Organic Chemistry Laboratory Report. Experiment 1: Crystallization. Objectives: 1. To study the crystallization process. 2. To identify the best suitable solvent to use for the crystallization process. 3. Gain an experience in purifying an organic compound by the techniques of the crystallization. Introduction. Crystallization is a technique which chemists use to purify solid compounds. It is one of the fundamental procedures each chemist
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EXPERIMENT 1 IDENTIFICATION OF UNKNOWN COMPOUNDS Date of experiment : 7th August 2012 Venue: ME204 INTRODUCTION Chemical reactions are regularly categorized into 3 types: oxidation-reduction (redox reaction)‚ precipitation (double displacement) and acid-base reaction (double displacement) Type 1: Oxidation-Reduction Reactions Oxidation-reduction processes include the movements of electrons form oxidants to reductants‚ which lead to increases in oxidation
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Progress in Organic Coatings 50 (2004) 75–104 Review Antifouling technology—past‚ present and future steps towards efficient and environmentally friendly antifouling coatings Diego Meseguer Yebra‚ Søren Kiil∗ ‚ Kim Dam-Johansen Department of Chemical Engineering‚ Technical University of Denmark‚ Building 229‚ DK-2800 Kgs. Lyngby‚ Denmark Received 28 January 2003; accepted 15 June 2003 Abstract The imminent ban of environmentally harmful tributyltin (TBT)-based paint products has been the
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EXPERIMENT 5- A FISCHER ESTERIFICATION SYNTHESIS OF PEAR OIL One of the major uses of Esterification (C(=O)-C-R) is forming compounds with different odors which can be used to make different synthetic and natural flavorings. Using the chemical and physical properties of organic compounds‚ an ester was catalyzed from the reaction of a carboxylic acid with an alcohol‚ producing an odor similar to that of pear oil. The mechanism of this Fischer- Esterification process is outlined as follows:
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Four elements‚ hydrogen‚ carbon‚ oxygen and nitrogen‚ are the major components of most organic compounds. Consequently‚ our understanding of organic chemistry must have‚ as a foundation‚ an appreciation of the electronic structure and properties of these elements. The truncated periodic table shown above provides the orbital electronic structure for the first eighteen elements (hydrogen through argon). According to the Aufbau principle‚ the electrons of an atom occupy quantum levels or orbitals starting
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primaryrx.com/pdf/C-PHEN%20DM%20DROPS.pdf 1 Chapter 15: Benzene and Aromaticity Introduction: • All organic compounds can be derived into two broad classes 1. Aliphatic compounds Nonaromatic hydrocarbons such as alkanes‚ alkenes or alkynes 2. Aromatic compounds A series of cyclic unsaturated compounds with unusually high stability • The properties and reactions of aromatic compounds are exemplified / demonstrated by the properties and reactions of benzene (the “parent” aromatic molecule)
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Amanda Jornd Experiment 7- Synthesis and Reactivity of tert-Butyl Chloride Via an SN1 Reaction Introduction/Background: Alkyl halides are compounds in which a halogen atom replaces a hydrogen atom of an alkane. Alkyl halides are classified as primary‚ secondary or tertiary depending on the number of alkyl substituents directly attached to the carbon attached to the halogen atom. The purpose of this lab was to properly prepare t-butyl chloride from t-butyl-alcohol in a concentrated hydrochloric
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Question Mark 1.Expand the following compound nouns:1.Weather report.2.Hairpin bend.3.Rubber pad.4.Fast food Mr.JA/ENG 2 2.Write a paragraph comparing and contrasting mobile phone and landline. Mr.JA/ENG 16 Cause & Effect Expression 1.1. A nail has pierced the Tyre. The essential commodities have become costlier. 2. Safety precautions were not observed. There were many accidents in the factory. 3. Many prefer private vehicles for commutations. It is over crowding in public transport. 4
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commencing at age 60 and ending at age 67 Plan 1: V0 * (1.06)42 = 3‚000‚000 V0 = 3‚000‚000/(1.06)42 V0 = $259‚582.20 Plan 2: V0 * (1.06)17 = 3‚000‚000 V0 = 3‚000‚000/(1.06)17 V0 = $1‚114‚093.26 Plan 3: C * Annuity Compound Factor (6%‚ 37) = 3‚000‚000 C * [((1.06)37 – 1)/0.06] = 3‚000‚000 C *127.27 = 3‚000‚000 C = $23‚572.28 Plan 4: C * Annuity Compounding Factor (6%‚17) = 3‚000‚000 C * 28.21 = 3‚000‚000 C = $106‚334.41 Plan 5: C * Annuity
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LU 1 - Tutorial 1. Write the formulas for the following compounds: Compound Copper (II) chloride Iron(II) sulphate Sodium sulphate Ammonium nitrate Sodium nitrite Potassium sulphate Magnesium phosphate Formula Compound Lithium acetate Manganese (IV) nitrate Potassium permanganate Iron (III) oxide Potassium sulphite Potassium hydrogen carbonate Sodium iodate Formula 2. Express the following numbers in scientific notation: a) 0.000000027 b) 356 c) 47 764 d) 0.096 3. Express the following numbers
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