barium sulphate‚ BaSO4‚ by gravimetric analysis. For the experiment‚ a dilute solution of barium chloride was slowly added to a hot unknown sulphate solution slightly acidified with concentrated HCl. The precipitate is then filtered off by suction filtration‚ washed with water and dried in the oven at 150 degree Celsius for about half an hour and cooled in the desiccator‚ and weighed as barium sulphate and the percentage of sulphate was calculated from the weight of barium sulphate. The Molarity of
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the mixture flask. This will stop the water escaping as it touches the cold surface of the condenser. The condenser has a permanent water flow going in and out and this is why the water drip back to the flask as it touches the cold surface. Vacuum filtration was used isolate the solid from the liquid-solid mixture after the sulphuric acid is qqadded and that the precipitation has occurred and that the pH is down to 2. Recrystallization was then used to purify the salicylic acid formed earlier. Boiling
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Filter Innovations Inc. Filter Innovations Inc. (FII) was created by John Dragasevich in 1992. It was based out of Toronto‚ Ontario and supplied small industrial water filtration systems and environmental products. It focused mostly on projects that required processing below 1 million litres per day. This was because competing companies were more interested in projects that required processing above 1 million litres per day. Problem Statement The problem Dragasevich faces with FII is that
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Glucansucrase to build long‚ sticky chains of sugars called glucans. These glucans glue the bacteria to the surfaces of the teeth and form a biofilm which is difficult to remove2. Glucansucrase has two reactions2. Firstly‚ the enzyme breaks sucrose into glucose and fructose. The fructose will be released‚ and the glucose will go to glucan chain. Glucansucrases are large proteins with many domains. These reactions are performed in the central portion of the enzyme‚ which is included in this structure
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total of 76 points but will be converted to a percentage grade when registered in your “Gradebook”. |ACTIVITY 1: The Effect of Arteriole Radius on Glomerular Filtration |Answers | |According to your lab manual‚ in humans the glomerular filtration rate (GFR) ranges from _____ to _____ |80 to 140 | |ml/minute.
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generate the pressure‚ because it does not diffuse. ACTIVITY 5: Filtration What were the results of your initial membrane analysis? Solute Residue for all solutes (Na/Cl‚ Urea‚ Glucose‚ Powdered charcoal) were present. Does the membrane MWCO affect filtration rate? Yes‚ the larger the MWCO the faster the filtration Does the amount of pressure applied affect the filtration rate? Yes‚ the greater the pressure‚ the faster the filtration Did all
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Activity 1: Simple diffusion Introduction: Simple diffusion is the net movement of substances from a region of high concentration to a region of low concentration so its overall net movement is along the concentration gradient‚ simple diffusion does not require energy therefore it is ’passive’‚ substances are diffused across the membrane between the phospholipids. Materials and methods: * 20 mwco dialysis membrane * 50 mwco dialysis membrane * 100 mwco dialysis membrane * 200 mwco
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Differences Between Ionic and Covalent Compounds Abstract In chemistry‚ a compound is a substance formed from two or more elements‚ with a fixed ratio determining the composition. A defining characteristic of a compound is that it has a chemical formula. Formulas describe the ratio of numbers of atoms in a substance. Compounds may have a number of possible phases. For a compound to be a liquid or a gas and still be called a compound‚ atoms from the various elements must be stuck together in
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Mole Fraction Return to Solutions Menu The mole fraction is: moles of target substance divided by total moles involved The symbol for the mole fraction is the lower-case Greek letter chi‚ χ. You will often see it with a subscript: χsolute is an example. Example #1: 0.100 mole of NaCl is dissolved into 100.0 grams of pure H2O. What is the mole fraction of NaCl? Solution: 100.0 g / 18.0 g mol¯1 = 5.56 mol of H2O Add that to the 0.100 mol of NaCl = 5.56 + 0.100 = 5.66 mol total Mole fraction
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NO I/C : 940129085210 LECTURER’S NAME : DR MOHD NOR BIN OTHMAN CONTENT INTRODUCTION……2 BACKGROUND……3 WATER TRAETMENT PROCESS 1) Coagulation process……4 2) Flocculation process……4-5 3) Sedimentation process……5 4) Filtration process……5 5) Disinfection process……6 CONCLUSION……6 INTRODUCTION A site visit to Sg Johor Water Treatment Plant was held on 20th April 2015 for water treatment class ‚ SKAA 2912. A total of approximately __ students take part in the site
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