Address 26 What s on the Final .


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Lecture #26 What’s on the Final?. Chemistry 142 B Autumn Quarter, 2004 J. B. Callis, Instructor. General. Date as Announced in Syllabus 44 Multiple-Choice Questions - Chapters 1-8 22 questions from chapters 1-5 22 questions from chapters 6-8
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Address #26 What\'s on the Final? Science 142 B Autumn Quarter, 2004 J. B. Callis, Instructor

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General Date as Announced in Syllabus 44 Multiple-Choice Questions - Chapters 1-8 22 questions from sections 1-5 22 questions from parts 6-8 Practice exam accessible on the class site – key on class notice board.

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Chapter 2 Dalton\'s Atomic Hypothesis and Beyond Law of various extents as clarified by Dalton Naming mixes

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Problem 26-1: Multiple Proportions A scientific expert studies three oxides of iodine, and finds their % oxygen as takes after: 20.14 %, 23.97% and 22.10 %. (a) Calculate the mass of oxygen per gram of iodine in every compound. Express the outcome as a proportion. At that point frame the proportion of proportions by partitioning by the proportion of the primary compound. (b) How do the numbers to some extent (a) bolster Dalton\'s nuclear hypothesis?

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Problem 26-1: Multiple Proportions

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Chapter 3 - Stoichiometry Convert between moles, mass and number of atoms. Decide experimental equation from natural creation. Adjust a synthetic response utilizing the particle adjust conditions. Tackle stoichiometric issues including a restricting reagent utilizing the reactant proportion technique.

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Problem 26-2: Bromoform is 94.85% bromine, 0.40% hydrogen and 4.75% carbon by mass. Decide its observational equation.

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Problem 26-3: Balance the accompanying compound condition: x NaCl + y SO 2 + z H 2 O + w O 2 => u Na 2 SO 4 + v HCl

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Problem 26-4: Sulfuric corrosive (H 2 SO 4 ) frames in the accompanying response: 2 SO 2 + O 2 + 2 H 2 O => 2 H 2 SO 4 Suppose 400. g SO 2 , 175. g O 2 and 125. g H 2 O are blended and the response continues until one of the reactants is spent. Recognize the constraining reactant and figure out what masses of alternate reactants remain.

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Chapter 4 Chemical Reactions Classify a given response and distinguish items, for example, encourages. Compose the net ionic response. Decide the oxidation number of a component in a compound. Adjust a redox condition utilizing the particle adjust and charge adjust conditions. Perform stoichiometric figurings on arrangements

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Problem 26-5: Balance the accompanying synthetic condition: x H + y H 2 O 2 + z Fe 2+ => u Fe 3+ + v H 2 O

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Chapter 5 – The Gas Phase Use the perfect gas law to register thickness from other free factors, e.g. weight, temperature. Work with halfway weights. Figure the measure of gas gathered over water. Utilize the perfect gas law in stoichiometric issues.

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Problem 26-6: A 20.6 L test of air is gathered in Greenland at –20. o C at a weight of 1.01 atm and constrained into a 1.05 L bottle for shipment to Europe for examination. Process the weight inside the jug as it is opened in the research center at 21. o C.

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Chapter 6 – Chemical Equilibrium Write the harmony consistent for an adjusted substance response. Take care of harmony issues given beginning fixations and K. Apply Le Chatelier\'s rule to decide bearing of move of response in light of an adjustment in response conditions.

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Problem 26-7: At 5000 K, even the nitrogen particle (N 2 ) separates (into 2 iotas of nitrogen). At this temperature, when the aggregate weight of nitrogen is 1.00 atm, N 2 ( g ) is a = 0.65% separated at balance: N 2 ( g ) = 2 N( g ). Figure the balance consistent at 5000K.

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Problem 26-7 (cont.)

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Chapter 7 – Acid/Base Equilibria Understand the pH scale Calculate the pH of solid, powerless and exceptionally frail acids and bases

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Problem 26-8: What is the pH of 0.15 M methylammonium bromide, CH 3 NH 3 Br? (K b of CH 3 NH 2 = 4.4 x 10 - 4

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Chapter 8 – Applications of Aqueous Equilibria Calculate the pH of support arrangements Give formulas for making up cradles Calculate pH at various focuses in a titration bend Perform solvency counts Understand the essential thoughts of complex particle equilibria

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Problem 9: You have available to you an adequate amount of an answer of 0.0500 M NaOH and 500 mL of an answer of 0.100 M formic corrosive (HCOOH, K a = 1.77 x 10 - 4 ). What amount of the NaOH arrangement ought to be added to the corrosive answer for create a cushion of pH 4.00? Ans: Use the base to create an adequate measure of the formate particle to give a cushion of the fancied pH. Do the issue at the Henderson-Hasselbalch level, overlooking the ionization of water. Take into consideration weakening of the first corrosive arrangement with the base, pretty much as in a titration. The applicable response is:

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Problem 9 (cont.):

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Problem 9 (cont.):

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Problem 10: A soaked arrangement of Mg(OH) 2 at 25 o C is set up by equilibrating strong Mg(OH) 2 with water. Concentrated NaOH is then included until the solvency of Mg(OH) 2 is 0.001 times that in H 2 O alone. (Disregard the adjustment in volume coming about because of the expansion of NaOH.) The solvency item K sp of Mg(OH) 2 is 1.2 x 10 - 11 at 25 o C. Ascertain the centralization of hydroxide particle in the arrangement after the expansion of the NaOH. Ans: First ascertain the solvency of Mg(OH) 2 in water. At that point ascertain the grouping of [Mg 2+ ] after expansion of OH - place this into the mass activity expression and settle for [OH - ].

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Problem 10 (cont.): Addition of OH - shifts the balance to one side and the centralization of Mg 2+ must reduce to keep up the mass-activity expression at a consistent esteem. After the expansion of base, the new centralization of Mg 2+ is

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Answers to Questions from Lecture 26 1, 1.25, 1.125 CHBr 3 4 NaCl + 2 SO 2 + 2 H 2 O + O 2 => 2 Na 2 SO 4 + 4 HCl 2 H + H 2 O 2 + 2 Fe 2+ => 2 Fe 3+ + 2 H 2 O 23.0 atm K = 1.67 x 10 - 4 pH = 5.73 639 mL 9.1 x 10 - 3

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