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    Polyvinylchloride Case Study

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    effect of unmodified and titanate treated nano-CaCO3 on the microstructure as well as on the mechanical‚ barrier‚ optical properties of PVC films‚ prepared by solution casting method were studied. The tensile strength and barrier properties of nano-CaCO3/PVC increase with increasing loading of CaCO3 nanoparticles due to increased interfacial contact area and enhanced interfacial adhesion between CaCO3 nanoparticles and PVC matrix. Titanate treated nano-CaCO3/PVC composites had superior tensile and barrier

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    Stoichiometry Lab

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    Stoichiometry Lab Name Questions A. From your balanced equation‚ what is the theoretical yield of your product? Theoretical yield of the CaCO3 is expected to be .69g. B. According to your data table‚ what is the actual yield of the product? The mass of the filter paper was 1.1g‚ and the total mass of the filter paper when dried with the CaCO3 was 1.8 total. Thus the actual yield of the product was .70g. C. What is the percent yield? Percentage yield is actual yield over the theoretical

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    Introduction: In this lab a total of six titrations are to be performed. Three of them will be done using a known Ca2+ solution‚ (1.000 g CaCO3 /L solution) and three of them will be done using an unknown solution obtained from the stock room. The objective of this lab is to determine the hardness of water‚ using the data collected from each titration performed with the unknown sample. Since the hardness of water arises from the presence of metal ions‚ we can use disodium salt of EDTA and the

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    Observations: CaCl2.H2O = m/M = 1/147 = 0.0068 mol CaCO3 = 0.0068*1/1 = 0.0068 mol CaCO3 = ? CaCO3 = CaCO3 mol *CaCO3 g =0.0068 mol*100.01 g =0.68 g Theoretical Yield: .0068 mol of CaCO3 *100.06 g CaCO3/1mol of CaCO3 =0.6804 g of CaCO3 Percentage Yield: = (0.8/0.68)*100 =117% Na2CO3(aq) + CaCl2. 2H2O(aq) → CaCO3(s) + 2NaCl(aq) + 2H2O Questions: A. Theoretical yield: 0.6804 g of CaCO3 B. Actual yield: 0.8g CaCO3 C. Percent yield: 117% D. My percentage yield was far more

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    ABSTRACT The average concentration of CaCO3 obtained was 212 ppm‚ with a standard deviation of 1 ppt. The results indicate that the unk B tap water can be considered as hard water. INTRODUCTION Hard water is due to metal ions (minerals) that are dissolved in the ground water. These minerals include Ca 2+‚ Mg2+‚ Fe3+‚ SO42-‚ HCO3-. When this water evaporates or boils‚ the difficult to dissolve metal salts remain as a scaly residue. Hard water inhibits the effectiveness of soap and detergents

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    Chemistry of Natural Waters

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    Very hard 180 + mg/l It is important to study the hardness of water because of the problems it can cause when left untreated. Hard water can cause “scale‚” or build-up in plumbing fixtures and water heaters.3 Scale occurs when calcium carbonate‚ CaCO3 ‚ is deposited as calcite crystals on the inner surfaces of pipes‚ and evaporator surfaces. This occurs when water with a high hardness value is evaporated or heated. This buildup can block pipes‚ make heat transfers in boilers inefficient‚ and

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    James|Date:3/10/13| Exp 9: Stoichiometry of a Precipitation Reaction|Lab Section: 73426| Data Tables: Step 3: Show the calculation of the needed amount of Na2CO3 CaCl2.H2O(aq)= m/M =1/147 =0.0068 mol CaCO3(s)=0.0068*1/1 =0.0068 mol CaCO3(s)= CaCO3 (s)= CaCO3 mol *CaCO3 g =0.0068 mol*100.01 g =.68 g Step 4: Mass of weighing dish _0.6_g Mass of weighing dish and Na2CO3 .72_g Net mass of the Na2CO3 .12_g Step 6: Mass of filter paper __1.0__g

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    were streaked on slants and incubated at 37 C. Change in color was visible at 5 day of incubation. Sealing of Concrete: Cultures precipitating CaCO3‚ were inoculated in LB and incubated for overnight at 37C. One stock of CaCO3 precipitating bacteria were inoculated in precipitation media also‚ to get the CaCO3 precipitation. Bacteria start precipitating CaCO3 after 1-2 days‚ at 37C. Media having cultures were pelleted down and cultures with media were injected between the concrete and kept at 37C for

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    hardening ions in the water sample. 1 mol of Na2H2Y = 1 mol of CaCO3 If mol of Na2H2Y = 0.0109 M x 0.01633 L = 1.78 x 10-4 mol then there are 1.78 x 10-4 mol of CaCO3 b. Express the hardness concentration in mg CaCO3 / L sample. (1.78 x 10-4mol x 100.1 x 1000) / (50ml / 1000) = 17.8 / 0.5  356 mg CaCO3 / L c. What is the hardness concentration express in ppm CaCO3? The ratio is 1mg CaCO3 / L = 1 ppm  356.3 ppm of CaCO3 d. Classify the hardness of this water according to table 9

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    Danielle Hall|Date: 10.15.2012| Exp 9: Stoichiometry of a Precipitation Reaction|Lab Section: | Data Tables: Step 3: Show the calculation of the needed amount of Na2CO3 CaCl2.H2O(aq)= m/M =1/147 =0.0068 mol CaCO3(s)=0.0068*1/1 =0.0068 mol CaCO3(s)= CaCO3 (s)= CaCO3 mol *CaCO3 g =0.0068 mol*100.01 g =.68 g Step 4: Mass of weighing dish _0.6___g Mass of weighing dish and Na2CO3 __0.72__g Net mass of the Na2CO3 ___0.12_g Step 6: Mass of filter paper __1.0__g

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