JK-TK/Newdata huggingface dataset:
6.Formation of synthetic rubber Synthetic rubber is able to resist action of oil,abrasion and organic solvents which rubber cannot. Common synthetic rubber is a polymer of 2-chlorobut-1,3-diene ; H Cl H H CH2=C (Cl CH = CH2 H - C = C – C = C - H During polymerization to synthetic rubber, one double C=C bond is broken to self add to another molecule. The double bond remaining move to carbon “2” thus; H Cl H H H Cl H H - C - C = C - C - C - C = C - C - H H H H Generally the structure of rubber is thus; Sulphur atoms make cross link between polymers
114 H Cl H H -(- C - C = C - C -)n- H H Rubber is thus strengthened through vulcanization and manufacture of synthetic rubber. (b)Condensation polymerization Condensation polymerization is the process where two or more small monomers join together to form a larger molecule by elimination/removal of a simple molecule. (usually water). Condensation polymers acquire a different name from the monomers because the two monomers are two different compounds During condensation polymerization: (i)the two monomers are brought together by high pressure to reduce distance between them. (ii)monomers realign themselves at the functional group. (iii)from each functional group an element is removed so as to form simple molecule (of usually H2O/HCl) (iv)the two monomers join without the simple molecule of H2O/HCl Examples of condensation polymerization 1.Formation of Nylon-6,6 Method 1: Nylon-6,6 can be made from the condensation polymerization of hexan1,6-dioic acid with hexan-1,6-diamine.Amines are a group of homologous series with a general formula R-NH2 and thus -NH2 as the functional group. During the formation of Nylon-6,6: (i)the two monomers are brought together by high pressure to reduce distance between them and realign themselves at the functional groups.
(ii)monomers realign themselves at the functional group. (iii)from each functional group an element is removed so as to form simple molecule (of usually H2O/HCl) (iv)the two monomers join without the simple molecule of H2O/HCl Examples of condensation polymerization 1.Formation of Nylon-6,6 Method 1: Nylon-6,6 can be made from the condensation polymerization of hexan1,6-dioic acid with hexan-1,6-diamine.Amines are a group of homologous series with a general formula R-NH2 and thus -NH2 as the functional group. During the formation of Nylon-6,6: (i)the two monomers are brought together by high pressure to reduce distance between them and realign themselves at the functional groups. O O H H H- O - C – (CH2 ) 4 – C – O - H + H –N – (CH2) 6 – N – H
115 (iii)from each functional group an element is removed so as to form a molecule of H2O and the two monomers join at the linkage . O O H H H- O - C – (CH2 ) 4 – C – N – (CH2) 6 – N – H + H 2O . Polymer bond linkage Nylon-6,6 derive its name from the two monomers each with six carbon chain Method 2: Nylon-6,6 can be made from the condensation polymerization of hexan1,6-dioyl dichloride with hexan-1,6-diamine. Hexan-1,6-dioyl dichloride belong to a group of homologous series with a general formula R-OCl and thus -OCl as the functional group. The R-OCl is formed when the “OH” in R-OOH/alkanoic acid is replaced by Cl/chlorine/Halogen During the formation of Nylon-6,6: (i)the two monomers are brought together by high pressure to reduce distance between them and realign themselves at the functional groups.
Polymer bond linkage Nylon-6,6 derive its name from the two monomers each with six carbon chain Method 2: Nylon-6,6 can be made from the condensation polymerization of hexan1,6-dioyl dichloride with hexan-1,6-diamine. Hexan-1,6-dioyl dichloride belong to a group of homologous series with a general formula R-OCl and thus -OCl as the functional group. The R-OCl is formed when the “OH” in R-OOH/alkanoic acid is replaced by Cl/chlorine/Halogen During the formation of Nylon-6,6: (i)the two monomers are brought together by high pressure to reduce distance between them and realign themselves at the functional groups. O O H H Cl - C – (CH2 ) 4 – C – Cl + H –N – (CH2) 6 – N – H (iii)from each functional group an element is removed so as to form a molecule of HCl and the two monomers join at the linkage . O O H H Cl - C – (CH2 ) 4 – C – N – (CH2) 6 – N – H + HCl . Polymer bond linkage The two monomers each has six carbon chain hence the name “nylon-6,6” The commercial name of Nylon-6,6 is Nylon It is a a tough, elastic and durable plastic. It is used to make clothes, plastic ropes and carpets. 2.Formation of Terylene
116 Method 1: Terylene can be made from the condensation polymerization of ethan1,2-diol with benzene-1,4-dicarboxylic acid. Benzene-1,4-dicarboxylic acid a group of homologous series with a general formula R-COOH where R is a ring of six carbon atom called Benzene ring .The functional group is -COOH. During the formation of Terylene: (i)the two monomers are brought together by high pressure to reduce distance between them and realign themselves at the functional groups.
2.Formation of Terylene
116 Method 1: Terylene can be made from the condensation polymerization of ethan1,2-diol with benzene-1,4-dicarboxylic acid. Benzene-1,4-dicarboxylic acid a group of homologous series with a general formula R-COOH where R is a ring of six carbon atom called Benzene ring .The functional group is -COOH. During the formation of Terylene: (i)the two monomers are brought together by high pressure to reduce distance between them and realign themselves at the functional groups. O O H- O - C – C6H5 – C – O - H + H –O – CH2 CH2 – O – H (iii)from each functional group an element is removed so as to form a molecule of H2O and the two monomers join at the linkage . O O H- O - C – C6H5 – C – O – (CH2) 6 – N – H + H 2O . Polymer bond linkage of terylene
117 Method 2: Terylene can be made from the condensation polymerization of benzene-1,4-dioyl dichloride with ethan-1,2-diol. Benzene-1,4-dioyl dichloride belong to a group of homologous series with a general formula R-OCl and thus -OCl as the functional group and R as a benzene ring. The R-OCl is formed when the “OH” in R-OOH is replaced by Cl/chlorine/Halogen During the formation of Terylene (i)the two monomers are brought together by high pressure to reduce distance between them and realign themselves at the functional groups. O O Cl - C – C5H5 – C – Cl + H –O – CH2 CH2 – O - H (iii)from each functional group an element is removed so as to form a molecule of HCl and the two monomers join at the linkage . O O Cl - C – C5H5 – C – O – CH2 CH2 – O – H + HCl . Polymer bond linkage of terylene
118 The commercial name of terylene is Polyester /polyster It is a a tough, elastic and durable plastic.
O O Cl - C – C5H5 – C – Cl + H –O – CH2 CH2 – O - H (iii)from each functional group an element is removed so as to form a molecule of HCl and the two monomers join at the linkage . O O Cl - C – C5H5 – C – O – CH2 CH2 – O – H + HCl . Polymer bond linkage of terylene
118 The commercial name of terylene is Polyester /polyster It is a a tough, elastic and durable plastic. It is used to make clothes, plastic ropes and sails and plastic model kits. 119 Practice questions Organic chemistry 1. A student mixed equal volumes of Ethanol and butanoic acid. He added a few drops of concentrated Sulphuric (VI) acid and warmed the mixture (i) Name and write the formula of the main products Name…………………………………. Formula…………………………………….. (ii) Which homologous series does the product named in (i) above belong? 2. The structure of the monomer phenyl ethene is given below:- a) Give the structure of the polymer formed when four of the monomers are added together b) Give the name of the polymer formed in (a) above 3. Explain the environmental effects of burning plastics in air as a disposal method 4. Write chemical equation to represent the effect of heat on ammonium carbonate 5. Sodium octadecanoate has a chemical formula CH3(CH2)6 COO-Na+, which is used as soap. HC = CH2 O
120 Explain why a lot of soap is needed when washing with hard water 6. A natural polymer is made up of the monomer: (a) Write the structural formula of the repeat unit of the polymer (b) When 5.0 x 10-5 moles of the polymer were hydrolysed, 0.515g of the monomer were obtained. Determine the number of the monomer molecules in this polymer. (C = 12; H = 1; N = 14; O =16) 7. The formula below represents active ingredients of two cleansing agents A and B O CH3CH2CH C OH
121 Which one of the cleansing agents would be suitable to be used in water containing magnesium hydrogen carbonate? Explain
122 8.
(C = 12; H = 1; N = 14; O =16) 7. The formula below represents active ingredients of two cleansing agents A and B O CH3CH2CH C OH
121 Which one of the cleansing agents would be suitable to be used in water containing magnesium hydrogen carbonate? Explain
122 8. Study the polymer below and use it to answer the questions that follow: (a) Give the name of the monomer and draw its structures (b) Identify the type of polymerization that takes place (c) State one advantage of synthetic polymers 9. Ethanol and Pentane are miscible liquids. Explain how water can be used to separate a mixture of ethanol and pentane 10. (a) What is absolute ethanol? GLUCOSE SOLUTION CRUDE ETHANOL 95% ETHANOL ABSOLUTE ETHANOL G H H H H H C C C C
123 (b) State two conditions required for process G to take place efficiently 11. (a) (i) The table below shows the volume of oxygen obtained per unit time when hydrogen peroxide was decomposed in the presence of manganese (IV) Oxide. Use it to answer the questions that follow:- Time in seconds Volume of Oxygen evolved (cm3) 0 30 60 90 120 150 180 210 240 270 300 0 10 19 27 34 38 43 45 45 45 45 (i) Plot a graph of volume of oxygen gas against time
124 (ii) Determine the rate of reaction at time 156 seconds (iii) From the graph, find the time taken for 18cm3 of oxygen to be produced (iv) Write a chemical equation to show how hydrogen peroxide decomposes in the presence of manganese (IV) Oxide (b) The diagram below shows how a Le’clanche (Dry cell) appears:- (i) What is the function of MnO2 in the cell above? (ii) Write the equation of a reaction that occurs at the cathode
125 (iii) Calculate the mass of Zinc that is consumed when a current of 0.1amperes flows through the above cell for 30minutes (1F =96500c Zn =65) 12.
(a) (i) The table below shows the volume of oxygen obtained per unit time when hydrogen peroxide was decomposed in the presence of manganese (IV) Oxide. Use it to answer the questions that follow:- Time in seconds Volume of Oxygen evolved (cm3) 0 30 60 90 120 150 180 210 240 270 300 0 10 19 27 34 38 43 45 45 45 45 (i) Plot a graph of volume of oxygen gas against time
124 (ii) Determine the rate of reaction at time 156 seconds (iii) From the graph, find the time taken for 18cm3 of oxygen to be produced (iv) Write a chemical equation to show how hydrogen peroxide decomposes in the presence of manganese (IV) Oxide (b) The diagram below shows how a Le’clanche (Dry cell) appears:- (i) What is the function of MnO2 in the cell above? (ii) Write the equation of a reaction that occurs at the cathode
125 (iii) Calculate the mass of Zinc that is consumed when a current of 0.1amperes flows through the above cell for 30minutes (1F =96500c Zn =65) 12. (a) Give the IUPAC names of the following compounds: (i) CH3COOCH2CH3 * (ii) (b) The structure below shows some reactions starting with ethanol. Study it and answer the questions that follow: CH2 = C – CHCH3 CH3COOH CH3COONa CH3CH2OH CH2=CH2 CH3CH3 CH2 CH2 n S P CH4 T Na Metal Compound U Step II Step I Step III CH3COOH Reagent R NaOH(aq) Heat Excess Cl2/U V
126 (i) Write the formula of the organic compounds P and S * (ii) Name the type of reaction, the reagent(s) and condition for the reactions in the following steps :- (I) Step I * (II) Step II * (III) Step III * (iii) Name reagent R …………………………………………………………… * (iv) Draw the structural formula of T and give its name * (v) (I) Name compound U………………………………………………………..
(ii) Write the equation of a reaction that occurs at the cathode
125 (iii) Calculate the mass of Zinc that is consumed when a current of 0.1amperes flows through the above cell for 30minutes (1F =96500c Zn =65) 12. (a) Give the IUPAC names of the following compounds: (i) CH3COOCH2CH3 * (ii) (b) The structure below shows some reactions starting with ethanol. Study it and answer the questions that follow: CH2 = C – CHCH3 CH3COOH CH3COONa CH3CH2OH CH2=CH2 CH3CH3 CH2 CH2 n S P CH4 T Na Metal Compound U Step II Step I Step III CH3COOH Reagent R NaOH(aq) Heat Excess Cl2/U V
126 (i) Write the formula of the organic compounds P and S * (ii) Name the type of reaction, the reagent(s) and condition for the reactions in the following steps :- (I) Step I * (II) Step II * (III) Step III * (iii) Name reagent R …………………………………………………………… * (iv) Draw the structural formula of T and give its name * (v) (I) Name compound U……………………………………………………….. (II) If the relative molecular mass of U is 42000, determine the value of n (C=12, H=1) (c) State why C2H4 burns with a more smoky flame than C2H6 * 13. a) State two factors that affect the properties of a polymer b) Name the compound with the formula below : CH3CH2CH2ONa c) Study the scheme below and use it to answer the questions that follow:- CH3CH2CH3 P Step CH3CH = CH2 Step CH3CH2CH2OH Step R H H C – C n K
127 i) Name the following compounds:- I. Product T ………………………… II.
(II) If the relative molecular mass of U is 42000, determine the value of n (C=12, H=1) (c) State why C2H4 burns with a more smoky flame than C2H6 * 13. a) State two factors that affect the properties of a polymer b) Name the compound with the formula below : CH3CH2CH2ONa c) Study the scheme below and use it to answer the questions that follow:- CH3CH2CH3 P Step CH3CH = CH2 Step CH3CH2CH2OH Step R H H C – C n K
127 i) Name the following compounds:- I. Product T ………………………… II. K ……… ii) State one common physical property of substance G iii) State the type of reaction that occurred in step J
128 iv) Give one use of substance K v) Write an equation for the combustion of compound P vi) Explain how compounds CH3CH2COOH and CH3CH2CH2OH can be distinguished chemically vii) If a polymer K has relative molecular mass of 12,600, calculate the value of n (H=1 C =12) 14. Study the scheme given below and answer the questions that follow:- H2 (g) Ni High temp Polymer Q Polymerization Compound P CH3CH2CH3 CH3CH2CH2ONa + H2 Na(s) Propan-l-ol Step I Propylethanoate CH3CH2COOH Solution T + CO2 (g) Step III Na2CO3(aq) Conc. H2SO4 180oC Step II
129 (a) (i) Name compound P …………………………………………………………………… (ii) Write an equation for the reaction between CH3CH2COOH and Na2CO3 (b) State one use of polymer Q (c) Name one oxidising agent that can be used in step II ………………………………….. (d) A sample of polymer Q is found to have a molecular mass of 4200. Determine the number of monomers in the polymer (H = 1, C = 12) (e) Name the type of reaction in step I ………………………………………………………….. (f) State one industrial application of step III (g)State how burning can be used to distinguish between propane and propyne.
(d) A sample of polymer Q is found to have a molecular mass of 4200. Determine the number of monomers in the polymer (H = 1, C = 12) (e) Name the type of reaction in step I ………………………………………………………….. (f) State one industrial application of step III (g)State how burning can be used to distinguish between propane and propyne. Explain your answer (h) 1000cm3 of ethene (C2H4) burnt in oxygen to produce Carbon (II) Oxide and water vapour. 130 Calculate the minimum volume of air needed for the complete combustion of ethene (Air contains 20% by volume of oxygen) 15. (a) Study the schematic diagram below and answer the questions that follow:- (i) Identify the following: Substance Q .............................................................................................................. Substance R............................................................................................................... Gas P.......................................................................................................................... CH3CH2COOCH2CH2CH3 CH3CHCH2 CH3CH2CH2ONa + Gas P CH3CH2CH2OH X V HCl Step 5 Step 1 R Na H+ Step 3 Q + H2O MnO4 Step 4Ni H2
131 (ii) Name: Step 1................................................................................................. Step 4................................................................................................. (iii) Draw the structural formula of the major product of step 5 (iv) State the condition and reagent in step 3 16. Study the flow chart below and answer the questions that follow (a) (i) Name the following organic compounds: M……………………………………………………………..…….. L………………………………………………………………….. M KMnO4/H+ CH2CH2 Ethyl Ethanoate CH2CH2OH L J K CO2 (g) STEP 2 Reagent Q St3KMnO4/H+(aq) Ni/H2(g) Step 4 Reagent P
132 Products C2H5COONa Step V ClbiCH CH Step I Step II CH2 = CH2 Step III C2H6 Step IV + Heat (ii) Name the process in step: Step 2 ………………………………………………………….…. Step 4 ………………………………………………………….… (iii) Identify the reagent P and Q (iv) Write an equation for the reaction between CH3CH2CH2OH and sodium 17.
L………………………………………………………………….. M KMnO4/H+ CH2CH2 Ethyl Ethanoate CH2CH2OH L J K CO2 (g) STEP 2 Reagent Q St3KMnO4/H+(aq) Ni/H2(g) Step 4 Reagent P
132 Products C2H5COONa Step V ClbiCH CH Step I Step II CH2 = CH2 Step III C2H6 Step IV + Heat (ii) Name the process in step: Step 2 ………………………………………………………….…. Step 4 ………………………………………………………….… (iii) Identify the reagent P and Q (iv) Write an equation for the reaction between CH3CH2CH2OH and sodium 17. a) Give the names of the following compounds: i) CH3CH2CH2CH2OH …………………………………………………………………… ii) CH3CH2COOH ………………………………………………………………… iii) CH3C – O- CH2CH3 …………………………………………………………………… 18. Study the scheme given below and answer the questions that follow;
133 n i) Name the reagents used in: Step I: ……………………………………………………………………… Step II …………………………………………………………………… Step III ……………………………………………………………………… ii) Write an equation to show products formed for the complete combustion of CH = CH iii) Explain one disadvantage of continued use of items made form the compound formed in step III 19. A hydrated salt has the following composition by mass. Iron 20.2 %, oxygen 23.0%, sulphur 11.5%, water 45.3% i) Determine the formula of the hydrated salt (Fe=56, S=32, O=16, H=11) ii) 6.95g of the hydrated salt in c(i) above were dissolved in distilled water and the total volume made to 250cm3 of solution. Calculate the concentration of the resulting salt solution in moles per litre. (Given that the molecula mass of the salt is 278) 20. Write an equation to show products formed for the complete combustion of CH = CH
134 iii) Explain one disadvantage of continued use of items made form the compound formed in step III
135 21.
Calculate the concentration of the resulting salt solution in moles per litre. (Given that the molecula mass of the salt is 278) 20. Write an equation to show products formed for the complete combustion of CH = CH
134 iii) Explain one disadvantage of continued use of items made form the compound formed in step III
135 21. Give the IUPAC name for each of the following organic compounds; i) CH3 - CH - CH2 - CH3 OH ii)CH3 – CH – CH2 – CH2 - CH3 C2H5 iii)CH3COOCH2CH2CH3 22. The structure below represents a cleansing agent. O R – S – O-Na+ O a) State the type of cleansing agent represented above b) State one advantage and one disadvantage of using the above cleansing agent. 23. The structure below shows part of polymer .Use it to answer the questions that follow. CH3 CH3 CH3   
136 ― CH - CH2 – CH- CH2 - CH – CH2 ― a) Derive the structure of the monomer b) Name the type of polymerization represented above 24. The flow chart below represents a series of reactions starting with ethanoic acid:- (a) Identify substances A and B (b) Name the process I 25. a) Write an equation showing how ammonium nitrate may be prepared starting with ammonia gas (b) Calculate the maximum mass of ammonium nitrate that can be prepared using 5.3kg of ammonia (H=1, N=14, O=16) 26. (a) What is meant by the term, esterification? Ethanol B Ethanoic acid Na2CO3 Salt A + CO2 + H2O
137 (b) Draw the structural formulae of two compounds that may be reacted to form ethylpropanoate 27. (a) Draw the structure of pentanoic acid (b) Draw the structure and give the name of the organic compound formed when ethanol reacts with pentanoic acid in presence of concentrated sulphuric acid
138 28. The scheme below shows some reactions starting with ethanol.
Ethanol B Ethanoic acid Na2CO3 Salt A + CO2 + H2O
137 (b) Draw the structural formulae of two compounds that may be reacted to form ethylpropanoate 27. (a) Draw the structure of pentanoic acid (b) Draw the structure and give the name of the organic compound formed when ethanol reacts with pentanoic acid in presence of concentrated sulphuric acid
138 28. The scheme below shows some reactions starting with ethanol. Study it and answer the questions that follow:- (i) Name and draw the structure of substance Q Q P CH3COONa Ethanol CH3CH2ONa C H2SO4(l) Cr2O7(aq) / H+(aq) Na(s) Step 2 Step 4 CH3CH2OH/H2SO4 Step 3 2-
139 (ii) Give the names of the reactions that take place in steps 2 and 4 (iii) What reagent is necessary for reaction that takes place in step 3 29. Substances A and B are represented by the formulae ROH and RCOOH respectively. They belong to two different homologous series of organic compounds. If both A and B react with potassium metal: (a) Name the common product produced by both (b) State the observation made when each of the samples A and B are reacted with sodium hydrogen carbonate (i) A (ii) B 30. Below are structures of particles. Use it to answer questions that follow. In each case only electrons in the outermost energy level are shown key P = Proton N = Neutron X = Electron W U V 19P Z Y
140 (a) Identify the particle which is an anion 31. Plastics and rubber are extensively used to cover electrical wires. (a) What term is used to describe plastic and rubbers used in this way? (b) Explain why plastics and rubbers are used this way
141 32. The scheme below represents the manufacture of a cleaning agent X (a) Draw the structure of X and state the type of cleaning agent to which X belong (b) State one disadvantage of using X as a cleaning agent 33. Y grams of a radioactive isotope take 120days to decay to 3.5grams.
(b) Explain why plastics and rubbers are used this way
141 32. The scheme below represents the manufacture of a cleaning agent X (a) Draw the structure of X and state the type of cleaning agent to which X belong (b) State one disadvantage of using X as a cleaning agent 33. Y grams of a radioactive isotope take 120days to decay to 3.5grams. The halflife period of the isotope is 20days (a) Find the initial mass of the isotope (b) Give one application of radioactivity in agriculture 34. The structure below represents a polymer. Study and answer the questions that follow:- R Conc. R Cleaning agent X H H -C – C - n
142 (i) Name the polymer above.................................................................................. (ii) Determine the value of n if giant molecule had relative molecular mass of 4956 35. RCOO-Na+ and RCH2OSO3-Na+ are two types of cleansing agents; i) Name the class of cleansing agents to which each belongs ii) Which one of these agents in (i) above would be more suitable when washing with water from the Indian ocean. Explain iii) Both sulphur (IV) oxide and chlorine are used bleaching agents. Explain the difference in their bleaching properties 36. The formula given below represents a portion of a polymer H H H H C C C C O H O H n
143 (a) Give the name of the polymer (b) Draw the structure of the monomer used to manufacture the polymer
15.0.0 NITROGEN AND IT'S COMPOUNDS (30 LESSONS) Introduction to oxides of nitrogen Nirogen has a position in second period of group V in the modern periodic table. It has molecular formula N2. It has atomic number 7 and atomic weight 14.08 and its electronic configuration of 2,5. Besides combining with hydrogen and forming NH3 , nitrogen combines with oxygen in different ratios and forms five different oxides.