THE MINISTRY OF EDUCATION AND SCIENCE OF UKRAINE NATIONAL TECHNICAL UNIVERSITY "DNIPROVSKA POLYTECHNIC" O.Y. Svietkina, O.B. Netyaga, H.V. Tarasova CHEMISTRY METHODICAL INSTRUCTIONS AND TASKS FOR SELF-WORK ON THE DISCIPLINE for students of all specialties Part 1 Dnipro 2018 THE MINISTRY OF EDUCATION AND SCIENCE OF UKRAINE NATIONAL TECHNICAL UNIVERSITY "DNIPROVSKA POLYTECHNIC" FACULTY OF GEOLOGICAL PROSPECTING Departament of Chemestry O.Y. Svietkina, O.B. Netyaga, H.V. Tarasova CHEMISTRY METHODICAL INSTRUCTIONS AND TASKS FOR SELF-WORK ON THE DISCIPLINE for students of all specialties Part 1 Dnipro NTU “DP” 2018 Svietkina O.Y. Chemistry. Methodical instructions and tasks for self-work on the discipline for students of all specialties in two parts. 2 p. P. 1 / O.Y. Svietkina, O.B. Netyaga, H.V. Tarasova; Ministry of eduk. and sien of Ukrain, National Technical University "Dniprovska Polytechnic". – Dnipro : NTU «DP», 2018. – 21 p. Автори: О.Ю. Свєткіна, проф., д-р техн. наук (вступ, розділ 1, 2, 3); О.Б. Нетяга, старш. викл. (розділ 4, 5); Г.В. Тарасова, асист. (розділ 6,7). Затверджено до видання методичною комісією з галузі знань 10. Природничі науки (протокол № 3 від 15.03.2018) за поданням кафедри хімії (протокол № 9 від 14.02.2018). The methodical instructions contain individual tasks in the course of chemistry, which are compiled on the topics of the theoretical course. Методичні рекомендації містять індивідуальні завдання відповідно до тем теоретичного курсу хімії. Відповідальна за випуск завідувач кафедри хімії, д-р техн. наук, проф. О.Ю. Свєткіна. INTRODUCTION The current methodological guidelines include the tasks on conducting ongoing student knowledge check of the chemistry course. Part one contains tasks on the topic of the theoretical part of the course. Each topic has a number: 1, 2, 3, etc. Inside the topics there are sections and subsections, which are numbered by adding one or two numbers respectively to the topic number. Thus, in topic 2 there are sections 2.1, 2.2, 2.3, etc. and subsections 2.1.1, 2.1.2, 2.1.3, etc. One section (subsection) there is 30 tasks. If there are no tasks in the middle of the section, it is indicated that particular examples should be taken from one of the previous sections. For example, to complete the task 10 of section 1.2 means: to compile empirical and graphic formulas for possible oxides of an element which are specified in section 1.1 under No. 10, that is manganese. Each student of the group performs assignments on various topics during the semester, indicated by numbers from 01 to 30. 1. CLASSES OF INORGANIC COMPOUNDS 1.1. List the stable oxidation steps of the element: 01 – Nitrogen; 02 – Magnesium; 03 – Sodium; 04 – Chlorine; 05 – Sulfur; 06 – Titanium; 07 – Carbon; 08 – Zinc; 09 – Copper; 10 – Manganese; 11 – Chromium; 12 – Bromine; 13 – Calcium; 14 – Aluminum; 15 – Potassium; 16 – Silver; 17 – Barium; 18 – Silicon; 19 – Cadmium; 20 – Lithium; 21 – Nickel; 22 – Hydrogen; 23 – Lithium; 24 – Tin; 25 – Phosphorus; 26 – Antimony; 27 – Boron; 28 – Iron; 29 – Mercury; 30 – Iodine. 1.2. Make empirical and graphical formulas for possible oxides of the element given in section 1.1 and name these oxides. 1.3. Determine the type of oxide (solvent, non–soluble, basic, acid or amphoteric) specified in section 1.2. 1.4. Write the equation of the salt formation reactions, proving the nature of the oxides (basic, acid or amphoteric), defined in section 1.3. 1.5. Write the empirical, graphic formulas and the name of the hydroxides of oxides, compiled in section 1.2. 3 1.6. Name the compounds in the international nomenclature: 01 – 3KHCO ; 02 –  23HCOCa ; 03 –   32 NOOHFe ; 04 – 2CaOHNO ; 05 – 3CaSiO ; 06 –  243 POBa ; 07 – 4CaHPO ; 08 –  24HSOBa ; 09 – 42SOLi ; 10 – 4KHSO ; 11 – 32SONa ; 12 – PbS; 13 –  24HSOCa ; 14 – 2BiOHCl ; 15 – 4CaCrO ; 16 – 722 OCrK ; 17 – 4NaMnO ; 18 – 4MnSO ; 19 – 2FeOHCl ; 20 – CdOHCl ; 21 – MgOHCl ; 22 –   32 COMgOH ; 23 – 4AlOHSO ; 24 –    422 SOOHAl ; 25 – CaOHCl; 26 –   32CONiOH ; 27 – CoOHBr ; 28 – 3LiHCO ; 29 – 3AgNO ; 30 – 22ZnONa . 1.7. Determine the degree of oxidation of the elements forming the compounds specified in section 1.6 1.8. Write the empirical and graphic formulas for acids: 01 – boric; 02 – carbonic; 03 – nitrous ; 04 – phosphoric; 05 – sulphuric; 06 – nitric; 07 – hydrosulphuric; 08 – acetic; 09 –silicic; 10 – sulphuric; 11 – hydrochloric acid; 12 – nitric; 13 – phosphoric; 14 – silicic; 15 – carbonic; 16 – hydrobromic; 17 – hydrofluoric acid; 18 – chromic; 19 – boric; 20 – permanganic; 21 – sulphuric; 22 – phosphoric; 23 – hydrobromic; 24 – sulphurous; 25 – nitrous; 26 – hydroiodic; 27 – chromic; 28 – nitric; 29 – acetic; 30 – permanganic. 1.9. Write the empirical and graphic base formulas: 01 – aluminum(III) hydroxide; 02 – calcium hydroxide; 03 – iron(III) hydroxide; 04 – strontium hydroxide; 05 – sodium hydroxide; 06 – iron(II) hydroxide; 07 – bismuth(III) hydroxide; 08 – copper(I) hydroxide; 09 – nickel(III) hydroxide; 10 – cobalt(III) hydroxide; 11 – potassium hydroxide; 12 – cesium hydroxide; 13 – barium hydroxide; 14 – cadmium hydroxide; 15 – magnesium hydroxide; 16 – manganese(II) hydroxide; 17–francium hydroxide; 18 – copper(II) hydroxide; 19 – bismuth(III) hydroxide; 20 – nickel(III) hydroxide; 21– mercury(II) hydroxide; 22 – lithium hydroxide; 23 – barium hydroxide; 24 – radium hydroxide; 25 – zirconium(III) hydroxide; 26 – copper(II) hydroxide; 27 – chromium(II) hydroxide; 28 – yttrium(III) hydroxide; 29 – rubidium hydroxide; 30 – thallium(III) hydroxide. 1.10. Write the empirical and graphic salt formulas: 01 – lithium nitrate; 02 – calcium carbonate; 03 – potassium sulphate; 04 – sodium nitrate; 05 – zinc sulfite; 06 – lithium chloride; 07 – sodium silicate; 08 – beryllium fluoride; 09 – barium sulfide; 10 – sodium dichromate; 11 – magnesium hydrogen sulfide; 12 – calcium dihydrogenphosphate; 13 – iron(III) hydrogen sulfate; 4 14 – aluminum hydroxosulfate; 15 – chromium(III) dihydroxonitrate; 16 – potassium zincate; 17 – aluminum dihydrogenphosphate; 18 – strontium dichromate; 19 – aluminum dihydroxosulphate; 20 – potassium hydrogencarbonate; 21 – iron(III) hydroxychloride; 22 – calcium hydrogenphosphate; 23 – aluminum dihydroxychloride; 24 – sodium hydrogenphosphate; 25 – cadmium carbonate; 26 – strontium hydrogensulfate; 27 – mercury(II) nitrate; 28 – barium chloride; 29 – tin(II) sulfate; 30 – silver chromate. 2. MAIN CONCEPTS AND LAWS OF CHEMISTRY 2.1. Atomic mass, molecular weight, mole, molar mass 2.1.1. Determine the mass in g and in a.m.u of an atom of the element: 01 – Titanium; 02 – Carbon; 03 – Zinc; 04 – Copper; 05 – Manganese; 06 – Chromium; 07 – Bromine; 08 – Calcium; 09 – Aluminum; 10 – Potassium; 11 – Silver; 12 – Barium; 13 – Silicon; 14 – Lithium; 15 – Nickel; 16 – Mercury; 17 – Iodine; 18 – Hydrogen; 19 – Lithium; 20 – Tin; 21 – Phosphorus; 22 – Antimony; 23 – Boron; 24 – Iron; 25 – Nitrogen; 26 – Magnesium; 27 – Sodium; 28 – Chlorine; 29 – Sulfur; 30 – Cadmium. 2.1.2.Determine the mass in g and in a.m.u of a molecule of the substance: 01 – carbonic acid; 02 – sulphuric acid; 03 – nitrous acid; 04 – ammonium hydroxide; 05 – zinc hydroxide; 06 – potassium sulfate; 07 – lithium sulfate; 08 – nitric acid ; 09 – barium sulfide; 10 – calcium hydrogensulfate; 11 – aluminum hydrogensulfate; 12 – sodium hydrogencarbonate; 13 – silver hydroxide; 14 – sulfurous acid; 15 – hydrosulphuric acid; 16 – barium hydroxide; 17 – copper(II) hydroxide; 18 – sodium silicate; 19 – magnesium hydrogensulfide; 20 – silver chromate; 21 – sodium chloride; 22 calcium nitrate; 23 – potassium sulfite; 24 – iron(III) hydrogensulfate; 25 – ammonium nitrite; 26 – aluminum phosphate; 27 – chromium(III) dihydroxochloride; 28 – sodium chromate; 29 – aluminum sulfate; 30 – calcium carbonate. 2.1.3. Calculate how many atoms fit into 01 – 15 g of titanium; 02 – 0,3 g of carbon; 03 – 17 g of zinc; 04 – 24 g of copper; 05 – 0,01 g manganese; 06 – 7 g of chromium; 07 – 5 g of bromine; 08 – 10 g of calcium; 09 – 50 g of aluminum; 10 – 15 g of potassium; 11 – 24 g of silver; 12 – 22 g of barium; 13 – 16 g of silicon; 14 – 0,9 g of lithium; 15 – 27 g of nickel; 16 – 41,2 g mercury; 17 – 12,7 g of iodine; 18 – 25,7 g of hydrogen; 19 – 24,3 g of lithium; 20 – 16,8 g of tin; 21 – 35,1 g of phosphorus; 22 – 33,5 g of stibium; 23 – 20,6 g boron; 24 – 33,9 g of iron; 25 – 36,8 g of nitrogen; 26 – 43,6 g of magnesium; 27 – 0,66 g of sodium; 28 – 1,23 g of chlorine; 29 – 2,27 g of sulfur; 30 – 23 g of cadmium. 5 2.1.4. Calculate how many molecules are contained in: 01 – 100 g of calcium carbonate; 02 – 54 g of chromium(III) dihydroxychloride; 03 – 22 g sodium chromate; 04 – 82 g aluminum sulfate; 05 – 4,8 g of iron(III) hydrogensulfate; 06 – 30 g of ammonium nitrite; 07 – 58 g of aluminum phosphate; 08 – 0,64 g of sodium chloride; 09 – 53 g of calcium nitrate; 10 – 21 g of potassium sulfite; 11 – 87 g of silver chromate; 12 – 5,21 g copper(II) hydroxide; 13 – 15,8 g sodium silicate; 14 – 5,4 g of magnesium hydrogensulfide; 15 – 73 g of sulphurous acid; 16 – 10,2 g of hydrosulphuric acid; 17 – 16,7 g barium hydroxide; 18 – 22,5 g aluminum hydrogensulfate; 19 – 36,2 g sodium hydrogencarbonate; 20 – 42,8 g of silver hydroxide; 21 – 51,3 g of nitric acid; 22 – 66,3 g barium sulfide; 23 – 72,2 g of calcium hydrogensulfate; 24 – 23,6 g of ammonium hydroxide; 25 – 44,6 g zinc hydroxide; 26 – 85,3 g of potassium sulfate; 27 – 44,9 g of lithium hydroxide; 28 – 1,3 g of carbonic acid; 29 – 0,66 g of sulphuric acid; 30 – 36,8 g of nitrite acid. 2.1.5. How many moles are the amount of substance specified in the section 2.1.4. 2.1.6. Determine the mass in g: 01 – 2 mol. of sodium chromate; 02 – 0,3 mol. of aluminum sulfate; 03 – 3,4 mol. of calcium carbonate; 04 – 6,3 mol. of aluminum phosphate; 05 – 5,1 mol. of chromium(III) dihydroxochloride; 06 – 3 mol. of potassium sulfite; 07 – 2,5 mol. of iron(III) hydrogensulfate; 08 – 3,2 mol. of ammonium nitrite; 09 – 6 mol. of sodium chloride; 10 – 2,8 mol. of calcium nitrate; 11 – 10 mol. of silver chromate; 12 – 7,3 mol. of copper(II) hydroxide; 13 – 0,43 mol. of sodium silicate; 14 – 3,8 mol. of magnesium hydrogensulfide; 15 – 11,5 mol. of sulfite acid; 16 – 6,22 mol. of hydrosulphuric acid; 17 – 0,7 mol. of barium hydroxide; 18 – 3 mol of aluminum hydrogensulfate; 19 – 1,4 mol. of sodium hydrogencarbonate; 20 – 22 mol. of silver hydroxides; 21 – 1,8 mol. of – nitrous acid; 22 – 2,27 mol. barium sulfide; 23 – 1,3 mol. of calcium hydrogensulfate; 24 – 2,9 mol. of ammonium hydroxide; 25 – 0,22 mol. of zinc hydroxide; 26 – 112 mol. of potassium sulfate; 27 – 44 mol. of lithium sulfate; 28 – 39 mol. of carbonic acid; 29 – 0,88 mol. of sulphuric acid; 30 – 5 mol. of – nitrous acid. 2.2. Equivalent, molar mass of equivalent of a substance 2.2.1. Calculate the equivalent and molar mass of the equivalent of the element specified in section 1.1. 2.2.2. Calculate the equivalent and molar mass of the oxide equivalent in section 1.2. 2.2.3. Calculate the equivalent and molar mass of the acid equivalent given in section 1.8. 6 2.2.4. Calculate the equivalent and molar mass of the base equivalent given in section 1.9. 2.2.5. Calculate the equivalent and the molar mass of the salt equivalent given in section 1.10. 2.2.6. Determine the mass in g of the specified number of equivalents of the substance: 01 – 3,0 – sulphuric acid; 02 – 0,4 – barium hydroxide; 03 – 6,0 – sodium hydroxide; 04 – 0,8 – hydrochloric acid; 05 – 5,1 – sodium carbonate; 06 – 2,1 – tin(IV) chloride; 07 – 1,4 – water; 08 – 6,2 – aluminum chloride; 09 – 18 –2,9 – phosphoric acid; 10 – 10,2 – calcium chloride; 11 – 5,0 – zinc oxide; 12 – 6,2 – nitrous acid; 13 – 0,41 – strontium hydroxide; 14 – 4,5 – ammonium nitrate; 15 – 5,5 – potassium chromate; 16 – 0,28 – sodium sulfite; 17 – 0,38 – barium sulfide; 18 – 2,9 – potassium sulfate; 19 – 3,6 – lead(II) bromide; 20 – 2,2 – lead(II) sulfate; 21 – 8,2 – silver carbonate; 22 – 1,3 – nickel(II) iodide; 23 – 0,84 – sodium sulfide; 24 – 11,5 – iron(III) hydroxide; 25 – 0,2 – zinc hydroxide; 26 – 7,2 – aluminum sulfate; 27 – 5,2 – zinc nitrite; 28 – 3,2 – magnesium chloride; 29 – 0,7 – sulfur trioxide; 30 – 0,5 – nitrogen dioxide. 2.2.7. How many equivalents are contained: 01 – 10 g of iron(III) hydroxide; 02 – 20,3 g of zinc hydroxide; 03 – 3,6 g of lead(II) bromide; 04 – 22,8 g of lead (II) sulfate; 05 – 13 g of sulphuric acid; 06 – 24 g of barium hydroxide; 07 – 6,5 g of sodium hydroxide; 08 – 28,6 g of chloride acid; 09 – 52,1 g of sodium carbonate; 10 – 21,2 g of the tin(IV) chloride; 11 – 1,4 g of water; 12 – 16,2 g of aluminum chloride; 13 – 11,8 g of – phosphoric acid; 14 – 1,2 g of calcium chloride; 15 – 15,3 g of zinc oxide;16 – 6,2 g of nitrite acid; 17 – 2,41 g of strontium hydroxide; 18 – 14,5 g of ammonium nitrate; 19 – 25,5 g of potassium chromate; 20 – 22,9 g of sodium sulfite; 21 – 3,38 g of barium sulfide; 22 – 12,9 g of potassium sulfate; 23 – 28,2 g of silver carbonate; 24 – 8,6 g of nickel(II) iodide; 25 – 84 g of sodium sulfide; 26 – 7,9 g of aluminum sulfate; 27 – 25,2 g of zinc nitrite; 28 – 13,2 g of magnesium chloride; 29 – 2,7 g of sulfur trioxide; 30 – 3,5 g nitrogen dioxide. 7 2.3. The law of equivalents 2.3.1. Determine the molar mass of the equivalent and the atomic mass of metal trivalent, when combustion m1 g, to which m2 g of metal oxide is formed. The values of m1 and m2 are respectively equal to: 01 – 7,50; 14,2 11 – 2,25; 4,25 21 – 7,75; 14,60 02 – 3,50; 6,60 12 – 3,75; 7,08 22 – 8,25; 15,58 03 – 3,00; 5,66 13 – 3,25; 6,14 23 – 8,75; 16,52 04 – 1,50; 2,83 14 – 1,75; 3,30 24 – 11,0; 20,77 05 – 8,00; 15,1 15 – 1,10; 2,08 25 – 12,0; 22,66 06 – 5,00; 9,44 16 – 1,60; 3,02 26 – 0,60; 1,13 07 – 1,00; 1,88 17 – 2,60; 4,90 27 – 13,0; 24,54 08 – 4,50; 8,50 18 – 2,10; 3,96 28 – 4,10; 7,74 09 – 6,50; 12,3 19 – 3,60; 6,80 29 – 14,0; 26,43 10 – 1,25; 2,36 20 – 3,10; 5,85 30 – 15,0; 28,32 2.3.2. Determine the molar mass of the equivalent and the atomic mass of the metal (II), if m2 g of salt was obtained when m1 g of metal and chlorine were connected, The molar mass of chlorine equivalent is 35,45 g/mol, The values of m1 and m2 are respectively: 01 – 2,16; 8,46 11 – 19,44; 76,14 21 – 24,48; 95,88 02 – 0,72; 2,86 12 – 20,88; 81,78 22 – 22,32; 87,42 03 – 25,2; 98,7 13 – 21,60; 84,60 23 – 18,72; 73,32 04 – 23,04; 90,24 14 – 9,36; 36,66 24 – 16,56; 64,86 05 – 17,28; 67,68 15 – 7,92; 31,02 25 – 13,68; 53,58 06 – 12,96; 50,76 16 – 10,08; 39,48 26 – 5,76; 22,56 07 – 7,20; 28,20 17 – 8,64; 33,84 27 – 6,48; 25,38 08 – 3,60; 14,10 18 – 14,40; 56,40 28 – 5,04; 19,74 09 – 20,16; 78,96 19 – 15,84; 62,04 29 – 4,32; 16,92 10 – 8,35; 32,46 20 – 15,12; 59,22 30 – 2,88; 11,28 2.3.3. Determine the molar mass of the equivalent and the atomic mass of the metal (II) if it is displaced from the sulfate acid of V l of hydrogen measured under normal conditions. The mass of the metal m and the volume of hydrogen V respectively are: 01 – 4,90; 1,68 11 – 9,48; 3,25 21 – 7,52; 2,58 02 – 5,23; 1,79 12 – 9,15; 3,14 22 – 7,19; 2,46 03 – 5,66; 1,90 13 – 8,82; 3,02 23 – 6,86; 2,36 04 – 1,96; 0,67 14 – 4,58; 1,57 24 – 6,54; 2,24 05 – 2,29; 0,78 15 – 4,25; 1,46 25 – 6,20; 2,13 06 – 2,61; 0,90 16 – 3,92; 1,34 26 – 5,88; 2,02 8 07 – 7,84; 2,69 17 – 3,59; 1,25 27 – 3,27; 1,12 08 – 8,17; 2,80 18 – 0,98; 0,34 28 – 2,94; 1,01 09 – 8,50; 2,91 19 – 0,65; 0,22 29 – 1,63; 0,56 10 – 9,80; 3,36 20 – 0,33; 0,11 30 – 1,31; 0,45 2.3.4. Determine the molar mass of the metal equivalent that is in contact with the element if it is known that this compound contains A % of the element, the molar mass equivalent of which is Mе g/mol, The values of А and Mе for the corresponding elements are: 01 – 06 – Sulfur; 48,04; 16,00 07 – 12 – Chlorine; 79,78; 35,45 13 – 18 – Florine; 45,24; 19,00 19 – 24 – Bromine; 80,00; 80,00 25 – 30 – Iodine; 94,84; 126,90 2.4. Avogadro's Law 2.4.1. Determine the mass of a substance that fits under normal conditions in: 01 – 3 l of ammonia; 02 – 3,8 l of a neon; 03 – 3 l of oxygen; 04 – 0,9 l of sulfur dioxide; 05 – 0,4 l of carbon dioxide; 06 – 112 l of oxygen; 07 – 0,5 l nitrogen dioxide; 08 – 1 l of nitrogen(IV) oxide; 09 – 0,6 l carbon(IV) oxide; 10 – 3,2 l of fluorine; 11 – 1,8 l of hydrogen; 12 – in 2,9 l of nitrogen(II) oxide; 13 – 1,4 l of helium; 14 – 1,5 l of nitrogen; 15 – 2,5 l of methane; 16 – 6,72 l of argon; 17 – 21,5 l of a neon; 18 – 2,8 l of hydrogen fluoride; 19 – 3,7 l of xenon; 20 – 2,3 l of chlorine; 21 – 2,5 l of hydrogen chloride; 22 – 0,6 l of sulfur dioxide; 23 – 2,6 l of hydrogen sulphide; 24 – 2,8 l of methane; 25 – 5 l of nitrogen dioxide; 26 – 15 l of hydrogen; 27 – 10,7 l of ammonia; 28 – 20 l of oxygen; 29 – 15 l of helium; 30 – 0,8 l of neon. 2.4.2. Determine the volume occupied under normal conditions: 01 – 36 g of nitrogen dioxide; 02 – 4,9 g of hydrogen sulphide; 03 – 42 g carbon(II) oxide; 04 – 5 g of hydrogen; 05 – 2,3 g of hydrogen fluoride; 06 – 3,4 g of oxygen; 07 – 6,8 g of nitrogen; 08 – 5,2 g of nitrogen(II) oxide; 09 – 7,3 g of helium; 10 – 8,6 g of argon; 11 – 72 g of sulfur trioxide; 12 – 9,2 g of hydrogen sulphide; 13 – 78 g of ammonia; 14–5,4 g of carbon dioxide; 15 – 5,8 g of hydrogen; 16–68 g of hydrogen sulphide; 17 – 98 g helium; 18 – 12 g of nitrogen dioxide; 19–25 g of hydrogen iodide; 20–67 g of hydrogen bromide; 21 – 8 g of sulfur dioxide; 22 – 16 g of acetylene; 23 – 15 g of nitrogen; 24 – 20 g of oxygen; 25 – 11,6 chlorine; 26 – 18 g of methane; 27 – 17 g of nitrogen oxide; 28 – 25 g of xenon; 29 – 0,7 g of nitrogen(I) oxide; 30 – 5 g of carbon(II) oxide. 9 3. CHEMICAL EQUILIBRIUM 3.1. Calculate the equilibrium constant and the initial concentrations of the reactants in the system (equilibrium concentrations, mol/l of substances indicated by the corresponding formula in the equation of reaction; the initial concentrations of the reaction products are 0): 3.1.1. 2 N O + O 2  2 N O 2 . 3.1.2. С О + C l 2  C O C l 2 . 01 0,2 0,1 0,1 06 1,0 1,5 1,4 02 0,3 0,2 0,1 07 0,3 0,4 0,2 03 0,4 0,1 0,2 08 2,0 1,8 1,6 04 1,5 0,7 0,6 09 3,0 2,0 2,0 05 2,0 1,4 1,6 10 0,5 0,4 0,1 3.1.3. N2O4  2NO2 . 3.1.4. H 2 + I 2  2 H I . 11 2,0 6,0 16 2,4 1,6 0,8 12 0,3 0,4 17 0,04 0,02 0,01 13 0,5 0,7 18 1,5 1,2 1,1 14 1,2 1,4 19 0,1 0,2 0,3 15 2,5 1,6 20 3,2 1,6 1,8 3.1.5. 2 C O + O 2  2 C O 2 . 3.1.6. 2NO + Cl 2  2NOCl . 21 1,2 1,4 1,4 26 0,03 0,02 0,01 22 0,3 0,5 0,7 27 0,4 0,6 0,6 23 0,01 0,02 0,03 28 0,6 0,2 0,1 23 2,2 1,8 1,4 29 2,6 1,8 1,6 25 1,6 0,3 0,4 30 4,0 2,2 2,6 3.2. Calculate the equilibrium concentrations of all substances and the constant of the equilibrium of the reaction (initial concentrations, mol/l, reactants and equilibrium concentration, mol/l of one of the products indicated by the corresponding formulas): 3.2.1. 2 N O 2  2 N O + O 2 . 3.2.2. 2 S О 2 + O 2  2 S O 3 . 01 0,5 0,1 06 0,8 0,6 0,2 02 0,4 0,01 07 2,4 1,8 0,6 03 2,5 1,4 08 0,8 0,4 0,1 04 2,8 0,6 09 1,2 0,8 0,4 05 0,08 0,02 10 3,6 2,2 0,8 10 3.2.3. 2 NO2  N2O4. 3.2.4. CO + Cl 2  2COCl 2 . 11 2,2 0,4 16 4,8 2,2 1,2 12 1,8 0,2 17 0,08 0,04 0,01 13 0,08 0,01 18 1,3 0,8 0,3 14 1,6 0,3 19 6,2 4,4 2,8 15 2,8 1,4 20 0,2 0,08 0,04 3.2.5. 2 H I  H 2 + I 2 . 3.2.6. 2CO + O 2  2CO 2 . 21 0,8 – 0,2 26 1,4 0,6 0,2 22 0,4 0,1 – 27 2,7 0,5 0,1 23 1,5 0,8 – 28 0,8 0,3 0,1 23 2,6 – 1,2 29 4,2 2,2 1,6 25 3,6 2,2 – 30 3,4 2,1 1,8 3.3. Calculate the equilibrium concentrations of the reactants (initial concentrations, mol/l of substances indicated by the corresponding formulas in the equation of reaction): 3.3.1. А + В  А В . 3.3.2. H2 + I2  2HI. 01 0,6 0,2 06 1,5 0,7 02 2,2 0,8 07 2,4 1,6 03 1,6 1,2 08 1,6 0,6 04 1,4 0,6 09 1,4 0,4 05 1,2 0,8 (Кр = 1) 10 1,6 0,2 (Кр = 2) 3.3.3. 2HCl  H2 + Cl 2 . 3.3.4. CO +Cl2  2COCl2. 11 0,6 16 1,4 8,8 12 0,8 17 0,4 0,1 13 1,4 18 1,2 0,4 14 0,2 19 1,8 1,6 15 1,8 (Кр = 0,1) 20 2,4 3,2 (Кр= 0,02) 3.3.5. А + В  C + D. 3.3.6. 2 N O + O 2  2 N O 2 . 21 2,0 4,0 26 1,2 0,8 22 3,0 6,2 27 0,6 0,4 23 4,2 4,4 28 3,8 2,2 24 1,2 1,0 29 4,6 1,2 25 6,0 2,2 (Кр = 3) 30 2,6 0,2 (Кр = 4) 11 3.4. Shift of chemical equilibrium In what direction the equilibrium is shifted at the specified pressure changes (P), concentration (C) of one of the reactants, temperature (Tо) for the following reciprocal reactions: 3.4.1. 2SО2 + O2 2SO3; Но298 = – 196,6 кДж. 01 Р will rise, [SO3] will decrease 02 Р will decrease, [SO2] will rise 03 [O2] will increase, То will rise 04 [O2] will decrease, То will decrease 05 [SO3] will increase, [SO2] will decrease 3.4.2. N2 + O2 2NO; Но298 = – 180,6 кДж. 06 [N2] will increase, То will decrease 07 [NО] will decrease, То will rise 08 [O2] will increase, Р will rise 09 [O2] will decrease, Р will decrease 10 [NО] will increase, [N2] will decrease 3.4.3. COCl2  CO + Cl 2 ; Но298 = – 112,5 кДж. 11 [CОCl2] will increase, То will rise 12 [CОCl2] will decrease, То will decrease 13 [CО] will increase, Р will decrease 14 [Cl2] will decrease, Р will rise 15 [CО] will decrease, [Cl2] will increase 3.4.4. 2H2 + O2 2H2O; Но298 = – 483,6 кДж. 16 [H2] will decrease, То will rise 17 [O2] will increase, То will decrease 18 [H2] will increase, [H2O] will decrease 19 [O2] will decrease, Р will rise 20 [H2O] will increase, Р will decrease 3.4.5. 2CO + O2  2CO2; Но298 = – 566 кДж. 21 [CО] will increase, [СО2] will decrease 22 [CО] will decrease, То will rise 23 [O2] will decrease, То will decrease 24 [СО2] will decrease, Р will decrease 25 [O2] will increase, Р will rise 3.4.6. N2 + 3Н2  2NH3; Но298 = – 92,4 кДж. 26 [N2] will increase, Р will decrease 27 [H2] will decrease, То will rise 28 [N2] will decrease, Р will rise 29 [NH3] will increase, То will decrease 30 [H2] will increase, [NH3] will decrease 12 4. CONCENTRATION OF SOLUTIONS 4.1. How many grams, moles, and equivalents of the dissolved substance are contained in: 01 – 250 g 8 % solution of potassium carbonate; 02 – 200 g 20 % solution of sulphuric acid; 03 – 150 g 15 % solution of hydrochloric acid; 04 – 180 g 17,5 % solution of potassium hydroxide; 05 – 100 g 3 % solution of silver nitrate; 06 – 270 g 10 % solution of sodium dichromate; 07 – 50 g of 5 % potassium permanganate solution; 08 – 300 g 10 % solution of lithium chloride; 09 – 120 g 7 % solution of manganese(II) sulfate; 10 – 400 g of 12 % sodium chloride solution; 11 – 160 g 20 % solution of ammonium hydroxide; 12 – 170 g 2 % solution of barium nitrate; 13 – 210 g 11 % solution of sodium nitrate; 14 – 180 g 5 % solution of sodium hydroxide; 15 – 190 g of 11 % potassium iodide solution; 16 – 110 g 30 % solution of hydrochloric acid; 17 – 75 g of 5 % potassium permanganate solution; 18 – 210 g 27 % solution of sulphuric acid; 19 – 700 g 3 % solution of phosphate acid; 20 – 90 g 45 % solution of hydrobromic acid; 21 – 40 g 32 % solution of iodide acid; 22 – 70 g 90 % solution of sulfate acid; 23 – 170 g 11 % solution of potassium carbonate; 24 – 400g of 53 % hydrochloric acid solution; 25 – 160 g 3 % solution of hydrosulphuric acid; 26 – 30 g 5 % solution of nitric acid; 27 – 180 g 3% solution of bismuth(III) nitrite; 28 – 145 g 8 % solution of magnesium chloride; 29 – 250 g of 1 % – boric acid solution; 30 – 115 g 3 % solution of sodium sulfate. 4.2. How many grams, moles, and equivalents of the dissolved substance are contained in: 01 – 250 ml 20% solution of ammonium hydroxide (*= 0,923); 02 – 175 ml 8 % solution of aluminum chloride ( = 1,071); 03 – 310 ml 50 % solution of sodium hydroxide ( = 1,525); 04 – 160 ml 40 % solution of calcium chloride ( = 1,395); 05 – 220 ml 4 % solution of sodium carbonate ( = 1,039); 06 – 50 ml 8 % solution of acetic acid ( = 1,010); 07 – 300 ml 5 % solution of potassium hydroxide ( = 1,045); 08 – 190 ml 27 % solution of hydrochloric acid ( = 1,135); 09 – 40 ml 50 % solution of potassium carbonate ( = 1,540); 10 – 420 ml 8 % solution of potassium dichromate ( = 1,055); 11 – 165 ml 20 % solution of silver nitrate ( = 1,194); 12 – 69 ml 4,2 % solution of cooper(II) sulfate ( = 1,040); 13 – 170 ml 60 % solution of zinc chloride ( = 1,568); 14 – 150 ml 35,5 % solution of phosphoric acid ( = 1,220); 15 – 115 ml 8 % solution of sodium sulfate ( = 1,072); 16 – 56 ml 50 % solution of potassium iodide ( = 1,545); 17 – 250 ml 4 % solution of iron(III) sulfate( = 1,033); 18 – 620 ml 4 % solution of barium chloride ( = 1,034); 19 – 160 ml 27 % solution of nitric acid ( = 1,160); 20 – 170 ml 8 % solution of cooper(II) sulfate ( = 1,084); 21 – 360 ml 50 % solution of iron(III) chloride ( = 1,551); 22 – 112 ml 4 % solution of lithium hydroxide ( = 1,043); 23 – 170 ml 5 % solution of cadmium sulfate ( = 1,047); 24 – 370 ml 13,5 % ________________________________________________________________ * = g/ml 13 solution of hydrochloric acid ( = 1,066); 25 – 450 ml 40 % solution of potassium bromide ( = 1,374); 26 – 107 ml 8 % solution of zinc chloride ( = 1,071); 27 – 210 ml 50 % solution of silver nitrate ( = 1,608); 28 – 650 ml 20 % solution of aluminum sulfate ( = 1,226); 29 – 340 ml 5 % solution of lithium hydroxide ( = 1,047); 30 – 700 мл 4 % solution of potassium dichromate ( = 1,026). 4.3. How many grams, moles, and equivalents of the dissolved substance are contained in: 01 – 0,5 l 0,3 M solution of hydrochloric acid; 02 – 4,2 l of 0,16 M ammonium chloride solution; 03 – 6,2 l 0,35 M solution of strontium nitrate; 04 – 0,25 l of 0,18 M potassium dichromate solution; 05 – 3,2 l 0,21 M solution of lithium hydroxide; 06 – 1,5 l of 0,8 M sodium carbonate solution; 07 – 10,0 l of 0,01 M boron acid solution; 08 – 10,5 l of 0,1 M sodium sulfide solution; 09 – 0,5 l of 0,25 M potassium bromide solution; 10 – 1,5 l 0,7 M solution of chloride acid; 11 – 8,0 l 1 M solution of cobalt nitrate; 12 – 4,6 l 0,5 M solution of silver nitrite; 13 – 6,0 l 12 M solution of sulphuric acid; 14 – 20,0 l of 2 M potassium iodide solution; 15 –3,0 l 1,25 M solution of ammonium hydroxide; 16 – 16,0 l 10,5 M solution of nitrite acid; 17 – 14,0 l 2 M solution of potassium carbonate; 18 – 15,0 l 2,5 M solution of ammonium carbonate; 19 – 7,0 l 0,18 M solution of sodium sulfate; 20 – 2,4 l 0,15 M solution of sulphurous acid; 21 – 1,25 l 0,4 M solution of aluminum chloride; 22 – 1,75 l 2 M solution of hydrosulphuric acid; 23 – 1,96 l 0,17 M solution of magnesium sulfate; 24–2,6 l of 0,002 M potassium bromide solution; 25 – 0,5 l 0,7 M solution of copper(II) sulfate; 26 – 3,7 l 0,75 M solution of cesium sulfate; 27 – 5,2 l 0,2 M solution of lithium iodide; 28 – 3,73 l 0,14 M solution of nickel (II) chloride; 29 – 2,8 l 0,7 M solution of potassium sulfate; 30 – 16,0 l 0,3 M solution of iron(III) sulfate. 4.4. Determine the molar, normal and molar concentration of the solution, the percentage concentration and density of which are given in section 4.2. 4.5. Determine the molar concentration of the solution: 01 – 2 N iron(III) chloride; 02 – 1,5 N – phosphoric acid; 03 – 6 N ammonium hydroxide; 04 – 1,2 N sodium carbonate; 05 – 0,02 N barium chloride; 06 – 0,17 N zinc sulfate; 07 – 1,5 N sulphuric acid; 08 – 3 N sodium hydroxide; 09–4 N hydrochloric acid; 10 – 1,7 N ammonium chloride; 11 – 0,2 N sulphuric acid; 12 – 0,11 N ammonium sulfate; 13 – 0,6 N potassium chloride; 14 – 0,04 N potassium nitrate; 15 – 0,15 N sodium phosphate; 16 – 0,05 N copper(II) sulfate; 17 – 0,01 N sodium hydroxide; 18 – 0,3 N of iron(II) chloride; 19 – 0,5 N calcium chloride; 20 – 0,2 N aluminum nitrate; 21 – 0,7 N – nitrous acid; 22 – 0,25 N sulphurous acid; 23 – 2,2 N calcium bromide; 24 – 2,8 N lithium iodide; 25 – 0,3 N barium hydroxide; 26 – 2,4 N aluminum sulfate; 27 – 0,5 N nitric acid; 28 – 0,01 N lithium carbonate; 29 – 1,75 N rubidium sulfate; 30 – 0,82 N of potassium permanganate, 14 4.6. Determine the normal concentration of the solution: 01 – 2 M sulphuric acid; 02 – 1,5 M phosphoric acid; 03 – 0,21 M iron(III) chloride; 04 – 0,5 M aluminum sulfate; 05 – 7,4 M nitric acid; 06 – 0,32 M zinc chloride; 07 – 1,7 M potassium bromide; 08–3M acetic acid; 09 – 0,14 M hydrofluoric acid; 10 – 0,2 M hydrosulphuric acid; 11 – 0,2 M iron(II) sulfate; 12 – 0,4 M potassium permanganate; 13 – 0,4 M potassium sulfate; 14–2,1 M ammonium chloride; 15–3 M – phosphoric acid; 16 – 1,7 M nitric acid; 17 – 0,002 M lead(II) chloride; 18 – 0,02 M aluminum sulfate; 19 – 1,7 M sodium chloride; 20 – 0,3 M potassium carbonate; 21 – 0,75 M ammonium hydroxide; 22 – 1,8 M calcium iodide; 23 – 0,5 M copper(II) sulfate; 24 – 0,1 M cadmium sulfate; 25 – 0,35 M sodium chromate; 26 – 0,8 M tin(II) chloride; 27 – 0,8 M barium nitrate; 28 – 0,33 M sodium sulfide; 29 – 0,13 M zinc sulfate; 30 – 0,7 M magnesium bromide. 5. PROPERTIES OF SOLUTIONS OF NON–ELECTRICITY 5.1. Determine the osmotic pressure of the solution, which is contained in V l m g non-electrolyte at t oC, The values m, V, and t respectively correspond to: 5.1.1. (glucose С6Н12О6 ) 01 – 24; 1,5; 20,2 04 – 36; 1,2; 14,0 02 – 44; 2,2; 37,0 05 – 12; 0,8; 22,9 03 – 62; 4,6; 32,6 06 – 16; 1,1; 18,0 5.1.2. (glycerin С3Н8О3) 07 – 12,0; 1,2; 16,0 10 – 3,2 1,2; 14,0 08 – 8,4; 0,8; 5,6 11 – 4,8; 2,2; 18,0 09 – 4,6; 0,6; 12,9 12 – 5,2; 4,2; 20,0 5.1.3. (sugar С12Н22О11) 13 – 2,22; 0,8; 16,2 16 – 2,44; 1,4; 20,9 14 – 3,28; 0,6; 18,0 17 – 4,22 1,6; 14,9 15 – 1,46; 1,2; 22,6 18 – 3,12; 0,8; 12,2 5.1.4. (aniline С6Н5NH2) 19 – 10,2; 0,8; 20,0 22 – 8,6; 1,26; 14,0 20 – 12,6; 1,2; 10,6 23 – 9,8; 1,6; 18,2 21 – 14,4; 1,4; 12,0 24 – 13,2; 1,8; 16,9 5.1.5. (methyl alcohol СН3ОН) 25 – 3,2; 1,0; 18,2 28 – 5,2; 0,8; 10,2 26 – 4,6; 1,2; 12,3 29 – 5,6; 1,4; 15,2 27 – 2,8; 1,6; 14,5 30 – 4,2; 1,8; 17,4 15 5.2. Determine the boiling and freezing point of A % solution of the non– electrolyte in a suitable solvent, The value of A equals: 5.2.1. niatrobenzene С6Н5NО2 in benzene, Кз = 5,1 grade/mol; Кк = 2,57 grade/mol, tз= –5,4 оС; tк = 80,2 оС. 001 –– 5,0 04 –– 12,4 007 –– 8,6 10 –– 5,2 13 –– 2,4 002 –– 7,2 005 –– 8,3 008 –– 6,8 11 –– 4,8 14 –– 3,6 003 –– 10,8 006 –– 9,6 009 –– 7,6 12 –– 3,6 15 –– 4,2 5.2.2. glycerine С3Н8О3 in acetone, Кз = 2,4 grade/mol; Кк = 1,48 grade/mol; tз= –94,6 оС; tк = 56,0 оС. 16 –– 1,2 19 –– 4,2 22 –– 7,8 25 –– 14,4 28 –– 6,8 17 –– 4,8 20 –– 5,8 23 –– 10,2 26 –– 16,8 29 –– 7,2 18 –– 6,6 21 –– 3,6 24 –– 12,4 27 –– 10,8 30 –– 8,4 6. PROPERTIES OF SOLUTIONS OF HIGH ELECTROLYTES, IZOTONIC COEFFICIENT, DISSOCIATION DEGREE OF ELECTROLYTE 6.1. Calculate the isotonic coefficient См of the electrolyte solution, whose osmotic pressure at t ° C is P kPa, The values of См, P and t, respectively, are: 6.1.1. ZnSO4 ; См = 0,05 М. 6.1.2. Ca(NO3)2; См = 0,1 М. 01 – 159; 0 04 – 167; 14 06 – 569; 3 09 – 598; 17 02 – 160; 2 05 – 165; 10 07 – 581; 9 10 – 575; 6 03 – 162; 5 08 – 590; 13 6.1.3. HNO3; См = 1 М. 6.1.4.Ca(OH)2; См = 0,05 М. 11 – 4130; 0 14 – 4267; 9 16 – 299; 8 19 – 303; 12 12 – 4367; 15 15 – 4372; 16 17 – 295; 4 20 – 310; 19 13 – 4237; 7 18 – 294; 3 6.1.5. BaCl2; См = 0,01 М. 6.1.6. HCl; См = 0,5 М. 21 – 328; 13 24 – 326; 11 26 – 2238; 15 29 – 2184; 7 22 – 335; 19 25 – 320; 6 27 – 2254; 17 30 – 2200; 10 23 – 318; 4 28 – 2192; 8 6.2. Calculate the isotonic coefficient of solution, which contains а g of electrolyte in1000 g of water and boil at t °C (Kb = 0,156 grade/mol), the values of а and t tare respectively: 16 6.2.1. NaOH 6.2.2. KBr 01 – 8; 100,184 06 – 180; 101,330 02 – 13; 100,299 07 – 210; 101,550 03 – 17; 100,390 08 – 150; 101,110 04 – 25; 100,574 09 – 140; 101,030 05 – 41; 100,940 10 – 130; 103,960 6.2.3. Ba(NO3)2 6.2.4. KCl 11 – 20; 100,076 16 – 210; 102,530 12 – 24; 100,091 17 – 215; 102,590 13 – 38; 100,144 18 – 225; 102,710 14 – 45; 100,170 19 – 230; 102,770 15 – 32; 100,122 20 – 248; 102,980 6.2.5. ZnCl2 6.2.6. KOH 21 – 6; 100,063 26 – 8; 100,137 22 – 12; 100,037 27 – 15; 100,257 23 – 4; 100,068 28 – 20; 100,342 24 – 10; 100,093 29 – 17; 100,291 25 – 15; 100,140 30 – 6; 100,103 6.3. Calculate the degree of dissociation of the electrolyte, whose isotonic coefficient of solution is equal to і, The value і is equal to: 01 – HBr; 1,89 11 – NaOH; 1,73 21 – CaCl2; 2,76 02 – HCl; 1,78 12 – KCl; 1,85 22 – Lі2SO4; 2,90 03 – HF; 1,07 13 – KI; 1,93 23 – CsCl; 1,85 04 – HI; 1,90 14 – K2S; 1,98 24 – RbCl; 1,93 05 – HNO3; 1,80 15 – BaCl2; 2,76 25 – NaI; 1,85 06 – H3PO4; 1,51 16 – LiCl; 1,85 26 – NaBr; 1,97 07 – H2SO4; 2,00 17 – Na2SO4; 2,76 27 – ZnCl2; 2,38 08 – Ba(OH)2; 2,38 18 – MgSO4; 1,43 28 – CrCl3; 2,95 09 – Ca(OH)2; 2,56 19 – MnSO4; 1,66 29 – Rb2SO4; 2,74 10 – KOH; 1,77 20 – K3PO4; 2,92 30 – Cs2SO4; 2,86 6.4. Calculate the degree of dissociation of the electrolyte, in which the water solution contains n mole of substance per V l of water, if the freezing point of the solution is t oC (Kf = 1,86 grade/mol), the values of V and t respectively are: 6.4.1. HNO3; n = 0,25. 6.4.2. MgSO4; n = 0,1. 01 – 1,7; (–0,51) 04 – 3,5; (–0,25) 06 – 3,4; (–0,04) 09 – 2,0; (–0,13) 02 – 2,2; (–0,40) 05 – 4,6; (–0,20) 07 – 2,6; (– 0,10) 10 – 3,0; (–0,09) 03 – 2,5; (–0,35) 08 – 1,8; (– 0,15) 17 6.4.3. Ba(OH)2; n = 1,5. 6.4.4. LiCl; n = 0,2. 11 – 2,7; (–2,45) 14 – 1,5; (– 4,4) 16 – 3,6; (– 0,19) 19 – 4,4; (–0,16) 12 – 3,8; (– 1,74) 15 – 3,2; (–2,06) 17 – 5,0; (– 0,14) 20 – 3,1; (–0,22) 13 – 1,2; (– 5,50) 18 – 2,0; (– 0,34) 6.4.5. K3PO4; n = 0,3. 6.4.6. Na2SO4; n = 0,04. 21 – 2,5; (–0,65) 24 – 5,2; (–0,31) 26 – 0,8; (– 0,26) 29 – 4,0 (– 0,05) 22 – 4,5; (–0,36) 25 – 3,8; (– 0,43) 27 – 1,2; (– 0,17) 30 – 2,2 (– 0,09) 23 – 3,0; (–0,56) 28 – 0,5; (– 0,41) 6.5. Exchange reactions between electrolytes in a solution. Write in the molecular and molecular-ionic form the reaction interaction equation for the following substances: 01 – KCN + HCl  16 – Zn(OH)2 + KOH  02 – Na2S + FeSO4  17 – Ba(OH)2 + HCl  03 – NaCN + HNO3  18 – Al(OH)3 + NaOH  04 – CH3COONa + HNO3  19 – Cu(OH)2 + H2SO4  05 – Na2S + HCl  20 – CH3COOK + HCl  06 – H2SO4 + KOH  21 – LiCN + H2SO4  07 – Pb(NO3)2 + NaI  22 – Na2CO3 + HNO3  08 – Cu(NO3)2 + Na2SO4  23 – K2S + H2SO4  09 – BaCl2 + K2SO4  24 – (NH4)2SO4 + KOH  10 – KNO3 + NaCl  25 – Ag NO3 + Na2S  11 – AgNO3 + KCl  26 – CdCO3 + HNO3  12 – CaCO3 + HCl  27 – AlCl3 + NaOH  13 – Ba(OH)2 + HNO3  28 – CuSO4 + Na2CO3  14 – SrSO4 + BaCl2  29 – CuCl2 + K2S  15 – NH4Cl + Ca(OH)2  30 – H2SO4 + Ca(OH)2  6.6. Ionic product of water, Hydrogen index (pН) 6.6.1. Calculate the pH of the solution of strong acid (= 1), The value of См mol/l, is equal to: 6.6.1.1. H2SO4 6.6.1.2. HNO3 01 – 2,210 –2 04 – 1,210 –3 06 – 1,610 –3 09 – 2,310 –1 02 – 5,610 –4 05 – 7,210 –4 07 – 2,410 –2 10 – 5,810 –3 03 – 4,210 –3 08 – 1,710 –3 18 6.6.1.3. HCl 6.6.1.4. HClO4 11 – 2,910 –3 14 – 1,410 –2 16 – 7,710 –3 19 – 3,510 –2 12 – 5,710 –4 15 – 7,810 –3 17 – 8,110 –6 20 – 4,410 –4 13 – 9,210 –5 18 – 2,510 –3 6.6.1.5. HІ 6.6.1.6. HBr 21 – 2,510 –5 24 – 5,710 –3 26 – 0,810 –6 29 – 4,010 –3 22 – 4,510 –3 25 – 3,810 –5 27 – 1,210 –7 30 – 2,210 –5 23 – 3,010 –4 28 – 0,510 –5 6.6.2. Calculate the pH and рОН of the solution of alkali (= 1), The value of См mol/l, is equal to: 6.6.2.1. NaOH 6.6.2.2. Ca(OH)2 01 – 5,210 –2 04 – 6,210 –3 06 – 1,910 –3 09 – 8,310 –1 02 – 9,610 –4 05 – 3,610 –4 07 – 6,410 –2 10 – 3,810 –3 03 – 4,310 –3 08 – 7,710 –3 6.6.2.3. LiOH 6.6.2.4. Вa(OH)2 11 – 2,610 –3 14 – 5,410 –2 16 – 7,810 –3 19 – 6,510 –2 12 – 3,710 –4 15 – 7,810 –3 17 – 5,110 –6 20 – 4,910 –4 13 – 3,210 –5 18 – 2,810 –3 6.6.2.5. KOH 6.6.2.6. RbOH 21 – 6,510 –5 24 – 5,310 –3 26 – 5,810 –6 29 – 4,310 –3 22 – 5,510 –3 25 – 3,410 –5 27 – 6,210 –7 30 – 2,610 –5 23 – 3,510 –4 28 – 4,510 –5 6.6.3. Calculate the concentration of H + and ОН  ions in a solution, whose pH is: 01 – 1,28 11 – 3,86 21 – 13,17 02 – 2,34 12 – 4,56 22 – 12,89 03 – 11,07 13 – 8,12 23 – 4,68 04 – 10,12 14 – 3,14 24 – 3,08 05 – 2,52 15 – 7,18 25 – 8,35 06 – 7,34 16 – 9,28 26 – 9,12 07 – 12,83 17 – 2,67 27 – 5,63 08 – 5,46 18 – 12,84 28 – 3,89 09 – 11,28 19 – 13,58 29 – 6,54 10 – 5,18 20 – 11,23 30 – 7,38 19 6.7. Hydrolysis of salts 6.7.1. Make up the molecular and ionic hydrolysis equation of a strong acid and weak base: 01 – CuSO4 11 – NH4Br 21 – CdCl2 02 – Fe(NO3)2 12 – CuCl2 22 – Ni(NO3)2 03 – (NH)2SO4 13 – Al2(SO4)3 23 – AlI3 04 – ZnCl2 14 – NiI2 24 – NiBr2 05 – CuI2 15 – ZnSO4 25 – Fe2(SO4)3 06 – Al(NO3)3 16 – Fe(NO3)3 26 – CdI2 07 – NH4NO3 17 – NH4Cl 27 – NiSO4 08 – ZnBr2 18 – CuBr2 28 – NiCl2 09 – CdSO4 19 – FeSO4 29 – AlBr3 10 – FeCl3 20 – FeCl2 30 – Co (NO3)2 6.7.2. Make a molecular and ionic hydrolysis equation of a weak acid and strong base: 01 – NaNO2 11 – Ca(CN)2 21 – Rb2CO3 02 – CaCl2 12 – Li2S 22 – CaSO3 03 – Ba(CN)2 13 – K2SO3 23 – CsNO2 04 – Li3PO4 14 – NaCN 24 – LiCN 05 – Na2SO3 15 – Rb2S 25 – Cs2S 06 – Li2CO3 16 – K3PO4 26 – Na2SO3 07 – KNO2 17 – K2S 27 – CsCN 08 – Cs2CO3 18 – RbCN 28 – Ca(NO2)2 09 – BaSO3 19 – Na3PO4 29 – LiNO2 10 – Na2S 20 – RbNO2 30 – Li2SO3 6.7.3. Make a molecular and ionic equation for the hydrolysis of a weak acid and weak base: 01 – 05 – Cr2S3 16 – 20 – (NH4)3PO4 06 – 10 – Al(CH3COO)3 21 – 25 – Al2S3 11 – 15 – Al2(CO3)3 26 – 30 – Fe2(SO3)3 7. Atomic Structure 7.1. Write the electronic formula of the atom and show WHICH s–, p–, d– or f–elements it belongs to: 01 – Radon 11 – Titanium 21 – Arsenic 02 – Bismuth 12 – Sulfur 22 – Erbium 03 – Lead 13 – Iron 23 – Germanium 04 – Hafnium 14 – Antimony 24 – Magnesium 05 – Osmium 15 – Niobium 25 – Prometheum 06 – Barium 16 – Selenium 26 – Nickel 07 – Platinum 17 – Rhodium 27 – Indium 08 – Yttrium 18 – Cobalt 28 – Xenon 09 – Chromium 19 – Bromine 29 – Zinc 10 – Strontium 20 – Tin 30 – Mendelevium 20 Свєткіна Олена Юріївна Нетяга Ольга Борисівна Тарасова Ганна Володимирівна ХІМІЯ МЕТОДИЧНІ РЕКОМЕНДАЦІЇ ТА ЗАВДАННЯ до самостійної роботи з дисципліни для студентів усіх спеціальностей (Частина 1) (Англійською мовою) Видано в редакції авторів Підп. до друку 01.03.2018. Формат 30 х 42/4 Папір офсет. Ризографія. Ум. друк. арк. 1,3. Обл.-вид. арк. 1,6. Тираж 30 пр. Зам. НТУ «Дніпровська політехніка» 49005, м. Дніпро, просп. Д. Яворницького, 19.