with potassium iodate in the presence of potassium iodide. Vitamin C‚ more properly called ascorbic acid‚ is an essential antioxidant needed by the human body (see additional notes). When iodate ions (IO3−) are added to an acidic solution containing iodide ions (I−)‚ an oxidation-reduction reaction occurs; - the iodate ions are reduced to form iodine Burette containing potassium iodate solution IO3− + 6 H+ + 5 e− → ½ I2 + 3 H2O - while the iodide ions are oxidised to form iodine. 2 I− → I2 +
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Water-We studied the difference between mineral oil and water using a hand lens. * Making iodine in water- We added one drop of potassium iodide and sodium hypochlorite (bleach) then mixed the two together producing iodine. * Making iodine in oil-We added two drops of mineral oil‚ one drop of water on the mineral oil‚ then added one drop of potassium iodide and sodium hypochlorite. Producing Iodine. Results: In class we conducted four experiments on
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the best methods is the iodine-thiosulfate titration procedure. The iodide ion‚ I-‚ is easily oxidized by almost any oxidizing agent. In acid solution‚ hypochlorite ions oxidize iodide ions to form iodine‚ I2. The iodine that forms is then titrated with a standard solution of sodium thiosulfate. The analysis takes place in a series of steps: 1. Acidified iodide ion is added to hypochlorite ion solution and the iodide is oxidized to iodine. 2 H+(aq) + ClO-(aq) + 2 I-(aq)
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Chemistry F332 Notes Ions in solids and solutions: Structure of an ionic lattice (Sodium Chloride): * Consists of sodium ions (Na+) surrounded by six chloride ions (Cl-) * Chloride ions also surrounded by six sodium ions. * Held together by attraction of oppositely charged ions. * Giant ionic lattice. * Electrostatic bonds hold lattices together. * Structure is simple cubic. * Some ionic crystals contain water. * Known as water of crystallisation. * These crystals
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Discussion and Conclusion: Preparation of 1-Bromobutane The purpose of this experiment was to demonstrate the conversion of a primary alcohol‚ 1-butanol‚ to a primary bromoalkane‚ 1-bromobutane‚ a SN2 reaction. The conversion of 1-butanol to 1-bromobutane relies on sulfuric acid which plays two important roles. First‚ it protonates the alcohol of 1-butanol to form an oxonium ion which is a good leaving group. Secondly‚ it produces the hydrobromic acid‚ the nucleophile‚ which attacks 1-butanol
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J for ice‚ 2.02 J for steam‚ g°C g°C J for liquid 4.18 g°C 2.05 Specific heat of water C p (water) Metal Specific Heat J g°C Density (g/cm3) Melting Point (°C) Aluminum Copper Gold Iron Lead Magnesium Mercury Nickel Titanium Zinc 0.897 0.385 0.129 0.449 0.129 1.023 0.140 0.444 0.523 0.388 2.702 8.92 19.31 7.86 11.3437 1.74 13.5939 8.90 4.5 7.14 660 1083 1064 1535 328 649 —39 1455 1660 420 NCDPI Reference Tables for Chemistry (October 2006 form A-v1) Page 1
Free Hydrogen Oxygen Chlorine
To carry this out‚ our method combines sodium iodide and common bleach as the oxidizing agent in aqueous alcohol as the solvent. Balanced Chemical Equations: Physical Properties: Name of Chemical Chemical Structure Molar Mass (g/mol) BP/MP (ºC) Density (g/mL) Mass/Vol. Used Purpose 3-methoxy-4-hydroxybenzaldehye C8H8O3 152.1494 MP=81-83‚ BP=285 NA 0.1g Solute Ethanol CH3CH2OH 46.0688 MP=-114.1‚ BP=78.3 0.789 2.0mL Solvent Sodium Iodide NaI 149.8943 MP=661‚ BP=1300 3.667 0.117g Catalyst
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(NaOCl)‚ is determined by redox titration. The method used is iodine-thiosulfate titration. It is a very useful method‚ since the iodide ion‚ I-‚ is easily oxidized by almost any oxidizing agent. The analysis takes place in a series of steps as follows: 1. A diluted sample of the bleach will be allowed to react with potassium iodide in acidic solution. The iodide ion will be oxidized to iodine while the hypochlorite ion will be reduced to chloride (Equation 1). 2 H+(aq) + OCl-(aq) + 2 I-(aq)
Free Oxidizing agent Sodium Iodine
Phenyl Salicylate per parts A‚ B‚ C * 2 g Potassium Iodide per parts A‚ B‚ C * 2 g Sodium Chloride per parts A‚ B‚ C * 2 g Sucrose per parts A‚ B‚ C * 6 Bunsen Burners * 6 Test Tubes with 25 ml Ethanol each * 6 Test Tubes with 25 ml Water each * 6 100 ml beakers with 50 ml Water each * 10 g Calcium Chloride per parts D * 10 g Citric Acid per parts D * 10 g Phenyl Salicylate per parts D * 10 g Potassium Iodide per parts D * 10 g Sodium Chloride per parts D
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Using the Iodine clock method to find the order of a reaction Introduction When peroxodisulfate (VI) ions and iodide ions react together in solution they form sulfate (VI) ions and iodide. This reaction is shown below: S2O82-aq+ 2I-aq SO42-aq+ I2(aq) The reactants and the sulfate (VI) ions are colourless however the Iodine is a yellow/brown colour. This allows you to measure the progress of the reaction through the colour change when the iodine is produced. In order to determine the order of
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