"Iodide and persulfate ions" Essays and Research Papers

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    Questions 1. The charge on an iodide ion is -1. It is determined by the formula for potassium iodide in which neither element has subscripts. This signifies that they have equal amount charges and cancel each other out. The charge of the lead ion is +2‚ as shown on the periodic

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    the effect of varying the concentration of iodide ions on the rate of reaction between hydrogen peroxide and an acidified solution of potassium iodide: H2O2(aq) + 2H+(aq) + 2I⁻ → 2H2O(l) + I2(aq) The course of this reaction can be followed by carrying it out in the presence of small quantities of starch and sodium thiosulfate solutions. As the iodine molecules are produced they immediately react with the thiosulfate ions and are converted back to iodide ions: I2(aq) + 2S2O32⁻(aq) → 2I⁻(aq) + S4O62⁻(aq)

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    Development of an Equation

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    1210 Laboratory Manual. Vol. 2013-2014. Plymouth: Hayden-McNeil. 32-35. Data/Results: Part A: In the potassium iodide solution‚ I think there were potassium atoms as well as iodine atoms. In the lead nitrate solution‚ I think there were lead and nitrate ions. The potassium atoms and the lead atoms can be classified as cations‚ since they are metals. The iodine atoms and the nitrate ion can be classified as anions since they are nonmetals. Upon mixing‚ the solution turned into a cloudy yellow color

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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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    The redox arena 1

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    and the importance of the electron in the formation of these new bonds. Materials Boileezers (boiling chips) Desiccant 0.17M acetic acid (acidified water) Solution of Sodium thiosulfate Granular zinc Iodine crystals Zinc ion and iodine-iodide-triiodide ion in water Solid zinc iodide Mineral oil Silver nitrate Magnesium turnings 3M Hydrochloric acid (HCl) solution 2 boiling tubes 1 large tube (we used a small) Pipet Beakers (to hold tubes) Bunsen burner PART 1 Procedure Get a boiling tube and label

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    Titration

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    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) ! Cl-(aq) + I2(aq) + H2O(l)

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    important for plant fertilizers since potassium helps plant growth. The metal halides form white ionic crystalline solids. They are all soluble in water except LiF because of a high lattice enthalpy due to the electrostatic attraction between Li+ and F- ions. All halogens are quite reactive‚ and in the natural world they always occur combined with other elements. Fluorine reacts so readily with almost any substance it contacts that chemists were not successful in isolating pure fluorine until. Chlorine

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    Nucleophilic Substitution

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    halogen atom becomes a halide ion. | | | | Some typical nucleophiles are the hydroxy group (−OH)‚ the alkoxy group (RO−)‚ and the cyanide ion (−C N). Reaction of these nucleophiles with an alkyl halide (R—X) gives the following reactions and products: | | | | The halogen ion that is displaced from the carbon atom is called the leaving group‚ and the overall reaction is called a nucleophilic substitution reaction. Procedure: 1. Sodium Iodide in Acetone. Acetone‚ with

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    What Is The Winkler Method

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    throughout the procedure are outlined below: Mn2+ + 2OH– + 1/2 O2 oxygen-manganese complex + H2O (1) oxygen-manganese complex + 4H+ + 2I– I2 + Mn2+ + 2H2O (2) I2 + 2Na2S2O3 Na2S4O6 + 2NaI (3) Addition of the manganous sulfate and the alkaline-iodide results in the formation of an insoluble oxygen-manganese complex (1)‚ the precipitate in step 2. The oxygen is stable in this form for several days. Both the

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    completely dry. Later the dried substance along with a colored solution ‚ the grey solid leftover‚ white product in H2O‚ standard zinc metal‚ standard iodine solid‚ standard zinc Iodide dissolved in water and standard zinc ion- Iodide ion- Iodine- Triiodide ion in H2O were tested for Iodine(I2)‚ Iodide‚ Triiodide‚ Zinc ion(Zn 2+) and Zinc metal. Then we dissolved some of the white substance with deionized water and made an electrolysis apparatus with a 9.38V battery. We then dried the rest of

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