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Calculate the volume of Oxygen required to completely react with 50cm3 of Hydrogen. Chemical equation: 2H2 (g) + O2 (g) -> 2H2O(l) Volume ratios 2 : 1 : 0 Reacting volumes 50cm3 : 25cm3 50cm3 of Oxygen is used 2. Calculate the volume of air required to completely reacts with 50cm3 of Hydrogen.(assume Oxygen is 21% by v... | {
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Chemical equation: CxHy (g) + O2 (g) -> H2O(g) + CO2(g) Volumes 5cm3 : 15cm3 : 10cm3 : 10cm3 Volume ratios 5cm3 : 15cm3 : 10cm3 : 10cm3 (divide by lowest volume) 5 5 5 5 Reacting volume ratios 1volume 3 volume 2 volume 2 volume Balanced chemical equation: CxHy (g) + 3O2 (g) -> 2H2O(g) + 2CO2(g) If “4H” are in 2H2O(g) t... | {
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Obtain the equation for the reaction and name the gaseous product. Chemical equation: NO (g) + O2 (g) -> NOx Volumes 100cm3 : 50cm3 : 100 Volume ratios 100cm3 : 50cm3 : 100cm3 (divide by lowest volume) 50 50 50 Reacting volume ratios 2volume 1 volume 2 volume Balanced chemical equation: 2 NO (g) + O2 (g) -> 2NO x(g) Th... | {
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5. Rewrite the chemical equation. It is an ionic equation. Practice (a)Precipitation of an insoluble salt
23 23 All insoluble salts are prepared in the laboratory from double decomposition /precipitation. This involves mixing two soluble salts to form one soluble and one insoluble salt 1. When silver nitrate(V) solutio... | {
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Balanced stoichiometric equation AgNO3(aq) + NaCl(aq) -> AgCl (s) + NaNO3 (aq) Split reactants product existing in aqueous state as cation/anion Ag+(aq) + NO3- (aq) + Na+(aq) + Cl-(aq) -> AgCl(s) + Na+(aq)+ NO3- (aq) Cancel out ions appearing on reactant and product side Ag+(aq) + NO3- (aq) + Na+(aq) + Cl-(aq) -> AgCl(... | {
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When barium nitrate(V) solution is added to copper(II)sulphate(VI) solution, copper(II) nitrate(V) solution and a white precipitate of barium sulphate(VI) are formed. Balanced stoichiometric equation Ba(NO3)2(aq) + CuSO4(aq) -> BaSO4 (s) + Cu(NO3) 2 (aq) Split reactants product existing in aqueous state as cation/anion... | {
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Balanced stoichiometric equation Ba(NO3)2(aq) + CuSO4(aq) -> BaSO4 (s) + Cu(NO3) 2 (aq) Split reactants product existing in aqueous state as cation/anion Ba2+(aq) + 2NO3- (aq) + Cu2+(aq) + SO42-(aq) -> BaSO4 (s) + 2NO3- (aq)+ Cu2+(aq) Cancel out ions appearing on reactant and product side Ba2+(aq) + 2NO3- (aq) +Cu2+ (a... | {
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Balanced stoichiometric equation Pb(NO3)2(aq) + 2KI (aq) -> PbI2 (s) + 2KNO3 (aq)
24 24 Split reactants product existing in aqueous state as cation/anion Pb2+(aq) + 2NO3- (aq) + 2K +(aq) + 2I - (aq) -> PbI2 (s) + 2NO3- (aq)+ 2K +(aq) Cancel out ions appearing on reactant and product side Pb2+(aq) + 2NO3- (aq) + 2K +(aq... | {
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(i)Reaction of alkalis with acids 1.Reaction of nitric(V)acid with potassium hydroxide Balanced stoichiometric equation HNO3(aq) + KOH (aq) -> H2O (l) + KNO3 (aq) Split reactants product existing in aqueous state as cation/anion H+(aq) + NO3- (aq) + K +(aq) + OH - (aq) -> H2O (l) + NO3- (aq)+ K +(aq) Cancel out ions ap... | {
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The SI unit of concentration is Molarity denoted M. Molarity may be defined as the number of moles of solute present in one cubic decimeter of solution. One cubic decimeter is equal to one litre and also equal to 1000cm3. The higher the molarity the higher the concentration and the higher/more solute has been dissolved... | {
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x 10-3moles 1000 1000 Mass of HNO3 =Moles x molar mass => 2.5 x 10-3 x 40 = 0.1 g 3. Calculate the volume required to dissolve : (a)(i) 0.25moles of sodium hydroxide solution to form a 0.8M solution Volume (in cm3) = moles x 1000 => 0.25 x 1000 = 312.5cm3 Molarity 0.8 (ii) 100cm3 was added to the sodium hydroxide solut... | {
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Volume (in cm3) = moles x 1000 => 0.01 x 1000 = 1000 cm3 Molarity 0.01 (ii) Determine the quantity of water which must be added to the sodium hydroxide solution above to form a 0.008M solution. C1 x V1 = C2 x V2 where: C1 = molarity/concentration before diluting/adding water C2 = molarity/concentration after diluting/a... | {
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0.388g of a monobasic organic acid B required 46.5 cm3 of 0.095M sodium hydroxide for complete neutralization. Name and draw the structural formula of B Moles of NaOH used = molarity x volume 1000 => 0.095 x 46.5 = 0.0044175 /4.4175 x 10-3moles 1000 Mole ratio B : NaOH = 1:1
34 34 Moles B = 0.0044175 /4.4175 x 10-3mole... | {
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10.5 g of an impure sample containing ammonium sulphate (VI) fertilizer was warmed with 250cm3 of o.8M sodium hydroxide solution.The excess of the alkali was neutralized by 85cm3 of 0.5M hydrochloric acid. Calculate the % of impurities in the ammonium sulphate (VI)fertilizer. (N=14.0,S=32.0,O=16.0, H=1.0) Equation for ... | {
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07875moles Molar mass (NH4) 2SO4= 132 gmole-1
35 35 Mass of in impure sample = moles x molar mass =>0. 07875 x 132 = 10.395 g Mass of impurities = 10.5 -10.395 = 0.105 g % impurities = 0.105 x 100 = 1.0 % 10.5 Practically volumetric analysis involves titration. Titration generally involves filling a burette with known/... | {
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The results of titration are recorded in a titration table as below
36 36 Sample titration table Titration number 1 2 3 Final burette reading (cm3) 20.0 20.0 20.0 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of solution used(cm3) 20.0 20.0 20.0 As evidence of a titration actually done examining body requires the ca... | {
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37 37 (a)Sample Titration Practice 1 (Simple Titration) You are provided with: 0.1M sodium hydroxide solution A Hydrochloric acid solution B You are required to determine the concentration of solution B in moles per litre. Procedure Fill the burette with solution B. Pipette 25.0cm3 of solution A into a conical flask. T... | {
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How many moles of: (i)solution A were present in 25cm3 solution. Moles of solution A = Molarity x volume = 0.1 x 25 = 2.5 x 10-3 moles 1000 1000 (ii)solution B were present in the average volume. Chemical equation: NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O(l) Mole ratio 1:1 => Moles of A = Moles of B = 2.5 x 10-3 moles (iii... | {
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Table 1:Sample results Titration number 1 2 3 Final burette reading (cm3) 20.0 20.0 20.0 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of solution A used(cm3) 20.0 20.0 20.0 Sample worked questions 1. Calculate the average volume of solution A used Average titre = Titre 1 + Titre 2 +Titre 3 => ( 20.0 +20.0 +20.0 ) =... | {
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Pipette 25cm3 of solution C into a conical flask, Titrate solution C with solution A until a permanent pink colour just appears. Complete table 2. Table 2:Sample results Titration number 1 2 3 Final burette reading (cm3) 20.0 20.0 20.0 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of solution A used(cm3) 20.0 20.0 2... | {
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Moles of solution A = Molarity x volume = 0.025 x 20 = 5.0 x 10-4 moles 1000 1000 (ii)solution C in 25cm3 solution given the equation for the reaction: MnO4- (aq) + 8H+(aq) + 5Fe2+ (aq) -> Mn2+(aq) + 5Fe3+(aq) + 4H2O(l) Mole ratio MnO4- (aq): 5Fe2+ (aq) = 1:5 => Moles of 5Fe2+ (aq) = Moles of MnO4- (aq) = 5.0 x 10-4 mo... | {
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You are required to determine the concentration of N in moles per litre and the % of calcium carbonate in mixture M. Procedure 1 Pipette 25.0cm3 of solution L into a conical flask. Add 2-3 drops of phenolphthalein indicator. Titrate with dilute hydrochloric acid solution N and record your results in table 1(4mark) Samp... | {
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Titrate with dilute hydrochloric acid solution N and record your results in table 1(4mark) Sample Table 1 1 2 3 Final burette reading (cm3) 6.5 6.5 6.5 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of N used (cm3) 6.5 6.5 6.5 Sample questions (a) Calculate the average volume of solution N used 6.5 + 6.5 + 6.5 = 6.5 ... | {
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Mole ratio NaOH : HCl from stoichiometric equation= 1:1 Moles HCl =Moles NaOH => 0.0125 moles (ii)the molarity of hydrochloric acid solution N. Molarity = moles x 1000 => 0.0125 moles x 1000 =1.9231M/moledm-3 6.5 6.5 Procedure 2 Place the 4.0 g of M provided into a conical flask and add 25.0cm3 of the dilute hydrochlor... | {
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Swirl the contents of the flask vigorously until effervescence stop.Using a 100ml measuring cylinder add 175cm3 distilled waterto make up the solution up to 200cm3.Label this solution K.Using a clean pipettetransfer 25.0cm3 of the solution into a clean conical flask and titrate with solution L from the burette using 2-... | {
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Sample Table 2 1 2 3 Final burette reading (cm3) 24.5 24.5 24.5 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of N used (cm3) 24.5 24.5 24.5 Sample calculations (a)Calculate the average volume of solution L used(1mk) 24.5 + 24.5 + 24.5 = 24.5cm3 3 (b)How many moles of sodium hydroxide are present in the average volu... | {
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Moles = molarity x average burette volume => 0.5 x 24.5 1000 1000 = 0.01225 /1.225 x 10-2 moles
42 42 (c) How many moles of hydrochloric acid are present in the original 200cm3 of solution K? Mole ratio NaOH: HCl = 1:1 => moles of HCl = 0.01225 /1.225 x 10-2 moles Moles in 200cm3 = 200cm3 x 0.01225 /1.225 x 10-2moles 2... | {
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Mole ratio NaOH: HCl = 1:1 => moles of HCl = 0.01225 /1.225 x 10-2 moles Moles in 200cm3 = 200cm3 x 0.01225 /1.225 x 10-2moles 25cm3(volume pipetted) =0.49 /4.9 x 10-1moles (d)How many moles of hydrochloric acid were contained in original 25 cm3 solution N used Original moles = Original molarity x pipetted volume => 10... | {
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Moles that reacted = original moles –moles in average titre => 0.04807/4.807 x 10-2moles - 0.01225 /1.225 x 10-2moles = 0.03582/3.582 x 10 -2 moles (f)Write the equation for the reaction between calcium carbonate and hydrochloric acid. CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + CO2(g) + H2O(l) (g)Calculate the number of moles ... | {
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From the equation CaCO3(s):2HCl(aq) = 1:2 => Moles CaCO3(s) = 1/2moles HCl = 1/2 x 0.03582/3.582 x 10 -2 moles = 0.01791 /1.791 x 10-2moles
43 43 (h)Calculate the mass of calcium carbonate in 4.0g of mixture M (Ca=40.0,O = 16.0,C=12.0) Molar mass CaCO3 = 100g Mass CaCO3 = moles x molar mass => 0.01791 /1.791 x 10-2mole... | {
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Pipette 25.0cm3 of solution L into a conical flask. Heat this solution to about 70oC(but not to boil).Titrate the hot solution L with solution M until a permanent pink colour just appears .Shake thoroughly during the titration. Repeat this procedure to complete table 1. Sample Table 1 1 2 3 Final burette reading (cm3) ... | {
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Moles = molarity x pipette volume => 0.5 x 25 = 0.0125/1.25 x10 -2 moles 1000 1000 (d)Given the mole ratio manganate(VII)(MnO4-): acid H2X is 2:5, calculate the number of moles of manganate(VII) (MnO4-) in the average titre. Moles H2X = 2/5 moles of MnO4- => 2/5 x 0.0125/1.25 x10 -2 moles = 0.005/5.0 x 10 -3moles (e)Ca... | {
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Moles per litre/molarity = moles x 1000 average burette volume =>0.005/5.0 x 10 -3moles x 1000 = 0.2083 molesl-1/M 24.0 Procedure 2 With solution M still in the burette ,pipette 25.0cm3 of solution N into a conical flask. Heat the conical flask containing solution N to about 70oC.Titrate while hot with solution M.Repea... | {
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Heat the conical flask containing solution N to about 70oC.Titrate while hot with solution M.Repeat the experiment to complete table 2. Sample Table 2 1 2 3 Final burette reading (cm3) 12.5 12.5 12.5
45 45 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of N used (cm3) 12.5 12.5 12.5 Sample calculations (a)Calculate t... | {
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Moles = molarity of solution M x average burette volume 1000 => 0.2083 molesl-1/ M x 12.5 = 0.0026 / 2.5 x 10-3 moles 1000 (ii)The reaction between manganate(VII)ions and ethanedioate ions that reacted with is as in the equation: 2MnO4- (aq) + 5C2O42- (aq) + 16H+ (aq) -> 2Mn2+(aq) + 10CO2(g) + 8H2O(l) Calculate the num... | {
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Pipette 25cm3 of solution N into a conical flask and add 2-3 drops of phenolphthalein indicator. Titrate this solution N with solution P from the burette. Repeat the procedure to complete table 3. Sample Table 2 1 2 3 Final burette reading (cm3) 24.9 24.9 24.9 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of N used ... | {
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Sample Table 2 1 2 3 Final burette reading (cm3) 24.9 24.9 24.9 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of N used (cm3) 24.9 24.9 24.9 Sample calculations (a)Calculate the average volume of solution L used (1mk) 24.9 + 24.9 + 24.9 = 24.9 cm3 3 (b)Calculations: (i)How many moles of sodium hydroxide solution P w... | {
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Moles = molarity of solution P x average burette volume 1000 => 0.1 molesl-1 x 24.9 = 0.00249 / 2.49 x 10-3 moles 1000 (ii)Given that NaOH solution P reacted with the ethanedioate ions from the acid only and the equation for the reaction is: 2NaOH (aq) + H2C2O4 (aq) -> Na2C2O4(g) + 2H2O(l) Calculate the number of moles... | {
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(iii)How many moles of ethanedioic acid were contained in 250cm3 of solution N? 47 47 25cm3 pipette volume -> 0.001245/1.245 x10-3 moles 250cm3 -> 0.001245/1.245 x10-3 moles x 250 = 0.01245/1.245 x10-2 moles 25 (iii)Determine the % by mass of sodium ethanedioate in the micture (H= 1.0,O=16.0,C=12.0 and total mass of mi... | {
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47 47 25cm3 pipette volume -> 0.001245/1.245 x10-3 moles 250cm3 -> 0.001245/1.245 x10-3 moles x 250 = 0.01245/1.245 x10-2 moles 25 (iii)Determine the % by mass of sodium ethanedioate in the micture (H= 1.0,O=16.0,C=12.0 and total mass of mixture =2.0 g in 250cm3 solution) Molar mass H2C2O4 = 90.0g Mass of H2C2O4 in 250... | {
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12.5+12.5+12.5 = 12.5cm3 3 Theoretical Practice examples 1. 1.0g of dibasic acid HOOC(CH2)xCOOH was dissolved in 250cm3 solution. 25.0 cm3 of this solution reacted with 30.0cm3 of 0.06M sodium hydroxide solution. Calculate the value of x in HOOC(CH2)xCOOH. | {
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46.0g of a metal carbonate MCO3 was dissolved 160cm3 of 0.1M excess hydrochloric acid and the resultant solution diluted to one litre.25.0cm3 of this solution required 20.0cm3 of 0.1M sodium hydroxide solution for complete neutralization. Calculate the atomic mass of ‘M’ Equation Chemical equation NaOH(aq) + HCl(aq) ->... | {
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Calculate the atomic mass of ‘M’ Equation Chemical equation NaOH(aq) + HCl(aq) -> KCl(aq) + H2O(l) Moles of NaOH = Molarity x volume=> 0.1 x20 = 0.002 moles 1000 1000 Mole ratio HCl; NaOH = 1:1 Excess moles of HCl = 0.002 moles 25cm3 -> 0.002 moles 1000cm3 -> 1000 x 0.002 = 0.08moles 25cm3 Original moles of HCl = Molar... | {
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25.0cm3 of a mixture of Fe2+ and Fe3+ ions in an aqueous salt was acidified with sulphuric(VI)acid then titrated against potassium manganate(VI).The salt required 15cm3 ofe0.02M potassium manganate(VI) for complete reaction. A second 25cm3 portion of the Fe2+ and Fe3+ ion salt was reduced by Zinc then titrated against ... | {
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Mole ratio Fe2+ :Mn04- = 5:1 Moles Mn04- used = 0.02 x 15 = 3.0 x 10-4 moles 1000 Moles Fe2+ = 3.0 x 10-4 moles = 6.0 x 10-5 moles 5 Molarity of Fe2+ = 6.0 x 10-4 moles x 1000 = 2.4 x 10-3 moles l-1 25 Since Zinc reduces Fe3+ to Fe2+ in the mixture: Moles Mn04- that reacted with all Fe2+= 0.02 x 19 = 3.8 x 10-4 moles 1... | {
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(ii)carbon can covalently bond to form a single, double or triple covalent bond with itself. (iii)carbon atoms can covalently bond to form a very long chain or ring. When carbon covalently bond with Hydrogen, it forms a group of organic compounds called Hydrocarbons A.HYDROCARBONS (HCs) Hydrocarbons are a group of orga... | {
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When carbon covalently bond with Hydrogen, it forms a group of organic compounds called Hydrocarbons A.HYDROCARBONS (HCs) Hydrocarbons are a group of organic compounds containing /made up of hydrogen and carbon atoms only. Depending on the type of bond that exist between the individual carbon atoms, hydrocarbon are cla... | {
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The carbon atoms are linked by single bond to each other and to hydrogen atoms. 2 They include: n General/ Molecular formula Structural formula Name 1 CH4 H H C H H Methane 2 C2H6 H H H C C H H H Ethane 3 C3H8 H H H H C C C H H H H Propane 4 C4H10 H H H H H C C C C H H H H H Butane 5 C5H12 H H H H H H C C C C C H CH3 (... | {
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2.The structural formula shows the arrangement/bonding of atoms of each element making the compound e.g Decane has the structural formula as in the table above ;this means the 1st carbon from left to right is bonded to three hydrogen atoms and one carbon atom. The 2nd carbon atom is joined/bonded to two other carbon at... | {
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(iv)the physical properties (e.g.melting/boiling points)show steady gradual change) 8.The 1st four alkanes have the prefix meth_,eth_,prop_ and but_ to represent 1,2,3 and 4 carbons in the compound. All other use the numeral prefix pent_,Hex_,hept_ , etc to show also the number of carbon atoms. 9.If one hydrogen atom i... | {
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Practice on IUPAC nomenclature of alkanes (a)Draw the structure of: (i)2-methylpentane Procedure 1. Identify the longest continuous carbon chain to get/determine the parent alkane. Butane is the parent name CH3 CH2 CH2 CH3 2. Number the longest chain form the end of the chain that is near the branches so as the branch ... | {
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Identify the longest continuous carbon chain to get/determine the parent alkane. Butane is the parent name CH3 CH2 CH2 CH3 2. Number the longest chain form the end of the chain that is near the branches so as the branch get the lowest number possible The methyl group is attached to Carbon “2 and 3” 3. Determine the pos... | {
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As the carbon chain increase, the boiling point, viscosity (ease of flow) and colour intensity increase as flammability decrease. Hydrocarbons in crude oil are not pure. They thus have no sharp fixed boiling point. Uses of different crude oil fractions Carbon atoms in a molecule Common name of fraction Uses of fraction... | {
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Methane is prepared from the heating of a mixture of sodium ethanoate and soda lime and collecting over water Sodium ethanoate + soda lime -> methane + Sodium carbonate CH3COONa(s) + NaOH(s) -> C H4 + Na2CO3(s) The “H” in NaOH is transferred/moves to the CH3 in CH3COONa(s) to form CH4. 2. Ethane is prepared from the he... | {
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They are slightly soluble in water. The solubility decrease as the carbon chain and thus the molar mass increase The melting and boiling point increase as the carbon chain increase. This is because of the increase in van-der-waals /intermolecular forces as the carbon chain increase. The 1st four straight chain alkanes ... | {
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Summary of physical properties of alkanes Alkane General formula Melting point(K) Boiling point(K) Density gcm-3 State at room(298K) temperature and pressure atmosphere (101300Pa) Methane CH4 90 112 0.424 gas
12 Ethane CH3CH3 91 184 0.546 gas Propane CH3CH2CH3 105 231 0.501 gas Butane CH3(CH2)2CH3 138 275 0.579 gas Pen... | {
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Alkane + Air -> carbon(IV) oxide + water (excess air/oxygen) Alkanes burn with a blue/non-luminous no-sooty/non-smoky flame in limited air to form carbon(II) oxide and water. Alkane + Air -> carbon(II) oxide + water (limited air) Examples 1.(a) Methane when ignited burns with a blue non sooty flame in excess air to for... | {
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Ethane + Air -> carbon(IV) oxide + water (excess air/oxygen) 2C2H6(g) + 7O2(g) -> 4CO2(g) + 6H2O(l/g) (b) Ethane when ignited burns with a blue non sooty flame in limited air to form carbon(II) oxide and water. 13 Ethane + Air -> carbon(II) oxide + water (excess air/oxygen) 2C2H6(g) + 5O2(g) -> 4CO(g) + 6H2O(l/g) 3.(a)... | {
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(v)substitution stops when all the hydrogen in alkanes are replaced with halogens. Substitution reaction is a highly explosive reaction in presence of sunlight / ultraviolet light that act as catalyst. Examples of substitution reactions Methane has no effect on bromine or chlorine in diffused light/dark. In sunlight , ... | {
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Chemical equation 1.(a)Methane + chlorine -> Chloromethane + Hydrogen chloride
14 CH4(g) + Cl2(g) -> CH3Cl (g) + HCl (g) H H H C H + Cl Cl -> H C Cl + H Cl H H (b) Chloromethane + chlorine -> dichloromethane + Hydrogen chloride CH3Cl (g) + Cl2(g) -> CH2Cl2 (g) + HCl (g) H H H C Cl + Cl Cl -> H C Cl + H Cl H Cl (c) dich... | {
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In sunlight , a mixture of bromine and ethane explode to form colourless mixture of bromoethane and hydrogen chloride gas. The red/brown colour of bromine gas fades. Chemical equation (a)Ethane + chlorine -> Chloroethane + Hydrogen chloride CH3CH3(g) + Br2(g) -> CH3CH2Br (g) + HBr (g) H H H H H C C H + Br Br -> H C C H... | {
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(ii) Alkenes (a)Nomenclature/Naming These are hydrocarbons with a general formula CnH2n and C C double bond as the functional group . n is the number of Carbon atoms in the molecule. The carbon atoms are linked by at least one double bond to each other and single bonds to hydrogen atoms. They include: n General/ Molecu... | {
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They include: n General/ Molecular formula Structural formula Name 1 Does not exist 2 C2H6 H H H C C H Ethene
17 CH2 CH2 3 C3H8 H H H H C C C H H CH2 CH CH3 Propene 4 C4H10 H H H H H C C C C H H H CH2 CH CH2CH3 Butene 5 C5H12 H H H H H H C C C C C H H H H CH2 CH (CH2)2CH3 Pentene 6 C6H14 H H H H H H H C C C C C C H H H... | {
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2.Since Hydrogen is monovalent ,each atom of hydrogen in the alkene MUST always be bonded using one covalent bond/one shared pair of electrons. 3.One member of the alkene ,like alkanes,differ from the next/previous by a CH2 group.They also form a homologous series. e.g Propene differ from ethene by one carbon and two H... | {
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2.Number the longest chain form the end of the chain which contains the =C = C= double bond so he =C = C= double bond lowest number possible. 3 Indicate the positions by splitting “alk-positions-ene” e.g. but-2-ene, pent-1,3diene. 4.The position indicated must be for the carbon atom at the lower position in the =C = C=... | {
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Like alkanes ,an alkyl group can be attached to the alkene. Chain/branch isomers are thus formed. 10.Chain/branch isomers are molecules/compounds having the same general formula but different structural formula e.g Butene and 2-methyl propene both have the same general formualr but different branching chain. Practice o... | {
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I2C CICI CI2 1,1,2,3,4,4-hexaiodobut -1,3- diene 18. H2C C(CH3)C(CH3) CH2 2,3-dimethylbut -1,3- diene (c)Occurrence and extraction
23 At indusrial level,alkenes are obtained from the cracking of alkanes.Cracking is the process of breaking long chain alkanes to smaller/shorter alkanes, an alkene and hydrogen gas at high... | {
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1.(a)At about 180oC,concentrated sulphuric(VI)acid dehydrates/removes water from ethanol to form ethene. The gas produced contain traces of carbon(IV)oxide and sulphur(IV)oxide gas as impurities. It is thus passed through concentrated sodium/potassium hydroxide solution to remove the impurities. Chemical equation CH3CH... | {
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24 Ethanol vapour passes through the hot aluminium (III) oxide which catalyses the dehydration. Activated aluminium(III)oxide has a very high affinity for water molecules/elements of water and thus dehydrates/ removes water from ethanol to form ethene. Chemical equation CH3CH2OH (l) --(Al2O3/strong heat--> CH2=CH2(g) +... | {
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Activated aluminium(III)oxide has a very high affinity for water molecules/elements of water and thus dehydrates/ removes water from ethanol to form ethene. Chemical equation CH3CH2OH (l) --(Al2O3/strong heat--> CH2=CH2(g) + H2O(l) 2(a) Propan-1-ol and Propan-2-ol(position isomers of propanol) are dehydrated by conc H2... | {
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Chemical equation CH3CH2OH (l) --(Al2O3/strong heat--> CH2=CH2(g) + H2O(l) 2(a) Propan-1-ol and Propan-2-ol(position isomers of propanol) are dehydrated by conc H2SO4 at about 180oC to propene(propene has no position isomers). Chemical equation CH3CH2 CH2OH (l) -- conc H2SO4/180oC --> CH3CH2=CH2(g) + H2O(l) Propan-1-ol... | {
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Chemical equation CH3CH2 CH2OH (l) -- conc H2SO4/180oC --> CH3CH2=CH2(g) + H2O(l) Propan-1-ol Prop-1-ene CH3CHOH CH3 (l) -- conc H2SO4/180oC --> CH3CH2=CH2(g) + H2O(l) Propan-2-ol Prop-1-ene (b) Propan-1-ol and Propan-2-ol(position isomers of propanol) are dehydrated by heating strongly aluminium(III)oxide(Al2O3) form ... | {
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Chemical equation CH3CH2 CH2 CH2OH (l) -- Heat/Al2O3 --> CH3 CH2CH2=CH2(g) + H2O(l) Butan-1-ol But-1-ene
25 CH3CHOH CH2CH3 (l) -- Heat/Al2O3 --> CH3CH=CH CH2(g) + H2O(l) Butan-2-ol But-2-ene Laboratory set up for the preparation of alkenes/ethene Caution (i)Ethanol is highly inflammable (ii)Conc H2SO4 is highly corrosi... | {
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This is because of the increase in van-der-waals /intermolecular forces as the carbon chain increase. The 1st four straight chain alkenes (ethene,propane,but-1-ene and pent-1-ene)are gases at room temperature and pressure. The density of straight chain alkenes,like alkanes, increase with increasing carbon chain as the ... | {
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A homologous series with C = C double or C C triple bond is said to be unsaturated. A homologous series with C C single bond is said to be saturated.Most of the reactions of the unsaturated compound involve trying to be saturated to form a C C single bond . Examples of burning alkenes 1.(a) Ethene when ignited burns wi... | {
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Propene + Air -> carbon(IV) oxide + water (excess air/oxygen) C3H6(g) + 3O2(g) -> 3CO(g) + 3H2O(l/g) (b)Addition reactions An addition reaction is one which an unsaturated compound reacts to form a saturated compound.Addition reactions of alkenes are named from the reagent used to cause the addtion/convert the double =... | {
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Examples 1.When Hydrogen gas is passed through liquid vegetable and animal oil at about 180oC in presence of Nickel catalyst,solid fat is formed. Hydrogenation is thus used to harden oils to solid fat especially margarine. During hydrogenation, one hydrogen atom in the hydrogen molecule attach itself to one carbon and ... | {
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Chemical equation H2C=CH2 + H2 -Ni/Pa-> H3C - CH3 H H H H C = C + H – H - Ni/Pa -> H - C – C - H H H H H 2.Propene undergo hydrogenation to form Propane
29 Chemical equation H3C CH=CH2 + H2 -Ni/Pa-> H3C CH - CH3 H H H H H H H C C = C + H – H - Ni/Pa-> H - C – C - C- H H H H H H 3.Both But-1-ene and But-2-ene undergo hy... | {
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But-1,3-diene + Hydrogen –Ni/Pa-> Butane H2C CH CH=CH2 + 2H2 -Ni/Pa-> H3C CH2CH - CH3 H H H H H H H H H C C - C = C -H + 2(H – H) - Ni/Pa-> H - C- C – C - C- H
30 H H H H (ii) Halogenation. Halogenation is an addition reaction in which a halogen (Fluorine, chlorine, bromine, iodine) reacts with an alkene to form an alk... | {
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One bromine atom bond at the 1st carbon in the double bond while the other goes to the 2nd carbon. Examples 1Ethene reacts with bromine to form 1,2-dibromoethane. Chemical equation H2C=CH2 + Br2 H2 Br C - CH2 Br H H H H C = C + Br – Br Br - C – C - Br H H H H Ethene + Bromine 1,2-dibromoethane 2.Propene reacts with chl... | {
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Chemical equation H3C CH=CH2 + Cl2 H3C CHCl - CH2Cl Propene + Chlorine 1,2-dichloropropane H H H H H H H C C = C + Cl – Cl H - C – C - C- Cl H H H Cl H H H H H H H H H H C C - C = C + I – I H - C- C – C - C- I
31 H H H H H H H H 3.Both But-1-ene and But-2-ene undergo halogenation with iodine to form 1,2diiodobutane and... | {
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The main compound is one which the hydrogen atom bond at the carbon with more hydrogen . Examples 1. Ethene reacts with hydrogen bromide to form bromoethane. Chemical equation H2C=CH2 + HBr H3 C - CH2 Br H H H H
32 C = C + H – Br H - C – C - Br H H H H Ethene + Bromine bromoethane 2. Propene reacts with hydrogen iodide... | {
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But-1,3-diene react with hydrogen iodide to form 2,3- diiodobutane. The reaction uses two moles of hydrogen iodide molecules/two iodine atoms and two hydrogen atoms to break the two double bonds. But-1,3-diene + iodine 2,3-diiodobutane Carbon atom with more Hydrogen atoms gets extra hydrogen
33 H2C= CH CH=CH2 + 2HI2 H3... | {
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During the addition reaction .the halogen move to one carbon and the OH to the other carbon in the alkene at the =C=C= double bond to form a halogenoalkanol. Bromine water + Alkene -> bromoalkanol Chlorine water + Alkene -> bromoalkanol Examples 1Ethene reacts with bromine water to form bromoethanol. Chemical equation ... | {
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Bromine water + Alkene -> bromoalkanol Chlorine water + Alkene -> bromoalkanol Examples 1Ethene reacts with bromine water to form bromoethanol. Chemical equation H2C=CH2 + HOBr H2 Br C - CH2 OH H H H H C = C + Br – OH Br - C – C - OH H H H H Ethene + Bromine water bromoethanol 2.Propene reacts with chlorine water to fo... | {
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Chemical equation I.H3C CH=CH2 + HOCl H3C CHCl - CH2OH Propene + Chlorine water 2-chloropropane H H H H H H H C C = C + HO – Cl H - C – C - C- OH
34 H H H Cl H II.H3C CH=CH2 + HOCl H3C CHOH - CH2Cl Propene + Chlorine chloropropan-2-ol H H H H H H H C C = C + HO – Cl H - C – C - C- Cl H H H OH H 3.Both But-1-ene and But... | {
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The reaction uses two moles of bromine water molecules to break the two double bonds. But-1,3-diene + bromine water 2,4-dibromobutan-1,3-diol H2C= CH CH=CH2 + 2HOBr H2COH CHBrCHOH CHBr H H H H H H H H
35 H C C - C = C -H + 2(HO – Br) H - C- C – C - C- H HO Br HO Br (v) Oxidation. Alkenes are oxidized to alkanols with d... | {
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Propene is oxidized to propan-1,2-diol by acidified potassium/sodium manganate(VII) solution/ acidified potassium/sodium dichromate(VI) solution. The purple acidified potassium/sodium manganate(VII) solution is decolorized. The orange acidified potassium/sodium dichromate(VI) solution turns to green. Chemical equation ... | {
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This takes place in two steps: (i)Alkenes react with concentrated sulphuric(VI)acid at room temperature and pressure to form alkylhydrogen sulphate(VI). Alkenes + concentrated sulphuric(VI)acid -> alkylhydrogen sulphate(VI) (ii)On adding water to alkylhydrogen sulphate(VI) then warming, an alkanol is formed. alkylhydro... | {
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Alkenes + concentrated sulphuric(VI)acid -> alkylhydrogen sulphate(VI) (ii)On adding water to alkylhydrogen sulphate(VI) then warming, an alkanol is formed. alkylhydrogen sulphate(VI) + water -warm-> Alkanol. Examples (i)Ethene reacts with cold concentrated sulphuric(VI)acid to form ethyl hydrogen sulphate(VII) Chemica... | {
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Examples (i)Ethene reacts with cold concentrated sulphuric(VI)acid to form ethyl hydrogen sulphate(VII) Chemical equation H2C=CH2 + H2SO4 CH3 - CH2OSO3H H H H O-SO3H C = C + H2SO4 H - C – C - H
37 H H H H Ethene + H2SO4 ethylhydrogen sulphate(VI) (ii) Ethylhydrogen sulphate(VI) is hydrolysed by water to ethanol Chemica... | {
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Propene reacts with cold concentrated sulphuric(VI)acid to form propyl hydrogen sulphate(VII) Chemical equation CH3H2C=CH2 + H2SO4 CH3CH2 - CH2OSO3H H H H H H O-SO3H C = C - C - H + H2SO4 H - C - C – C - H H H H H H H Propene + H2SO4 propylhydrogen sulphate(VI) (ii) Propylhydrogen sulphate(VI) is hydrolysed by water to... | {
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Addition polymers are named from the alkene/monomer making the polymer and adding the prefix “poly” before the name of monomer to form a polyalkene During addition polymerization (i)the double bond in alkenes break (ii)free radicals are formed (iii)the free radicals collide with each other and join to form a larger mol... | {
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Examples of addition polymerization 1.Formation of Polyethene Polyethene is an addition polymer formed when ethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduc... | {
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During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting paticles) H H H H H H H H C = C + C = C + C = C + C = C + … H H H H H H H H Ethene + Ethene + Ethene + Ethene + … (ii)the double bond joining the ethane molecule break to free rea... | {
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Polyethene molecule can be represented as: H H H H H H H H extension of molecule/polymer - C – C - C – C - C – C - C – C- H H H H H H H H Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H H Where n is the number of monomers in the polymer. The number of monomers in the polymer can... | {
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The number of monomers in the polymer can be determined from the molar mass of the polymer and monomer from the relationship: Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer Examples Polythene has a molar mass of 4760.Calculate the number of ethene molecules in the polymer(C=12.0, ... | {
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40 It is an elastic, tough, transparent and durable plastic. Polythene is used: (i)in making plastic bag (ii)bowls and plastic bags (iii)packaging materials 2.Formation of Polychlorethene Polychloroethene is an addition polymer formed when chloroethene molecule/monomer join together to form a large molecule/polymer at ... | {
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Polythene is used: (i)in making plastic bag (ii)bowls and plastic bags (iii)packaging materials 2.Formation of Polychlorethene Polychloroethene is an addition polymer formed when chloroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)man... | {
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During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + … H Cl H Cl H Cl H Cl chloroethene + chloroethene + chloroethene + chloroethene + … (ii)the double bond joining the chlo... | {
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