Pressure Changing altitudes affect pressure. 4

Pressure Units of 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 5

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. 11

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 …. 12

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. 13

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)

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 16

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 and Heat Temperature (energy) is a measure of the random motions of the components of a substance. Hot water Cold water (90. oC) (10. oC) 19

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

Gas  Liquid Condensation 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-). 30

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: 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. 31

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. 32

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 34

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. 1.00g NaOH x 10^6 = 9900 ppm NaOH 100.0g H2O + 1.00g NaOH 36

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 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. 41

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. 42

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. 43

Example: HCl + NaOH --> HOH + NaCl 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. 44

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 Bronsted-Lowry Model Acid – proton (H+) donor Base – proton (H+) acceptor The general reaction for an acid dissolving in water is 46

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 48

Acid Strength For example with a weak acid (HA) – most of the acid molecules remain intact. 49