CHEM 210 1nd Edition Lecture 14 Outline of Last Lecture I. Dilute SolutionsOutline of Current Lecture II. Migration of Ions in a SolutionIII. Determining Moles of Ions in Aqueous SolutionsIV. Writing Equations for Aqueous Ionic ReactionsCurrent Lecture** It is very crucial that everyone understands that whatever Dr. Hosmane teaches, is the correct material for the chemistry class. 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 of these are water (H2O), some solids such as sugar (C12H22O11), and alcohols. Strong electrolytes have 100 percent dissociations in water, and are substances that exist as ions in solutions. Examples of these include strong acids,strong bases, and ionic compounds. Finally, weak electrolytes produce relatively few ions and partially dissociate in solutions. Examples of these are weak acids and weak bases.Moles of Ions in Aqueous Ionic SolutionsEx; 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 These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.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.Writing Equations for Aqueous Ionic ReactionsMolecular Equation Includes the symbols from the elements, subscripts, and the states of matterTotal Ionic Equation Includes everything the molecular equation has, including the charges for the elements or compoundsNet ionic Equation Simplified version of the total ionic equation. Net equations get rid of spectator ions, 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