CH 16 - the name of this functional group class is historically derived from the pleasant aromas associated with many of its constituent compounds discovered in past centuries - although not all aromatic compounds known today share the same distinction (many are in fact‚ odorless or even unpleasant)‚ they are the principle components in such pleasant smelling substances as cinnamon‚ licorice‚ wintergreen‚ clove & vanilla CLASSIFICATION – - also known as arenes‚ many aromatic compounds possess
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functions of society C. Biology is a multidisciplinary adventure II. III. OBJECTIVES After reading this chapter and attending lecture‚ the student should be able to: 1. Briefly describe unifying themes that pervade the science of biology. 2. Diagram the hierarchy of structural levels in biology. 3. Explain how the properties of life emerge from complex organization. 4. Describe seven emergent properties associated with life. 5. Distinguish between holism and reductionism. 6. Explain how technological
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6/6/6 CHY2F 2 Answer all questions in the spaces provided. Do not write outside the box 1 The diagram shows how the atoms are bonded in ammonia. H N H H 1 (a) 1 (a) (i) Draw a ring around the correct answer. The formula of an ammonia molecule is NH3 N3H NH3 (1 mark) 1 (a) (ii) The particles represented by the dots () and crosses (x) in the diagram are electrons neutrons protons (1 mark) 1 (a) (iii) The bonds between the nitrogen and hydrogen atoms in ammonia
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Some of the materials are red litmus paper‚ blue litmus paper‚ hydrion paper‚ pipets‚ and 12 well spot plates. The other materials that are chemicals or solutions are red cabbage juice‚ drain cleaner (NaOH)‚ detergent‚ baking soda (NaHCO3)‚ ammonia (NH3)‚ soft drink‚ distilled water‚ vinegar (CH3CO2H)‚ and lemon juice. For Part A the first step is on a paper towel lay-out seven pieces of red litmus paper‚ seven pieces of blue litmus paper‚ and seven pieces of hydrion paper. Second step is use the
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English Proficiency This test is designed to assess the test takers’ general proficiency in the use of English language as a means of self-expression in real life situations and specifically to test the test takers’ knowledge of basic grammar‚ their vocabulary‚ their ability to read fast and comprehend‚ and also their ability to apply the elements of effective writing. 1. Grammar 1. Agreement‚ Time and Tense‚ Parallel construction‚ Relative pronouns 2. Determiners‚ Prepositions‚ Modals
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|Chemistry Review 1 |TEACHER ANSWER KEY | | |June 06‚ 2011 | | | |’see explanation below’
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Chapter 1 1. Which is the most electronegative atom in the compound below? OH Br N H A) Carbon Ans: C B) Nitrogen C) Oxygen D) Bromine 2. Which of the following correctly describes the electrons of a carbon atom in its ground state? A) 3 s electrons; 3 p electrons B) 2 1s electrons; 4 2p electrons C) 2 1s electrons; 2 2s electrons; 2 2p electrons D) 2 1s electrons; 2 2s electrons; 4 2p electrons E) None of these choices is correct. Ans: C 3. Which of the following statements correctly
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Representative Elements I. Elements A. Representative Elements: Groups 1A through 8A ← Filling s and p orbitals ← First member of a group forms the strongest ( bonds; results in diatomic molecules B. Transition Elements: center of the table ← Filling d orbitals C. Lanthanides & Actinides: listed separately‚ on the bottom on the table ← Filling 4f and 5f orbitals II. Metals/Nonmetals/Metalloids A. Metals: ← Lose valence electrons to form cations
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The Grignard Synthesis of Triphenylmethanol Abstract: Grignard synthesis of triphenylmethanol was achieved by use of the Grignard reagent phenyl magesium brominde. The organometallic grignard reagent was synthesized by use of a reflux apparatus recrystallization techniques. Once synthesized it was used in a Grignard reaction that involved nucleophilic addition to a carbonyl in order to make triphenylmethanol. The final product was solidified and recrystallized and spectral data was
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CHEM 3411‚ Fall 2010 Solution Set 5 In this solution set‚ an underline is used to show the last significant digit of numbers. For instance in x = 2.51693 the 2‚5‚1‚ and 6 are all significant. Digits to the right of the underlined digit‚ the 9 & 3 in the example‚ are not significant and would be rounded off at the end of calculations. Carrying these extra digits for intermediate values in calculations reduces rounding errors and ensures we get the same answer regardless of the order of arithmetic
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