Chemistry Gen: Course Description Year 2 PART - II CGT 21a Unit I. Basic physical chemistry I * Gaseous state: Gas laws‚ kinetic theory of gas‚ collision and gas pressure derivation of gas laws from kinetic theory‚ average kinetic energy of translation. Boltzmann constant and absolute scale of temperature‚ Maxwell’s distribution law of molecular speeds (without derivation)‚ most probable‚ average and root mean square speed of gas molecules‚ principle of equipartition of energy (without
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previously made‚ the absorbency of each concentration solution will help find the pennies % of copper in your post-1982 penny. Equation for the reactions between the post-1982 penny and nitric acid: Blue color Cu(s) + HNO3(aq) → Cu(NO3)2(aq) + NO2(g) + 2H2O(l) colorless Zn(s) + HNO3(aq) →
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NAME: EDWARD AFUTU INDEX NO: 8030212 EXPERIMENT: I.2.2.2. DEMONSTRATOR: MISS NAOMI KABIRI
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The intent of this paper is to educate individuals about acid rain‚ define its chemical makeup‚ and describe its impact on the environment. Rain is an important part of all life; it is the source of water for almost everything. In some areas rain is actually a threat to life because of pollution from cars‚ factories‚ and power plants. This pollution releases gases into the atmosphere to create acid rain. Acid rain is rain‚ or any other form of precipitation that is uncharacteristically acidic
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Trinitrotoluene‚ TNT‚ or 2‚4‚6-trinitrotoluene‚ is a nitro chemical compound with the formula C₆H₂(NO₂)₃CH₃. TNT is a yellow-colored solid and contains in its structure three NO2 groups and one –CH3 group all bonded to carbon atoms on the aromatic ring (Juhasz and Naidu 2007‚ Pichtel 2012). Oxygen and nitrogen are the two highly electronegative elements where oxygen possesses the highest electro-negativity. Subsequently‚ the polarized N–O bonds within the nitro groups makes them easily reducible
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AROMATIC 1. Draw structures corresponding to the names given. a. m-fluoronitrobenzene b. p o-chlorophenol c. o-chlorophenol d. 3‚5-dimethylbenzoic acid e. 1-phenyl-3-methylpentane 2. Provide correct IUPAC names for each of the following compounds. a. NO2 CH3 b. c. CH3 d. NH2 H2N CH3 e. Ph C H3 C C CH2CH2CH3 Ph 3. For each molecule below‚ predict whether the molecule would be expected to show aromatic character or not. Explain your answer in each case. a. O N N H
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| 2 | Action of heat:The given salt is heated in a dry test tube. | (i)Reddish brown vapours(ii)Crackling sound(iii)Yellow when salt and white when cold(iv)Colourless gas giving pungent smelling gas(v) No characteristic change | Presence of NO3-‚ NO2-Presence of nitrates of lead and bariumPresence of Zn2+Presence of NH4+Absence of nitrate‚ nitrite‚ zinc and ammonium ions | 3 | Solubility: a) In water b) In dil. HCl | SolubleSoluble in dil. HCl but insoluble in water | Water soluble saltsPresence
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Chemistry 12: Reaction Kinetics Review 1. Looking at the expressions for reaction rate‚ write expressions with which you could express rates for the following reactions. (Hint: look at what happens to reactants and products.) Recall that solid or liquids can lose or gain mass‚ gases can lose or gain volume and aqueous solutions can increase or decrease in concentration. ("a" is done as an example.) a) Mg(s) + 2HCl(aq) H2(g) + MgCl2(aq) reaction rate = mass of Mg consumed unit time b) c) or
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UNIVERSITY TUNKU ABDUL RAHMAN FACULTY OF SCIENCE BACHELOR OF SCIENCE(HONS)BIOCHEMISTRY YEAR 1 SEMESTER 1 UDEC1164 ORGANIC CHEMISTRY Name: Cheong Wai See Student ID: 1404057 Practical Group: 1 Title of Experiment: Name of Tutor: Dr. Mohammod Aminuzzaman Date: 24 February 2016 Tiltle: Nitration of Methyl Benzoate Objectives: -To prepare methyl-3-nitrobenzoate from nitration of methyl benzoate by electropilic aromatic substitution. -To calculate the percentage yield and get the melting point of methyl-3-nitrobenzoate
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Chapter 2 22. A student heats 0.5585 g of iron with 0.3550 g of sulfur. She reports that she obtains 0.8792 g of iron sulfide and recovers 0.0433 g of unreacted sulfur. Show by calculation whether or not her results obey the law of conservation of mass. Total mass initial = 0.5585 g + 0.3550 g = 0.9135 g Total mass final = 0.8792 g + 0.0433 g = 0.9225 g These two values should be equal by the law of conservation of mass. These results do not obey the law of conservation of mass. Possibly she
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