澳大利亚全国化学竞赛试题(1998-2010)-FQEB003
1998年全国高中学生化学竞赛(省级赛区)试题.doc
1998年全国化学竞赛(预赛)模拟试卷(二)时间:3小时满分:100分一.(24分)1.(6分)磷的氯化物有PCl3和PCl5,氮的氯化物只有NCl3,为什么没有NCl5?。
白磷在过量氯气(其分子有三种不同的相对分子质量)中燃烧时,其产物共有种不同分子。
2.(6分)如果CH4分子中的氢原子能被F、Cl、Br、I四种卤原子取代,那么所得卤代烃有种,其中具有光学活性的有种。
3.(4分)某绝热容器中有饱和石灰水溶液,再往溶液中投入少量CaO粉末,下列判断中正确的是A 一定有晶体析出B Ca(OH)2浓度增大C pH不变D [H+][OH-]的积不变E [Ca2+][OH-]2的积不变F [H+]一定增大4.(4分)按各组指定内容,前者一定大于后者的是A 熔点:金刚石与石墨碘与水银B 沸点:甲醇与甲醛乙醇与乙硫醇C 密度:甲醛与乙烷(标况)铜与铁D 常温常压在水中的溶解度:O3与O2丙酮与甲醚E 足量镁粉与等体积等浓度的醋酸和盐酸反应后两溶液的pH(不考虑反应后溶液中[Mg2+]的差异)5.(4分)以下各组指定微粒构成正八面体顶点的是A 乙烷分子中的氢原子B XeF6分子中的F原子C NaCl晶体中与一个Na+最近的Cl-D NaCl晶体中与一个Na+最近的Na+E CsCl晶体中与一个Cs+最近的Cl-F CsCl晶体中与一个Cs+最近的Cs+G P4在少量O2中燃烧产物分子中的O原子H 高碘酸根离子中的O原子二.(22分)1.(3分)炸药TNT发生化学反应的方程式:2.(3分)完成反应方程式:在100%H2SO4中,1mol H3BO3和6mol H2SO4反应生成6mol(离子):3.(6分)配平Cl2+AgF+H2O-AgClO3+AgCl+HF+O2时有无穷多组系数,如5,10,5-1,9,10,1或7,14,7-1,13,14,2;……。
若H2O前面的系数为11,写出配平后的系数若a,b,c……为任意自然数,试用字母配第 1 页共6 页平方程式,并能反映出所有情况:4.(4分)最近几年,人们从各方面得到证据,表明苯的结构仍然是凯库勒式,最重要的证据之一是,邻二甲苯的臭氧化产物,只能用苯环是凯库勒式来解释,写出邻二甲苯与臭氧反应配平的方程式:5.(6分)将蓝矾放在一个可以称Array出质量的容器内加热,固体质量随温度变化的关系如右图所示,纵坐标是固体的相对质量,写出固体质量发生突变时的化学反应方程式:(相对原子质量:H. 1.0 O. 16.0 S. 32.0 Cu. 63.5)三.(16分)保险粉是Na2S2O3的工业名称,是最大宗的无机盐之一,年产量30万吨,大量用于印染业,并用来漂白纸张、纸浆和陶土。
1998年全国高中学生化学竞赛_决赛_理论试题
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澳大利亚全国化学竞赛试题(1998-2010)-FQEB002
AUSTRALIAN CHEMISTRY OLYMPIADFINAL PAPERPART B1989Instruction to candidates(1)You are allowed 10 minutes to read this paper, and 3 hours to complete the questions.(2)You are not permitted to refer to books, notes or periodic tables but you may use an electroniccalculator and molecular models.(3)You must attempt all questions, so be sure to allocate your time appropriately.(4)Answers must be written in the blank space provided immediately below each question in theexam booklet. Rough working must be on the back of pages.(5)Answers must provide clearly laid out working and sufficient explanation to show how youreached your conclusions.(6)Your name must be written in the appropriate place on each page of your answers.(7)Use only black or blue ball point pen for your written answers, pencil or other coloured pensare not acceptable.Question 1A white crystalline ionic solid A, because of its thermal instability, is used extensively as adirect explosive in mining and quarrying operations. At high temperatures (>300 °C) itcan explode violently giving off two colourless and odourless gases B and C and H2O.Under controlled thermolysis at about 250 °C, a gas D is liberated with H2O also as aby-product. D is a moderately unreactive gas comprised of linear unsymmetricalmolecules, as expected for a 16-electron triatomic species. Like A, D is alsothermodynamically unstable and decomposes exothermically upon heating to give Band C. With atomic oxygen, D gives E which exists in equilibrium with F, which isparamagnetic. At high pressures and moderate temperatures F rapidlydisproportionates to D and a brown gas G. In the presence of C, F is also quicklyconverted to G, which exists in equilibrium with colourless H. In a medium of highdielectric constant such as nitromethane, H undergoes a self-ionization equilibriumwith I and J. I is isoelectronic with CO and J is the anionic constituent of A. H can beoxidised by ozone to give a highly deliquescent light sensitive solid K. X-ray diffractionstudies show that solid K consists of an ionic array of linear L and planar J. In the gasand solution state, K is probably molecular, L is isoelectronic with CO2.Identify all the lettered compounds and write down all the equations described.Question 2After attending a Chemistry Olympiad Summer School, an enterprising and environmentally aware student from a mining town decided that it should be possibleto measure the concentration of cadmium in the city's river water if the followingexperimental apparatus was used.River water0.100M NaClThe student intended to fill the right-hand beaker with 0.100M NaCl solution and the left-hand beaker with river water and then to measure the cell potential with a very highresistance voltmeter. Knowing the cell potential, she figured that it ought to be a simplematter to calculate [Cd2+] in the river-water using Nernst's equation.Conveniently, the temperature of the cell components and river water was 25°C. Her trusty 2nd edition of Zumdahl told her that useful standard reduction potentials at 25°C were:Cd2++2e- Cdε0 = -0.40 VAg++e- Agε0 = +0.80 Vand that K sp(AgCl) = 1.6 x 10-10M2.-1;R = 8.314 JK-1mol-1.Data: F = 96,485 Cmol(a)Write an expression which allows you to calculate [Cd2+] from the potential of theelectrochemical cell which is set up in the apparatus.(b)If the measured cell potential was +0.90 V, what was the concentration of cadmium in theriver water?(c)The student was so enthused by the success of this experiment (she might be able to sellthe idea??) that she wondered whether the same setup could be used to measure thecadmium concentration in seawater.In fact she wondered if the same cell potential would be obtained when the left-hand beaker contained seawater with the same total concentration of cadmium as the riverwater?Being an experimentalist rather than a theorist, she decided to test her suspicions by adding just enough NaCl to the river water in the cell to make the chlorideconcentration identical to seawater, 0.55M.What cell potential did she measure?Data: In seawater only 0.95% of cadmium is uncomplexed and the major complexes are the chloro complexes. You can ignore the concentrations of hydroxy and carbonatocomplexes.2(d) The student noted that the cell potential (for the particular sample of river water in part (b))changed from 0.90V in the morning when the ambient temperature was 25°C, to 0.95Vin the afternoon when the temperature had risen to 35°C.Is the equilibrium2 Ag+ + Cd Cd2+ + 2 Agexothermic or endothermic? Explain your answer.You can assume that the solubility of silver chloride does not change from 25°C to 35°C.(e) Explain why it is better to have 0.1M NaCl and some solid AgCl in the right-hand beaker,rather than 0.1M AgNO3. Why is the AgCl necessary?Question 3The industrial conversion of an organic compound X to its isomer Y is carried out at constant temperature and pressure by passing X at 5 cm.s-1 down a reactor tubepacked with catalyst. The gas flows smoothly through the tube without any back mixingand its composition at any distance along the tube is uniform across the tube. Thetable below gives the gas analyses at various distances along the pipe.Distance (cm)% Y00.025 5.0509.510018.217028.929044.140055.2a)Assuming that the rate of the backward reaction is much less than the forward reactionshow the reaction order in X.b)For the process to operate economically, 80% of X should be converted to Y. How longshould the reactor tube be made;i)using the data given above;ii) if the flow rate is halved;iii)if a new catalyst is discovered which exactly doubles the rate constant ?c)Steam pipes are available for heating 3 metres of the reactor tube. Which 3 metres shouldbe so heated ? Explain you answer briefly.Question 4Simple chemical tests were once the backbone of structuredetermination in organic chemistry. For instance the reaction ofa molecule with "sodium hypoiodite" (NaOH/I2 in water) toyield a pale yellow precipitate of iodoform (CHI3) was a test fora methyl ketone or a group easily oxidised to a methyl ketone.The following problem is an example of a classical structuredetermination based on simple chemistry.The toxic principal in buttercups is an unstable oil called protoanemonin, C5H4O2, which is readily hydrolysed by acid or base to A, C5H6O3. Unlike protoanemonin, A dissolveswith effervescence in sodium bicarbonate solution and can be converted to a 2,4-dinitrophenylhydrazone. On treatment with hydrogen in the presence of platinum,protoanemonin gives B, C5H8O2. B does not dissolve in sodium bicarbonate solutionbut does so in sodium hydroxide solution; acidification of this solution gives an acid C,C5H10O3, which on standing reverts to B. When C is treated with sodium hydroxideand iodine, it gives iodoform (triiodomethane) and D, C4H6O4. D evolves a gas fromsodium bicarbonate solution. D also reacts with aqueous sodium hydroxide, 147.5mg3of D being required to neutralise 10mL of 0.125M barium hydroxide. D can be obtainedby treating 1,3-cyclohexadiene with KMnO4 under vigorous conditions.Explain the above reactions and deduce a probable structure for protoanemonin.Data: Relative atomic masses: C = 12.01; O = 16.00; H = 1.008;Ba = 137.3.4。
1988年-2013年全国高中学生化学竞赛(省级赛区)试题答案及解析
1988 年全国高中学生化学竞赛(初赛)试题参考答案
一、(1)3Na2CO3 + Al2(SO4)3 + 3H2O=Al(OH)3↓+ 3Na2SO4 + 3CO2↑;现象:白色沉淀和气泡逸出。 Na2CO3 + 2HAc=2NaAc + CO2↑+ H2O;现象:气体逸出。Na2CO3 + C17H33COOH=C17H33COONa + NaHCO3;现象:乳状液。产物 NaHCO3 写成 Na2CO3,Na2CO3 和 C17H33COOH 产物为 CO2,Na2CO3 和 Al2(SO4)3产物写成 NaHCO3 均给分 (2)CO2:Na2CO3 + CO2 + H2O=2NaHCO3; Al(OH)3:Al(OH)3 + 3HAc=Al(Ac)3 + 3H2O; C17H33COONa:C17H33COONa + HAc=C17H33COOH + NaAc 二、卡车在起动、刹车时排出黑色废烟者,是以柴油为燃料的。柴油分子中碳原子数目多于汽油分子
,所以沸点高。7 无氢键,沸点低于 6。
十、1.α-氰基,α-丁烯酸乙酯;2.
1990 年全国高中学生化学竞赛(初赛)试题参考答案
一、(1)Cu、Zn 或其他;(2)C、P 或其他;(3)Al、Fe 或其他;(4)Au、Pt 或其他 二、H2SO3+1/2O2=H2SO4;H2SO4 酸性比 H2SO3 强
五、A.C5H10O4 B.C4H8O4 C.C4H6O4 略
六、(1)
Si-O 交替成六、八元坏都给 2 分;(2)
写出
交联结构给 3 分。 七、(1)6FeC2O3 + 3H2O2 + 6K2C2O4=2Fe(OH)3 + 4K3Fe(C2O4)3(由于溶液显碱性所以有部分
澳大利亚全国化学竞赛试题(1998-2010)-FSE7B
Name:
FINAL SELECTION EXAMINATION for the
2008 AUSTRALIAN CHEMISTRY OLYMPIAD TEAM
PAPER B
2007
Please note that this answer book will be photocopied when returned and then
O N2
OSO OR
diazonanaphthaquinone derivative
carbene intermediate
+ N2
CO2H
rearranged intermediate H2O
OSO OR
(i) Draw three Lewis structures of diazoacetaldehyde (see structure below) the simplest compound containing the same active functional group as diazonaphthaquinone. Indicate the formal charge on each non-hydrogen atom.
1. You are allowed 10 minutes to read this paper, and 3 hours to complete the questions. 2. You are not permitted to refer to books or notes but you may use a non programmable electronic
The story continues: The solubility of photoresistors of Novolak polymers is affected by acid. The acid component can be produced photochemically from diazonaphthaquinone. “Novolak”, shown below, is one example of a photoresistor important in the microelectronic revolution.
澳大利亚全国化学竞赛试题(1998-2010)-FQE6B001
reactive, simple molecule plays a key role as a neurotransmitter in a wide range of
biochemical systems. As with all biologically active chemical species a number of important
compounds) with the aid of a radical-generating initiator. The reaction mechanism is:
I2 → 2I• I• + M →
R•1
Initiation
R1• + M →
R
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.
.
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the second initiation step is instantaneous and consumes negligible monomer)
(b)
Derive an expression for the average length of the polymer chains formed (explain your
(5) Answers must be written in the blank space provided immediately below each question in the exam booklet. Rough working must be on the backs of pages. Only material presented in the answer boxes will be assessed.
1998年全国高中学生化学竞赛(初赛)试题.doc
1998年全国高中学生化学竞赛(初赛)试题1.某芳香烃A,分子式C9H12。
在光照下用Br2溴化A得到两种一溴衍生物(B1和B2),产率约为1 : 1。
在铁催化下用Br2溴化A也得到两种一溴衍生物(C1和C2);C1和C2在铁催化下继续溴化则总共得到4种二溴衍生物(D1、D2、D3、D4)。
(1)写出A的结构简式。
(5分)(2)写出B1、B2、C1、C2、D1、D2、D3、D4的结构简式。
(8分)2.100.0g无水氢氧化钾溶于100.0g水。
在T温度下电解该溶液,电流强度I = 6.00安培,电解时间10.00小时。
电解结束温度重新调至T,分离析出的KOH·2H2O 固体后,测得剩余溶液的总质量为164.8g。
已知不同温度下每100g溶液中无水氢氧化钾的质量为:求温度T注:法拉第常数F = 9.65X104 C/mol,相对原子质量:K 39.1 O16.0 H 1.01 (15分)3.迄今已合成的最重元素是112号,它是用Zn 7030高能原子轰击Pb 20882的靶子,使锌核与铅核熔合而得。
科学家通过该放射性元素的一系列衰变的产物确定了它的存在,总共只检出一个原子。
该原子每次衰变都放出一个高能 粒子,最后得到比较稳定的第100号元素镄的含153个中子的同位素。
(7分) (1)112号是第几周期第几族元素?(2)它是金属还是非金属?(3)你认为它的最高氧化态至少可以达到多少?(4)写出合成112元素的反应式(注反应式中的核素要用诸如31H 、Pb 20882等带上下标的符号来表示,112号元素符号未定,可用M 表示)。
4.100.00mL SO 32–和S 2O 32–的溶液与80.00mL 浓度为0.0500mol/L 的K 2CrO 4的碱性溶液恰好反应, 反应只有一种含硫产物和一种含铬产物; 反应产物混合物经盐酸酸化后与过量的BaCl 2溶液反应, 得到白色沉淀, 沉淀经过滤、洗涤、干燥后称量,质量为0.9336g 。
1998年全国高中学生化学竞赛决赛实验试题
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澳大利亚全国化学竞赛试题(1998-2010)-FQEB005
2
Question 1
(a)
The following gas phase reactions are known to occur at varying temperatures.
n-C4H10 → (CH3)2C = CH2 + H2
C2H4 + Cl2 → CH2ClCH2Cl CH4 + 12O2 → CH3OH
respective initial partial pressures of NH3 were found.
Half life (minutes)
Initial partial pressure of NH3 (atmospheres)
7.6 0.349
3.7 0.171
1.7 0.076
(i) Determine the order of the reaction (ii) Determine the rate constant of decomposition.
(5) Answers must provide clearly laid out working and sufficient explanation to show how you reached your conclusin the appropriate place on each page of your answers.
No
(259)
71
Lu 175.0
103
Lr
(260)
Atomic mass values given here are to four significant figures. A value given in parentheses denotes the mass of the longest-lived isotope.
1998年第十二届全国高中学生化学竞赛(初赛)试题及答案
1998年全国高中学生化学竞赛(初赛)试题1.某芳香烃A,分子式C9H12。
在光照下用Br2溴化A得到两种一溴衍生物(B1和B2),产率约为1 : 1。
在铁催化下用Br2溴化A也得到两种一溴衍生物(C1和C2);C1和C2在铁催化下继续溴化则总共得到4种二溴衍生物(D1、D2、D3、D4)。
(1)写出A的结构简式。
(5分)(2)写出B1、B2、C1、C2、D1、D2、D3、D4的结构简式。
(8分)2.100.0g无水氢氧化钾溶于100.0g水。
在T温度下电解该溶液,电流强度I = 6.00安培,电解时间10.00小时。
电解结束温度重新调至T,分离析出的KOH·2H2O 固体后,测得剩余溶液的总质量为164.8g。
已知不同温度下每100g溶液中无水氢氧化钾的质量为:求温度T注:法拉第常数F = 9.65X104 C/mol,相对原子质量:K 39.1 O16.0 H 1.01 (15分)3.迄今已合成的最重元素是112号,它是用Zn 7030高能原子轰击Pb 20882的靶子,使锌核与铅核熔合而得。
科学家通过该放射性元素的一系列衰变的产物确定了它的存在,总共只检出一个原子。
该原子每次衰变都放出一个高能 粒子,最后得到比较稳定的第100号元素镄的含153个中子的同位素。
(7分) (1)112号是第几周期第几族元素?(2)它是金属还是非金属?(3)你认为它的最高氧化态至少可以达到多少?(4)写出合成112元素的反应式(注反应式中的核素要用诸如31H 、Pb 20882等带上下标的符号来表示,112号元素符号未定,可用M 表示)。
4.100.00mL SO 32–和S 2O 32–的溶液与80.00mL 浓度为0.0500mol/L 的K 2CrO 4的碱性溶液恰好反应, 反应只有一种含硫产物和一种含铬产物; 反应产物混合物经盐酸酸化后与过量的BaCl 2溶液反应, 得到白色沉淀, 沉淀经过滤、洗涤、干燥后称量,质量为0.9336g 。
中国化学会1998年全国高中学生化学竞赛(省级赛区)试题与答案(Word版)
1998年全国高中学生化学竞赛(初赛)试题评析1.某芳香烃A,分子式C9H12。
在光照下用Br2溴化A得到两种一溴衍生物(B1和B2),产率约为1︰1。
在铁催化下用Br2溴化A也得到两种一溴衍生物(C1和C2);C1和C2在铁催化下继续溴化则总共得到4种二溴衍生物(D1、D2、D3、D4)。
(13分)(1)写出A的结构简式。
(5分)(2)写出B1、B2、C1、C2、D1、D2、D3、D4的结构简式。
(8分)【评析】A是“芳香烃”,至少要有一个苯环。
按分子式C9H12,已不可能有比苯环更高级的芳香环,例如萘的分子式是C10H8,已超过9个碳。
因而,C9H12是有烃基侧链的苯的衍生物是无疑的了。
问题是:侧链有几个?侧链的组成?粗想起来,以下所有结构似乎都是可能的:但分析试题提供的信息之一——光照下得到2种一溴衍生物,不必细想,就可以立即排除掉上列的2和7,因为2、7无疑只能得到一种一溴衍生物。
3也可排除,因为当溴取代三个甲基中的一个氢就可得到3种一溴衍生物,超过2种。
如果选手以为7的一溴衍生物不止一种,就说明没有一溴代只发生在侧链的 -碳上的基本知识。
这个基本知识比中学有机化学深,被命题人假设为完全可以通过课外活动达到的最低限度的高一级的有机化学知识。
究竟有多少参赛学生可以达到这个最低限度的高一级的有机化学知识?这需要作统计。
如果统计的结果,超过85% 的学生达不到这个高一级的有机化学知识,说明命题人过高地估计参赛选手的知识水平了,学生解答该题其余部分也就困难了。
1式也可以得到2种一溴代衍生物,但1-溴代甲基-2,3-二甲基苯的产率跟2-溴代甲基-1,3-二甲基苯的产率不会相等,不符合2种一溴衍生物的产率相等的信息。
于是只剩下4、5、6三种可能性。
然后再分析另一个信息——芳香环上的氢被溴取代得到2种一溴衍生物。
这又需要选手有芳香环上的氢被取代的定位知识——烃基是一种邻对位定向基团。
选手有这个超过中学化学知识的高一级的化学知识吗?命题人又一次面临对选手的化学知识的估计的考验。
1998年全国高中学生化学竞赛_决赛_理论试题答案与评分标准
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1998年全国初中奥林匹克化学竞赛试题参考答案
1998年全国初中奥林匹克化学竞赛试题参考答案一、选择题1.B2.BD3.B4.B D5.CD6.BC7.A8.D9.AC 10.A 11.CD 12.C二、填空题 点燃13.2C 21H 30O 2 + 55O 2 ==== 42CO 2 + 30H 2O14.CaH 2 + 2H 2O + Na 2CO 3==== CaCO 3↓ + 2NaOH + 2H 2↑15.(1)C ,A ,D ,B 或D ,A ,C ,B (2)Ca(OH)2 自来水或蒸馏水(3)200- x-3/2y (4)100x/22.4 100y/22.4(5)44.8 121.616.(1)CaO·3MgO·4SiO 2 (2)K 2O·3Al 2O 3·6SiO 2·4H 2O三、实验题 17.18.第一步: 高温TiO 2 +2C+2Cl 2 ====TiCl 4 + 2CO 2 还原剂:C第二步:(1)2Mg+ TiCl 4Ar2MgCl 2+Ti (2)常温下,用盐酸除去过量的镁,然后过滤,并用蒸馏水洗涤固体,即得金属钛 Mg+2HCl==MgCl 2+H 2↑19.(1)略(2)80mL 56mL (3)0.65g 14.6%四、20.(1)X :Ca(2)A :CaCO 3 B :Ca(OH)2 C :CaO D :Ca(NO 3)2 E :CaCl 2 F :HCl G :AgNO 3 H :Na 2CO 3(3)D→A :Ca(NO 3)2+Na 2CO 3=CaCO 3↓+2NaNO 3E→D :CaCl 2 +2AgNO 3=2AgCl↓+Ca(NO 3)2A→E :CaCO 3+2HCl=CaCl 2+H 2O+CO 2↑21.(1)硫的化合价下降,碳的化合价升高 (2)BaCl 2(3)BaCl 2+Na 2CO 3=BaCO 3↓+2NaCl方案1不正确,pH 试纸使用方法不对,Na 2CO 3,NaOH 溶液都显碱性 方案2不正确,NaOH 与CuSO 4反应生成蓝色沉淀 方案3 正确,BaCl 2与Na 2CO 3反应生成白色沉淀,所以原NaOH 中混有Na 2CO 3高温BaCO 3==BaO+CO 2↑ BaO+H 2O == Ba(OH)2(4)不加入盐酸而加入Na 2CO 3,不会产生H 2S 气体 BaS+ Na 2CO 3=BaCO 3↓+2NaSBaCO 3+2HCl=BaCl 2+H 2O+CO 2↑ BaCl 2+ Na 2SO 4== BaSO 4↓+ 2NaCl22.固体C 增加的质量即为SO 42-的质量,得SO 42-的质量为(b-a)g 。
澳大利亚全国化学竞赛试题(1998-2010)-FSE7A
FINAL SELECTION EXAMINATIONf or th e2008 AUSTRALIAN CHEMISTRYOLYMPIAD TEAMPAPE R A2007Please note that this answer book will be photocopied when returned and thensplit so that answers are sent to the appropriate markers.For this reason it is extremely important that you observe instructions 6 to 8.In st r u ct i o n s t o S c h o l ar s1.You are allowed 10 minutes to read this paper, and 3 hours to complete the questions.2.You are not permitted to refer to books or notes but you may use a non programmable electroniccalculator.3.All questions to be attempted. A guide for time allocation is supplied at the beginning of each question.4. A periodic table with atomic masses and the values of some physical constants are provided on thefollowing page. Data are supplied, where necessary, with each question.5.Answers must provide clearly laid out working and sufficient explanation to show how you reachedyour conclusions.6.Answers must be written in the blank space provided immediately below each question. Rough workingmust be on the backs of pages. Only material presented in the answer boxes will be assessed.7.Ensure that your name is written in the appropriate place on ALL of the pages (even those you mayhave left blank) in this examination booklet.e only black or blue pen for your written answers, pencil or other coloured pens are notacceptable.Supervisor DeclarationI certify that the final selection examination was carried out under strict examination conditions and that no improper actions occurred during the examination period.Name of Exam Supervisor: (please print) ……………………………………………………………..…… Signed: ……………………………………………………………… Date: ……………………………Please use the enclosed pre-addressed Express Post Envelope to return all examination papers to: Mr R W Switzer, ASO Chemistry Program, 14 Liverpool Street, Golden Crest Manors, NERANG QLD 4211. EXAMINATIONS SHOULD BE RETURNED BY EXPRESS POST ON WEDNESDAY 12th MARCH SO THAT THEY ARE RECEIVED BY FRIDAY 14th MARCH 2008.CONSTANTSspeed of light in vacuum c = 2.998 × 108 m s –1Planck’s constant h = 6.626 × 10–34 J selementary charge e = 1.602 × 10–19 Celectron mass m e = 9.109 × 10–31 kgAvogadro constant N A = 6.022 × 1023 mol –1Faraday constant F = 9.6485 × 104 C mol –1ideal gas constant R = 8.3145 J K –1 mol –1Boltzmann constant k = 1.381 × 10–23 J K –1atomic mass unit u= 1.661 × 10-27 kgpi π = 3.14159Ångström 1 Å = 10–10 mQuestion 1 (16 minutes)(a) (i) Write the ground state electron configuration for a chlorine atom, C l.(ii) Write an excited state electron configuration for a boron atom, B.(iii) Draw a Lewis structure for Br2O5 (which contains a Br–O–Br bridge). Describe the shape around each of the Br atoms.(b) Consider the molecule, CO2.(i) Use VSEPR theory to rationalise the shape of CO2.(ii) Draw resonance structures for CO2 and indicate which structure will make the major contribution to the bonding.(iii) Construct a well-labelled hybrid orbital diagram for CO2 and include a description of the formation of the sigma and pi bonds.(c) Comment on each of the following observations.(i) XeF6 has a pentagonal pyramidal (distorted octahedral) structure.(ii) N5+ is bent at the central N atom.(iii) NOF3 is well established with N–O and N–F single bonds.(iv) BH3 can accept a pair of electrons to form compounds such as H3BN(CH3)3 in which the B atom is tetrahedral.(d) The dipole moment of (Z)-N2F2 in the gas phase is 0.16 Debyes, but (E)-N2F2 is non-polar. Explainhow this difference arises.Question 2 (36 minutes)(a) Data:!p Ka(CH3COOH)=4.76(i) Calculate the pH of each of the following solutions:A. 10.0 mL 1.00 mol L–1 HC l.B. 10.0 mL 1.00 mol L–1 CH3COOHC. 10.0 mL 1.00 mol L–1 CH3COO–(ii) For each of the solutions above calculate the absolute change in pH when 5.00 mL of 1.00 mol L–1 NaOH is added.A.B.C.A 20.00 mL aliquot of an unknown monoprotic acid of unknown concentration is titrated with5.00 × 10–1mol L–1sodium hydroxide solution. The p H was measured after the addition of10.00 mL and 20.00 mL of base and was found to be 8.50 and 9.50 respectively.(iii) Give approximate ranges for reasonable values of original concentration and p K a for the unknown acid given the information above.(iv) Calculate the original concentration and p K a of the monoprotic acid.(v) What will be the p H of the titration at equivalence?(b) Data:!E°(VO2+/VO2+)=1.00 V!E°(VO2+/V3+)=0.34 V!E°(V3+/V2+)="0.26 V!T=298 KBatteries use electrochemical reactions to store energy in the form of potential chemical energy. Redox batteries are a type of galvanic cell that only utilizes aqueous species. An example is the vanadium redox battery (VRB) which only has vanadium species involved in the reaction. In this battery VO2+ is converted to VO2+ while V2+ is oxidised to V3+ in the presence of sulphuric acid.(i) Write half equations and a balanced full equation for the VRB.(ii) Calculate E0cell, ΔG0 and K for the main reaction that occurs in a VRB.(iii) Given that the starting solutions for the battery have a concentration of 2.00 mol L–1with respect to sulphuric acid and all other species are present at a concentration of 1.00 mol L–1what is the starting voltage, E cell?(iv) What is the voltage, E cell, of the battery if no sulfuric acid was present initially? (Assume a neutral solution).One of the advantages of the VRB over other redox batteries is that if a contamination was to occur the battery could still be used as vanadium is present in both halves of the cell.(v) What would be the resulting concentrations of!VO2+ and!VO2+ in the cathode solution if 100 mL ofthe anode solution was to contaminate 1.00 L of the cathode solution? (assume no change in p H) (vi) Hence what would be the new E cell?Question 3 (20 minutes)The hypochlorite anion (C l O–) is found in many strong oxidising agents. NaC l O is the active ingredient in bleach, and HC l O is the 'chlorine' in salt-based swimming pools.(a) (i) Which element, oxygen or chlorine, is more electronegative?(ii) Draw a molecular orbital energy level diagram of the hypochlorite anion, showing how the atomic orbitals combine.(iii) Draw a 95% inclusion surface for each of the following MOs:(iv) What is the bond order of C l O–, and is your answer consistent with VB theory?(b) The particle-in-a-box model, despite its simplicity, makes surprisingly accurate predictions of theabsorption spectra of conjugated organic molecules. The thiacyanine dyes are one such set of molecules, where n = {0, 1, 2, ...} specifies the length of the conjugated chain. Take β = 140 pm to be the average bond length in the conjugated chain.(i) Draw two resonance forms of the dye with n = 0, showing how the electrons are delocalised alongthe π-system.(ii) Take the length of the 'box' to be the distance over which the π-electrons are delocalised. What is the length of the box when n = 0 (ignore any correction due to bond angles)?(iii) On the same graph, draw the potential energy function V(x) and the second lowest energy wavefunction Ψ2(x) of the system.(iv) Derive formulae for L, the length of the box, and p, the number of electrons in the box in terms of n.(v) Draw an energy level diagram for the thiacyanine dye with n = 3. Label the HOMO and LUMO. (vi) Derive a formula for the energy levels of a particle in a box.(vii) Calculate the wavelength of light absorbed in the HOMO-LUMO transition of the dye with n = 3.What colour is the dye?Question 4 (20 minutes)(a) Name each of the following transition metal complexes:(i) [Rh(en)BrCO](ii) K2[Fe(CN)4(H2O)2](iii) Na[Co(ox)2(NH3)2](iv) Na[Au(CN)4]Data: en = NH2CH2CH2NH2ox = C2O42–(b) For each of the following compounds, draw all possible isomers, designating each isomer with anappropriate label:(i) [Co(en)2ClBr]+(ii) [Pd(NH2CH2CH(CH3)NH2)2]2+ (racemic mixture of ligands) (iii) [Fe(NH2CH2C(CH3)2NH2)3]2+(c) For each of the following compounds, construct a well-labelled crystal field energy diagram,showing electron occupancy. Also indicate:• Predicted geometry of the ligands• Spin-only magnetic momentum (in units of Bohr Magneton (BM))• Overall crystal field stabilisation energy (CFSE) of the complex(i) [Mn(NH3)6]2+(ii) [Os(CN)6]4–(iii) [NiBr4]2–(d) Explain whether you would expect Jahn-Teller distortion to play a significant role in any of thecomplexes in part (c).Question 5 (20 minutes)Data: Gas Constant R = 8.3145 J K–1 mol–10° C = 273.15 KSulfonamides are an important group of drugs known for their effect on inhibiting bacterial growth, amongst other things. Because cell membranes consist mainly of lipid, the lipid/water partition coefficient of these drugs is a major determinant of properties of the drug, such as its absorption and penetration into bacteria.We will study the thermochemistry of the n-octanol/water partition coefficients of three different sulfonamides as a close approximation of how sulfonamide drugs behave.Partition coefficient K D is defined as the equilibrium constant of the partition process (assume the drug is not ionizable in aqueous phase):[drug]aq octK D (at 298.15 K) K D (at 333.15 K)sulfamethazine 1.8423 0.7893sulfamethoxazole 7.5057 0.5320sulfachloropyrazine 1.7927 0.3121(a) Write an equation for the molar standard ΔG of the partition process from water to n-octanol, interms of the partition coefficient K D and other relevant variables/constants.(b) Hence, calculate the molar standard ΔG of the partition process of the three given drugs (from waterto n-octanol) at the two temperatures 298.15 K and 333.15 K.(c) Hence, calculate the molar ΔH and ΔS of the partition process (from water to n-octanol) of the threesulfonamides given.(d) What does the sign of ΔS and ΔH tell us about what drives the solvation of sulfonamides fromwater to n-octanol? Hence explain, using the Second Law of Thermodynamics, why the partition coefficients of all three drugs decrease with increased temperature.(e)Calculate the concentration of each drug in the n-octanol phase in equilibrium with an aqueous phase concentration of 5.00 × 10–5 mol L –1 of each sulfonamide at body temperature (310.15 K). Which drug would be expected to have the best penetration into cell membrane lipids at body temperature?Drug with best penetration into cell membrane lipids: (f)Derive an expression that relates the ratio of partition coefficients at different temperatures(2)(1)D D K K !"#$#$%&solely in terms of ΔH, T 1 and T 2.Question 6 (10 minutes)Draw and name the products when (Z)-2-butene is treated with the following reagents. Indicate stereochemical and regiochemical outcomes where appropriate.(a) O3 followed by Zn/H+ or DMS(b) hot, concentrated KMnO4(c) cold, dilute KMnO4(d) gaseous HBr(e) liquid Br2(f) aqueous Br2Question 7 (10 minutes)Draw and name the products of the following reactions. Indicate stereochemical and regiochemical outcomes where appropriate.(a)(b)(c)(d)(e)(f)Question 8 (6 minutes)A single product forms when sodium ethoxide in ethanol is added to benzaldehyde and then acetoneadded dropwise to the mixture.(a) Draw the structure of this product.(b) Heating the initial product results in a product of molecular formula C10H10O. Draw its structureand explain its formation.(c) If the acetone is not added dropwise another product competes. Draw the structure of thiscompeting product and explain why the acetone must be added dropwise.Question 9 (10 minutes)p-aminobenzoic acid (PABA) is a common additive to sunscreens. When PABA is treated with bromine in the presence of iron(III) bromide, a single isomer of formula C7H6NO2Br is produced.Draw a mechanism including Whelan intermediates for attack at two different positions to predict the structure of this isomer. Indicate any particularly good or bad resonance contributors in your answer.The stoichiometric equation of the thermal decomposition of azomethane (CH3NNCH3) isH3CNNCH3 (g)N2(g) + C2H6(g)The total pressure of the reaction mixture was measured at different times and at different temperatures as follows.T = 571.6 KT (min)10.0 20.0 33.0 InfinityP (torr)491.9 548.0 609.7 861.6T = 593.6 KT (min)9.118.2InfinityP (torr)318.5371.5424.6Determine the reaction order, rate constant k and the parameters of Arrhenius equation for this reaction.The measurement of 14C disintegrations is a powerful and widely used technique for determining the age of carbon-containing specimens.(a) Obtain the law of radioactive disintegrations, assuming it follows first order kinetics.(b) The specific activity of 14C, in equilibrium with the atmosphere and the earth-skin, is18.3 disintegations min–1 g–1. The minimum measurable value is 0.03. The half-life of 14C is 5720years. Calculate the oldest measurable age.(c) In Zagreb, an Egyptian mummy, enveloped in a cloth with Etruscan inscriptions, was found. Inorder to ascertain the Etruscan origin of the cloth, the 14C method was used. By combustion of a splinter a gas was obtained which contained CO2. An activity of 12.9 disintegrations min–1was counted in 2 dm3of purified CO2at 1 atmosphere and 25˚C. Calculate the age of the specimen.(1 dm = 10 cm)Insulin isolated from meat has been found to have molecular weight 6000 and to contain two peptide chains bound together by disulfide bridges. The two chains were separated after oxidation of the disulfide bridges. One of the chains, A-chain, was found to contain 21 amino acids units in the following proportions:Gly, Ala, 2 x Val, 2 x Leu, Ile, 4 x Cys, 2 x Asp, 4 x Glu, 2 x Ser and 2 x TyrN-terminal assay by the dinitrofluorobenzene method (DNFB) gave DNP-Gly, and C-terminal assay using carboxypeptidase liberated Asp.After partial acid hydrolysis the following peptides were isolated and determined using the method of Edman Degradation.1. Cys-Asp 5. Glu-Cys-Cys9. Ser-Leu-Tyr2. Tyr-Cys 6. Glu-Glu-Cys10. Ser-Val-Cys3. Cys-Cys-Ala7. Glu-Leu-Glu11. Gly-Ile-Val-Glu-Glu4. Glu-Asp-Tyr8. Leu-Tyr-GluFurthermore, it was found that the A chain in its native state contained an intramolecular disulfide bridge giving a dodeca-ring.Give the amino acid sequence for the A chain.Isoelectric point (pI) is defined as the pH at which a particular molecule or surface carries no net electrical charge. Consider a generic naturally occurring amino acid whose side group R is neutral.(a) Draw the structure in Fischer projection form for such an amino acid.(b) By considering the two equilibria associated with the C terminus and N terminus of the aminoacids, and where the respective equilibrium constants are k a1 and k a2, derive an expression for pI.Glutamatic acid (Glu) is very important in cellular metabolism. Its other important function is to serve as the precursor for the synthesis of GABA (an inhibitory neurotransmitter) in a reaction that involves glutamic acid decarboxylase (GAD), which is very abundant in the cerebellum and pancreas.(c) What class of biomolecules does GAD belong to? What is the function of GAD?(d) Where would you expect to find glutamic acid in the 3D structure of GAD in an aqueousenvironment?(e) Calculate the pI for glutamic acid given its pK a(N-terminus) = 9.67, pK a(C-terminus) = 2.19,pK a (side group R-COOH) = 4.25.(f) Calculate the pH at which two thirds of glutamic acid’s γ-COOH group has been deprotonated.Lysine is a basic amino acid with side group of (CH)4NH3+ at pH 7. It has p K a (C-terminus) = 2.18, p K a (N-terminus) = 9.04 and p K a (side group R'–NH3+) = 8.95.(g) Write down (in short hand) all possible dipeptides that can be formed from Lys and Glu.(h) If we extend the idea of isoelectric point to peptides, calculate the isoelectric point of the dipeptideLys-Glu. [Assume that there is no difference in the pK a of each amino acid functional group before and after they form peptides.](i) Draw the dipeptide Lys-Glu at p I. You may represent the respective side groups asR–COOH/R–COO– and R'–NH2/R'–NH3+。
1998年全国化学竞赛预赛模拟试卷参考答案
1998 年全国化学比赛(初赛)模拟试卷(一)参照答案说明:本套试卷依据“全国比赛初赛纲领”和“降低原理水平,提升能力要求”命题新动向进行拟题。
本试题在剖析能力上有很高的要求,并波及许多的数学、物理等交错学科的知识,合用于对优异学生的选拔与训练。
1998(一)一.(共15 分)1. 2/9 (5 分,答 4/9 给 2 分)此题经过对O3的剖析解决问题,打破“题海”中单质间的转变成非氧化复原反响的约束。
O3分子中各原子电荷为,即 O3为,每摩尔O3转变成O2转移电子数为2/3mol ,此题还对物理中的气态方程进行了考察。
2.、(各1分);(3分)此题主要考察了异构体和数列的思想(或摆列组合或概括法的思想)。
本答案仅供给一种方法,该物质可代替的 H 为 2m+4,当 m为奇数时,二氯代替物为 (2m+3) +(2m- 1) ++ 5+ 1(共项),等差数列乞降,; m为偶数时, S=(2m+3) ++ 7+3,可得同样答案。
此外我们能够依据m=2,3,4,5概括出通式;也能够用摆列组合思想解答(详见《中学生化学报》第413 期。
3. 2240(4 分)此题主要考察复杂氧化复原反响中电子转移的本质性问题。
在多步氧化复原中,可忽视中间反响,只要抓住起始物质和最后产物的氧化数变化。
此题中,Fe→Fe(Ⅲ) ,N(Ⅴ) →N(Ⅱ) 。
1.NO2++H3O++2HSO4-(4 分 )联系硝基苯的制取。
此题为在 100%H2SO4中的酸碱反响,并考察对数据的办理。
在100%H2SO4+-++-和水溶液中的-相当。
中, H2SO4释出 H和 HSO4,H能和 H2O形成 H3O, HSO4OH2.PCl5+9NH3=P(NH)(NH2) 3 +5NH4Cl (3 分)NH3与 H2 O的类比,同时应注意同中有异。
3.Mg+4HNO3=Mg(NO3) 2+2NO↑+2H2 O ( 2 分)40Mg+100HNO3==5NO↑+H2↑+NO2↑+3N2↑+4NHNO3+40Mg(NO3) 2+41H2 O(5 分)此题考察识图能力,图象的定量剖析,氧化复原反响得失电子守恒,方程式比率系数确实定。
澳大利亚全国化学竞赛试题(1998-2010)-FQEB004
(iii) If several forms are possible, choose one, taking into account the fact that the repulsion between pairs decreases as follows:
free-pair / free-pair > bonded-pair / free-pair > bonded-pair / bonded-pair
(ii) Draw the possible structure (or structures) of each molecule.
(iii) If several forms are possible, choose one, taking into account the fact that the repulsion between pairs decreases as follows:
AUSTRALIAN CHEMISTRY OLYMPIAD
FINAL PAPER PART B 1991
Instruction to candidates
(1) You are allowed 10 minutes to read this paper, and 3 hours to complete the questions.
(ii) Give the maximum value of the X—P—X bond angle.
(iii) With reference to the volume occupied by the bonding pairs around the phosphorus, predict whether the X—P—X bond angles will be increased or decreased when X is F, Cl, Br and I.
澳大利亚全国化学竞赛试题(1998-2010)-FQE5B001
b)
What is the value of the quantum number L of the highest occupied level in the ground
electronic state of the C60 molecule?
c)
How many unpaired electrons should the C60 molecule have according to the model of free
dihydrate of the sodium salt, Na2[H2Y].2H2O are available in highly pure form and are often
used as primary standards.
The acid dissociation constants of H4Y at 25°C are: Ka1 = 1.02 x 10–2 Ka2 = 2.14 x 10–3 Ka3 = 6.92 x 10–7 Ka4 = 5.50 x 10–11
AUSTRALIAN CHEMISTRY OLYMPIAD
FINAL PAPER PART B 1995
Please note that this answer book will be photocopied when returned and then split so that answers are sent to the appropriate markers. For this reason it is extremely important that you observe instructions 5 to 7.
L(L
+
澳大利亚全国化学竞赛试题(1998-2010)-FQEA009
F I N A L S E L E C T I O N E X A M I N A T I O Nf o r t h e2002A U S T R A L I A N C H E M I S T R Y O L Y M P I A D T E A MPA R T A2001Please note that this answer book will be photocopied when returned and thensplit so that answers are sent to the appropriate markers.For this reason it is extremely important that you observe instructions 6 to 8.Ins tructions to Stude nt1.You are allowed 10 minutes to read this paper, and 3 hours to complete the questions.2.You are not permitted to refer to books, notes or periodic tables but you may use a non programmableelectronic calculator.3.All questions to be attempted. A guide for time allocation is supplied at the beginning of eachquestion.4.Data is supplied, where necessary, with each question.5.Answers must provide clearly laid out working and sufficient explanation to show how you reachedyour conclusions.6.Answers must be written in the blank space provided immediately below each question in the exambooklet. Rough working must be on the backs of pages. Only material presented in the answer boxes will be assessed.7.Ensure that your name is written in the appropriate place on each page of your examination booklet.e only black or blue pen for your written answers, pencil or other coloured pens are notacceptable.Question 1 (20 minutes)Molecular orbital theory is used to reconcile concepts such as resonance and magnetic properties of molecules which cannot be adequately explained by atomic orbital theory.a)Draw the molecular orbital energy diagram for He2+, He2(ground state) and He2(1st excitedstate). State the bond order and the electron configuration of each species. Which ones would you expect to be stable?b)The homodinuclear species B2 and O2are known to be paramagnetic. Use an MO orbitalenergy diagram to illustrate why. Are there any differences between the splitting of orbital energies in these two species? If so, why?c)Write out the electron configurations of B 2 and O 2. Based on your answer in (b), predict whichone would have a lower first ionisation energy compared to the corresponding atom’s firstionisation energy.d)We have considered the effects of reducing a dimension on chemistry from Flatlandia. Let usnow investigate the effects of adding a dimension. In the hypothetical 4th -landia, electrons aredescribed using five, not four, quantum numbers. n, l, m l and m s remain as they are. The fifthnumber m h (the “hyper” magnetic quantum number) behaves like m l . The values of mlDO NOT affect m h . Eg if l = 3, –3 ≤ m l , m h ≤ –3; if l = 4, m l = 2, –4 ≤ m h ≤ 4.Predict the total number of elements expected in the first four periods.e)Consider SO 42–. Using both atomic and molecular orbital theory account for its structure. Includein your answer a hybrid orbital diagram and a molecular orbital energy diagram. Whichorbitals does the sulfur atom use to form molecular orbitals? How many electrons areinvolved in bonding orbitals?Question 2 (25 minutes)a)For each of the following species draw the Lewis structure and predict the electron pairgeometry and molecular shape: XeF2, ICl 3 and the diprotic acid H 3P O 3. What is thehybridisation of the central atom in each of these species? Which of these species would be expected to have a dipole moment?b)Like 1,2-diaminoethane (en) the following related ligands containing phosphorus donor atomsare also potent chelating agents: (2-aminoethyl)dimethylphosphine (PN)[H 2NCH 2C H 2P(CH 3)2] and 1,2-bis(diphenylphosphino)ethane (d ppe). How manystereoisomers are possible for each of the following species containing these ligands?Sketch and name them.(i)[Pt(SCN)2(dppe)](ii)[RuCl 2(dppe)2](iii)[Co(PN)3]3+(iv)Which, if any, of these species would be expected to exhibit linkage isomerism? Sketch all ofthe possible linkage isomers.c)Consider the bidentate ligand (2-aminoethyl)methylphenylphosphine (PN*). What type ofisomerism, if any, could be exhibited by a square planar complex of the type [M(PN*)2]?Sketch the isomers.Question 3 (45 minutes)Data for this question can be found at the end of the question.Our adventure into the world of analytical chemistry begins on a different planet, Ammonica.The world of Ammonica is similar in many ways to our own, except for the fact that theirvarious life forms are based around ammonia (which exists as a liquid under Ammonica’senvironmental conditions) instead of water.a)Just as the autodissociation of water (and its dissociation constant, Kw ) is so crucial to thepeople of Earth, so too is the autodissocation of ammonia to the Ammonites. The process of autodissociation now involves the transfer of a proton from one molecule of ammonia toanother: ie,NH 3 (l ) + NH 3 (l ) NH 4+ (am ) + NH 2– (am )with equilibrium constant K am = 1.0 x 10–22 at ambient conditions.(The notation (am) indicatesthat the species is present as an ammoniacal solution). The special symbols pH and pNH2are given to the values of -log 10[NH 4+] and -log 10[NH 2–] respectively. The definition of acidityand basicity constants in Ammonica is analogous to our own system, with the hydrogen ion and hydroxide ion concentrations replaced by ammonium and amide ion concentrationsrespectively.(i)Calculate the pH of neutral ammonia.Water (a solid the Ammonites find has an unbearable odour when dissolved) was dissolved in liquid ammonia to give a solution with [H 2O]total = 0.10 M.(ii)Adapt the Arrhenius definition of acids and bases on Earth to Ammonica.(iii)Why would water be considered an acid in Ammonica?(iv)Give the overall equation for the reaction between water and ammonia, and the expression forits equilibrium constant K a (H 2O).(v)Calculate the values of pH and pNH 2 for the solution above under ambient conditions.(vi)Calculate the change in pH effected by the addition of 1.00 x 10–3 moles of ammonium chlorideto 1.0 L of this solution.Solid sodium hydroxide (2.0 g) was added to 1.0 L of the solution to create a buffer based on the H 2O / OH – acid / base pair.(vii)Calculate the pH of this solution and the change in pH that would occur upon addition of 1.00 x10–3 moles of ammonium chloride. Comment on your answer with respect to your answer topart (vi) above.b)The partial pressure of ammonia gas, p(NH 3), in the atmosphere of Ammonica is 0.020 atm. Bycoincidence, the atmospheric pressure in Ammonica is identical to that on Earth.(i)Calculate the equilibrium constant, Kvap , for the vaporisation of ammonia at the ambienttemperature of Ammonica.(ii)Calculate the ambient temperature on Ammonica.c)Liquid ammonia has a number of interesting and unusual properties, not least of which is itsability to support electrons in solution. Sodium (among other metals) will dissolve in liquidammonia to give a solution containing both sodium ions and electrons in ammoniacalsolution:Na (s ) → Na (am )Na (am ) Na + (am ) + e – (am )p K = -6.22Organic chemists make use of this extraordinary fact here on Earth to reduce alkynes to alkenes; for example, 3-hexyne can be reduced to trans -3-hexene by adding it to a solutionof sodium in liquid ammonia. A chemist on Ammonica dissolves 17.4 mg of sodium metal in1.0 L of ammonia. To this solution, they carefully add 30 mg of 3-hexyne.(i)Give a balanced equation for the reduction of 3-hexyne in ammonia by ammoniacal electrons.(ii)Calculate the initial concentration of electrons and ammoniacal sodium atoms in the solution.(iii)Calculate the potential of the hexyne / hexene couple at the point where half of the 3-hexynehas been converted to trans -3-hexene.(iv)Calculate the amount of 3-hexyne remaining in solution once the system has come toequilibrium and hence the maximum yield of the reduction product, trans -3-hexene underthese conditions. Give your answer to as many significant figures as are necessary toillustrate a deviation from 100%.Data:p K a (H2O (am )) = 4.75∆G o vap (NH 3) = +5.84 kJ mol –1FW(Na) = 22.99 g mol –1FW(NaOH) = 40.00 g mol –1FW(C 6H 10) = 82.15 g mol –1E o (C 6H 10 (am ) / C 6H 12 (am )) = +0.90 VQuestion 4 (45 minutes)a)90232Th is one of the longest-lived radioactive isotopes, with a half-life of 1.40 x1010 years.(i)The end product of the 90232Th decay chain is 82208Pb . A sample of ore is analysed and is foundto contain 56.5mg of 208-lead for every gram of 232-thorium. Assuming that all other chainhalf-lives are much shorter and can therefore be ignored, how old is the rock?(ii)The longest-lived intermediate decay product of the thorium chain is the first member of theseries, 88228Ra , with a half-life of 5.76 years. How many grams of Ra-228 will be found in asample of ore containing 1.000g of thorium?b)Data: C v (O 2(g)) = 24.0 J K –1 mol –1∆f H o (H 2O (g)) = -242.00 kJ mol –1.Gaseous Species Average C p (J K –1 mol –1)H 230N 233H 2O 41A perfect gas has internal energy U = 3/2(nRT), and PV = nRT.(i)Show that molar constant-volume heat capacity, C v v dq dT =, is equal to 32R for a perfect gas.(ii)Determine a numerical ratio of molar constant pressure heat capacity C p to C v for a perfect gas.Qualitatively explain the reason for the difference in the two quantities. The ratio found holdsapproximately for many real gases. Using the ratio find C p for gaseous O 2.Hydrogen is burnt in air (20%:80% O 2:N2) in a torch. It is assumed the system is adiabatic.Assume that the reaction takes place completely, and that 200% more air than the stoichiometric ratio is present.(iii)What is the composition of the exit mixture of gases?(iv)Using any data given or previously found, determine the maximum temperature reached by theflame (inlet temperature is 298.15K)?c)Given:ln ()pp H R T T vap 212111=−°−∆Data: 1 atm = 101.3kPa N A = 6.022 x 1023 mol –1R = 8.314 J K –1 mol –1 = 0.08206 L atm K –1 mol –1∆diss S o (Br 2(g)) = 104.5 J K –1 mol –1.[Assume all enthalpies and entropies constant over the given temperature ranges, and that gases may be treated as ideal.]Bromine is extremely toxic, and it presents a special hazard due to a very high volatility. At normal atmospheric pressure the boiling point of bromine is 59.2o C, and even at 9.3o C thepressure of vapours in a closed system is 13.33kPa.(i)Use these data to estimate the standard molar enthalpy of vaporisation of elementary bromine.The dissociation equilibrium of bromine is significant because of the low enthalpy of dissociation (∆diss H o = 192.9kJ mol –1), and because very low concentrations of bromine radical are enoughto initiate chain reaction mechanisms.(ii)Evaluate the number of bromine atoms contained in saturated bromine vapour at boiling pointand normal atmospheric pressure in a 1.000L vessel, and the percentage of atomic brominein bromine vapours in these conditions.(iii)Consider the reactionBr 2(l) 2Br (g).What is the equilibrium constant at 50o C? What is the vapour pressure of Br atoms above the liquid?ORGANIC SECTION (45 minutes)Question 5 (35 minutes)The structure of an optically active hydrocarbon A was deduced from the following:Treatment of A with hot concentrated acidified KMnO4 gave a single product, B, which upon heating withiodine in alkaline solution gave C (C9H16O4) and a yellow precipitate of iodoform. C can alsobe obtained from the following reaction scheme: 4-isopropyl-1,2-dihydroxybenzene wascatalytically hydrogenated with a palladium on carbon catalyst under a pressure of 138 atm(2000 psi), forming D, and D treated with periodic acid in aqueous THF to form C.a)Give two possible structures for A consistent with this information, and draw out the reactionschemes, clearly labeling products A-D (do not show stereochemistry).Addition of HBr to A in CCl4results in an achiral product E.b)Draw the single possible structure for A (do not show stereochemistry). Give the structure of Eand explain why it is not chiral.The main component of the essential oil of oranges, known as limonene (F) (C10H16), isstructurally very similar to A, and in fact hydrogenation of both limonene and A over a palladium on carbon catalyst gives the same product, p-menthane (G). Treatment of limonene with ozone followed by dimethyl sulfide produces formaldehyde (methanal) and H.Treatment of H with I2/NaOH gives J and twice the amount of iodoform as B gave under thesame conditions.c)Draw out this reaction scheme and clearly label products A, B, F-H and J (do not showstereochemistry).Limonene (F) is a member of a family of coupounds known as terpenes. In fact a number of different terpenoids (derivatives of terpenes) can be synthesized from limonene.Hydration of limonene with an acid catalyst may result in the formation of two constitutionally isomeric bicyclic alcohols. One of them (K) upon treatment with sodium dichromate isoxidised to camphor (structure below), but the other (L) cannot be oxidized with sodiumdichromate.Camphord)Draw the structures of K and L (do not show stereochemistry), and show a mechanism for theirformation from limonene under an acid catalyst.Dehydration of L with concentrated sulfuric acid can result in either of the two isomeric terpenes that are the main constituents of wood turpentine: α- and β-pinene. When treated with hotconcentrated KMnO4, only β-pinene gives off carbon dioxide, and α-pinene forms a keto acid.Interestingly although dehydration of L can give α-pinene, if α-pinene is treated with dilutesulfuric acid, K, and not L is formed.e)Draw the structures of α- and β-pinene, and propose a mechanism for the formation of K fromα-pinene in the presence of dilute sulfuric acid.f)Is camphor chiral? Draw a diagram clearly showing either the reason you think camphor is notchiral, or all possible stereoisomers of camphor, assigning any stereocentres R or S.g)What would be the organic product or products formed when A is treated with the followingreagents? If more than one product is possible, indicate which, if any, will be the majorproduct, and be sure to note any regiochemical and stereochemical consequences of these reactions.(i)bromine in CCl 4.(ii)iodine monochloride.(iii)CHCl 3 in concentrated KOH.(iv)m -chloroperbenzoic acid followed by sodium hydroxide and acidic workup.(v)m -chloroperbenzoic acid followed by phenyl magnesium bromide and acidic workup.Question 6 (5 minutes)Give the organic product or products of the following reactions, noting any stereochemical or regiochemical consequences:a)1-butyne 1. BH 3.THF22–?b)1-butyne NaNH 2???c)1-butyne H 2/Lindlar catalyst ?Question 7 (5 minutes)An inexperienced young chemist wanted to make compound B , shown below, and found a bottle of the structurally similar compound A , also shown below. Trying to convert A into B in one step and not really sure of how to do so, our adventurous young chemist decided to apply some classical trial and error, simply heated A with a catalytic amount of dilute acid,and was delighted when Bwas formed, as well as 2-methylpropene.H 322H Compound A Compound B+ 2-methylpropenePropose a mechanism for this reaction.。
1998年全国高中学生化学竞赛(初赛)试题答案
1998年全国高中学生化学竞赛(初赛)试题答案1. (13分)(1)A CH 32H 5(5分)(2)B 1CH 2Br 2H 5 B 2CH 33 C 1CH 3Br2H 5C 2 CH 32H 5Br(8分)2.电解反应是水分解为氢气和氧气。
10.00小时6.00安培总共提供电量Q =It =216×103C ,相当于2.24mol 电子,每电解1mol 水需电子2mol, 故有1.12mol 水,即20.1g 水被电解。
从剩余溶液的总质量(200g -164.8g )知,有35.2g 物质离开溶液,其中有20.1g 水被电解,故结晶的KOH ·2H 2O 的质量为15.1g 。
结晶的KOH 的质量为:(15.1g/92.10g/mol)·M(KOH)=9.2g结晶水的质量为:(15.1g/92.10g/mol)×2M(H 2O)=5.9g剩余溶液的溶质质量分数为:m(KOH)/[m(H 2O)+m(KOH)]=55.1%根据溶解度数据,T 应在20~30o C 之间,设此期间溶解度与温度呈线性关系,则 T =[273+20+(55.1-52.8)/(55.8-52.8)×10] K =301K (答:28o C )3.(7分) (1)第七周期第IIB 族元素 (2)是金属(3)最高氧化态至少可以达到+2。
(4)Zn 7030+Pb 20882=M 277112+n 14.(10分) (1)SO 42- Cr(OH)4-或KCr(OH)4或Cr(OH)63-或K 3Cr(OH)6或Cr(OH)3,(2)设n(SO 32-)=x ,n(S 2O 32-)=y , x +2y =0.9336g/233.3g/mol2/3x +8/3y =80.00mL ×0.0500mol/L解得:x =2.0×10-3mol ;y =1.0×10-3mol (4分)原始溶液的浓度:c(SO 32-)=2.0×10-2mol/L ;c(S 2O 32-)=1.0×10-2mol/L (4分)。
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Question 2
The reduction of a 4-substituted cyclic ketone A yields a mixture of liquid isomeric compounds B and C (with composition of 76.00% C and 12.75% H and density d430 = 0.9092 gcm-3) which during dehydration form one product D.
AUSTRALIAN CHEMISTRY OLYMPIAD
FINAL PAPER PART B 1990
Instruction to candidates
(1) You are allowed 10 minutes to read this paper, and 3 hours to complete the questions.
(iii) Give the energetically favoured conformation of molecules of the compounds A - C taking into consideration spatial repulsive interaction.
Question 3
-200
2 1 2 2 1 12 2 2 1 2 2 2 3
1 2 2 2 1 11 2 2 1 2 1 1 2
2 2 1 1 2 22 2 1 1 2 2 1 2
1 2 1 1 2 11 1 2 1 1 2 2 2
-300
2 1 1 2 2 21 3 1 2 2 3 3 1
An element A is mixed with fluorine in a molar ratio 1:9 at room temperature and high pressure (ca. 1 MPa). Upon heating to 900oC three compounds B, C, and D are formed. All three products are crystalline solids with melting points below 150oC. The fluorine content of C is
(7) Use only black or blue ball point pen for your written answers, pencil or other coloured pens are not acceptable.
Question 1
The formation of phosgene, COCl2 by the reaction of carbon monoxide and chlorine in the
(ii) Determine the composition of the mixture of compounds B and C, assuming additivity, when the
densities d430, of the more stable isomer B and less thermodynamically stable isomer C are 0.9073 and 0.9120 gcm-3 respectively.
(2) You are not permitted to refer to books, notes or periodic tables but you may use a non-programable electronic calculator and molecular models.
(3) You must attempt all questions.
found to be 36.7% and that of D 46.5% (by weight).
When B is treated with anhydrous HOSO2F at -75oC, a compound E is formed:
B + HOSO2F → E + HF
E is a crystalline solid which is stable for weeks at 0oC, but decomposes in days at room temperature.
the identities of the atoms in E.
2
Answer on diagrams
The electron density distribution of E on the X-Y plane Y
2 2 1 1 2 22 1 1 1 2 2 2 2
1 2 1 1 2 11 2 2 2 1 2 1 1
(1) E
O2 , ∆
(2) Η3Ο+ , ∆
F+G
Product G undergoes the iodoform reaction and shows only one signal in 1H NMR spectrum.
(i) Give the formulas of products A - G and all possible constitutional isomers of the compound G.
Dehydrogenation of compound D in the presence of palladium gives the product E. Product E
forms the basis of an important two-stage industrial synthesis to form compounds F (M = 94.114 gmol-1) and G, the reaction conditions are shown in the following equation.
(5) Answers must provide clearly laid out working and sufficient explanation to show how you reached your conclusions.
(6) Your name must be written in the appropriate place on each page of your answers.
(a)
Show clearly where the maxima lie by drawing the contour curves around the maxima, that is
connecting points of approximately equal electron densities. Label each maximum to show
(i) on the assumption that the equilibria in steps (a) and (b) are very fast compared with step (c).
(ii) by making the steady state approximation for the reactive intermediates.
(b)
Using the rate law obtained in part (i) apply the isolation method to obtain the half life for the
consumption of chlorine (Cl2) as a function of chlorine (Cl2) concentration.
The electron density distribution of E (the electron density map of E), which is obtained through X-ray diffraction studies, is shown on two intersecting, mutually perpendicular planes (see enclosed). These planes are arbitrarilly marked as X-Y and X-Z planes. The numbers indicated on these maps relate to the electron density in the neighbourhood of the atoms of E as a function of the spatial coordinates. The maxima of the electron density found in these maps coincide with the locations of the atoms. The values of the maxima are approximately proportional to the number of electrons in the atom in question. The cartesian coordinate axes are indicated on both planes. All significant electron density is shown on these maps.
presence of light is believed to proceed by the following mechanism;