CHEM 210 1nd EditionExam # 2 Study Guide Lectures: 12- 22Lecture 12 (February 11)Using balanced equation to solve word problemsHow do you use a balanced equation to solve word problems?Before you being a word problem, and a balanced equation is given to you, you must first check to see if it’s balanced. For example,2Cu2S + 302  2Cu2O + 2SO2There are 4 moles of copper on the left side as well as the right side. There are 2 moles of sulfuron both sides of the equations as well, and finally there are 6 moles of oxygen on each side, so the equation is balanced.If the question from the problem asks you how many moles of O2 are required to roast 10.0 moles of copper (I) sulfide, you simply multiply by the mole ratio of oxygen and copper (I) sulfide, to get the answer.10.0 mol CuS2 X (3 mol O2) / (2 mol Cu2S) = 15.0 mol O2. If another problem asks you for the amount of grams of sulfur dioxide that forms when 10.0 molof copper (I) sulfide is roasted, you would multiply by the mole ratio of SO2 and Cu2S, and then multiply that by the molar mass of SO2 to end up with an answer in grams.10.0 mol Cu2S X (2 mol SO2) / (2 mol Cu2S) X (64. 07g SO2) / (mol SO2) = 641 g SO2. Finally, if the problem asks you how many kilograms of oxygen are required to form 2.86 kg of copper (I) oxide? You do the same steps to end up with an answer in kilograms.2.86 g Cu2O X (10^3 g Cu2O) / (1 kg Cu2O) X (1 mol Cu2O) / (143.10 g Cu2O) = 20.0 mol Cu2OThen,20.0 mol Cu2O X (3 mol O2) / (2 mol Cu2O) X (32.00 g O2) / (1 mol O2) X (1 kg O2) / (10^3 g O2) = 0.959 kg O2Lecture 13 (February 16)What is dilution and how do you solve dilution problems in chemistry?Dilution is used to lessen concentrations of different solutions by adding more volume to the solution. The lighter the color of the solution, the more diluted it is. The chemical equation used to solve dilutions is M1 V1 = M2 V2. The equation can be rearranged to solve for different components, depending on what the problem asks you to solvefor. For example,If you dilute 175 mL of a 1.6 M solution of LiCl to 1.0 L, determine the new concentration of the solution.First we find what the problem is looking for, which in this case is the final concentration or M2.Next we rearrange our equation to solve for M2M1 V1 / V2 = M2Finally, we plug in the values. Make sure the units are consistent. For this case, I will convert the 1.0 L into mL, to be constant with the initial volume in mL.(1.6 mol/L) (175 mL) / (1000 mL) = M2M2 = 0.28 mol / L or 0.28 M.Lecture 14 (February 18)How do Ions migrate in a solution? How do we determine the moles of ions in aqueous solutions? How do we write equations for aqueous ionic reactions?Conduction is the movement of electrically charged particles. Opposite charges attract while like charges repel. Electrolytes are substances that produce ions in solutions.Nonelectrolytes do not produce any ions. Examples are water, sugar and alcohols.Strong electrolytes dissociate in water 100 percent whole weak electrolytes partially dissociate in water. Strong electrolytes include strong acids and strong bases while weak electrolytes include weak acids and weak bases.Examples on how to solve equations using moles of ions in aqueous solutions:How many moles of each ion are in the following solutions?1. 5.0 moles of ammonium sulfate dissolved in H20.First, we look at how many ions are in ammonium sulfate. NH4 or ammonium has a plus 1 charge, giving it one ion, while sulfate (SO4 ^-2) has a 2- charge, resulting in 2 ions. This totals to 3 ions. 1mole is of ammonium sulfate is equal to 3 ions. Next, you just multiply the 5.0 moles given to the 3 ions, which results in 15 moles.2. 78.5 grams of cesium bromide dissolved in H20.First we have to write a balanced equation.CsBr  Cs (+1) + Br (-1)Next, we convert from grams to moles.78.5 g CsBr x (1 mol CsBr) / (212.8 g CsBr) = 0.369 mol Cs and 0.369 mol Br.Molecular equations include the symbols from the elements, subscripts and the states of matter.Total Ionic Equations include everything the molecular equation has, including the charges for the elements or compounds.Net ionic equations are simplified versions of the total ionic equations. They get rid of spectatorions or ions that appear on both sides of the equation. Ex: NaOH + HCl  H2O + NaClNa(+ charge)(aq) + OH (- charge)(aq) + H(+charge)(aq) + Cl (- charge)  H2O (l) + Na(+ charge)(aq) + Cl (- charge)(aq)Notice how we separate the compounds into ions. Also notice that Na (+) and Cl (-) appear the same on both sides of the equation. We can cross those out.We are left with OH (- charge)(aq) + H (+charge)(aq)  H2O (l)Lecture 15 (February 20)What are precipitates? Which ionic compounds are soluble and which ionic compounds are insoluble? What are strong/ weak acids and bases?Precipitates are formed when two aqueous reactants form a solid as the product in a reaction. Soluble Ionic Compounds are compounds that dissolve in water or aqueous solutions.- Groups 1A in the periodic table and NH4+ are all soluble.- Common nitrides, acetates, and most perchlorates are soluble.- Chlorides, bromides, and iodides are soluble, except those of Ag+, Pb 2+, Cu+, and Hg.Insoluble Ionic Compounds are compounds that cannot dissolve in water or aqueous solutions.- All common metal hydroxides, except those of group 1A in the periodic table, and larger members of group 2A, beginning with Ca+, are insoluble.- Common carbonates and phosphates, except those of group 1A and NH4+ are insoluble.- Common sulfides, except those of group 1, group 2 and NH4+ are insoluble.There are also strong/weak acids and bases in chemistry.Strong acids include;- HCl (hydrochloric acid)- HNO3 (nitric acid)- HBr (hydrobromic acid)- H2SO4 (sulfuric acid)- HI (hydroiodic acid)- HClO4 (perchloric acid)Weak acids are;- CH3COOH (acetic acid)- HF (hydrofluoric acid)- HN02 (nitrous acid)Strong bases include;- NaOH (sodium hydroxide)- KOH (potassium hydroxide)- Ba (OH)2 (barium hydroxide)Weak bases are;- NH3 (ammonia)- CH3NH2 (methylamine)- C5H5N (pyridine)Lecture 16 (February 23)What is oxidation and reduction and when do they occur? What are the rules for finding oxidation numbers in compounds? Oxidation occurs when the oxidation number in the compound in the reactant increases when itreaches the product. Reduction is the opposite, where the oxidation number in the product is less than the oxidation number in the reactant. There are a set of rules used to find