Chemistry 261 Practical Report Practical 2 Thermos Thurman 211526893 PRACTICAL 2: Preparation of sodium hexanitrocobaltate (III) Date: 12 March 2014 INTRODUCTION Sodium hexanitrocobaltate (III) is a coordination complex which has the formula Na3[Co(NO2)6]. This compound is yellow in colour. It consists of a central Co3+ ion surrounded by six nitro ligands [1]. This compound is used as a qualitative test for potassium and ammonium ions (so long as certain
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809g/1mL)(1mol/88.15g) = 0.02294mol Volume of glacial acetic acid (d=1.06g/mL; MW=60g/mol): 3.5mL Moles of glacial acetic acid: (3.5mL)(1.06g/1mL)(1mol/60g) = 0.06183mol Therefore‚ the acetic acid is in excess‚ which makes the isopentyl alcohol the limiting reagent. The theoretical yield is: Molecular weight of isopentyl acetate is 130.19g (0.02294mol isopentyl alcohol)(1 mol isopentyl acetate/1mol isopentyl alcohol)(130.19g/1mol)=2.9866g The observed yield is: Weight of 3mL conical vial:
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WJEC CBAC AS/A LEVEL GCE in Chemistry REVISION AID UNIT 1 AS UNIT CH1 – Controlling and Using Chemical Changes (in order to make things‚ produce energy and solve environmental problems) Preamble This unit begins with some important fundamental ideas about atoms and the use of the mole concept in calculations. Three key principles governing chemical change are then studied‚ viz. the position of equilibrium between reactants and products‚ the energy changes associated with a
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Lab Protocol #8 - Dehydration of 3‚3-dimethyl-2-butanol KEYWORDS: alkenes‚ E2‚ E1‚ carbocation stability‚ elimination A. Introduction. The dehydration of alcohols is an ELIMINATION reaction that is commonly used to form alkene molecules. The mechanism for the dehydration of alcohols consists of transforming the –OH functional group into a better leaving group by using a strong acid to protonate the alcohol. After –OH has been transformed into –OH2+‚ it becomes easier to break the σ-bond between the
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UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level *0281936391* CHEMISTRY Paper 5 Planning‚ Analysis and Evaluation Candidates answer on the Question Paper. No Additional Materials are required. READ THESE INSTRUCTIONS FIRST Write your Centre number‚ candidate number and name on all the work you hand in. Write in dark blue or black pen. You may use a soft pencil for any diagrams‚ graphs or rough working. Do not use staples‚ paper clips‚ highlighters
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Page 1 of 10 5.111 Principles of Chemical Science EXAM # 3 ============================================================================ Write legibly your name and your TA’s name below. Do not open the exam until the start of the exam is announced. The exam is closed notes and closed book. You have 50 minutes (1 academic hour) to complete it. ● Read each part of each problem carefully. ● Write your answers legibly in the corresponding spaces of the attached sheets. ● For problems requiring
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prepared varying concentration of SCN- and constant concentration of H+ and Fe+ that are in stoichiometric excess. The excess of H+ ions will ensure that Fe3+ has no engage in the reaction and excess of Fe3+ ions will make the SCN- ion which is a limiting reagent. In part ll the solution of unknown [SCN-] was analyze using the same procedure as in part
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9.5 – Industrial Chemistry (Option): Δ. Construct word and balanced formulae equations of all chemical reactions as they are encountered in this module: The EQUILIBRIUM Constant: For the basic reversible reaction‚ depicted as: aA + bB cC + dD; Uses of SULFURIC ACID: Fertiliser Production: Creating ammonium sulfate: sulfuric acid + ammonia ammonium sulfate H2SO4 (aq) + 2NH3 (aq) (NH4)2SO4 (aq) Dehydrating Agent: Dehydration of ethanol: ethanol ethylene + water C2H5OH (l) C2H4 (g) + H2O (l) Dehydration
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Results: Limiting Reactant: Eq 1 Limiting reactant = Benzoin Theoretical yield of Benzil: Eq 2 Theoretical Yield Benzil | 0.296 g | Mass of Crude Benzil | 0.188 g | Mass of Final Benzil | 0.127 g | % Yield | 43% | % Recovery | 66% | Table 1: Mass of crude/final Benzil‚ % yield‚ and % recovery Percent Yield: % Yield = (Final product/Theoretical product) x 100 Eq 3 = (0.127 g/0.296 g) x 100 = 43% yield Percent Recovery
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Lab 8: Isopentyl Acetate CHM 2211L October 22‚ 2014 Abstract Many esters are naturally occurring compounds that are responsible for the fragrance of fruits and flowers. Isopentyl Acetate is the compound responsible for the smell of bananas. However‚ when Isopentyl acetate is treated with H3O+ it will hydrolyze back into the carboxylic acid‚ which will change the fragrance. In this experiment 3.11g of Isopentyl Acetate were produced with a 51.92 percent yield. Introduction Esters are carboxylic
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