IChO(国际化学奥林匹克竞赛)考试大纲
国际化学奥林匹克竞赛基本要求

国际化学奥林匹克竞赛基本要求说明:☆化学论题的分类(下表各知识点后面的数字,前者是它归属的组别,后者是知识点序号)·组1:这些论题是绝大多数中学化学大纲里有的。
·组2:这些论题是许多中学化学大纲里有的;若大纲里没有,预期各国应向参赛学生介绍。
·组3:这些论题是大多数中学化学大纲里没有的。
☆主办国出的预备题不必涉及第一和二组论题,但对第二组论题应开具细目。
所有竞赛将涉及到的第三组论题必须包括在预备题中。
译注:☆1:本大纲制订的基础是对参赛队领队的大量问卷调查与综合分析,制订工作历时8年,三度修订。
通读本大纲可得出如下认识:本大纲是采取举例的方式来明确知识点的层次,重在体现知识点的不同层次,不能说大纲里没有列入的知识点就不属于竞赛内容。
也可以有如下认识:本大纲的一级知识水平大致与Nuffield O-水平吻合,二级知识大致与Nuffield A-水平吻合,三级知识大致为大学本科低年级知识水平。
☆2:知识点的分级旨在明确竞赛前参赛选手需要记忆多少知识作为解题的基础,不等于在竞赛试题中不得出现超纲的供参赛选手现学现用的化学知识信息以及考核参赛选手把握新知识内涵的能力(学习能力)与应用新知识的能力(应用能力)以及创造性思维的能力。
☆3:本大纲是98年10月在斯洛伐克首都布拉津斯拉伐工作会议上修订的。
在该次工作会议上与会代表用6个单元时间逐字逐句地对原在华沙会议制订的大纲进行分组审核,提出修改意见,然后又用1个单元时间进行讨论,最后由指导委员会主席携回加工定稿。
第31、32届IChO预备题均将此大纲作为附件,预备题的编制以此大纲为依据。
☆4:全国化学竞赛基本要求应根据此大纲作出相应调整。
全国初赛应与本大纲一级知识水平大致吻合,决赛(冬令营)应于本大纲一、二级知识水平大致吻合,对国家集训队进行的讲座与国家队四名选手的选拔赛应依据本大纲一、二级知识和预备题中明确的三级知识,属于本大纲三级知识而预备题未涉及的不应作为国家集训队选手已经掌握的(记忆了的以及能够熟练应用的)知识来考核。
国际化学奥林匹克竞赛-国际化学奥林匹克竞赛-第38届ICHO理论试题(中文版)答案

1-1.The mass of a water droplet:m = V ρ = [(4/3) π r3] ρ = (4/3) π (0.5x10-6 m)3 (1.0 g/cm3)= 5.2x10-16 kg⇒10 marksAverage kinetic energy at 27o C:KE = mv2/2 = (5.2x10-16 kg) (0.51x10-2 m/s)2/2= 6.9x10-21 kg m2/s2= 6.9 x10-21 J ⇒15 marks*.The average kinetic energy of an argon atom is the same as that of a water droplet.KE becomes zero at –273 o C.From the linear relationship in the figure, KE = aT (absolute temperature)where a is the increase in kinetic energy of an argon atom per degree.a = KE/T = 6.9x10-21 J/(27+273K) = 2.3x10-23 J/K⇒25 marksS: specific heat of argon N: number of atoms in 1g of argonS = 0.31 J/g K = a x NN = S/a = (0.31 J/g K) / (2.3x10-23 J/K)= 1.4x1022 ⇒30 marksAvogadro’s number (N A) : Number of argon atoms in 40 g of argonN A = (40)(1.4x1022)= 5.6 x1023⇒20 marks2-1. ⇒ 30 marksmass of a typical star = (4/3)(3.1)(7x108 m)3(1.4 g/10-6 m 3) = 2×1033 g mass of protons of a typical star = (2×1033 g)(3/4 + 1/8) = 1.8×1033 g number of protons of a typical star = (1.8×1033 g)(6×1023/g) = 1×1057number of stellar protons in the universe = (1×1057)(1023) = 1×1080Partial credits on principles:Volume = (4/3)(3.14)radius 3×density; 4 marks 1 mole = 6×1023; 4 marksTotal number of protons in the universe = number of protons in a star ×1023; 2 marks Mass fraction of protons from hydrogen = (3/4)(1/1); 5 marks Mass fraction of protons from helium = (1/4)(2/4); 10 marks2-2. ⇒ 30 marks∆E(2→3) = C(1/4 - 1/9) = 0.1389 C λ(2→3) = 656.3 nm ∆E(1→2) = C(1/1 - 1/4) = 0.75 Cλ(1→2) = (656.3)(0.1389/0.75) = 121.5 nmNo penalty for using Rydberg constant from memory. 15 marks penalty if answered in a different unit (Hz, etc.)2-3.T = (2.9×10-3 m K)/1.215×10-7 m = 2.4×104 K ⇒ 10 marks2-4..⇒ 20 marksλ = 3 × 108 m/1.42 × 109 = 0.21 mT = (2.9 × 10-3 m K)/0.21 m = 0.014 K2-5. ⇒ 10 marks14N + 4He → (17O ) + 1HO-17, O acceptable1783-1.k des = A exp(-E des/R T)= (1x1012 s-1)(5x10-32) = 5x10-20 s-1 at T = 20 K ⇒10 markssurface residence time, τresidence = 1 / k des = 2x1019 s = 6x1011 yr ⇒20 marks(full credit for τhalf-life = ln2 / k des = 1x1019 s = 4x1011 yr)residence time = 2x1019s3-2.The distance to be traveled by a molecule: x = πr = 300 nm.k mig = A exp(-E mig/R T)= (1x1012 s-1)(2x10-16 ) = 2x10-4 s-1 at T = 20 K ⇒ 5 marksaverage time between migratory jumps,τ = 1 / k mig = 5x103 sthe time needed to move 300 nm= (300 nm/0.3 nm) jumps x (5x103 s/jump) = 5x106 s = 50 days ⇒15 marks(Full credit for the calculation using a random-walk model. In this case:t = τ (x/d) 2 = 5 x 109 s = 160 yr. The answer is still (b).)(a) (b)(c) (d) (e)10 marks3-3.k(20 K) / k(300 K) = exp[(E/R) (1/T1 - 1/T2)]= e-112 = ~ 10-49 for the given reaction ).) ⇒15 marks The rate of formaldehyde production at 20 K= ~ 10-49 molecule/site/s = ~ 10-42 molecule/site/ yr⇒10 marks(The reaction will not occur at all during the age of the universe (1x1010 yr).)rate = 10-42molecules/site/yr3-4. circle one(a) (b) (c) (a, b) (a, c) (b,c)(a, b, c)(15 marks, all or nothing)4-1.H PNumber of atoms ( 11.3 ) 1⇒ 10 marksTheoretical wt % ( 3.43 )⇒ 10 marks4-2.adenineN NN NN H H guanineNN N NO N HH HNN O N H H cytosineNN H O O thymine(10 marks on each)4-3. 7 marks each, 20 marks for threeadenineNNNNNHHguanine NN NNON HHH NNH OOthymineNNONHH cytosine NNH OOthymineguanine NN NNON HHHcytosineNNONHHcytosineNNON HHNNHOO thyminethymineNNHOONNH OOthyminethymine NNHOONNONHH cytosineadenineNNNNNHH adenineNNNNNHHadenine NNNNNHHguanineguanine NNNNON HHHNNNNONHHH4-4. 2.5 marks for each bracketadenineN NN N HNH 2guanine N NH N N HO NH 2Uracil N H NH O cytosineN H N NH 2OOHCN ( 5 ) ( 5 ) ( 4 )( 4 )H 2O ( 0 ) ( 1 ) ( 2 ) ( 1 )5-1.(20 marks)1st ionization is complete: H2SO4→ H+ + HSO4-[H2SO4] = 02nd ionization: [H+][SO42-]/[HSO4-] = K2 = 1.2 x 10-2 (1)Mass balance: [H2SO4] + [HSO4-] + [SO42-] = 1.0 x 10-7 (2)Charge balance: [H+] = [HSO4-] + 2[SO42-] + [OH-] (3)Degree of ionization is increased upon dilution.[H2SO4] = 0Assume [H+]H2SO4 = 2 x 10-7From (1), [SO42-]/[HSO4-] = 6 x 104 (2nd ionization is **plete)[HSO4-] = 0From (2), [SO42-] = 1.0 x 10-7 [5 marks]From (3), [H+] = (2 x 10-7) + 10-14/[H+][H+] = 2.4 x 10-7(pH = 6.6) [8 marks][OH-] = 10-14/(2.4 x 10-7) = 4.1 x 10-8[2 marks]From (1), [HSO4-] = [H+][SO42-]/K2= (2.4 x 10-7)(1.0 x 10-7)/(1.2 x 10-2) = 2.0 x 10-12[5 marks]Check charge balance:2.4 x 10-7≈ (2.0 x 10-12) + 2(1.0 x 10-7) + (4.1 x 10-8)Check mass balance:0 + 2.0 x 10-12 + 1.0 x 10-7≈ 1.0 x 10-7Species Concentration** x 10-12HSO4-** x 10-7SO42-** x 10-7H+** x 10-8 OH-5-2. (20 marks)mmol H3PO4 = 0.85 ⨯ 3.48 mL ⨯ 1.69g/mL ⨯ 1 mol/98.00 g ⨯ 1000 = 51.0 [5 marks]The desired pH is above p K2.A 1:1 mixture of H2PO4- and HPO42- would have pH = p K2 = 7.20.If the pH is to be 7.40, there must be more HPO42- than H2PO4-.We need to add NaOH to convert H3PO4to H2PO4-and to convert to the right amount of H2PO4-to HPO42-.H3PO4 + OH-→ H2PO4- + H2OH2PO4- + OH-→ HPO42- + H2OThe volume of 0.80 NaOH needed to react with to to convert H3PO4 to H2PO4- is:51.0 mmol / 0.80M = 63.75 mL [5 marks]To get pH of 7.40 we need:H2PO4- + OH-→ HPO42-Initial mmol 51.0 x 0Final mmol 51.0-x 0 xpH = p K2 + log [HPO42-] / [H2PO4-]7.40 = 7.20 + log {x / (51.0-x)}; x = 31.27 mmol [5 marks]The volume of NaOH needed to convert 31.27 mmol is :31.27 mmol / 0.80 M = 39.09 mLThe total volume of NaOH = 63.75 + 39.09 =102.84 mL , 103 mL [5 marks]Total volume of 0.80 M NaOH (mL) 103 mL5-3. (20 marks)p K = 3.52pH = pK a + log ([A-]/[HA])[A-]/[HA] = 10(pH-pKa) [5 marks]In blood, pH =7.40, [A-]/[HA] = 10(7.40-3.52) = 7586Total ASA = 7586 +1 = 7587 [5 marks]In stomach, pH = 2.00, [A-]/[HA] = 10(2.00-3.52) = 3.02x10-2Total ASA = 1+ 3.02x10-2 = 1.03 [5 marks]Ratio of total aspirin in blood to that in stomach = 7587/1.03 = 7400 [5 marks]** ( 103Ratio of total aspirin in blood to that in stomach6-1. (5 marks)4 H2O + 4 e-→ 2 H2(g) + 4 OH- (or 2 H2O + 2 e-→ H2(g) + 2 OH-)6-2. (5 marks)2 H2O → O2 + 4 H+ + 4 e-(or H2O → 1/2 O2 + 2 H+ + 2 e- )6-3. (5 marks)Cu → Cu2+ + 2e-6-4. (20 marks)Reduction of sodium ion seldom takes place.It has a highly negative reduction potential of –2.710 V.Reduction potential for water to hydrogen is negative (water is very stable).But, it is not as negative as that for sodium ion. It is –0.830 V.Reduction of both copper ion and oxygen takes place readily and the reduction potentials for both are positive.In the present system, the reverse reaction (oxidation) takes place at the positive terminal. Copper is oxidized before water.Reduction potential for hydrogen ion is defined as 0.000 V.6-5. (15 marks)pOH = 14.00 – 4.84 = 9.16[OH-] = 6.92 x 10-10K sp = [Cu2+][OH-]2 = 0.100 x (6.92 x 10-10) = 4.79 x 10-206-6.E = E o Cu2+/Cu + (0.0592/2) log [Cu2+]= +0.340 + (0.0592/2) log [Cu2+]= +0.340 + (0.0592/2) log (K sp / [OH-]2)= +0.340 + (0.0592/2) log (K sp) - (0.0592/2) log [OH-]2= +0.340 + (0.0592/2) log (K sp) - 0.0592 log [OH-],3 marksBy definition, the standard potential for Cu(OH)2(s) + 2e-→ Cu(s) + 2OH- is the potential where [OH-] = 1.00.E = E o Cu(OH)2/Cu = +0.340 + (0.0592/2) log (K sp)= +0.340 + (0.0592/2) log (4.79 x 10-20)= +0.340 - 0.5722 marks= -0.232 V10 marks-------------------------------------------------------------------------------------------------------------- One may solve this problem as following.Eqn 1: Cu(OH)2(s) + 2e -→ Cu + 2OH-E+o = E o Cu(OH)2/Cu = ?Eqn 2: Cu(OH)2(s) → Cu2+ + 2OH-E o = (0.05916/n) logK sp= (0.05916/2) log(4.79×10-20)= -0.5715 V3 marksEqn 1 – Eqn 2 : Cu2+ + 2e-→ CuE-o = E+o - E o = E o Cu2+/Cu = 0.34 VTherefore, E+o = E-o + E o = + 0.34 + (-0.5715)2 marks= -0.232 V10 marks-0.232 V6-7.Below pH = 4.84, there is no effect of Cu(OH)2 because of no precipitation.Therefore,E = E Cu2+/Cu = +0.340 + (0.0592/2) log [Cu2+]= +0.340 + (0.0592/2) log 0.1003 marks= +0.340 – 0.0296 = +0.310 V7 marks** V6-8.** g graphite = 0.0833 mol carbon6 mol carbon to 1 mol lithium; 1 g graphite can hold 0.0139 mol lithiumTo insert 1 mol lithium, 96487 coulombs are needed.Therefore, 1 g graphite can charge 96487 × 0.0139 = 1340 coulombs. 5 marks1340 coulombs / g = 1340 A sec / g = 1340 x 1000 mA × (1 / 3600) h = 372 mA h / g 5 marks372 mA h / g7-1. (10 marks)n/V = P/RT = (80 x 106 / 1.013 x 105 atm)/[(0.082 atm L/mol/K)(298K)] = 32 mol/L5 marksdensity = mass/volume = d = 32 x 2 g/L = 64 kg/m 3 5 marks64 kg/m 37-2.** or 0.23H 2(g) + 1/2 O 2(g) → H 2O(l); ∆H rexn-1 = ∆H f [H 2O(l)] = -286 kJ/mol = -143 kJ/g 7 marksC(s) + O 2(g) → CO 2(g); ∆H rexn-2 = ∆H f [CO 2(g)] = -394 kJ/mol = -33 kJ/g 7 marks(-∆H rexn-1) / (-∆H rexn-2) = 4.3 or (-∆H rexn-2) / (-∆H rexn-1)= 0.236 marks7-3. (a) (-)1.2 x 105 kJ, (b) (-)6.9 x 104 kJ** x 108 sec or 3.3 x 104 hr or 1.4 x 103 days or 46 month or 3.8 yrI = 0.81 AH 2(g) + 1/2 O 2(g) → H 2O(l)∆H c = -286 kJ/mol = -143 kJ/g = -143 x 103 kJ/kg 5 marksΔG = ΔH – T ΔSΔS c= 70 – 131 – 205/2 = -163.5 J/K/mol5 marksΔG c = -286 kJ/mol + 298K x 163.5 J/K/mol = -237 kJ/mol = -1.2 x 105 kJ/kg 5 marks(a) electric motor W max = ΔG c ⨯ 1 kg = - 1.2 x 105 kJ 5 marks (b) heat engine W max = efficiency x ∆H c 5 marks= (1 – 298/573) x (-143 x 103 kJ) = -6.9 x 104 kJ 5 marks119 x 103 kJ = 1 W x t(sec)t = 1.2 x 108 sec = 3.3 x 104 hr = 1.4 x 103 days = 46 month = 3.8 yr 5 marksΔG = -nFE n = # of electrons involved in the reaction F = 96.5 kC/molH 2(g) + 1/2 O 2(g) → H 2O(l) n = 2 5 marksE = - ΔG/nF = 237 kJ/mol / 2 / 96.5 kC/mol = 1.23 V5 marksI = W/E = 0.81 A5 marks8-1-1. (5 marks on each)①C②C③CO8-1-2.③ Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) 5marks① C(s) + O2(g) → CO2(g) ΔH①◦ = -393.51 kJ = ΔH f◦(CO2(g))② CO2(g) + C(s) → 2CO(g) ΔH②◦ = 172.46 kJFrom ① and ②,ΔH f◦(CO(g)) = (1/2){172.46 + (-393.51)} = -110.525 kJΔH f◦(Fe2O3) = -824.2 kJΔH③◦ = 3ⅹΔH f◦(CO2(g)) - ΔH f◦(Fe2O3) - 3ⅹΔH f◦(CO(g))= 3ⅹ(-393.51) – (-824.2) - 3ⅹ(-110.525) = -24.8 kJ 7 marks ΔS③°=2ⅹ27.28+3ⅹ213.74-87.4-3ⅹ197.674=15.36 J/K 3 marks ΔG③°=ΔH°-TΔS°=-24.8kJ-15.36J/Kⅹ1kJ/1000Jⅹ1473.15K=-47.43 kJ5 marksK = e(-ΔG°/RT)= e(47430J/(8.314J/Kⅹ1473.15K)) = 48 5 marksBalanced equation of ③:K = 48Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)8-2-1. (20 marks)One AB2O4 unit has available 4 (= 1 + (1/4)ⅹ12) octahedral sites.48-2-2. (20 marks)Since one face-centered cube in AB2O4 represents one Fe3O4 unit in this case, it has 8 available tetrahedral sites. In one Fe3O4 unit, 1 tetrahedral site should be occupied by either one Fe2+ (normal-spinel) or one Fe3+ (inverse-spinel). Therefore, in both cases, the calculation gives (1/8) ⅹ100% = 12.5% occupancy in available tetrahedral sites.**%8-2-3. (10 marks for d-orbital splitting, 10 marks for elec. distribution)9-1-1. 1 answer for 8 marks, two for 15 marksH 3CN NNH 3CNNN :::+_+::_:9-1-2. ( 10 marks)H 3CN::9-1-3.H 3CNCH 2CH 2:H 3CN HH CCH 2:(10 marks) (10marks )9-2-1. 5 marks eachHONN +_::ONN:H+:HH_O NN:H+:H_::::::9-2-2.( 10 marks)CH 2CO ::9-3-1.(40 marks)CH 3H 3CH 3C+BC H 2CCH 3CH 3CO 2DEOOO_9-3-2.(10 marks)O OH O n+F10-1. 10 marks eachNMLCH 2OHCH 2OHMeOOMeH HH HOMeMeO CHOCHOCH 2OHCH 2OHHHH H OHOMeMeO OH10-2. 8 marks each for correct structuresNumber of possible structures24 marks12OH(OH)OH(H)HH HHOMeOMeOH COOMeOH(OH)OH(H)HH HHOMeOMeOHCOOMe34OH(OH)OH(H)OH(OH)OHe(H)10-3. 10 marks eachGICH 2OHCH 2OHHHHHMeOOMeOHOMeCH 2OHCH 2OHHHHOMeOMeOMe10-4. 10 marksNumber of the correct structure for C from 10-2110-5.BOH(OH)OH(H)HHHH OHCOOHOHOH10 marks eachDJOH(OH)OH(H)HHHHOMeOMeCOOMeOMeOH(OMe)OMe(H)HHHHOMeOMeOMeCOOMe10-6. 20 marksHOOCOHHH OOOHOOH COOHOOHOHOH COOH11-1. 10 marks311-2. 30 marksCOOHHOOCOOH11-3. 2.5 marks eacha, c, d11-4 30 marksOOCOCOOOHTransition State11-5.For the enzyme-catalyzed reaction, Arrehnius equation could be applied.k cat/k uncat = A exp (-E a, cat/ RT) / A exp (-E a, uncat / RT)= exp [-∆E a, cat-uncat/ RT]= exp [-∆E a, cat-uncat(J/mol) / (2,480 J/mol)] = 106Therefore, -∆E a, cat-uncat = 34,300 J/mol 15 marksk uncat, T/k uncat, 298 = exp (-∆H≠ uncat/ RT) / exp (-∆H≠uncat / 298R)= exp [(-∆H≠ uncat/R)(1/T-1/298)]ln(k uncat, T/k uncat, 298 )= 13.8 = [(-86900/8.32)(1/T-1/298)]Therefore, T = 491 K, or 218o C 15 marks-E a, cat-uncat = 34,300 J/molT = 491 K, or 218o C。
国际化学奥林匹克竞赛-40thIchotheoreticalanswers

Yes, but only in quite dilute solutions can this happen. 1 pt for ticking yes
c = [HA] + [A–] = [H+]
(1 pt)
[H+] = [A–] + [OH–]
(1 pt)
This means that [HA] = [OH–]
(1 pt for reasonable guess – between 6 and 7)
A good approximation is: [H+ ] = 3 (KKw )
The full equation can be solved through iteration: [H+ ] = 3 (K + [H+ ])Kw
In equilibrium constant calculations all concentrations are referenced to a standard concentration of 1 mol/dm3. Consider all gases ideal throughout the exam.
4
Name:
Code: XXX-
c)
Could it be possible that the solution contained acetic acid?
Acetic acid: pKa = 4.76
Yes No
If yes, calculate the pH (or at least try to estimate it) and show your work.
Official English version
全国化学竞赛决赛考试大纲

决赛大纲1.原子结构——四个量子数、原子轨道(波函数的角向分布图像与轨道名称)。
2.分子结构——用波函数的角向分布图像说明π键和σ键。
分子轨道的基本概念(成键、反键及非键)。
第二周期同核双原子分子的分子轨道能级图。
氟化氢的分子轨道能级图。
最高占有轨道(HOMO)与最低未占有轨道(LUWO)的概念。
键级的概念。
分子轨道想论对氧的磁性的解释。
3.晶体结构——点阵概念与14种点阵单位、7大晶系。
简单点阵单位和带心点阵单位的概念(如点阵单位中点阵点的数目与位置)。
结构基元。
晶胞中的等同原子。
堆积模型和堆积-填隙模型。
4.范德华力——取向力、诱导力和色散力。
氢键。
5.键能。
6.化学热力学基础——热力学能、焓、自由能和熵。
热化学计算〔等温过程的盖斯定律、(标准摩尔)生成焓、(标准摩尔)生成自由能、标准熵以及有关计算;假设焓变与熵变不是温度的函数的计算;假设为绝热体系的火焰最高温度计算〕。
自由能与反应的方向性。
吉布斯一亥姆霍兹方程及其应用(对反应方向与温度的关系的解释;对放热反应与吸热反应的反应方间性的说明;假设焓变与熵变不随温度变化的近似计算)。
范特霍夫方程及其应用。
标准自由能与平衡常数。
平衡常数与温度的函数关系。
热力学循环。
热力学分解温度(标态和非标态——压力对分解温度的影响)。
反应熵变的正负号的判断。
7.化学动力学基础——反应速率的基本概念;反应级数的概念;用实验数据推求反应级数。
一级反应(的积分式)的有关计算(速率常数、半衰期、C-1 4法推断年代等);阿累尼乌斯方程及其有关计算(阿累尼乌斯活化能的概念与计算;速率常数的计算;温度对速率常数的影响的计算等);活化能与反应热的关系;反应机理的基本概念;用稳态近似推求速率方程。
催化剂对反应的影响的本质。
化学反应的动力学制约和热力学制约。
8.酸碱质子理论。
缓冲溶液。
利用多元酸的平衡常数的计算。
盐的水解度和溶液的pH值的计算。
溶度积原理以及利用溶度积的计算。
氢氧化物沉淀与溶解的pH值。
2004年第36届ICHO预备试题及答案(中文)课件

准备题及解答Kiel, Germany January, 2004序言准备题目是为挑战和激励参加在德国Kiel举行的第36届国际化学奥林匹亚的学生所设计。
题目含盖广大范围的主题,但是大部份能以基本化学知识解答。
详细的题目和解答使学生能够学习和了解题目的背景。
你也可以在网站上得到此题目:www.icho.de...Chemistry(没有解答)。
解答将于2004年五月公告。
在同一网页(...Participation),你可以找到-IChO的规章-实作和理论部分的大纲-国际评审委员会制定的安全规则及建议-危险警告符号,名称,说明,R-警语,S-警语给教练的话:在理论测验时,会提供元素周期表和与准备题相类似的常数及公式。
假如须要额外的资料,请来信。
在考试时,学生会有一本试题和与试题分开的答案卷。
只有写在格内的答案才会被评分。
在考试时,大会准备了英文版的试题,但是只有在被要求的情形下才会提供。
对于须要计算的答案,只有在有写计算过程的情况才会被评分。
我们认为安全是很重要的。
在准备题的实作题目之前,你可以找到将在Kiel遵循的有关安全预防和步骤的规则表。
在Kiel注册时,我们将要求总教练签证关于他的学生队知道这些规则,且被训练和能够遵循的声明。
最后,虽然审慎的检视题目,也许您还是会发现一些没有被发觉的错误。
我们欢迎您对题目的意见。
(Hampe@t-online.de)欢迎参加第36届IChO,欢迎到德国,欢迎到Kiel。
Wolfgang Hampe内容页码题目/解答1 燃烧能(Combustion Energy)2 哈柏-柏西法Haber-Bosch Process)3 生物化学之热力学(Thermodynamics in Biochemistry)4 热导(Heat Conductivity)5 “绿色”化学---超临界流体CO2的应用(“Green”Chemistry – The Use of Supercritical CO2)6 过氧双硫酸盐离子的化学动力学(Chemical Kinetics of thePeroxodisulfate)7 乙烯氢化反应的催化(Catalytic Hydrogenation of Ethylene)8 酵素反应动力学(Kinetics of an Enzymatic Reaction)9 CaCN2, 一个旧但是重要的肥料(CaCN2– An Old but stillImportant Fertilizer)10 最密堆积的结构(Closed-Packed Structure)11 碳化钛,一个高科技的固体(Titanium Carbide – AHigh-Tech Solid)12 金属的奈米聚合物(Metal Nanoclusters)13 分子的光吸收(Absorption of Light by Molecules)14 观察单一分子(Observing Single Molecules)15 四面体分子的红外线光谱(Infrared Spectroscopy ofTetrahedral Molecules)16 生物有机化学光谱(Spectroscopy in Bioorganic Chemistry)17 去氧核醣核酸,核醣核酸和蛋白质(DNA, RNA, Proteins)18 脂肪酸的降解(Fatty Acid Degradation)19 脂质(Lipids)20 凯库勒的苯环结构(Kekulé, Benzene, and the Problem ofAromaticity)21 苯与环己烷(Benzene and Cyclohexane)22 非苯系芳香族化合物(Non-Benzoid Aromatic System)23 止痛剂(Pain Reliefers)24 羰基化学(Carbonyl Chemistry)25 环己烷(Cyclohexanes)26 掌性化合物(Chiral Compounds)27 单醣(Monosaccharides)28 Epibatidine29 Crixivan30 立体选择性的还原反应(Stereoselective Reduction)31 接口活性微胞(Surfactant Micelles)32 两性共聚高分子的自我组装(Self-assembly of AmphiphilicBlock Copolymers)33 微乳化技术(Microemulsions)34 硅石奈米结构(Silica Nanostructures)35 八水合过氧化锶的制备和测定(Preparation and volumetricdetermination of strontium peroxide octahydrate)36 碘酸钾的制备和测定(Preparation and iodometricdetermination of potassium iodate)37 定性分析未知混合物中的阴离子(Qualitative analysis ofanions in an unknown mixture)38 聚甲基丙烯酸甲酯的回收再利用(Recycling ofPolymethylmethacrylate)39 对氯苯甲醇的合成-坎尼札罗反应的例子(Synthesis ofpara-chlorobenzyl alcohol – an example of CannizzaroReaction)40 活泼碳酸衍生物的氨解:氰基乙醯胺的合成(Ammonolysis ofan activated carbonic acid ester: synthesis of cyanoacetamide)题目1: 燃烧能(Combustion Energy)1.1 写出丙烷和丁烷在空气中完全燃烧的化学方程式。
第51 届国际化学奥林匹克试题(理论部分)

126
Vol. 35
大 学 化 学
总则
本理论考卷共 58 页。
“Start (开始)”命令发出后,即可开始答题。
考试时间共 5 小时。
所有结果和答案必须用笔清楚地书写在考卷指定的相应答题框中。写在答题框外的答案不予评
判。
若需打草稿,可以利用试卷的背面。但谨记:写在指定区域之外的解答不予评判。
:醛
:羰基 α 位
苄基 CHn—C6H5:
芳环上的:
11.0
10.0
9.0
8.0
7.0
6.0
50
. All
Rights
Reserved.
H―H
耦合常数(Hz)
氢的种类
|Jab| (Hz)
R2CHaHb
4–20
R2HaC―CR2Hb
2–12
如果 σ 键可以自由转动:6–8
这次比赛。经过各 5 小时的实验和理论考试,共决出 196 枚奖牌——金牌
37 枚,银牌 64 枚,铜牌 95 枚。中国代表队何流(浙江杭州学军中学)、邓
子杰(河北衡水第一中学)、杨景程(湖南长沙市第一中学)、陈劭炎(山东省
广饶县第一中学) 4 名选手发挥出色,获得三金一银的优异成绩。其中,何
流总分第一、实验(满分)与另两位选手并列第一,邓子杰总分第二、理论成绩第一,获得国际纯粹与应用化学联合
教授观摩比赛并为申办工作提供指导;南开大学朱守非教授、梁广鑫教授和郭东升教授观摩比赛并代表中国化学会
向国际化学奥赛组委会提出申请并进行答辩,为我国获得了 2022 年 IChO 的举办权。
兹将本届国际化学奥林匹克理论试题及实验试题中文版刊出,以飨读者。为最大程度展示试题原貌,编辑保持
化学学科的学科竞赛与奖项介绍

化学学科的学科竞赛与奖项介绍化学学科竞赛与奖项在学生学习和研究中起着重要的推动和鼓励作用。
通过参加学科竞赛,学生可以展示自己的专业知识和研究成果,提高学术水平,并获得荣誉和奖励。
本文将介绍一些著名的化学学科竞赛和相关奖项。
一、国际化学奥林匹克竞赛(IChO)国际化学奥林匹克竞赛是世界上最具声誉和影响力的化学学科竞赛之一。
该竞赛每年举办一次,吸引来自全球各个国家和地区的优秀化学学生参加。
竞赛中,学生们需在实验和理论两个方面展示他们的化学素养和解决问题的能力。
获奖者不仅能获得个人荣誉,还有机会获得奖学金和进入顶尖大学学习的机会。
二、全国中学生化学竞赛全国中学生化学竞赛是中国中学生学科竞赛中的重要组成部分。
该竞赛根据学年不同分为初赛和决赛两个阶段。
初赛主要通过考试形式进行,考察学生对基础化学知识的理解和应用能力。
决赛则通过笔试和实验实训等形式,对学生的创新思维和实践能力进行考察。
在全国化学竞赛中获奖的学生将受到学术界的关注,并有机会获得奖学金和科研机会。
三、国家级化学竞赛国家级化学竞赛是中国高中生的一项重要学科竞赛。
该竞赛由教育部主办,面向全国各地的高中学生。
竞赛分为初赛、复赛和决赛三个阶段。
初赛主要以选择题形式考察学生的记忆能力和基础知识掌握情况。
复赛则通过计算题和实验题考察学生的应用能力和实验操作技巧。
决赛则是对复赛成绩较好的学生进行更加深入的考察,涉及到更高层次的化学知识和问题解决方法。
四、地区化学竞赛地区化学竞赛是在省市一级组织的学科竞赛活动。
不同地区的化学竞赛有不同的名称和形式,如上海市化学竞赛、北京市化学竞赛等。
这些竞赛旨在推动地方教育发展,挖掘和培养优秀的化学学生。
获奖的学生将受到地方教育部门和高校的关注,有机会获得奖学金和升学机会。
除了这些竞赛外,其他化学学科领域还有许多奖项和科研机会可供学生展示才华和获得更多的学术发展机会。
例如,针对化学研究方向的国际学术会议,如国际化学联合会大会、美国化学学会年会等,提供了与化学界专家学者面对面交流的机会。
国际化学奥林匹克竞赛试题汇编-第38届ICHO理论试题(中文版)答案

1-1.The mass of a water droplet:m = V ρ = [(4/3) π r3] ρ = (4/3) π (0.5x10-6 m)3 (1.0 g/cm3)= 5.2x10-16 kg⇒10 marksAverage kinetic energy at 27o C:KE = mv2/2 = (5.2x10-16 kg) (0.51x10-2 m/s)2/2= 6.9x10-21 kg m2/s2= 6.9 x10-21 J ⇒15 marks*.The average kinetic energy of an argon atom is the same as that of a water droplet.KE becomes zero at –273 o C.From the linear relationship in the figure, KE = aT (absolute temperature)where a is the increase in kinetic energy of an argon atom per degree.a = KE/T = 6.9x10-21 J/(27+273K) = 2.3x10-23 J/K⇒25 marksS: specific heat of argon N: number of atoms in 1g of argonS = 0.31 J/g K = a x NN = S/a = (0.31 J/g K) / (2.3x10-23 J/K)= 1.4x1022 ⇒30 marksAvogadro’s number (N A) : Number of argon atoms in 40 g of argonN A = (40)(1.4x1022)= 5.6 x1023⇒20 marks2-1. ⇒ 30 marksmass of a typical star = (4/3)(3.1)(7x108 m)3(1.4 g/10-6 m 3) = 2×1033 g mass of protons of a typical star = (2×1033 g)(3/4 + 1/8) = 1.8×1033 g number of protons of a typical star = (1.8×1033 g)(6×1023/g) = 1×1057number of stellar protons in the universe = (1×1057)(1023) = 1×1080Partial credits on principles:Volume = (4/3)(3.14)radius 3×density; 4 marks 1 mole = 6×1023; 4 marksTotal number of protons in the universe = number of protons in a star ×1023; 2 marks Mass fraction of protons from hydrogen = (3/4)(1/1); 5 marks Mass fraction of protons from helium = (1/4)(2/4); 10 marks2-2. ⇒ 30 marks∆E(2→3) = C(1/4 - 1/9) = 0.1389 C λ(2→3) = 656.3 nm ∆E(1→2) = C(1/1 - 1/4) = 0.75 Cλ(1→2) = (656.3)(0.1389/0.75) = 121.5 nmNo penalty for using Rydberg constant from memory. 15 marks penalty if answered in a different unit (Hz, etc.)2-3.T = (2.9×10-3 m K)/1.215×10-7 m = 2.4×104 K ⇒ 10 marks2-4..⇒ 20 marksλ = 3 × 108 m/1.42 × 109 = 0.21 mT = (2.9 × 10-3 m K)/0.21 m = 0.014 K2-5. ⇒ 10 marks14N + 4He → (17O ) + 1HO-17, O acceptable1783-1.k des = A exp(-E des/R T)= (1x1012 s-1)(5x10-32) = 5x10-20 s-1 at T = 20 K ⇒10 markssurface residence time, τresidence = 1 / k des = 2x1019 s = 6x1011 yr ⇒20 marks(full credit for τhalf-life = ln2 / k des = 1x1019 s = 4x1011 yr)residence time = 2x1019s3-2.The distance to be traveled by a molecule: x = πr = 300 nm.k mig = A exp(-E mig/R T)= (1x1012 s-1)(2x10-16 ) = 2x10-4 s-1 at T = 20 K ⇒ 5 marksaverage time between migratory jumps,τ = 1 / k mig = 5x103 sthe time needed to move 300 nm= (300 nm/0.3 nm) jumps x (5x103 s/jump) = 5x106 s = 50 days ⇒15 marks(Full credit for the calculation using a random-walk model. In this case:t = τ (x/d) 2 = 5 x 109 s = 160 yr. The answer is still (b).)(a) (b)(c) (d) (e)10 marks3-3.k(20 K) / k(300 K) = exp[(E/R) (1/T1 - 1/T2)]= e-112 = ~ 10-49 for the given reaction ).) ⇒15 marks The rate of formaldehyde production at 20 K= ~ 10-49 molecule/site/s = ~ 10-42 molecule/site/ yr⇒10 marks(The reaction will not occur at all during the age of the universe (1x1010 yr).)rate = 10-42molecules/site/yr3-4. circle one(a) (b) (c) (a, b) (a, c) (b,c)(a, b, c)(15 marks, all or nothing)4-1.H PNumber of atoms ( 11.3 ) 1⇒ 10 marksTheoretical wt % ( 3.43 )⇒ 10 marks4-2.adenineN NN NN H H guanineNN N NO N HH HNN O N H H cytosineNN H O O thymine(10 marks on each)4-3. 7 marks each, 20 marks for threeadenineNNNNNHHguanine NN NNON HHH NNH OOthymineNNONHH cytosine NNH OOthymineguanine NN NNON HHHcytosineNNONHHcytosineNNON HHNNHOO thyminethymineNNHOONNH OOthyminethymine NNHOONNONHH cytosineadenineNNNNNHH adenineNNNNNHHadenine NNNNNHHguanineguanine NNNNON HHHNNNNONHHH4-4. 2.5 marks for each bracketadenineN NN N HNH 2guanine N NH N N HO NH 2Uracil N H NH O cytosineN H N NH 2OOHCN ( 5 ) ( 5 ) ( 4 )( 4 )H 2O ( 0 ) ( 1 ) ( 2 ) ( 1 )5-1.(20 marks)1st ionization is complete: H2SO4→ H+ + HSO4-[H2SO4] = 02nd ionization: [H+][SO42-]/[HSO4-] = K2 = 1.2 x 10-2 (1)Mass balance: [H2SO4] + [HSO4-] + [SO42-] = 1.0 x 10-7 (2)Charge balance: [H+] = [HSO4-] + 2[SO42-] + [OH-] (3)Degree of ionization is increased upon dilution.[H2SO4] = 0Assume [H+]H2SO4 = 2 x 10-7From (1), [SO42-]/[HSO4-] = 6 x 104 (2nd ionization is **plete)[HSO4-] = 0From (2), [SO42-] = 1.0 x 10-7 [5 marks]From (3), [H+] = (2 x 10-7) + 10-14/[H+][H+] = 2.4 x 10-7(pH = 6.6) [8 marks][OH-] = 10-14/(2.4 x 10-7) = 4.1 x 10-8[2 marks]From (1), [HSO4-] = [H+][SO42-]/K2= (2.4 x 10-7)(1.0 x 10-7)/(1.2 x 10-2) = 2.0 x 10-12[5 marks]Check charge balance:2.4 x 10-7≈ (2.0 x 10-12) + 2(1.0 x 10-7) + (4.1 x 10-8)Check mass balance:0 + 2.0 x 10-12 + 1.0 x 10-7≈ 1.0 x 10-7Species Concentration** x 10-12HSO4-** x 10-7SO42-** x 10-7H+** x 10-8 OH-5-2. (20 marks)mmol H3PO4 = 0.85 ⨯ 3.48 mL ⨯ 1.69g/mL ⨯ 1 mol/98.00 g ⨯ 1000 = 51.0 [5 marks]The desired pH is above p K2.A 1:1 mixture of H2PO4- and HPO42- would have pH = p K2 = 7.20.If the pH is to be 7.40, there must be more HPO42- than H2PO4-.We need to add NaOH to convert H3PO4to H2PO4-and to convert to the right amount of H2PO4-to HPO42-.H3PO4 + OH-→ H2PO4- + H2OH2PO4- + OH-→ HPO42- + H2OThe volume of 0.80 NaOH needed to react with to to convert H3PO4 to H2PO4- is:51.0 mmol / 0.80M = 63.75 mL [5 marks]To get pH of 7.40 we need:H2PO4- + OH-→ HPO42-Initial mmol 51.0 x 0Final mmol 51.0-x 0 xpH = p K2 + log [HPO42-] / [H2PO4-]7.40 = 7.20 + log {x / (51.0-x)}; x = 31.27 mmol [5 marks]The volume of NaOH needed to convert 31.27 mmol is :31.27 mmol / 0.80 M = 39.09 mLThe total volume of NaOH = 63.75 + 39.09 =102.84 mL , 103 mL [5 marks]Total volume of 0.80 M NaOH (mL) 103 mL5-3. (20 marks)p K = 3.52pH = pK a + log ([A-]/[HA])[A-]/[HA] = 10(pH-pKa) [5 marks]In blood, pH =7.40, [A-]/[HA] = 10(7.40-3.52) = 7586Total ASA = 7586 +1 = 7587 [5 marks]In stomach, pH = 2.00, [A-]/[HA] = 10(2.00-3.52) = 3.02x10-2Total ASA = 1+ 3.02x10-2 = 1.03 [5 marks]Ratio of total aspirin in blood to that in stomach = 7587/1.03 = 7400 [5 marks]** ( 103Ratio of total aspirin in blood to that in stomach6-1. (5 marks)4 H2O + 4 e-→ 2 H2(g) + 4 OH- (or 2 H2O + 2 e-→ H2(g) + 2 OH-)6-2. (5 marks)2 H2O → O2 + 4 H+ + 4 e-(or H2O → 1/2 O2 + 2 H+ + 2 e- )6-3. (5 marks)Cu → Cu2+ + 2e-6-4. (20 marks)Reduction of sodium ion seldom takes place.It has a highly negative reduction potential of –2.710 V.Reduction potential for water to hydrogen is negative (water is very stable).But, it is not as negative as that for sodium ion. It is –0.830 V.Reduction of both copper ion and oxygen takes place readily and the reduction potentials for both are positive.In the present system, the reverse reaction (oxidation) takes place at the positive terminal. Copper is oxidized before water.Reduction potential for hydrogen ion is defined as 0.000 V.6-5. (15 marks)pOH = 14.00 – 4.84 = 9.16[OH-] = 6.92 x 10-10K sp = [Cu2+][OH-]2 = 0.100 x (6.92 x 10-10) = 4.79 x 10-206-6.E = E o Cu2+/Cu + (0.0592/2) log [Cu2+]= +0.340 + (0.0592/2) log [Cu2+]= +0.340 + (0.0592/2) log (K sp / [OH-]2)= +0.340 + (0.0592/2) log (K sp) - (0.0592/2) log [OH-]2= +0.340 + (0.0592/2) log (K sp) - 0.0592 log [OH-],3 marksBy definition, the standard potential for Cu(OH)2(s) + 2e-→ Cu(s) + 2OH- is the potential where [OH-] = 1.00.E = E o Cu(OH)2/Cu = +0.340 + (0.0592/2) log (K sp)= +0.340 + (0.0592/2) log (4.79 x 10-20)= +0.340 - 0.5722 marks= -0.232 V10 marks-------------------------------------------------------------------------------------------------------------- One may solve this problem as following.Eqn 1: Cu(OH)2(s) + 2e -→ Cu + 2OH-E+o = E o Cu(OH)2/Cu = ?Eqn 2: Cu(OH)2(s) → Cu2+ + 2OH-E o = (0.05916/n) logK sp= (0.05916/2) log(4.79×10-20)= -0.5715 V3 marksEqn 1 – Eqn 2 : Cu2+ + 2e-→ CuE-o = E+o - E o = E o Cu2+/Cu = 0.34 VTherefore, E+o = E-o + E o = + 0.34 + (-0.5715)2 marks= -0.232 V10 marks-0.232 V6-7.Below pH = 4.84, there is no effect of Cu(OH)2 because of no precipitation.Therefore,E = E Cu2+/Cu = +0.340 + (0.0592/2) log [Cu2+]= +0.340 + (0.0592/2) log 0.1003 marks= +0.340 – 0.0296 = +0.310 V7 marks** V6-8.** g graphite = 0.0833 mol carbon6 mol carbon to 1 mol lithium; 1 g graphite can hold 0.0139 mol lithiumTo insert 1 mol lithium, 96487 coulombs are needed.Therefore, 1 g graphite can charge 96487 × 0.0139 = 1340 coulombs. 5 marks1340 coulombs / g = 1340 A sec / g = 1340 x 1000 mA × (1 / 3600) h = 372 mA h / g 5 marks372 mA h / g7-1. (10 marks)n/V = P/RT = (80 x 106 / 1.013 x 105 atm)/[(0.082 atm L/mol/K)(298K)] = 32 mol/L5 marksdensity = mass/volume = d = 32 x 2 g/L = 64 kg/m 3 5 marks64 kg/m 37-2.** or 0.23H 2(g) + 1/2 O 2(g) → H 2O(l); ∆H rexn-1 = ∆H f [H 2O(l)] = -286 kJ/mol = -143 kJ/g 7 marksC(s) + O 2(g) → CO 2(g); ∆H rexn-2 = ∆H f [CO 2(g)] = -394 kJ/mol = -33 kJ/g 7 marks(-∆H rexn-1) / (-∆H rexn-2) = 4.3 or (-∆H rexn-2) / (-∆H rexn-1)= 0.236 marks7-3. (a) (-)1.2 x 105 kJ, (b) (-)6.9 x 104 kJ** x 108 sec or 3.3 x 104 hr or 1.4 x 103 days or 46 month or 3.8 yrI = 0.81 AH 2(g) + 1/2 O 2(g) → H 2O(l)∆H c = -286 kJ/mol = -143 kJ/g = -143 x 103 kJ/kg 5 marksΔG = ΔH – T ΔSΔS c= 70 – 131 – 205/2 = -163.5 J/K/mol5 marksΔG c = -286 kJ/mol + 298K x 163.5 J/K/mol = -237 kJ/mol = -1.2 x 105 kJ/kg 5 marks(a) electric motor W max = ΔG c ⨯ 1 kg = - 1.2 x 105 kJ 5 marks (b) heat engine W max = efficiency x ∆H c 5 marks= (1 – 298/573) x (-143 x 103 kJ) = -6.9 x 104 kJ 5 marks119 x 103 kJ = 1 W x t(sec)t = 1.2 x 108 sec = 3.3 x 104 hr = 1.4 x 103 days = 46 month = 3.8 yr 5 marksΔG = -nFE n = # of electrons involved in the reaction F = 96.5 kC/molH 2(g) + 1/2 O 2(g) → H 2O(l) n = 2 5 marksE = - ΔG/nF = 237 kJ/mol / 2 / 96.5 kC/mol = 1.23 V5 marksI = W/E = 0.81 A5 marks8-1-1. (5 marks on each)①C②C③CO8-1-2.③ Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) 5marks① C(s) + O2(g) → CO2(g) ΔH①◦ = -393.51 kJ = ΔH f◦(CO2(g))② CO2(g) + C(s) → 2CO(g) ΔH②◦ = 172.46 kJFrom ① and ②,ΔH f◦(CO(g)) = (1/2){172.46 + (-393.51)} = -110.525 kJΔH f◦(Fe2O3) = -824.2 kJΔH③◦ = 3ⅹΔH f◦(CO2(g)) - ΔH f◦(Fe2O3) - 3ⅹΔH f◦(CO(g))= 3ⅹ(-393.51) – (-824.2) - 3ⅹ(-110.525) = -24.8 kJ 7 marks ΔS③°=2ⅹ27.28+3ⅹ213.74-87.4-3ⅹ197.674=15.36 J/K 3 marks ΔG③°=ΔH°-TΔS°=-24.8kJ-15.36J/Kⅹ1kJ/1000Jⅹ1473.15K=-47.43 kJ5 marksK = e(-ΔG°/RT)= e(47430J/(8.314J/Kⅹ1473.15K)) = 48 5 marksBalanced equation of ③:K = 48Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)8-2-1. (20 marks)One AB2O4 unit has available 4 (= 1 + (1/4)ⅹ12) octahedral sites.48-2-2. (20 marks)Since one face-centered cube in AB2O4 represents one Fe3O4 unit in this case, it has 8 available tetrahedral sites. In one Fe3O4 unit, 1 tetrahedral site should be occupied by either one Fe2+ (normal-spinel) or one Fe3+ (inverse-spinel). Therefore, in both cases, the calculation gives (1/8) ⅹ100% = 12.5% occupancy in available tetrahedral sites.**%8-2-3. (10 marks for d-orbital splitting, 10 marks for elec. distribution)9-1-1. 1 answer for 8 marks, two for 15 marksH 3CN NNH 3CNNN :::+_+::_:9-1-2. ( 10 marks)H 3CN::9-1-3.H 3CNCH 2CH 2:H 3CN HH CCH 2:(10 marks) (10marks )9-2-1. 5 marks eachHONN +_::ONN:H+:HH_O NN:H+:H_::::::9-2-2.( 10 marks)CH 2CO ::9-3-1.(40 marks)CH 3H 3CH 3C+BC H 2CCH 3CH 3CO 2DEOOO_9-3-2.(10 marks)O OH O n+F10-1. 10 marks eachNMLCH 2OHCH 2OHMeOOMeH HH HOMeMeO CHOCHOCH 2OHCH 2OHHHH H OHOMeMeO OH10-2. 8 marks each for correct structuresNumber of possible structures24 marks12OH(OH)OH(H)HH HHOMeOMeOH COOMeOH(OH)OH(H)HH HHOMeOMeOHCOOMe34OH(OH)OH(H)OH(OH)OHe(H)10-3. 10 marks eachGICH 2OHCH 2OHHHHHMeOOMeOHOMeCH 2OHCH 2OHHHHOMeOMeOMe10-4. 10 marksNumber of the correct structure for C from 10-2110-5.BOH(OH)OH(H)HHHH OHCOOHOHOH10 marks eachDJOH(OH)OH(H)HHHHOMeOMeCOOMeOMeOH(OMe)OMe(H)HHHHOMeOMeOMeCOOMe10-6. 20 marksHOOCOHHH OOOHOOH COOHOOHOHOH COOH11-1. 10 marks311-2. 30 marksCOOHHOOCOOH11-3. 2.5 marks eacha, c, d11-4 30 marksOOCOCOOOHTransition State11-5.For the enzyme-catalyzed reaction, Arrehnius equation could be applied.k cat/k uncat = A exp (-E a, cat/ RT) / A exp (-E a, uncat / RT)= exp [-∆E a, cat-uncat/ RT]= exp [-∆E a, cat-uncat(J/mol) / (2,480 J/mol)] = 106Therefore, -∆E a, cat-uncat = 34,300 J/mol 15 marksk uncat, T/k uncat, 298 = exp (-∆H≠ uncat/ RT) / exp (-∆H≠uncat / 298R)= exp [(-∆H≠ uncat/R)(1/T-1/298)]ln(k uncat, T/k uncat, 298 )= 13.8 = [(-86900/8.32)(1/T-1/298)]Therefore, T = 491 K, or 218o C 15 marks-E a, cat-uncat = 34,300 J/molT = 491 K, or 218o C。
- 1、下载文档前请自行甄别文档内容的完整性,平台不提供额外的编辑、内容补充、找答案等附加服务。
- 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
- 3、如文档侵犯您的权益,请联系客服反馈,我们会尽快为您处理(人工客服工作时间:9:00-18:30)。
SYLLABUS OF THEINTERNATIONAL CHEMISTRY OLYMPIADPart I Theoretical partLevel 1:school chemistry programs and need not be mentioned in the preparatory problems.Level 2: These topics are included in a substantial number of secondary school programs and maybe used without exemplification in the preparatory problems. Level 3: These topics are not included in the majority of secondary school programs and can only be used in the competition if examples are given in the preparatory problems.1. The atom1.1.Introduction1.1.1. Counting of nucleons 11.1.2. Isotopes 1 1.2.The hydrogen atom1.2.1. Concept of energy levels 11.2.2. Shape of s-orbitals 11.2.3. Shape and orientation of p-orbitals 11.2.4. Shape and orientation of d-orbitals 31.2.5. Understanding the simplest Schrodinger equation 31.2.6. Square of the wave function and probability 31.2.7. Quantum numbers (n, l, m l) 3 1.3.Radioactivity1.3.1. Types of radioactivity 11.3.2. Radioactive decay 11.3.3. Nuclear reactions 22. Chemical bonding2.1.VSEPR – Simple molecular structures with2.1.1. no more than four electron pairs about central atom 12.1.2. with central atom exceeding the “octet rule” 3 2.2.Delocalization and resonance 3 2.3.Hybrid orbital theory 3 2.4.Molecular orbital theory2.4.1. molecular orbital diagram (H2 molecule) 32.4.2. molecular orbital diagram (N2 and O2 molecules) 32.4.3. bond orders in O2, O2–, O2+ 32.4.4. unpaired electrons and paramagnetism 33. Chemical calculations3.1.1. Balancing equations 13.1.2. Stoichiometric calculations 13.1.3. Mass and volume relations (including density) 13.1.4. Empirical formula 13.1.5. Avogadro’s number 13.1.6. Concentration calculations 14. Periodic trends4.1.Electron configuration4.1.1. Pauli exclusion principle 14.1.2. Hund’s Rule 14.1.3. Main group elements 14.1.4. Transition metal elements 14.1.5. Lanthanide and actinide metals 3 4.2.Electronegativity 1 4.3.Electron affinity 2 4.4.First ionization energy 1 4.5.Atomic size 1 4.6.Ion size 14.7.Highest oxidation number 15. Inorganic Chemistry5.1.Introduction5.1.1. Trends in physical properties of elements (Main groups)5.1.1.1. melting point 15.1.1.2. boiling point 15.1.1.3. metal character 15.1.1.4. magnetic properties 35.1.1.5. electrical conductivity 25.1.2. Oxidation number 15.1.3. Nomenclature5.1.3.1. main group compounds 15.1.3.2. transition metal compounds 15.1.3.3. simple metal complexes 3 5.2.Groups 1 and 25.2.1. Trend in reactivity of (heavy elements more reactive) 15.2.2. Products of reaction with5.2.2.1. water 15.2.2.2. halogens 15.2.2.3. oxygen 25.2.3. Basicity of oxides 15.2.4. Properties of hydrides 35.2.5. Other compounds, properties and oxidation states 3 5.3.Groups 13 – 18 and Hydrogen5.3.1. Binary molecular compounds of hydrogen5.3.1.1. Formulae 15.3.1.2. Acid-base properties of CH4, NH3, H2O, H2S 15.3.1.3. Other properties 35.3.2. P block elementsGroup 13 (Boron group)15.3.2.1 The oxidation state of boron and aluminium in theiroxides and chlorides is +325.3.2.2. The acid-base properties of aluminiumoxide/hydroxide5.3.2.3. Reaction of boron(III) oxide with water 35.3.2.4. Reaction of boron(III) chloride with water 35.3.2.5. Other compounds, properties and oxidation states 35.3.3. Group 14 (Carbon group)5.3.3.1. The oxidation state of Si in its chloride and oxide is1+425.3.3.2. The +2 and +4 oxidation states of carbon, tin andlead, the acid-base and redox properties of theoxides and chlorides5.3.3.3. Other compounds, properties and oxidation states 35.3.4. Group 15 (Nitrogen group)25.3.4.1. Phosphorus(+5) oxide and chloride, and theirreaction with water5.3.4.2. Phosphorus(+3) oxide and chloride, and their2reaction with water5.3.4.3. Oxides of nitrogena. Reaction of NO to form NO2 1b. Dimerization of NO2 1c. Reaction of NO2 with water 15.3.4.4. Redox properties ofa. HNO3 and nitrates 1b. HNO2 and NH2NH2 35.3.4.5. Bi(+5) and Bi(+3) 35.3.4.6. Other compounds, properties and oxidation states 35.3.5. Group 16 (Oxygen group)5.3.5.1. The +4 and +6 oxidation states of sulfur, reaction1of their oxides with water, properties of their acids5.3.5.2. Reaction of thiosulfate anion with I2 35.3.5.3. Other compounds, properties and oxidation states 35.3.6. Group 17 (Halogens)5.3.6.1. Reactivity and oxidant strength decreases from F21to I25.3.6.2. Acid-base properties of the hydrogen halides 115.3.6.3. The oxidation state of fluorine in its compounds is–15.3.6.4. The –1, +1, +3, +5, +7 oxidation states of chlorine 15.3.6.5. Mononuclear oxoanions of chlorine 25.3.6.6. Reactions of halogens with water 35.3.6.7. Reaction of Cl2O and Cl2O7 with water 35.3.6.8. Other compounds, properties and oxidation states 35.3.7. Group 18 (Rare gases) 3 5.4. Transition elements5.4.1. Common oxidation states of common transition metals:1 Cr(+2), Cr(+3) Mn(+2), Mn(+4), Mn(+7) Ag(+1)Fe(+2), Fe(+3) Co(+2) Zn(+2)Hg(+1), Hg(+2) Cu(+1), Cu(+2) Ni(+2)5.4.2. Colours of ions listed above in aqueous solution 25.4.3. Insolubility of Ag, Hg and Cu in HCl 25.4.4. M2+ arising by dissolution of the other metals in HCl 225.4.5. Cr(OH)3 and Zn(OH)2 are amphoteric and the other +2oxides/hydroxides of the metals listed above are basic5.4.6. MnO4– and Cr2O72– are strong oxidants in acid solution 15.4.7. pH dependence of products of MnO4– acting as oxidant 25.4.8. Interconversion between CrO42– and Cr2O72– 35.4.9. Other compounds, properties and oxidation states 3 nthanides and actinides 3 5.6.Coordination chemistry including stereochemistry5.6.1. Definition of coordination number 115.6.2. Writing equations for complexation reactions given allformulae5.6.3. Formulae of common complex ions5.6.3.1. Ag(NH3)2+ 15.6.3.2. Ag(S2O3)23– 35.6.3.3. FeSCN2+ 35.6.3.4. Cu(NH3)42+ 15.6.3.5. Other complex ions 35.6.4. (6.5) Ligand field theory (e g and t2g terms, high and low spin) 35.6.5. Stereochemistry5.6.5.1. (6.7) cis and trans 35.6.5.2. enantiomers 3 5.7.Selected industrial processes5.7.1. Preparation of H2SO4 15.7.2. Preparation of NH3 15.7.3. Preparation of Na2CO3 25.7.4. Preparation of Cl2 and NaOH 25.7.5. Preparation of HNO3 26. Physical chemistry6.1.Gases6.1.1. Ideal gas law 16.1.2. van der Waal’s gas law 36.1.3. definition of partial pressure 26.1.4. Dalton’s Law 3 6.2. Thermodynamics6.2.1. First Law6.2.1.1. Concept of system and surroundings 26.2.1.2. Energy, heat and work 26.2.2. Enthalpy6.2.2.1. Relationship between internal energy and enthalpy 36.2.2.2. Definition of heat capacity 26.2.2.3. Difference between C p and C v (ideal gas only) 36.2.2.4. Enthalpy is a state propert y (Hess’s Law) 26.2.2.5. Born-Haber cycle for ionic compounds 36.2.2.6. Use of standard formation enthalpies 26.2.2.7. Enthalpies of solution and solvation 36.2.2.8. Bond enthalpies (definition and use) 26.2.3. Second Law (Entropy and Free Energy)6.2.3.1. Entropy definition (d q / T) 36.2.3.2. Entropy and disorder 36.2.3.3. Entropy definition (S = k ln W) 36.2.3.4. Gibbs energy definition (∆G = ∆H –T∆S) 36.2.3.5. Using ∆G to predict direction of natural change 36.2.3.6. Relationship between ∆G︒ and equilibrium constant K 3 6.3.Equilibrium6.3.1. Acid-base6.3.1.1. Arrhenius definitions of acids and bases 16.3.1.2. Bronsted-Lowry definitions 16.3.1.3. Conjugate acids and bases 16.3.1.4. pH definition 16.3.1.5. K w definition 16.3.1.6. K a and K b as a measure of acid and base strength 16.3.1.7. Acidity or basicity of ions 16.3.1.8. Calculation of pH from p K a(weak acid) 16.3.1.9. Calculation of pH of a simple buffer solution 26.3.2. Gas phase6.3.2.1. Equilibrium constant in partial pressures 36.3.2.2. Relating K p and K c 36.3.3. Solubility6.3.3.1. Solubility constant (product) definition (K sp) 26.3.3.2. Calculation of solubility in water from K sp 26.3.4. Compleximetric6.3.4.1. Complex formation constant (definition) 36.3.4.2. Problems involving compleximetric equilibria 36.3.4.3. Lewis acids and bases 36.3.4.4. Hard and soft Lewis acids and bases 36.3.5. Phase6.3.5.1. Temperature dependence of vapour pressure 36.3.5.2. Clausius-Clapeyron equation 36.3.5.3. Single component phase diagramsa. triple point 3b. critical point 36.3.5.4. liquid-vapour systema. ideal and nonideal systems 3b. diagram 3c. use in fractional distillation 36.3.5.5. Henry’s Law 36.3.5.6. Raoult’s Law 36.3.5.7. Deviation from Raoult’s Law 36.3.5.8. Boiling point elevation 36.3.5.9. Freezing point depression 36.3.5.10. Osmotic pressure 36.3.5.11. Partition coefficient 36.3.5.12. Solvent extraction 36.3.6. Multiple6.3.6.1. Calculation of pH for multiprotic acids 36.3.6.2. Calculation of pH for weak acid mixtures 3 6.4.Electrochemistry6.4.1. Electromotive force (definition) 16.4.2. First kind electrodes 16.4.3. Standard electrode potential 16.4.4. Nernst equation 36.4.5. Second kind electrodes 36.4.6. Relationship between G and electromotive force 37. Chemical kinetics (Homogeneous reactions)7.1.Introduction7.1.1. Factors affecting reaction rate 17.1.2. Reaction coordinates and the basic idea of a transition state 1 7.2.Rate law7.2.1. Differential rate law 27.2.2. Concept of reaction order 27.2.3. Rate constant definition 27.2.4. First order reactions7.2.4.1. Dependence of concentration on time 37.2.4.2. Concept of half life 37.2.4.3. Relationship between half life and rate constant 37.2.4.4. Calculation of first order rate constant froma. differential rate law 3b. integrated rate law 37.2.4.5. Rate constant for second and third order reactions 3 7.3.Reaction mechanisms7.3.1. Concept of molecularity 37.3.2. Rate-determining step 37.3.3. Basic concepts of collision theory 37.3.4. Opposing parallel and consecutive reactions 37.3.5. Arrhenius’s law 37.3.5.1. Definition of activation energy 37.3.5.2. Calculation of activation energy 38. Spectroscopy8.1.UV/visible8.1.1. Identification of aromatic compound 38.1.2. Identification of chromophore 38.1.3. Dyes: colour vs structure 38.1.4. Beer’s Law 3 8.2.Infrared8.2.1. Interpretation using a table of frequencies 38.2.2. Recognition of hydrogen bonds 3 8.3.x-Ray8.3.1. Bragg’s Law 38.3.2. Concept of8.3.2.1. coordination number 38.3.2.2. unit cell 38.3.3. Solid structures8.3.3.1. NaCl 38.3.3.2. CsCl 38.3.3.3. metals 3 8.4.NMR8.4.1. General Concepts8.4.1.1. chemical shift 38.4.1.2. spin-spin coupling and coupling constants 38.4.1.3. integration 38.4.2. Interpretation of a simple 1H spectrum (like ethanol) 38.4.3. Identification of o- and p-disubstituted benzene 338.4.4. Interpretation of simple spectra of 13C (proton decoupled) andother 1/2 spin nuclei8.5.Mass spectrometry8.5.1.1. Recognition of molecular ion 38.5.1.2. Recognition of fragments with the help of a table 38.5.1.3. Recognition of typical isotope distribution 39. Organic Chemistry9.1.Introduction9.1.1. (3.1.1) Alkane naming (IUPAC) 19.1.2. Trends in boiling points of9.1.2.1. (3.1.3) alkanes with structure 19.1.2.2. (3.7.1) alcohols vs ethers due to hydrogen-bonding 119.1.3. (3.3.1, 3.4.1) Geometry at singly, doubly, and triply bondedcarbon9.1.4. Identification of common functional groups 19.1.5. Isomerism of alkenes9.1.5.1. cis-trans 19.1.5.2. E/Z 39.1.6. Enantiomers9.1.6.1. Optical activity 29.1.6.2. R/S nomenclature 3 9.2.Reactivity9.2.1. Alkanes9.2.1.1. reaction with halogensa. products 1b. free radical mechanism (initiation, termination) 29.2.1.2. Cycloalkanesa. names 2b. Strain in small rings 3c. chair/boat conformations of cyclohexane 39.2.2. Alkenes9.2.2.1. Products from Br2, HBr and H2O/H+ 19.2.2.2. Markownikoff’s rule 29.2.2.3. Mechanism involving carbocation intermediates 39.2.2.4. Relative stability of carbocations 39.2.2.5. 1,4 addition to dienes 3 9.2.3. Alkynes9.2.3.1. Acidity relative to alkenes 39.2.3.2. Differences in chemical properties from alkenes 2 9.2.4. Benzene9.2.4.1. formula 19.2.4.2. stabilization by resonance 19.2.4.3. electrophilic substitution (nitration, halogenation)a. directing effect of first substituent 3b. effect of first substituent on reactivity 3c. explanation of substituent effects 39.2.5. Halogen compounds9.2.5.1. Nomenclature of monofunctional 19.2.5.2. Substitution reactionsa. giving alcohols 3b. in which halogen is exchanged 3c. reactivityi. primary vs secondary vs tertiary 3ii. aliphatic vs aromatic 3d. S N1 and S N2 mechanisms 39.2.5.3. Elimination reactions 29.2.5.4. Competition of elimination and substitution 2 9.2.6. Alcohols9.2.6.1. Nomenclature of monofunctional 19.2.6.2. Comparison of acidity of alcohols and phenols 29.2.6.3. Dehydration to alkenes 19.2.6.4. Esters with inorganic acid 29.2.6.5. Oxidation reactions 1 9.2.7. Aldehydes and ketones9.2.7.1. Nomenclature of monofunctional 19.2.7.2. Oxidation of aldehydes 19.2.7.3. Reduction to alcohols (LiAlH4, NaBH4) 39.2.7.4. Keto/enol tautomerism 39.2.7.5. Nucleophilic addition reactions witha. HCN 3b. RNH2 (R = alkyl, HO, NH2) 3c. enolate anions (aldol condensation) 3d. alcohols to form acetals/ketals 3e. Grignard reagents 39.2.8. Carboxylic acids and their derivatives29.2.8.1. Nomenclature of carboxylic acids and theirderivatives (esters, acid halides, amides)9.2.8.2. Acidity strength related to inductive effects 39.2.8.3. Preparation of carboxylic acids by hydrolysis ofa. esters (including soaps) 1b. amides 2c. nitriles 39.2.8.4. Reaction of carboxylic acidsa. with alcohols to form esters 1b. to form acid chlorides 3c. to form anhydrides 39.2.8.5. Reaction of acid chlorides to form amides 39.2.8.6. Mechanism of esterification 39.2.8.7. Multifunctional acids (hydroxyacids, ketoacids) 39.2.8.8. Polycarboxylic acids 39.2.9. Amines9.2.9.1. Nomenclaturea. simple amines 1b. recognition of primary, secondary, tertiary 19.2.9.2. Basicitya. As a property of an amine 1b. Comparison of basicity of aliphatic and aromatic 3c. Comparison of basicity of amines and amides 3d. Preparation of aminesi. from halides 3ii. from aromatic nitro compounds 3iii. from amides (by hydrolysis) 39.2.9.3. Diazotizationa. of aliphatic amines 3b. of aromatic amines 310. PolymersSynthetic10.1.10.1.1. Addition polymers10.1.1.1. polystyrene 210.1.1.2. polyethene 110.1.1.3. chain mechanism of formation 210.1.2. Condensation polymers10.1.2.1. polyesters 210.1.2.2. polyamides 210.1.3. Silicones 310.1.4. Concept of cross-linking and its affect on properties 3 10.2.Natural10.2.1. Silicates 310.2.2. Rubber 311. Biochemistry11.1.Carbohydrates11.1.1. Glucose and fructose11.1.1.1. chain formulae 111.1.1.2. Fischer projections 211.1.1.3. Haworth formulae 311.1.2. Difference between starch and cellulose 211.1.3. Difference between α- and β- D glucose 2 11.2.Fats11.2.1. Structure of fats in relationship to properties 211.2.2. Formula of glycerol 1 11.3.Nitrogen-containing Compounds of Biological Importance11.3.1. Amino acids11.3.1.1. Ionic structure 111.3.1.2. Isoelectric point 311.3.1.3. 20 amino acids (classification with structures2provided)11.3.1.4. Separation by electrophoresis 311.3.1.5. The peptide linkage 111.3.2. Proteins11.3.2.1. Primary structure 111.3.2.2. –S-S- bridges 311.3.2.3. Sequence analysis 311.3.2.4. Secondary structure 311.3.2.5. Details of α-helix structure 311.3.2.6. Tertiary structure 3211.3.2.7. Denaturation (change in pH, temperature,metals, ethanol)11.3.3. Nuclei Acids and Protein Synthesis11.3.3.1. Pyrimidine and purine 311.3.3.2. Nucleosides and nucleotides 311.3.3.3. Formulae of pyrimidine and purine bases 311.3.3.4. Difference between ribose and 2-deoxyribose 311.3.3.5. Base combination CG and AT (hydrogen-bonding) 311.3.3.6. Difference between DNA and RNA 311.3.3.7. Difference between mRNA and tRNA 3 11.4.Enzymes11.4.1.1. General properties, active centers 311.4.1.2. Nomenclature, kinetics, coenzymes, function of ATP 312. Analytical chemistry12.1.Titrations12.1.1. acid-base12.1.1.1. Titration curve; pH (strong and weak acid) 212.1.1.2. Choice of indicators for acidimetry 212.1.2. Redox titration 3 12.2.Qualitative analysis12.2.1. Ions (Inorganic)12.2.1.1. Identification of Ag+, Ba2+, Cl–, SO42– 212.2.1.2. Identification of other anions and cations 312.2.2. Organic functional groups12.2.2.1. Lucas reagent (1-, 2-, 3-alcohols) 312.2.2.2. Iodoform reaction 3312.2.2.3. Identification of primary, secondary, tertiary,quarternary amines in the laboratory12.3.Chromatographic methods of separation 3Part I Experimental partLevel 1: is assigned to the basic experimental activities which are supposed to be mastered by competitors very wellLevel 2: is assigned to the activities which are parts of school experimental exercises in developed countries and the authors of IChO tasks mayincorporate them into the tasks without being bounded to mention it inadvanceLevel 3: is assigned to such activities which are not in the chemistry syllabus in the majority of participating countries and the authors are obliged to mentionthem in the set of preparatory tasksIf the organizer wants to apply a technique which is not mentioned in the above syllabus, this technique is set to level 3 automatically.1. Synthesis of inorganic and organic compounds1.1. Heating with burners and hotplates 1 1.2. Heating of liquids 1 1.3. Handling the work with inflammable substances and materials 1 1.4. Measuring of masses (analytical balance) 11 1.5. Measuring of volumes of liquids (measuring cylinder, pipette,burette)1.6. Preparation of solutions from a solid compound and solvent 1 1.7. Mixing and dilution of solutions 11.8. Mixing and stirring of liquids 1 1.9. Using mixer and magnetic stirrer 2 1.10. Using a dropping funnel 1 1.11. Syntheses in flat bottom vessels – general principles 1 1.12. Syntheses in round bottom vessels – general principles 1 1.13 Syntheses in a closed apparatus – general principles 1 1.14. Using microscale equipment for synthesis 3 1.15. Apparatus for heating of reaction mixture under reflux 2 1.16. Apparatus for distillation of liquids at normal pressure 2 1.17. Apparatus for distillation of liquids at reduced pressure 2 1.18. Apparatus for steam distillation 3 1.19. Filtration through flat paper filter 1 1.20. Filtration through a folded paper filter 1 1.21. Handling a water vacuum pump 1 1.22. Filtration through a Büchner funnel 1 1.23. Suction through a glass filter 1 1.24. Washing of precipitates by decantation 1 1.25. Washing of precipitates on a filter 2 1.26. Drying of precipitates on a filter with appropriate solvents 2 1.27. Recrystallization of substances from aqueous solution 1 1.28. Recrystallization of substances from a known organic solvent 23 1.29. Practical choice of an appropriate solvent for recrystallization of asubstance1.30. Drying of substances in a drying box 2 1.31. Drying of substances in a desiccator 2 1.32. Connecting and using of a gas washing bottle 21.33. Extraction with an inmiscible solvent 12. Identification of inorganic and organic compounds:general principles2.1. Test-tube reactions 1 2.2. Technique of reactions performed in a dot dish and on a filter paper 12 2.3. Group reactions of some cations and anions specified by theorganizer2.4. Selective reactions of some cations and anions specified by the2 organizer2.5. Specific reactions of some cations and anions specified by the organizer 32 2.6. Identification of elements by flame coloration (using a platinumwire/MgO rod, Co-glass)2.7. Using a hand spectroscope/Bunsen spectroscope 3 2.8. Melting point determination with Kofler or similar type of apparatus 32.9. Qualitative evidence of basic functional groups of organic2 substances specified by the organizer3 2.10. Exploitation of some specific reactions for identification of organiccompounds (specified by the organizer)3. Determination of some inorganic and organic compounds:general principles3.1. Quantitative determinations using precipitation reactions 2 3.2. Igniting of a precipitate in a crucible 1 3.3. Quantitative volumetric determinations 1 3.4. Rules at titrating 1 3.5. Use of a pipetting ball 1 3.6. Preparation of a standard solution 2 3.7. Alkalimetric and acidimetric determinations 22 3.8. Color transitions of indicators at alkalimetric and acidimetricdeterminations3.9. Direct and indirect determinations (back titration) 3 3.10. Manganometric determinations 3 3.11. Iodometric determinations 3 3.12. Other types of determinations on basis of redox reactions 3 3.13. Complexometric determinations 3 3.14. Color transitions of solutions at complexometric determinations 3 3.15. Volumetric determinations on basis of precipitation reactions 33.16. Thermometric titration 34. Special measurements and procedures4.1. Measuring with a pH-meter 2 4.2. Chromatography on thin layers 3 4.3. Column chromatography 3 4.4. Separation on ion exchanger 3 4.5. Measuring of UV-VIS absorbances with a spectral photometer 34.6. Performing of conductivity measurements 35. Evaluation of results5.1. Estimation of experimental errors (significant figures, plots scales) 1。