CO3 + H2SO4 = bubbly reaction‚ little to no colour change. CO3 + 6M HCl = Barrium hydroxide began to go cloudy‚ indicating the presence of BaCO3(Carbonate anions) Chloride Solution + 0.1M AgNO3 =white precipitate formed‚ very fine texture. Iodide solution + 0.1M AgNO3 =yellow/white precipitate formed‚ cloudy texture. Silver Chloride + ammonium hydroxide = white precipitate forms‚ slowly begins to disappear. Adding HNO3 the reappearance of a white precipitate began‚ indicating the presence
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#1 LIMITING REAGENTS INVESTIGATION Aim: To determine the limiting reagent and percent yield of the reaction between potassium iodide with lead (II) nitrate solution. Apparatus required: Safety glasses‚ funnel stands‚ watch glass‚ oven‚ electronic balance‚ wash bottle with distilled water‚ test tubes‚ 10.0mL 0.50M lead (II) nitrate‚ 10.0mL 0.30M of potassium iodide solution‚ two 100.0 mL beakers‚ funnel‚ filter paper. Reaction Involved in Experiment: 2KI + Pb(NO3)2
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Lorraine Chen Chem 106 Lab South Street Seaweed Seaport Warm up: 1. Coffee Beans are crushed into small pieces‚ water is added to it and the mixture is heated over a flame. What do you think would happen to the coffee beans as it interacts with the water? Explain your prediction I think the heating of the water and coffee beans would cause the water to turn a light brown color because this process is most likely removing some kind of excess substance from the surface of the coffee beans. This
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1 Bromthymol Blue‚ 0.04% - 2 mL in Pipet 1 Copper (II) Sulfate‚ 0.2 M – 2 mL in Pipet 1 FDC Blue Dye #-1.0.1% - 2 mL in Pipet 1 Hydrochloric Acid‚ 1.0M-1 mL 1 Lead (II) Nitrate‚ 0.2 M- 2 mL in Pipet 1 PhenolphthaleinSolution1%1mL 1 Potassium Iodide‚ 0.1 M-2 mL in Pipet 1 Silver Nitrate‚ 0.1 N – 2 mL in white Dropper Bottle 1 Sodium Bicarbonate‚ 1 M – 2 mL in Pipet 1 Sodium Hydroxide‚ 1 M – 1 mL 1 Sodium Hypochlorite‚ 1% - 2 mL in Pipet 1 Starch Solution‚ 1% Stabilized - 2 mL in Pipet
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factors affecting the kinetics of reaction between peroxodisulfate (vi) and iodide d. del prado1 and j. belano2 1 department of food science and nutrition‚ college of home economics 2 department of food science and nutrition‚ college of home economics university of the philppines‚ diliman‚ quezon city 1101‚ philippines date submitted: january 7‚ 2013 ------------------------------------------------- ------------------------------------------------- ABSTRACT -------------------------------------------------
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present. To access the degree of the alkyl halide formed (1° or 2° halide) and to determine whether Markovnikov’s addition occurred‚ two qualitative tests can be used: (a) reaction with alcoholic silver nitrate solution and (b) reaction with sodium iodide in acetone. For the silver nitrate test‚ one drop of alkyl halide is added to 2 mL of 0.1 M solution of silver nitrate in
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rule‚ reactivity increases as you move down a group in the periodic table. This means in regards to solubility that the more you farther you move down the group the more insoluble the element is when combined with hydroxides‚ chlorides‚ bromides‚ iodides‚ sulfates‚ carbonates‚ and oxalates. My results were consistent with this theory in that the mixtures went from no reaction to forming a precipitate or from forming a light precipitate to a heavy one as the elements moved down the periodic
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oxidize potassium iodide as follow: Equation: K S 0 (aq) + 2KI (aq)->2K S0 (aq)+I (aq) 2 2 8 2 4 2 KI(aq) +I (aq) -> KI (aq) 2 3 _________________________________________________ K S 0 (aq) +3KI(aq) -> 2K S0 (aq)+ KI (aq) 2 2 8 2 4 2 The rate law of this reaction can be represented as follow: Rate=k[S208 2-]^a [I-]^b When the concentration of peroxydisulphate ions is fixed‚ the order of reaction with respect to iodide ion is formed. Alos‚ when the concentration of iodide ion is fixed‚ the
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Chemical equilibrium is said to be dynamic because‚ at equilibrium‚ there are reactions continually occurring. The rate of the forward reaction equals the rate of the reverse reaction. Equilibrium Constants At equilibrium as much hydrogen iodide is being decomposed as is formed and so the concentrations of all three substances remains constant. Kc is the equilibrium constant in terms of molar concentration. This is known as the Equilibrium Law. The equilibrium constant shows the position
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iodine. The solubility of iodine is increased by complexation with iodide to form triiodide I2 (aq) + I- ºI3- Triiodide then oxidizes vitamin C to dehydroascorbic acid C6H8O6+ I3-+ H2O → C6H6O6+ 3I-+ 2H+(2) Vitamin C dehydroascorbic acid The endpoint is indicated by the reaction of iodine with starch suspension‚ which produces a blue-black product. As long as vitamin C is present‚ the triiodide is quickly converted to iodide ion‚ and no blue-black iodine-starch product is observed. However‚ when
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