Flammable Table 6.2Physical Constant: Melting Point Parameter Experimental Theoretical (From Literature) Melting Point (oC) Trial 1 Trial 2 Average 88 88 86 86 Table 6.3Solubility Test Sample Solubility Solubility Class Water Ether 5% NaOH5% Na2CO3 5% HClConc. H2SO4 Unknown Sample + + + + + + Basic Compound Table 6.4Qualitative Tests for Elements Qualitative Tests Sample Treatment Observations Inference Test for Carbon Unknown Liquid Sample CuO + Lime Water White precipitate formation with
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SERANGOON JUNIOR COLLEGE General Certificate of Education Advanced Level Higher 2 CHEMISTRY JC2 Preliminary Examination Paper 1 Multiple Choice Additional Materials: Data Booklet Optical Mark Sheet (OMS) 9647/01 23 August 2011 1 hour READ THESE INSTRUCTIONS FIRST On the separate multiple choice OMS given‚ write your name‚ FIN/NRIC and class in the spaces provided. Shade correctly your class and FIN/NRIC number. Eg. If your NRIC is S9306660Z‚ shade S9306660Z for the item “index number”. There
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The Synthesis of Organic Aspirin ABSTRACT: The purpose of this experiment was to find out how a reaction undergoes for a globally known painkiller called aspirin‚ and to become familiar with achieving successful yields. Aspirin is synthesized from salicylic acid and acetic anhydride. Those two chemicals are mixed together along with sulfuric acid to form a crude solid. Filtration is used separate the impurities from the crude aspirin. To get purified aspirin‚ the precipitate was heated until all
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reaction arrow as on the products side. Details on balancing chemical equations are found in the units on Stoichiometry and Redox Reactions. The general format for writing a chemical equation can be written in a short-hand version as a A + b B + … → c C + d D + … where the lower case letters are the stoichiometric coefficients needed to balance a specific equation. The units on Stoichiometry‚ Redox Reactions‚ and Acid-Base Chemistry contain additional background reading‚ example problems‚ and
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Zn(NO3)2‚ 0.2 M‚ 100 mL* Hydrochloric acid‚ HCl‚ 6 M‚ 125 mL* Nitric acid‚ HNO3‚ 6 M‚ 125 mL* Ammonia‚ NH3‚ 6 M‚ 125 mL* Sulfuric acid‚ H2SO4‚ 6 M‚ 125 mL* Acetic acid‚ CH3COOH‚ 6 M‚ 125 mL* Sodium chloride‚ NaCl‚ 0.2 M‚ 100 mL* Sodium carbonate‚ Na2CO3‚ 0.2 M‚ 100 mL* Sodium hydroxide‚ 6 M‚ 100 mL* Sodium sulfate‚ Na2SO4‚ 0.2 M‚ 100 mL* Sodium nitrate‚ NaNO3‚ 0.2 M‚ 100 mL* Barium chloride‚ BaCl2‚ 0.1 M‚ 100 mL* Barium hydroxide‚ Ba(OH)2‚ 5 g* Potassium thiocynate‚ KSCN‚ 0.1 M‚ 100 mL* Potassium
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11:30-2:20 Abstract: In this experiment‚ phenylmagnesium bromide‚ a Grignard reagent was synthesized from bromobenzene and magnesium strips in a diethyl ether solvent. The Grignard reagent was then converted to triphenylmethanol‚ a tertiary alcohol with HCl. The reaction for phenylmagnesium bromide was: The reaction for Grignard to triphenylmethanol was: In the formation of the Grignard reagent‚ the limited reagent‚ magnesium was determined and 0.00617mol was calculated. In the second part
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enthalpy‚ heat of reaction‚ and the heat of the calorimeter is: Ccal=qcal∆T qrxn=nLR x ∆H The students are the ones responsible in preparing their own solution. Inside the test tube the students put the reagents used for calibration to determine the heat capacity of the calorimeter. These reagents are: sodium hydroxide and HCl. Then the students measure the temperature with 15 seconds intervals. The students made two trials and determine the change in temperature which is needed for the calculations
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CALCULATIONS Determining the amount Limiting Reagent used. nlimiting reagent = Molarity x Volume or Mass / Molar Mass Example: Limiting reagent is 5mL of 1.0 M HCl nlimiting reagent = Molarity x Volume nlimiting reagent = (1.0 [mol/L]) x 0.005 [L]) = 0.005 mol Determining the qrxn and qcal. qrxn + qcal = 0 -qrxn = qcal qrxn = ΔHrxn x nlimiting reagent qcal = Ccal ΔT qrxn = - Ccal ΔT + mcsolid ΔT (note: only if there is a precipitate formed in the reaction)
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The limiting reagent it CuCl2 because it is the one that will run out first compared to the .009 moles of Na2Cl3‚ which it the excess reagent because it is a higher amount of moles compared to the .007 moles of the CuCl2. The amount of excess reagent in grams that should remain in solution if the theoretical yield of CuCO3 is produced is: 1 mole of CuCl2 (63.55) +(2*35.45)
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Number particles = NA x n Mass % Determination of empirical formula and molecular formula Combustion analysis Balancing equations Chapter 4 Limiting reactant Mole to mole conversion from reaction stoichiometry Theoretical yield‚ Percent yield‚ Actual yield Solution concentration (molarity) M = n/V ‚ V always in L M1V1=M2V2 ( dilution calculations) Stoichiometry of reactions in solutions M1V1=M2V2 ( dilution calculations) Ionic reactions (formula unit equation‚ complete ionic and net ionic equation)
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