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MODULE FOR INTERNATIONAL STANDARD CLASS SMK NEGERI 1 TAROGONG KALER SUBJECT DATE – TIME TEACHER SUB. THEME : Chemistry : ………………………………….. : Undang Suryana, HS. : Hydrocarbons ( Isomerism & Reactions Of Hidrocarbons ) CLASS/SEMESTER,YEAR: X/2, 2005 – 2006 I. General Objectives The students can understand organic compounds and molecul macro, determine product of reaction and synthesise molecul macro as its using. II. Spesific Objectives After completing the lessons, students are expected to be able to : - explain isomerism concept & apply of classity of hydrocarbon compounds. - Write simple reaction of alkanes, alkenes, alkynes. III. Materials - ISOMERISM Isomerism is a hydrocarbon compound has same molecule formula and different structural. Example : C4H8 → a) CH3 – CH2 – CH2 – CH3 = butane b) CH3 – CH – CH3 = Isobutane ∣ CH3 - REACTIONS OF HYDROCARBONDS – 1. Oxidation CH4 (8) + O2 ( 8) → CO2 = 2H2O 2. Substitution Reactions CH4 + Cl2 → CH3Cl + HCl 3. Addition Reactions CH3 CH = CH2 + Br2 → CH3 CH – CH2 ∣ ∣ Br Br 4. Elimination Reactions CH3 – CH – CH2 + Zn → CH3 – CH = CH2 + Zn Cl2 ∣ ∣ propen Cl Cl IV. Method / Technique Demonstration : Use molymood V. Competence Test 1) Make isomerism from C6H14, C6H12, C6H10 2) Write the following equestion a. C3H8 + O2 → b. Propene + H2 → c. 2 – pentene + HCl → d. 2 – meter – 2 – pentene + HI → e. 2 cloro butane + KOH → Refferences Ebbing, General Chemistry, USA : Houghton Mifflin Company, 1987 Purba, Michael, Kimia untuk SMA 1b, Jakarta : Erlangga 2004. I. Give the IUPAC name for each of the following hydrocarbon 1) CH3 – CH – CH – CH – CH3 ∣ ∣ ∣ CH3 CH3 CH3 2) CH3 – CH – CH2 – CH2 ∣ ∣ C 2 H5 C2H5 3) CH3 – CH – CH2 = CH2 – CH – CH3 ∣ ∣ CH3H7 CH3 C2H5 ∣ 4) CH3 – C ≡ C - C – CH2 – CH – CH3 ∣ ∣ CH3 CH3 5) CH2 = CH – CH2 – C = CH – CH2 ∣ ∣ CH3 CH3 II. Give the condesed structural formula for each of the following compounds 6) 2, 36 – trimetil oktana 7) 2,4 – dimetil – 1 – pentena 8) 4 – metil – 2 – pentuna 9) 3,3 – dietil – 4 – propil – 1,5 – heksadiena 10) Isobutana III. Make isomerism from C6H10 MODULE FOR INTERNATIONAL STANDARD CLASS SMK NEGERI 1 TAROGONG KALER SUBJECT DATE – TIME TEACHER SUB. THEME : Chemistry : ………………………………… : Undang Suryana, HS. : Auto Redox I. General Objectives Students explain the development of redox concept and relation with name system of compound. II. Spesific Objectives After completing the lessons, students are expected to be able to : - Give the name IUPAC of compound - Explain the auto redox reaction III. Materials - Otoredox reaction was called : disproporsionasi. - Otoredox reaction : redoks eaction which the same substance as redactor and As oxidator. Example : 1. Cl2 (g) + 2 NaOH (g) → NaCl (ag) + NaClO (ag) + H2O reduksi -1 oksidasi +1 Past of Cl2 gas (biloks = 0) realize reduction to NaCl (biloks Cl = -1) and another part realize oxidation to NaCl O (BILOKS Cl = + 1) Cl2 = reduktor Cl2 = oksidator +4 2. H2S -2 + SO2 reduction o → oksidation 3S + 2 H2O o SO2 = oksidatur H2S = reductor - The name system of Ionic Compound by IUPAC Oxidation number can used for indicate “muatan” ionic metal in the ionic comp. Exp : CO2S → 2 CO + S2Biloks Cu = +1 Name : Tembaga (I) sulfide Fe2 (SO4)3 → 2 Fe 3+ + 3 SO4 2biloks Fe = + 3 name : besi (III) sulfat Pb (NO3)2 → Pb 2+ + 2NO3Biloks Pb = +2 Name : Timbal (II) nitrat IV. Competence Test Answer the question ! 1. Define auto redox or not, these reaction bellow ? a. 3 NaClO → 2NaCl + NaClO3 b. 5 KI + KIO3 + 3 H2SO4 → 3 K2SO4 + 3 I2 + 3 H2O 2. Give the name these compound bellow : a. SnO b. CO2O c. Al2 (SO4)3 d. HgCl2 e. MgO V. Reference - Michael Purba, Kimia SMA 1B, Michael Purba. VI. Student’s difficulties and suggestions MODULE FOR INTERNATIONAL STANDARD CLASS SMK NEGERI 1 TAROGONG KALER SUBJECT DATE – TIME TEACHER SUB. THEME : Chemistry : ……………………………………… : Undang Suryana, HS. : Name System Of The Complex Ionic I. General Objectives Students understand about property of the transisi element through obaservation and concept applied. II. Spesific Objectives After completing the lessons, students are expected to be able to : - Give name of compleks ionic. - Give name of coordinasi compound III. Materials The name system of complex ionic, differentiate : 1. Complex Positive Exp : coordinate number + ligan + name atomic centre + oxidation number Zn (NH3)4 2 + = tentra aming seng (II) Ag (NH3)2 + = di amin perak (I) Cu (H2O)4 2+ = tetra aguo tembaga (II) 2. Complex Negative Exp : Koordinate number + ligan + name atomic centre + “at” + oxidation number Ag (CN)2 - = disiano Argentant (I) Fe (CN)6 4 - = Heksa siano ferat (II) CoF6 3= Heksa Floura Cobaltat (III) IV. Competence Test Answer the question 1. Write the name of : a). Zn (NH3) Cl + b) Cr (H2O)4 Cl2 + c) Cu (NH3)4 2+ 2. Write chemistry formula of : a). Kalium Heksasiano marganat (III) b). Natrium tetrasianozinkat (II) C). Tetra amin dikloeakobalt III nitrat V. Reference - Michael Purba, Kimia 2000 3B, Erlangga VI. Student’s Difficulties and Suggestions MODULE FOR INTERNATIONAL STANDARD CLASS SMK NEGERI 1 TAROGONG KALER SUBJECT DATE – TIME TEACHEAR SUB. THEME CLASS / SEMESTER, YEAR : Chemistry : ………………………………… : Undang Suryana, HS. : Petroleum & Petrochemical : X / 2, 2005 - 2006 I. General Objectives Describe process of forming & technique of saparation of petroleum fraction with Its use. II. Spesific Objectives - Explain process of forming petroleum and natural gas. - Explain main components of composing petroleum. - Interpret of diagram of distilling stratified for explaining base and technique of saparation in petroleum fraction. - Differ gasoline quality base on octan number. - Explain of using of petroleum residue in petrochemical industry. - Analyze effect of fuel burning of environment. III. Materials 1. Petroleum composition 2. Petroleum refining Petroleum, as we noted earlier, is a mixture of hydrocarbons, principally alkanes and cycloalkanes. The object of petroleum refining is to obtain various hydrocarbon products from the mixture. In part, this is accomplished by fractional distillation of the petroleum, but the demand for certain products, particularly gasoline, is greater than what can be supplied by distillation. For the reson, petroleum refiners resort to various chemical processes is catalytic cracking. The antiknock characteristics of a gasoline are rated by the octane number scale. 3. Process of petroleum. 4. Quality of gasoline. IV. Method / Technique Demonstration : Use diagrams of petroleum. IV. Competence Tests Answer the questions below ! 1. How process of petroleum ? 2. What is meaning ? a. Stratified distilation b. Petroleum cracking c. Octan number 3. How to improve petroleum quality ? 4. Mention additive substance in fuel ? And mention profit and as unprofit ? References Purba, Michael, Kimia untuk SMA 1b, Jakarta : Erlangga 2004. Ebbing, General Chemistry USA : Houghton Mifflin Company, 1987. MODULE FOR INTERNATIONAL STANDARD CLASS SMK NEGERI 1 TAROGONG KALER SUBJECT : Chemistry DATE – TIME : ……………………………… TEACHER : Undang Suryana HS. SUB. THEME : Reaction of Stoichiometry CLASS/SEMESTER, YEAR: X (Science) / 2; 2005-2006 I. General Objectives The students can apply the law of Avogadro and Gay iussac with mol concept in solving stoichiometry. II. Specific Objectives After completing the lessons, students are expected to be able to : Determine limiting reagent in a reaction. III. Materials Stoichiometry is the study of mole, mass, energy, and volume relationship In chemical reactions. In stoichiometry, we usually look at the quantities of reactans that combine together to produce various amount of products. Example : 1) Calculate the theoretical maximum number of moles of NH3 that results when 0,55 mol H2 combines with excess N2 ? N2 + 3H2 → 2NH3 solutions Ar C = 1 mol C X125 mol H2O x 12 g C = 15000 C. 1 mol H2O Limiting Reagent Problems Is reactant limit in reaction and the maximum yield of product. Example: 1) What mass of SO3 is produced when 0,6 g SO2 is combined with 0,4 g 02 ? 2SO2 (g) + O2 (g) → 2SO3 (g) Solution 2SO2(G) + O2(g) → 2SO3(g), Mr SO3 = 80 mol So2 = 0,68 SO2 = 0,00938 mol SO2 → limiting reagent 64g SO2 mol O2 = 0,4g SO2 = 0,0125 mol O2 32g O2 mass of SO3 = 2mol SO3 X 0,00938 mol SO2 x 80 gr SO3 = 0,75 gr SO2 2 mol SO2 IV. Competence Test. Answer the question below ! 1) What mass of oxygen gas O2, is liberated when a 2,5 g sample of sodium Nitrate is heated ? 2 NaNO3(S) → 2 Na NO2(S) + O2(g) 2) What volume of SO2, is liberated when 884 g S8is combined with excess O2 ? S8(s) + 8 O2 (g) → 8 SO2(g) 3) Consider the equition : H2 + I2 → 2HI Calculate the mass of HI that forms when 1,05 g H2 combine with 122 g I2 ? V. References Drew H. Wholfe, introducing the college chemistry, US: Mc. Graw Hill, 1984. Purba, Michael, Kimia untuk anak SMA kelas X Jilid 1 b, Jakarta : Erlangga 2002 VI. Student’s Defficulties and Suggestions MODULE FOR INTERNATIONAL STANDARD CLASS SMK NEGERI 1 TAROGONG KALER SUBJECT : Chemistry DATE-TIME : ………………………………….. TEACHER : Undang Suryana, HS. SUB. THEME : Compounds of Stoichiometry CLASS/SEMESTER, YEAR: X (Science) / 2; 2005 – 2006 I. General Objectives The students can apply the law of Gay Lussac and Avogadro with mol concept in Solving stoichiometry. II. Spesific Objectives After completing the lessons, students are expected to be able to : Determine empirical formula from the composition and molecular formula from empirical formula and mass percentages from the formula. III. Materials 1. An Empirical Formula or Simplest Formula from the Composition. The empirical formula of a compound shows the ratios of number of atoms in the compound. Example : A compound of nitrogen and axygen is analyzed, and a sample waging 1,87 g is found to contain 0,483 grams N and 1,104 grams O. What is the empirical formula of the compound ? Solution : Mol N = 0,483grams = 0,0345 mol N 14grams N Mol O = 1,104grams = 0,06900 mol O 16gramsO N : O 0,0345 : 0,069 So the empirical Formula : NO2 1 : 2 2. Molecular Formula from Empirical Formula. The molecular formula of a compound is multiple of its empirical formula n = molecularweight empiricalformulaweight Example : We found the percentages compositions of butyric acid to be 54,2 % C, 9,2 % H, and 36,6 % O. Determine the empirical formula. The molecular weight of Butyric acid was determined by experiment to be 88 amu.What is the molecular formula ? Solutions : C : H 54,2 : 12 4,51 : 1,97 : 2 : (C2H4O) n = 88 (44) n = 88 n = 2 : O 9,2 1 9,2 4 4 : 36,6 16 : 2,29 : 1 : 1 So the molecular formula of butyric acid is (C2H4O)2 or C4h8o2 3. Percentages of Element in Compound % = x.Ar x 100 % Mr Example : Calculate the percentage of C and N in CO (NH2)2 ? Known: Ar H = 1, C = 12, O = 16 Solutions : Mr CO (NH2)2 = 60 % C = 1(12) X 100 5 = 20 % 60 %N = 2(14) X 100 % = 46,67 % 60 IV. Compeyence Tests 1. Benzoid acid is a white crystalline powder uae as a food preservative. The Compound contains 68,8 % C, 5,0 % H, nad 26,2 % O by mass. What is its Empirical formula ? 2. The percentage composition of etyl is 62 % C, 10,4 % H and 27,6 % O and The molecular weight is 116 amu. Obtain the molecular formula. V. References Drew H. Wolfe, IN TRODUCTION TO COLLEGE Chemistry, USA: pany, 1987. Purba, Michael, KIMIA UNTUK sma Kelas I Jilid 1b. Jakrta: Erlangga, 2002.