The Iodine Clock Investigation Introduction This is an investigation into the rate of a reaction and the factors that contribute to how fast a reaction will take place. Through the recording and analysis of raw data‚ this investigation also allows us to apply generally accepted scientific rules and to test them against results gained from accurate experimental procedures. Aim The aim of this experiment is to investigate the rate at which iodine is formed when the concentration
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area. Enzymes are proteins that speed up the rate of chemical reactions‚ which would otherwise progress more slowly.(Background Information; pg. 1) pH is a measurement of the acidity or alkalinity (base) of a solution. When the liver got mixed with H2O2 ‚ the first time the chemical reaction was fast‚ the second time the reaction was slow and the last try was very fast. Temperature is the degree or intensity of heat present in a substance or object. When the temperature of the liver changed from freezing
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enzymes speed up reduction reactions‚ in which oxygen is removed. Many other enzymes catalyse other types of reactions. Individual enzymes are named by adding ’ASE’ to the name of the substrate with which they react. The enzyme that controls urea decomposition is called urease; those that control protein hydrolyses are known as proteinases. However‚ some enzymes‚ such as the proteinases trypsin and pepsin‚ retain the names used before this nomenclature was adopted Enzymes are large proteins that speed
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100 units/ml catalase was put in to each of the three water baths‚ one at a time. While the tubes sat in the baths‚ the 4 H2O2 vials were prepared and the depth of the H2O2 was recorded. After the 0 degree celsius test tube incubated in it’s bath‚ it was taken out and a filter was dropped into it and drained on a paper towel. The filter was then dropped into one of the 3% H2O2 vials. The distance traveled and the time was recorded in the data chart. This process was repeated for each of the test tubes
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challenged with A. solani‚ respectively. The 2‚4-D was the most effective treatment followed by H2O2 and ABA. Data also indicated that the spray with the fungicide (difenoconazole as a single application) with elicitors slightly decreased the disease severity compared with the elicitors alone. PAL gene expression was significantly and rapidly up-regulated after treating potato plants with 2‚4-D followed by H2O2 with or without the fungicide compared to ABA treatment‚ very obvious effects in the heat map
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Decolorization and Chemical Oxygen Demand Reduction (COD) of Simulated Textile Wastewater using Fenton’s Reagent Submitted to: Eric Siy COE 5100 – Statistical Research and Design Chemical Engineering Department College of Engineering De La Salle University – Taft‚ Manila by MARIA KATRINA A. PULUTAN MS Chemical Engineering 1st Trimester AY 2010-2011 1. INTRODUCTION Nature is threatened by the environmental contamination caused by the wastewater produced and discharged every day
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People who get exposed to H2O2 increase the substrate concentration that enters the body and therefore run the risk of Hydrogen Peroxide poisoning. Scientists can use the research done in this experiment to help catalase work its best and prevent H2O2 poisoning. Catalase can also potentially be used to help cure H2O2 poisoning. If scientists know where Catalase work best (ph-7.2‚ temperature- 37°C ‚ 1:5 ratio from Enzyme
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In the lab‚ the purpose was to standardize a KMnO4 solution than using it to evaluate how close a H2O2’s concentration was to its labeled concentration. In this lab‚ it was decided to compare the concentration of new Publix brand hydrogen peroxide to an old sample of the same solution. From the data collected in the first part by titrating a solution of FeSO4*7H2O with the KMnO4 solution‚ it was determined that the concentration was .028M MnO4-. This was able to be done because a known amount of
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Making up Hydrogen Peroxide Volume required 250 cm3 and concentration required 0.1 moldm-3 Given concentration of H2O2 = 1.7 moldm¬-3 Number of moles (n¬¬¬¬¬¬¬¬¬¬¬) = Concentration (moldm-3) x Volume (dm-3) = 0.1 x 0.25 = 0.025 mol Volume (dm-3) = Number of Moles (n) X 1000 Concentration (moldm-3) = (0.025/1.7) x 1000 = 14.7 cm3 Distilled water required: 250 cm3 – 14.7
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takes 20 minutes for the reaction to drop from 1.0 M to 0.60 M. The time required for the concentration to drop from 0.60 M to 0.36 M will be (a) more than 20 minutes (b) 20 minutes (c) less than 20 minutes (d) none of these For a certain decomposition‚ the rate is 0.30 M sec–1 when the concentration of reactant is 0.20 M. If the reaction is second order‚ the rate (in M sec–1) when concentration is increased 3-fold is (a) 0.30 (b) 0.90 (c) 0.60 (d) 2.70 5. A 10º rise in temperature
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