Transcription of Chapter 4: Solution Stoichiometry – Cont.
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Chapter 4: Solution Stoichiometry – Cont.
1 Chapter 4: Solution Stoichiometry Solutions2 Molarity (dilution calculations, Solution Stoichiometry ); Solubility and Solubility Rules Molecular, Ionic and Net Ionic Equations Precipitation Reactions Acid-Base ReactionsReading: Sections , , Problems: 27 a&c, 29, 31, 35c, 37a, 45b, 49, 53, 55, 57, 59, 61, 63, 67, 69, 71, 73, 75, 77, 79 and Solutions Recall from Chapter 2 Homogeneous mixtures are called solutions The component of the Solution that changes state ( saltin water) is called the solute The component that keeps its state and does the dissolving is called the solvent If both components start in the same state, the major component is the solventTro: Chemistry: A Molecular Approach, 2/e3 To describe solutions accurately, we must describe how much of each component is presentSolution Concentration Dilute solutionshave a small amount of solute compared to solvent Concentrated solutionshave a large amount of solute compared to solventTro: Chemistry: A Molecular Approach, 2/e43 Concentrations.
Tro: Chemistry: A Molecular Approach, 2/e 1 L Pb(NO 3)2 = 0.175 mol, 1 L KCl = 0.150 mol, 1 mol Pb(NO 3)2: 2 mol KCl Solution: 16 Tro: Chemistry: A Molecular Approach, 2/e Example 4.8 – Cont. Answer: 0.350 L or 350 mL of 0.150 M KCl is needed for the reaction
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