stuck to the stir stick and beaker. Data‚ Calculations and results: .007L * (3mol HCl)/L * (1mol C6H5COOH)/1mol HCl =0.0210 moles C6H5COOH 2.00g C6H5COONa * (1mol C6H5COONa)/(144.1g)* (1mol C6H5COOH)/(1 mol C6H5COONa) =0.0139 moles C6H5COOH Limiting Reactant: C6H5COONa Theoretical Yield: 0.0139 mol C6H5COOH*122.1g/ mol C6H5COOH = 1.70 g mol C6H5COOH mol C6H5COOH Actual Yield: 1.59g C6H5COOH Percent Yield: (1.59g/1.70g)*100 = 93.5% Conclusion: Based on the net equation listed
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HKDSE CHEMISTRY – A Modern View (Second Edition) (Chemistry) Coursebook 4A Suggested answers Chapter 36 Rate of chemical reaction Page Number Class practice 1 Think about 5 Chapter exercise 5 Chapter 37 Factors affecting rate of reaction Class practice 11 Think about 14 Writing practice 14 Chapter exercise 14 Chapter 38 Molar volume of gases at room temperature and pressure (r.t.p.) Class practice 21 Self-test 24 Think about 26 Chapter exercise 27 Part exercise 34 Chapter 39 Dynamic equilibrium
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Assessment: Answer the following item with the appropriate response. 1. What is a mole ratio? 2. What piece of information is needed to convert grams of a compound to moles of the same compound? 3. Phosphine (PH3) can be prepared by the hydrolysis of calcium phosphide‚ Ca3P2: Ca3P2 + 6 H2O 3 Ca (OH)2 + 2 PH3 a) One mole of Ca3P2 produces 2 mol of PH3. b) One gram of Ca3P2 produces 2 g of PH3. c) 3 moles of Ca(OH)2 are produced for each 2 mol of PH3 produced
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2:1 N/A 83 0.7856 H2SO4 98.06 ~15drops N/A N/A N/A 327 1.84 Methanol 32.0 ~6mL N/A N/A N/A 64.7 0.7914 Water 18.0 5mL N/A N/A 0 100 1.0 1‚4-Di-t-butyl-2‚5-dimethoxybenzene 250.17 product product 1:1 102-105 N/A N/A Reaction and its Mechanism: The limiting reagent for this reaction is 1‚ 4 – dimethoxybenzene. See calculations section for explanation. Overall Reaction: C8H10O2+
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give a mixture of products arising from both substitution and elimination. As with substitution reactions‚ the rate at which elimination reactions proceed can be proportional to both the concentration of the base and the concentration of the reactant alkyl halide (an "E2" reaction (elimination bimolecular)‚ or the rate can be proportional only to the alkyl halide (an "E1" reaction (elimination unimolecular). The mechanism for the E2 reaction is best described as concerted with the reaction coordinate
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I) INTRODUCTION * Industrial Chemistry deals with the preparation of products from raw materials through the agency of chemical change * Chemistry is important to Industry by: * Regulating manufacturing processes * Quality Control * Research and Development * Sources of raw materials from the natural environment 1) Lithosphere –Earth’s Crust 2) Hydrosphere – Marine and Oceanic Environment 3) Atmosphere – Air 4) Plants – Biosphere * Classification of Natural
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chapter we start out by discussing rates of reactions and the rate law. The rate law indicates the affect that the concentration of the reactants has on the reaction rate. In general‚ adding more of a reactant speeds things up (rather like pushing the gas pedal to put more gas into the car engine ). But how much faster is the reaction if say the concentration of a reactant is doubled ? the rate law will help us answer such equations. Another way to affect the rate of a reaction is to change the temperature
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develop an Experiment Beginning to End * Find a good reaction * Good reaction is defined by high yield‚ green chemistry (less waste‚ less harmful chemicals) * Reaction found on Reaxys * Find MSDS Sheet for all Compounds (Reactants‚ Products and Side Products) * It is important to know what you are working with * The toxicity category numbers are there to gauge how toxic the chemical is‚ 4 being most severe and 1 being least. * Safety‚ no exposed skin
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linked together by peptide bonds through condensation reactions. They are an essential part of all living organisms. Enzymes are “biological catalyst used to speed up chemical reactions (Athanasopoulos‚ 2014). Normally‚ for chemical reactions; “reactant molecules must collide with enough force [energy] and with the correct geometric orientation for bond breaking to occur” (L. Morris‚ 2014). As for energy‚ because the structure of proteins can be denatured at high temperatures‚ living
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Then put 50 ml of distilled water on the beaker with sodium carbonate. Use a stirring rod to stir it and then we you removed it rinse it properly so that you do not lose the reactant. Put 1.00 g of calcium carbonate and place it to the labeled beaker. Then record the mass and after put 50 ml of distilled water to the beaker. Stir it to dissolve the solid material. Then to the same thing in step 3 when removing the stirring rod
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