Kémia | Középiskola » Kémia angol nyelven középszintű írásbeli érettségi vizsga megoldással, 2012

ÉRETTSÉGI VIZSGA 2012. május 16 Név: . osztály: KÉMIA ANGOL NYELVEN KÖZÉPSZINTŰ ÍRÁSBELI VIZSGA 2012. május 16 8:00 Az írásbeli vizsga időtartama: 120 perc Kémia angol nyelven Pótlapok száma Tisztázati Piszkozati NEMZETI ERŐFORRÁS MINISZTÉRIUM középszint írásbeli vizsga 1012 Kémia angol nyelven középszint Név: . osztály: Important informations • The examination test should be solved within 120 minutes, after 120 minutes the work should be finished. • The sequence of answering the questions is free. • For the solution of the problems calculators without text-storage capability and fourplace logarithm tables can be used. Use of other electronic or written help is forbidden • Read the introductory text of the questions carefully and keep its instructions. • Write the answers in ink. If you cancel an answer or part of an answer, the canceled work can not be evaluated. • For the calculations, you can get maximum number of

points only if the main steps of the calculation are indicated, too. • Please, don’t write anything into the gray squares. írásbeli vizsga 1012 2 / 16 2012. május 16 Kémia angol nyelven középszint Név: . osztály: 1. Panel question Choose the suitable compound from the given list below and write its structural formula into the empty cells of the table. Answer the questions in the cells beside in a well readable writing. (The formula of a given compound can be used at several places in the table) ethanol, propan-2-ol, acetaldehyde, acetic acid, diethyl ether, glycine Structural formula of the compound, showing bonding and nonbonding electron pairs 1. Further information It has a low boiling point, it is easily inflammable. Formerly, it was used for narcosis. Under mild oxidation conditions, it is converted to acetone. 3. 4. It can be prepared from one of the above listed compounds. What is the name of this compound? 5. 6. Classify the product into one group

of oxygen-containing compounds. It is giving a positive silver mirror test. 7. 8. Reaction equation of the silver mirror test: It is a solid substance (at 25 °C and standard pressure). 9. 10. Type of lattice of the compound: The two compounds react to form an ester. 11. 13. Structural formula of the ester formed: At 25 oC, it is a liquid. Its aqueous solution is acidic. 2. Name a household use (1 example): 12. 14. Its name: 15 points írásbeli vizsga 1012 3 / 16 2012. május 16 Kémia angol nyelven középszint Név: . osztály: 2. Essay Read the following text carefully and answer the questions. Iron production could be “greener” The following summary about the experiments of Donald Sadoway concerning the metallurgy of iron was published in summer 2006 and since then, the method has been continuously developed. Donald Sadoway, a material scientist at MIT (Massachusetts Institute of Technology) would produce iron by electrolysis and this way, the

environmental pollution caused by the conventional smelting method could be consigned to the past. The generally used conventional method releases one tonne of carbon dioxide into the atmosphere for every tonne of metal produced. In conventional smelting, iron ore is mixed with coke which is reacting with the iron ore, producing carbon dioxide and carbon monoxide, leaving pure iron behind. Electrolysis produces iron in a different way. The iron ore is dissolved in a solvent of silicon dioxide at a very high temperature of 1600 degree Celsius and an electric current is passed through it. Negatively charged oxide ions migrate to the positively charged anode and in this case, on the anode exclusively oxygen bubbles off. Correspondingly, positively charged iron(III) ions migrate to the negatively charged cathode where they are reduced to liquid elemental iron which collects in a pool at the bottom of the cell and is simply siphoned off. A similar process is routinely used to produce

aluminum, whose oxide is so stable that it cannot be reduced by coke as in the case of iron. The steel industry has never had any reason to switch to electrolysis, since iron oxide is easily reduced by carbon to molten iron. Carbon is cheap, electricity is expensive and up to now, carbon dioxide emission has never been associated with any extra costs. However, if governments begin to impose serious carbon emission taxes, the new process may become more attractive, according to the American Iron and Steel Institute which provided $500 000 in funds for the researcher of MIT. According to the Institute, there are other factors in favor of electrolysis: for one thing, it would cut out the energy-intensive coke-making process which is further increasing the carbon dioxide emission. Sadoway demonstrated that the idea of electrolysis can work in a laboratory. However, things are going to take 10 to 15 years to develop and get to an industrial scale, says Lawrence Kavanagh, vice president of

the Institute, because the technology has a number of hurdles to overcome till a commercial application. The first and biggest hurdle is finding a practical anode material. In the first experiments, Sadoway used an anode made of graphite, similar to those used for producing aluminum. However, it is also a sort of carbon, releasing about as much carbon dioxide as smelting. The most suitable material would be platinum as an anode, but it is too expensive for large scale manufacturing. Possibilities include either metal alloys that form an oxide on the outer layer but still conduct electricity, or conductive ceramics. írásbeli vizsga 1012 4 / 16 2012. május 16 Kémia angol nyelven középszint Név: . osztály: Another great problem is that the process uses a lot of electricity, about 2000 kilowatt hours per tonne of iron produced. If it can not be considerably decreased, the method would actually result in more pollution than the conventional procedure and so, there is a

great skepticism about the invention. Most people are also skeptical about the likelihood of a practical carbon-free anode, since the industry has been trying for years to develop one and until now without any success. (After a paper of Richárd Balázs on August 31, 2006, www.sghu, Information and Technology) a) Iron production uses iron ore which contains mainly iron(III) oxide. Write balanced equations of the reduction of iron(III) oxide by carbon and carbon monoxide, respectively. b) What is the mass of carbon dioxide released into the air - according to the text – for every tonne of iron produced? c) Calculate the mass of carbon dioxide released while producing 1.00 tonne of iron using carbon monoxide. (Ar(Fe) = 56.0, Ar(C) = 120, Ar(O) = 160) d) Give a short definition of electrolysis. e) Fill the table concerning the electrolytic cell used in the experimental procedure. Name of the pole Process at the electrodes * Type of occurring (electrode) process Negative pole 1.

3. 5. Positive pole 2. 4. 6. * the answer can be given in text or writing the chemical equation f) Which type of electrolysis is used to produce iron in the experimental procedure? Underline the correct answer: electrolysis of a solution electrolysis of a molten electrolyte g) List advantages (2 examples) and disadvantages (2 examples) of the experimental procedure. h) Electrolysis is also used to produce aluminum. Which substance is electrolyzed in this case? 16 points írásbeli vizsga 1012 5 / 16 2012. május 16 Kémia angol nyelven középszint Név: . osztály: 3. Simple choice Write the only correct letter mark into the empty cell on the right-hand side of the answers. 1. What is the maximum number of electrons on the third (M) shell? A) B) C) D) E) 6 10 14 18 32 2. 0200 mol NaCl is contained in 2500 cm3 solution What is the molarity of the solution? A) B) C) D) E) 0.800 mol/dm3 0.0500 mol/dm3 0.200 mol/dm3 0.0050 mol/dm3 0.0008 mol/dm3 3. Which statement

is not true? A) Ionic bonding is a primary chemical bonding. B) Hydrogen bonding is the strongest secondary bonding. C) In case of an element consisting of molecules, there is no possibility for dipole-dipole interaction between the molecules. D) Dispersion forces can only act between molecules consisting of like atoms. E) Ionic bonding can be formed between 2+ charged cations and 1- charged anions, too. 4. Which of the following compounds contains the lowest oxidation number of nitrogen atom? A) B) C) D) E) NO NH4Cl HNO3 NO2 Cu(NO3)2 írásbeli vizsga 1012 6 / 16 2012. május 16 Kémia angol nyelven középszint Név: . osztály: 5. Which of the following substances is giving a basic solution if dissolved in water? A) B) C) D) E) methanol chlorine glucose sulfur dioxide methyl amine 6. Which statement is true? A) In a soft water, there are more dissolved calcium and magnesium salts than in a hard one. B) Calcium chloride dissolved in water causes temporary hardness of

water. C) Limestone dissolved by rain water containing carbon dioxide is leading to acidifying of the water of rivers and lakes. D) Both stalactites and scaling are calcium carbonate and/or magnesium carbonate. E) Soft water used for washing decreases the effect of washing powders. 7. Which of the following substances is white/colorless? A) B) C) D) E) calcium hydroxide copper sulfur nitrogen dioxide bromine 8. How many constitutional isomers are existing for the compound C5H12? A) B) C) D) E) 1 2 3 4 5 8 points írásbeli vizsga 1012 7 / 16 2012. május 16 Kémia angol nyelven középszint Név: . osztály: 4. Four types of association Below, you have to compare two terms. Write the correct letter mark in the empty cells of the table. A) B) C) D) 1. 2. 3. 4. 5. 6. 7. 8. 9. Substances having atomic lattice Substances having molecular lattice Both of them None of them In their lattice points, there are ions. Covalent bonding is the force holding the lattice together. It

can be found both among elements and compounds. It can be found among the allotropic modifications of carbon, too. Alkenes can be classified into this group. In their solid aggregation state, they are conducting electricity. These substances are characterized by low melting and boiling point. In solid state, these substances might be characterized by sublimation. Very hard substances belong into this group, too. 1. 2. 3. 4. 5. 6. 7. 8. 9. 9 points 5. Analytical question and calculation problem Hydrogen chloride Hydrogen chloride is produced in a lot of industrial processes. It is for example the byproduct in the reaction of alkanes with chlorine, but this by-product contains impurities of the educts (chlorine and organic substances). Most amount of hydrogen chloride produced this way is further converted mainly to produce vinyl chloride or chlorinated solvents. If high purity substance is needed, hydrogen chloride is produced by direct synthesis. a) List the characteristic

physical properties of hydrogen chloride. color: aggregation state (at 25 °C and standard pressure): odor: írásbeli vizsga 1012 8 / 16 2012. május 16 Kémia angol nyelven középszint Név: . osztály: b) Write (balanced) equation of the reaction first mentioned in the text for the example of methane and chlorine. Give the type of the reaction and the name of the produced organic substance. c) The by-product hydrogen chloride can be used for the production of PVC from ethyne. type of reaction: type of reaction: . . + HCl ethyne . . name of the product PVC d) Hydrogen chloride can also be produced by the reaction of hydrogen and chlorine. Write (balanced) equation of the occurring reaction. How many grams of hydrogen and how many grams of chlorine are needed to produce 500.0 dm3 hydrogen chloride at 250 °C and standard pressure? (Ar(H) = 1.00, Ar(Cl) = 355) e) When hydrogen chloride is dissolved in water, will the solution be acidic, neutral or basic? 16 points

írásbeli vizsga 1012 9 / 16 2012. május 16 Kémia angol nyelven középszint Név: . osztály: 6. Alternative question In the following question – depending on your field of interest – you have to solve only one version. At the corresponding place of the examination paper, you have to indicate the letter mark of the chosen question (A or B). If it doesn’t happen and the fact of your choice doesn’t emerge unambiguously from the test-paper, in every case the solution of the first alternative question will be evaluated. Letter mark of the chosen question: A) Analytical question N2 + 3 H2 2 NH3, heat of formation: ΔfH(NH3(g)) = –46.1 kJ/mol In equilibrium state: t = 400 °C, p = 2⋅105 kPa Based on the above equation and data, answer the questions below. In a container of volume V, 1 mole of nitrogen and 3 moles of hydrogen are mixed and the mixture is let to come to equilibrium. How will the number of moles of ammonia change, if the conditions are changed as

given below? Write on the dotted line the letter mark of one of the answers given below and explain your answer (based on Le Chatelier’s principle). A. the amount of ammonia will decrease B. the amount of ammonia will be unchanged C. the number of moles of ammonia will increase 1. A catalyst is added to the gas mixture: . Explanation:. 2. At constant pressure, the temperature is changed to 100 °C: . Explanation:. 3. At a constant temperature, further 2 moles of nitrogen are added to the mixture: Explanation:. 4. At a constant temperature, the pressure is changed to 1⋅105 kPa: . Explanation:. 5. At a constant temperature, the volume of the initial mixture is decreased to half of the initial volume :. Explanation:. írásbeli vizsga 1012 10 / 16 2012. május 16 Kémia angol nyelven középszint Név: . osztály: 6. Write the expression for the equilibrium constant of the above reaction B) Calculation problem In a container, 12.0 moles of carbon monoxide gas and

400 moles of oxygen gas were mixed and by a spark ignition, the two gases reacted with each other. Ar(C) = 12.0, Ar(O) = 160, ΔfH(CO(g)) = –1134 kJ/mol, ΔfH(CO2(g)) = –3944 kJ/mol a) Calculate the composition of the initial gas mixture in % by mass. b) Calculate the composition of the obtained gas mixture in mole %. c) Calculate the energy change following the above process. 11 points írásbeli vizsga 1012 11 / 16 2012. május 16 Kémia angol nyelven középszint Név: . osztály: 7. Analysis of experiments Experiments with distilled water a) Is distilled water acidic, basic or neutral? . Phenolphtalein is added to distilled water. The color of the solution: The molar concentration of H3O+ and OH- ions in distilled water at 25 °C is: [H3O+] = . [OH–] = . b) A small piece of sodium is put into distilled water containing phenolphatalein. Is there any change in the system? If yes, what can be observed? Explanation of the observation (if a reaction occurs, write

the reaction equation, too): c) A small piece of white phosphorus is put into distilled water containing phenolphthalein. Is there any change in the system? If yes, what can be observed? Explanation of the observation (if a reaction occurs, write the reaction equation, too): d) A small piece of quicklime is put into distilled water containing phenolphatalein. Is there any change in the system? If yes, what can be observed? Explanation of the observation (if a reaction occurs, write the reaction equation, too): e) Distilled water and concentrated sulfuric acid are mixed. The solution is strongly warmed up. Explanation of the observation: What is the recommended rule to mixing water and concentrated sulfuric acid? 12 points írásbeli vizsga 1012 12 / 16 2012. május 16 Kémia angol nyelven középszint Név: . osztály: 8. Calculation problem feladat The density of a 0.50 molar (mol/dm3) sulfuric acid solution is 103 g/cm3 Ar(H) = 1.00, Ar(O) = 160, Ar(S) = 320 a) How

many cm3 0.75 molar potassium hydroxide solution can neutralize 100 cm3 of the above solution? b) What is the concentration of the 0.50 molar sulfuric acid solution in % by mass? c) 120 g sulfur trioxide is dissolved in 500 cm3 of the initial sulfuric acid solution. What will be the concentration of the resulting solution in % by mass? 13 points írásbeli vizsga 1012 13 / 16 2012. május 16 Kémia angol nyelven középszint Név: . osztály: írásbeli vizsga 1012 14 / 16 2012. május 16 Kémia angol nyelven középszint Név: . osztály: írásbeli vizsga 1012 15 / 16 2012. május 16 Kémia angol nyelven középszint Név: . osztály: maximum points 1. Panel question 2. Essay 3. Simple choice 4. Four types of assotiation 5. Analytical question and calculation problem 6. Alternative question 7. Analysis of experiments 8. Calculation problem Points of the written exam reached points 15 16 8 9 16 11 12 13 100 marking teacher date

Points (rounded up) Achieved to write in the points rounded up / software / Elért pontszám Programba beírt egész számra egész pontszám kerekítve Test/Feladatsor marking teacher/javító tanár underwriter/jegyző date/dátum date/dátum írásbeli vizsga 1012 16 / 16 2012. május 16 ÉRETTSÉGI VIZSGA 2012. május 16 Kémia angol nyelven középszint Javítási-értékelési útmutató 1012 KÉMIA ANGOL NYELVEN KÖZÉPSZINTŰ ÍRÁSBELI ÉRETTSÉGI VIZSGA JAVÍTÁSI-ÉRTÉKELÉSI ÚTMUTATÓ NEMZETI ERŐFORRÁS MINISZTÉRIUM Kémia angol nyelven középszint Javítási-értékelési útmutató Basic guidelines of the evaluation of written exercises Evaluation of the written test-papers should follow the distributed correction instructions. Evaluation of the theoretical questions • No deviation from the correction instruction is allowed. • ½ points can not be given, the questions can only be

evaluated according to the allowed part-points in the correction key. Evaluation of the calculation problems • Test-papers following the way of solution of the correction instruction, must be evaluated according to the part-points of the correction key. • Beside objectivity, attention must be paid to honesty. During the evaluation, punishment with a pedagogical intention can not be applied! • In a given – errorless – solution no points can be subtracted because of the lack of not required (but in the correction key given) subresults. (Those subresults help only the evaluation of partial solutions.) • Approaches differing from the correction key – if correct – get maximum points or part-points according to the nodes of the correction key. • For a bare result without any derivation or explanation only 1-2 points can be given as a maximum according to the points of that result in the correction key! • A calculation problem gets maximum points even if it

contains a theoretically incorrect reaction equation which is not necessary to the solution (and the question did not ask to write it). • In case of a problem containing several subproblems, part-points for a given subproblem can be given even if the candidate makes the calculation using an incorrect result of a beforegoing subproblem – if the solution doesn’t lead to a contradiction. • Relations which can be regarded as trivial can be used without any derivation in calculation problems of the maturity examination, and they get maximum points – even without detailed explanation. For example: • conversion of mass, number of moles, volume and number of particles, • trivial facts following from Avogadro’s law (equal stoichiometric ratios or volume ratios in case of gases under the same conditions, and so on), • using the mixing (dilution) equation, and so on. • For each calculation error maximum 1-2 points can be subtracted (if the candidate continues the

calculation correctly with the incorrect subresult, he or she should get all other part-points for the further part of the calculation)! • In case of a smaller error, the candidate doesn’t get points for the incorrect part of the calculation, but the following steps of the calculation using incorrect data should get the corresponding points. A smaller error is for example: • incorrect use of density in the conversion of volume and mass, • other incorrect simple mathematical procedure, írásbeli vizsga 1012 2/8 2012. május 16 Kémia angol nyelven középszint Javítási-értékelési útmutató • incorrectly balanced equation, which doesn’t lead to an obviously unrealistic result. • In case of a gross error, the candidate doesn’t get further points for the given subcalculation as allowed in the correction key even if he continues the calculation correctly with the incorrect subresult. A gross error is for example: • a calculation based on an incorrect

(e. g not occurring) reaction equation, • if the result estimated from the data is obviously unrealistic (for example if the mass of the solution calculated from the mass of the solute is smaller than the mass of the solute dissolved in it, and so on). (The solution of further subcalculations which can be regarded as independent calculation units can be evaluated of course also in this case according to the previously discussed principles. Points can be given – if calculating correctly with incorrect subresults - if the calculation doesn’t lead to unrealistic results.) írásbeli vizsga 1012 3/8 2012. május 16 Kémia angol nyelven középszint Javítási-értékelési útmutató 1. Panel question (15 points) 1. 2. 3. 4. 5. 6. 7. 8. Correct structural formula of acetic acid Seasoning to food, scaling remover (1 correct example) Correct structural formula of diethyl ether. Ethanol. Correct structural formula of propan-2-ol. Ketones. Correct structural formula of

acetaldehyde. CH3CHO + 2 Ag+ + 2 OH– = 2 Ag + CH3COOH + H2O (1 point for the equation, 1 point for balancing) 9. Correct structural formula of glycine (structural formula of the zwitterions can be accepted, too) 10. Ionic lattice 11. Correct structural formula of ethanol (or propan-2-ol) 12. Correct structural formula of acetic acid (Answers given in points 11. and 12 can interchange places) 13. Correct structural formula of ethyl acetate or isopropyl acetate 14. Ethyl acetate/ethyl esther of acetic acid or isopropyl acetate/isopropyl esther of acetic acid. 1 point 1 point 1 point 1 point 1 point 1 point 1 point 2 points 1 point 1 point 1 point 1 point 1 point 1 point 2. Essay (16 points) a) Fe2O3 + 3 C = 2 Fe + 3 CO . 1 point (the equation 2 Fe2O3 + 3 C = 4 Fe + 3 CO2 can be accepted, too.) Fe2O3 + 3 CO = 2 Fe + 3 CO2 . 1 point b) 1 tonne 1 point c) n(Fe) = 106 g / 56.0 g/mol =17857 mol 1 point n(CO2) = 1.5⋅n(Fe) 1 point 1 point n(CO2) = 26786 mol m(CO2) = n(CO2)⋅M(CO2) m(CO2)

= 26786 mol⋅44.0 g/mol = 118⋅106 g = 118 tonne 1 point d) Electrical energy causes a redox reaction to occur. 1 point e) 1. Cathode 2. Anode (for answers 1 and 2 together1 point can be given) 1 point 3. Iron ions are converted to elemental iron (Fe3+ + 3 e– = Fe or Fe2+ + 2 e– = Fe) (Based on the text or the figure, respectively both answers can be accepted.) 1 point 4. Oxide ions are converted to oxygen (O2– = ½ O2 + 2 e–) 1 point 5. Reduction 6. Oxidation (for answers 5 and 6 together1 point can be given) 1 point f) Electrolysis of a molten electrolyte. 1 point g) Advantages: less carbon dioxide/ carbon monoxide release, no need for coke. 1 point Disadvantages: a lot of electricity is needed and therefore expensive, There is no suitable anode (platinum is expensive, graphite: increases carbon dioxide release, too). (one point can be given for two correct answers in both cases) 1 point h) Aluminum is produced from anhydrous aluminum oxide (Al2O3). 1 point írásbeli

vizsga 1012 4/8 2012. május 16 Kémia angol nyelven középszint Javítási-értékelési útmutató 3. Simple choice (8 points) For every correct answer one point can be given. 1. D 2. A 3. D 4. B 5. E 6. D 7. A 8. C 4. Four types of assotiation (9 points) For every correct answer one point can be given. 1. D 2. A 3. C 4. C 5. B 6. D 7. B 8. B 9. A 5. Analytical question and calculation problem (16 points) a) Colorless, gas, pungent odor. 2 points (One point can be given for two correct answers) b) CH4 + Cl2 = CH3Cl + HCl 1 point (A correct equation with a product containing more chlorine can also be accepted.) Substitution 1 point Cloromethane (methyl chloride) 1 point (In case of another equation, correct naming of the product should get one point, too.) c) CH≡CH 1 point Addition 1 point CH2=CH–Cl – vinyl chloride (chloroethene) 1 point Polimerization 1 point d) H2 + Cl2 = 2 HCl 1 point n(HCl) = V/Vm = 20.4 mol 1 point n(Cl2) = n(H2) = n(HCl) /2 1 point n(Cl2) =

n(H2) = 10.2 mol 1 point 1 point m(Cl2) = n · M(Cl2) = 10.2 mol ⋅ 710 g/mol = 724 g m(H2) = n · M(H2) = 10.2 mol ⋅ 20 g/mol = 204 g 1 point e) The obtained solution is acidic. 1 point írásbeli vizsga 1012 5/8 2012. május 16 Kémia angol nyelven középszint Javítási-értékelési útmutató 6. Alternative question (11 points) A) Analytical question 1. B the amount of ammonia will be unchanged 1 point Catalysts do not shift the position of the equilibrium (they only increase the rate of reaction). 1 point 2. C the number of moles of ammonia will increase 1 point The process is exothermic in the direction of the formation of ammonia and therefore, decreasing the temperature favors the formation of the product (the equilibrium is shifted to the right). 1 point 3. C the number of moles of ammonia will increase 1 point The concentration of the educts is increasing, which favors the formation of the product (the equilibrium is shifted to the right). 1 point 4. A the

amount of ammonia will decrease 1 point Formation of ammonia is connected to a decrease in the stoichiometric numbers (a decrease in the number of the moles), decreasing the pressure favors the decomposition of the product, the equilibrium is shifted into the direction of an increase in the number of molecules (to the left). 1 point 5. C the number of moles of ammonia will increase 1 point Decreasing the volume of the gas to half of the original value increases the pressure of the gas mixture and therefore, the equilibrium is shifted into the direction of a decrease in the number of the molecules, the formation of the product is favored. 1 point In the explanations, every other formulation can be accepted, but only mentioning Le Chatelier’s principle is not enough. [NH 3 ]2 6. K = 1 point 3 [N 2 ] ⋅ [H 2 ] B) Calculation problem a) m(CO) = 12.0 mol ⋅ 2800 g/mol = 3360 g m(O2) = 4.00 mol ⋅ 3200 g/mol = 1280g m(mixture) = 336.0 + 1280 = 4640 g 336.0 composition in % by mass:

⋅100 = 72.4% CO 464.0 128.0 ⋅100 = 27.6% O2 464.0 b) The occurring reaction: 2 CO + O2 = 2 CO2 the amount of oxygen is limiting n(CO) = 8.00 mol is reacting n(CO2) = 8.00 mol is produced the excess: n(CO) = 4.00 mol n(mixture) = 8.00 + 400 = 120 mol írásbeli vizsga 1012 6/8 (or use of it) 1 point 1 point 1 point 1 point 1 point 1 point 1 point 2012. május 16 Kémia angol nyelven középszint composition in mole %: Javítási-értékelési útmutató 8.00 ⋅100 = 66.7% CO2 12.0 4.00 ⋅100 = 33.3% CO 12.0 1 point c) the heat of reaction related to 1 mole of CO: 1 point ΔrH = ΔfH(CO2(g)) –ΔfH(CO (g)) = = –394.4 kJ/mol – (–1134 kJ/mol) = –281 kJ/mol 1 point n(CO) = 8.00 mol, Qr = – 281 kJ/mol ⋅ 8.00 mol = –2248 kJ 1 point (Only the use of a correct sign/correct formulation should get a point.) (Every other correct explanation/deduction should get maximum number of points!) 7. Analysis of experiments (12 points) a) Neutral. Colorless. (for the

two correct answers) [H3O+] = 10–7 mol/dm3, [OH–] =10–7 mol/dm3 (If the unit is missing or not correct, no point should be given.) b) Sodium is moving on the surface of water, the color of the solution turns violet/purple/cyclamen, there is a gas evolution, the piece of sodium is molten to a sphere. (Giving at least three correct observations.) Explanation of the given observations. 2 Na + 2 H2O = 2 NaOH + H2 c) White phosphorus remains unchanged (its density is higher than that of water, it is settled down to the bottom of the container), the solution remains colorless White phosphorus is not dissolved in water, because it is apolar and it is not reacting with it. d) A small amount of quicklime is dissolved, the color of the solution turns violet/purple/cyclamen. (When more quicklime is put into water, precipitation occurs.) Quicklime (CaO) reacts with water, the produced Ca(OH)2 makes the solution basic. CaO + H2O = Ca(OH)2 e) The dissolution of sulfuric acid is an exothermic

process (its heat of dissolution is negative). The acid is added carefully/slowly into water while constantly stirring (use of safety glasses and gloves). 1 point 1 point 1 point 1 point 1 point 1 point 1 point 1 point 1 point 1 point 1 point 1 point 8. Calculation problem (13 points) a) H2SO4 + 2 KOH = K2SO4 + 2 H2O 1 mol sulfuric acid is neutralized by 2 mol sodium hydroxide (Use of the reaction equation or the correct molar ratio) írásbeli vizsga 1012 7/8 1 point 2012. május 16 Kémia angol nyelven középszint Javítási-értékelési útmutató n(H2SO4) = 0.0100 dm3 ⋅ 050 mol/dm3 = 00050 mol n(KOH) = 0.0100 mol V(KOH) = 0.0100 mol / 0750 mol/dm3 = 00133 dm3 = 133 cm3 1 point b) Let’s take 1.00dm3 solution: m(solution) = 1000 cm3 ⋅ 1.03 g/cm3 = 1030 g n(H2SO4) = 0.50 mol, m(H2SO4) = 050 mol ⋅ 980 g/mol = 490 g 49,0 The concentration of the solution is: ⋅ 100 = 4.76 mass% 1030 1 point 1 point 1 point c) The initial solution is: V(solution) = 500 cm3,

m(solution) = 500 cm3 ⋅ 1.03 g/cm3 = 515 g n((H2SO4) = 0.250 mol M(SO3) = 80.0 g / mol, n(SO3) = 120 g / 800 g/mol = 150 mol SO3 + H2O = H2SO4 after dissolution: n(H2SO4) = 1.50 mol + 025 mol = 175 mol m(H2SO4) = 1.75 mol ⋅ 980 g/mol = 1715 g m(solution) = 515 g + 120 g = 635 g 171,5 the sulfuric acid concentration of the solution is: ⋅ 100 = 27.0 mass% 635 1 point 1 point 1 point 1 point 1 point 1 point 1 point 1 point (Every other correct way of calculation should get maximum number of points!) írásbeli vizsga 1012 8/8 2012. május 16