week-4 - First Mid-quarter Examination MQ-1 on Monday Jan...

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First Mid-quarter Examination MQ-1 on Monday, Jan. 29 at 6:30 pm Covering Chapters 10, 11, and 13 room TBA
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For students with a scheduled- class conflict with the first Mid-Quarter Exam: (1) Send me an email today with a copy of your class schedule , and (1) Let me know when you want to take the makeup exam : a) during the last week of classes. b) early (5:00-6:18 pm) on Mon, Jan 29. c) late (8:00-9:18 pm) on Mon, Jan 29.
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MQ-1 on Monday, Jan. 29 at 6:30 pm Covering Chapters 10, 11, and 13 Help session on Saturday, Jan 27 at 4:00 – 6:00 pm in MP1000
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Week 4 Sections 13.1- 13.6—Properties of Solutions 13.1 The Solution Process Energy Changes, Entropy, Rxns 13.2 Saturated Solutions and Solubility 13.3 Factors Affecting Solubility Solute-Solvent Interactions Pressure Effects Temperature Effects 13.4 Ways of Expressing Concentration Mass Percentage, ppm, and ppb Mole Fraction, Molarity, and Molality Conversion of Concentration Units 13.5 Colligative Properties 13.6 Colloids
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endothermic endothermic exothermic Figure 13.3
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e.g. MgSO 4 e.g. NH 4 NO 3 Figure 13.4
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Realize there is an inherent tendency for the two isolated materials to form solution, regardless of the energetics!!! This represents an “entropy” factor.
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Factors that FAVOR solubility: 1. Strong solute-solvent interactions 2. Weak solute-solute interactions 3. Weak solvent-solvent interactions More often we’ll settle for the solute-solvent interactions being similar to the solute-solute and solvent-solvent interactions. A general rule: Like dissolves like. i.e. polar and polar non-polar and non-polar
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Dissolution : solute + solvent solution. Crystallization : solution solute + solvent. Saturation : crystallization and dissolution are in equilibrium. Solubility : amount of solute required to form a saturated solution. Supersaturation : a solution formed when more solute is dissolved than in a saturated solution.
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Fig 13.12 Structure of glucose—note red O atoms in OH groups which can interact nicely with water.
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Consider gas solubilities in water at 20 o C with 1 atm gas pressure (~Table 13.2) Solubility/M He 0.40 x 10 -3 N 2 0.69 x 10 -3 CO 1.04 x 10 -3 O 2 1.38 x 10 -3 Ar 1.50 x 10 -3 Kr 2.79 x 10 -3 CO 2 3.1 x 10 -2 NH 3 ~ 53
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Fig 13.14 Henry’s Law, C g = k P g Note Table 13.2 again for k
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Consider N 2 dissolved in water at 4.0 atm. Note k = 0.69 x 10 -3 mol/L-atm C g = k P g = (0.69 x 10 -3 mol/L-atm)(4.0 atm) = 2.76 x 10 -3 mol/L at normal atmospheric conditions, however, P g = 0.78 atm C g = (0.69 x 10 -3 mol/L-atm)(0.78 atm) = 0.538 x 10 -3 mol/L Note that (2.76 - 0.54) x 10 -3 mol/L = 2.22 x 10 -3 mol/l Thus for 1.0 L of water, 0.0022 mol of nitrogen would be released = 0.0022 x 22.4L = 0.049 L = 49 mL !
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This note was uploaded on 07/17/2008 for the course CHEM 122 taught by Professor Zellmer during the Winter '07 term at Ohio State.

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week-4 - First Mid-quarter Examination MQ-1 on Monday Jan...

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