Chemistry Study Packet for Exam 2 You MUST MEMORIZE the 6 Strong Acids Everything that is not a strong acid is a weak acid You should also MEMORIZE the 8 Strong Bases HCl Hydrochloric Acid HBr Hydrobromic Acid HI Hydroiodic Acid H2SO4 Sulfuric Acid HClO4 Perchloric Acid HNO3 Nitric Acid NaOH Sodium Hydroxide LiOH Lithium Hydroxide RbOH Rubidium Hydroxide KOH Potassium Hydroxide CsOH Cesium Hydroxide Ca OH 2 Calcium Hydroxide Ba OH 2 Barium Hydroxide Sr OH 2 Strontium Hydroxide Everything that is not a strong base is a weak base Strong Electrolytes an ionic compound that is completely ionized in water An acid is something that increases the concentration of hydronium ions in solution Bases increase concentration of OH Weak bases react with water to form conjugate acids and hydroxide ions Acid H donor Base H acceptor 3 Ways to Name Acids and Bases 1 If it is a Hydrogen a singular element ex Add a hydro prefix ide ending turns to ic HCl Hydrochloric Acid 2 If it is a Hydrogen a polyatomic ion ending in ate ate goes to ic H2SO4 Sulfuric Acid note there is no hydro prefix here 3 If it is a Hydrogen a polyatomic ion ending in ite ite goes to ous ex ex HNO3 Nitrous Acid note there is no hydro prefix here Acid and Base Net Ionics HCl aq NaF aq NaCl aq HF aq aq Cl aq Na aq F aq Na aq Cl aq HF aq Molecular Ionic H Net Ionic H aq F aq HF aq Hydrofluoric Acid is a weak acid so it doesn t ionize HCl NaF and NaCl all ionize because HCl is a strong acid and the other two are soluble ionic compounds Acids and Bases normally have concentrations which is also called Molarity Molarity mols Liters You can use molarity as a conversion factor Ex If you have 1 52g of Al OH 3 how many mL of 110 M HCl can be created 3 HCl aq Al OH 3 aq AlCl3 3 H2O 1 52 g Al OH 3 1mol Al OH 3 78g AL OH 3 3mols HCl 1mol AL OH 3 1 L HCl 1000 mL HCl 110 mol HCl 1 L HCl 531 5 mL HCl Redox Reactions Loss of electrons oxidation Gain of electrons reduction Oxidizing agent is reduced Oxidation number of Oxygen is 2 except when it s not The oxidation rules will be given to you in the packet Ex 12 5 1 0 4 2 1 2 C5H12 8 O2 5 CO2 6 H2O Carbon loses electrons so it is Oxidized Oxygen gains electrons so it is Reduced treat oxidation numbers almost like charges they have to be equal to the overall charge If you can t do the math in your head treat it like an algebraic equation ex CO2 So if you know that Oxygen is 2 and the overall charge is 0 then you can set up an equation X 2 2 0 X 4 So Carbon has an oxidation number is 4 Sometimes you have to use the Activity Series to finish equations Electrons will Flow from the metal above to the ions below Ex Does Ag want more electrons than Cu AgNO3 Cu Cu NO3 2 Ag Cu aq 2 Ag aq Cu2 aq Ag s No Ag wants to oxidize so it therefore loses electrons however Cu wants lose electrons more than Ag which is why the reaction works Pressure PV nRT P Pressure in atm V Volume in L n number of mols R Constant 0821 T Temperature K The first step to completing a PV nRT problem is to make sure everything is in the right units Then simply plug and chug Ex 30oC 8 5 x 105 L 751 mmHg 751mmHg 1atm 760 mmHg 985 atm 30oC 273 313 K 985 atm 32 581 mols Partial Pressure you need to remember that Pressuretotal Pressurewater Pressuregas This is the case only when a gas is bubbled through water PSI Pounds per square inch o Gauge pressure 15 absolute pressure You can also manipulate PV nRT to create the other gas laws Ex P1V1 P2V2 PV nRT can also be used to calculate density d m v 0 685 mols 0821 298 K V 1 675 L 1atm Then convert Liters to mL and follow the formula for density 100g 1675 mL 0597g mL Mass 100g P 1 atm n 0 685 mols T 298K R constant V Empirical Formula Multiply till Mol Divide by Small Multiply till Whole Ex 64 9 C 13 5 H 21 6 O Always assume 100g sample 64 9g C 13 5g H 21 6g O C 64 gC 1mol C 12gC 5 40 mol C 13 5 g H 1 mol H 1g H 13 5 mol H 21 6 gO 1 mol O 16g O 1 35 mol O H O Divide by 1 35 and you get the empirical formula of C4H10O If you need the molecular formula divide the molar mass of the molecular compound provided by the molar mass of the empirical formula Chp 6 o Heat does not pressure o Food calories 1000 calories 1 Kcal o Joules 4 184 j 1 cal o Calorie amount of heat energy it takes to raise 1g of H2O 1oC when the water is at 15oC o Enthalpy H o H is positive endothermic o H is negative exothermic o The Reaction being exothermic makes it more favorable o Bond breaking is endothermic o Work on the system by the surroundings w 0 o Work on the surroundings by the system w 0 o Remember that enthalpy in an equation works like a mol ratio See recitation exercise 7
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