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    Acid Base Questions

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    ACID/BASE You might need to know the following K values: CH3COOH Ka = 1.8 x 10–5 Benzoic Acid Ka = 6.5 x 10–5 HNO2 Ka = 4.5 x 10–4 NH3 Kb = 1.8 x 10–5 HF Ka = 7.2 x 10–4 H2S Ka = 5.7 x 10–8 HSO4– Ka = 1.2 x 10–2 HS– Ka = 1.2 x 10–13 HCOOH Ka = 1.8 x 10–4 HOCl Ka = 3.0 x 10–8 SIMPLE ACIDS AND BASES 1. According to the Brønsted–Lowry definition‚ which species can function both as an acid and as a base? (A) Cl– (B) SO42– (C) NH4+ (D) HCO3– (E) H3O+ 2. Which of the following

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    Biology UNIT 5 Essay Plans

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    disease‚ describe how bacteria may affect the lives of humans and other organisms. 19. Inorganic ions include those of sodium‚ phosphorus and hydrogen. Describe how these and other inorganic ions are used in living organisms. 20. Condensation and hydrolysis and their importance in biology. 21. Negative feedback and its importance in biology. 22. Ways in which different species of organisms differ from each other 23. The transfer of energy between different organisms and between these organisms and

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    The Five Major Compounds That Make Up the Human Body The human body is one of the most complex and fascinating things on this planet. There are five major groups of compounds that compose the human body. These are carbohydrates‚ lipids‚ proteins‚ nucleotides‚ and water. These compounds are all very important to humans and without them humans would not be able to survive. Compounds have many functions that encourage a human cell and a human body to function. Compounds are pure substances

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    Exam 1 Part 2 Study Guide

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    Human Physiology Exam 1 Part 2 Study Guide Chapter 1 Define homeostasis * Maintenance of constant conditions in the internal environment. Composition‚ temperature and volume of extracellular fluid do not change significantly under normal conditions Know the components of a homeostatic control mechanism * Regulated Variable- Regulated to stay within relatively narrow limits * Set Point- normal desire value * Error Signal- difference between the actual value and the set point

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    Biochemistry

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    SBI4U – CHAPTER ONE TEST – PART A – Multiple Choice 1. Which of the following is the essential characteristic of a polar molecule? a) contains double or triple bonds b) is formed at extremely low temperatures c) contains ions as part of the structure d) has an asymmetrical distribution of electrical charge e) contains the element oxygen 2. Isomers are molecules that a) react readily with one another b) have the same molecular formula c) have different molecular masses d) differ

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    The Biological Importance of Water for Living Organisms Water is made up of two elements‚ 2 positively charged hydrogen molecules and one negatively charged oxygen molecule. Water molecules have uneven charge distribution as one end of the molecule is slightly positive and the other slightly negative‚ this is called polar. Ionic substances such as sodium chloride dissolve easily in water because the positively and negatively charged ions are separated due to the dipole nature of water. As water

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    Esterification

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    calculate the percentage yield of substance by esterification. Introduction: When a carboxylic acid reacts with an alcohol‚ the products are water and ester. The overall reaction is reversible and slow‚ as soon as the products begin to form‚ hydrolysis a reverse reaction of esterification occurs. To ensure the reaction is complete an acid catalyst is commonly used. The catalyst is usually concentrated sulphuric acid. This experiment covered 4 techniques; they are distillation‚ refluxing‚ extraction

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    Melting Point

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    Melting Point Determination Identity and Purity of Solid Organic Compounds Objectives • To introduce the technique of melting point determination. • To use the concept of melting points for identification and characterization of organic compounds. • Properly fill and use a capillary melting point tube. • Determine accurate melting point ranges for a wide variety of organic substances. Introduction The melting point of a solid can easily and accurately be determined

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    Robinson Annulation 2

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    step‚ Micheal addition takes place to an α‚β-unsaturated ketone. Enolate is formed and tautomerization takes place for further reaction. Cyclization takes place. That is‚ aldol condensation reaction to form six-membered cyclic product. Further hydrolysis results in the formation of α‚β-unsaturated cyclic ketones For example 1‚ consider the following reaction : Example 1 Mechanism of the Robinson Annulation The first step in the process is the Michael Addition to an α‚β-unsaturated ketone

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    amul

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    These are usually sugars‚ but it is possible to buy sugar-free sweets‚ such as sugar-free peppermints. Most common is the disaccharide sucrose. Hydrolysis of sucrose gives a mixture called invert sugar‚ which is sweeter and is also a common ingredient. Finally confectioneries‚ especially commercial ones‚ are sweetened by a variety of syrups obtained by hydrolysis of starch‚ these include corn syrup.[5] (CADBURY) PACKING Cadbury SA’s chocolate products’ marketing seems to be an ongoing SA case study

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