Solubility and Functional Groups _______________________________________________________ You will recall from general chemistry that a solution has two components: the solvent‚ which is the substance present in greater amount‚ and the solute‚ which is dissolved in the solvent. Solubility is defined as the mass (in grams) of solute dissolved in 100 g of solute at saturation. Molar solubility is defined as the amount (in moles) of solute per liter of saturated solution. The solubility of one compound
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PRACTICAL # 3 Title: Solubility Product of Ca(OH)2 Aim: To find the solubility and the solubility product of calcium hydroxide. Theory: Define‚ with equation‚ the solubility product. Find‚ from literature‚ the solubility product of calcium hydroxide at 25oC. Experimental: Reagents: solid calcium hydroxide‚ water‚ 0.1 moldm-3 hydrochloric acid Apparatus: Procedure:
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Determination of g by Free Fall Raw Data: Time (ms) ± 0.01ms Height of release of ball from the sensor plate (cm) ±0.1cm Trial 1 Trial 2 Trial 3 Trial 4 Trial 5 0.0 0.00 0.00 0.00 0.00 0.00 15.0 180.05 179.36 178.74 180.26 180.23 30.0 244.33 244.21 244.71 243.88 245.87 45.0 300.72 301.29 300.59 301.43 301.70 60.0 348.68 348.39 348.77 349.12 348.35 75.0 390.27 390.77 389.58 391.19 390.43 This table below is the results obtained during the experiment in cm/ms. This table below is the results
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INTRODUCTION Many substances contain water molecules as a part of their crystal structure. We call such solids hydrates‚ and we call the bound water the water of hydration. A hydrate has a definite number of water molecules bound to each anhydrous salt unit. The formula of the hydrate copper(II) sulfate pentahydrate is CuSO4 · 5 H2O The dot indicates that the molecules of water are attached to the ions in CuSO4 by weak bonds. We can drive off the water of hydration by heating the hydrate. If blue
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Effect of Temperature on Solubility Lab Purpose: What is the solubility of minerals in water? What is the relationship between temperature and solubility? Hypothesis: If salt and sugar are each tested in water of varying temperatures‚ then salt and sugar’s solubility will increase as the temperature also increases. Materials: Two 250 mL beakers Tap water 100 mL graduated cylinder Hot plate Two petri dishes Glass stirring rod Salt Sugar Thermometer
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The primary focus of experiment 4 was to teach a student the basic solubility rules of salts in aqueous solutions. After developing and using a scheme for the qualitative analysis of three cations in an aqueous mixture‚ a student would use a centrifuge to identify‚ precipitate‚ and separate the three ions in the mixture. After becoming familiar with all three cations‚ the student would use the scheme again to identify at least two of the cations in an unknown solution. In order to begin the separation
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Determination of a Rate Law Megan Gilleland 10.11.2012 Dr. Charles J. Horn Abstract: This two part experiment is designed to determine the rate law of the following reaction‚ 2I-(aq) + H2O2(aq) + 2H+I2(aq) + 2H2O(L)‚ and to then determine if a change in temperature has an effect on that rate of this reaction. It was found that the reaction rate=k[I-]^1[H2O2+]^1‚ and the experimental activation energy is 60.62 KJ/mol. Introduction The rate of a chemical reaction often
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reactants and products are constant. In this experiment set up‚ the reaction between ethanoic acid and ethanol to produce ethyl ethanoate and water is a reversible reaction as shown by the following equation: CH₃COOH + CH₃CH₂OH↔CH₃COOCH₂CH₃ +H₂O The Equilibrium Law states that at equilibrium the ratio [CH₃COOCH₂CH₃][H₂O]/[CH₃COOH][CH₃CH₂OH] is constant at constant temperature. This ‚ Kc‚ is called the equilibrium constant for the reaction. AIM: To verify the Equilibrium Constant for an Esterification
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affecting solubility 1. Effect of Temperature For some substances to dissolve in a given solvent‚ heat is absorbed. The reaction is endothermic. In this case‚ an increase in temperature increases solubility. For some substances‚ heat is released when they dissolve in a given solvent. The reaction is called exothermic. In this case‚ an increase in temperature decreases solubility. Generally‚ an increase in temperature in the solubility of solids in liquids increases solubility. But for solubility of gas
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Separating Sand & Salt Purpose: The purpose of this experiment is to separate a mixture of sand and salt. Materials: Small beaker Electric balance Sand Salt Water Graduated cylinder Stirring rod Filter paper Flask Rubber policeman Wash bottle Funnel Bunsen burner complete with rubber tubing and a source of gas Hot hands Insulator pad Ring stand complete with a ring Wire gauze Striker Pre-lab Questions: 1. Water will be useful in separating the salt and sand because salt is soluble in water‚ which
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