CHEM 1211 Lecture 11Outline of Last Lecture I. StoichiometryII. Clicker QuestionsIII. Concept BuildingIV. Aqueous ReactionsA. MolarityOutline of Current Lecture I. Problem OneII. Problem TwoIII. Dilution of SolutionA. Problem OneIV. Chemical ThermodynamicsA. Heat Transfer1. Through a Liquid…Current Lecture: I. Problem One27.0mL of a solution that is 0.100M in nickel (II) nitrate reacts with 14.0mL of a solution that is 0.0800M in sodium phosphate. What mass of nickel phosphate is produced?3 Ni(NO3)2(aq) + 2Na3PO4(aq) 6NaNO3(aq) + Ni3(PO4)2(s)0.0270L 0.100 mole Ni(NO3)21 mole Ni(PO4)2366.02g Ni(PO4)2=0.329g L 3 mole Ni(NO3)21 mole Ni(PO4)20.0140L 0.0800 mole Na3PO41 mol Ni3(PO4)2366.02g Ni3PO4=0.205g L 2 mol NiNa3PO41 mole Ni3PO4II. Problem Two20.00mL of H2SO3 solution is diluted to total volume of 50.00mL. The Final solution titrated with 16.78mL of NaOH. What is the molarity of the original H2SO3 solution? (.1100M NaOH)H2SO3(aq) + 2NaOH NaSO3(aq) + 2H2O(l)0.01678L 0.1100mol NaOH 1 mole H2SO31 = 1.845 x 10-2M L 2 mol NaOH 0.05000LIII. Dilution of Solution- To dilute a solution, add solvent to a concentrated solution.i.e. Making tea or soda- The number of moles of solute in the two solutions remain constant M1V1 = M2V2: Used for dilutions, not chem. Rxns or titrantsA. Problem One20.00mL of H2SO3 solution is diluted to total volume of 50.00mL. The Final solution titrated with 16.78mL of NaOH.M1 = ?V1 = 20.00mLM2 = 1.845 x 10-2MV2 = 50.00mLIV. Chemical ThermodynamicsA. Heat Transfer1. How do we know that heat is transferred through a liquid?(1.845 x 10-2M)(50.00mL) = 4.613 x 10-2M (20.00mL)Hot metalTmetal TwaterHeat(q): Hot Cold until equilibriumqAu + qH2O =
View Full Document
Unlocking...