INVESTIGATING THE DIFFERENT RELATIONSHIPS THAT CONCENTRATION AND TEMPERATURE HAVE ON THE RATE OF REACTION Aim To study the effect that temperature and concentration of iodide ion solution have on the rate of iodide ion I⁻ oxidation by peroxodisulphate ion S₂O₈⁻‚ creating an iodine clock reaction. Introduction I decided to choose as the topic for my investigation the rate of reaction for its vital importance in the human body. Indeed I am really interested in Biology and especially physiology and
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For example‚ adults need about 150 micrograms of iodine in the form of iodide ion per day. However‚ an insufficient supply of iodide via diet and drinking water causes the non-cancerous enlargement of the thyroid gland known as goiter. Prolonged lack of iodide can lead to loss of thyroid function and the birth defect known as cretinism‚ which has been recognized since the Middle Ages. Iodide ion‚ in the form of potassium iodide‚ may be added to table salt to produce "iodized salt" in order to easily
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determine the amount of hypochlorite ion present in commercial bleach. Introduction Many commercial products‚ such as bleaches and hair coloring agents‚ contain oxidizing agents. The most common oxidizing agent in bleaches is sodium hypochlorite‚ NaClO (sometimes written NaOCl). Commercial bleaches are made by bubbling chlorine gas into a sodium hydroxide solution. Some of the chlorine is oxidized to the hypochlorite ion‚ ClO-‚ and some is reduced to the chloride ion‚ Cl-. The solution remains strongly
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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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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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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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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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