Areal Density Measurement Is a Convenient Method for the Determination of Porcine Islet Equivalents
2020智慧树知到《分析化学(山东联盟)》章节测试完整答案
2020智慧树知到《分析化学（山东联盟）》章节测试完整答案智慧树知到《分析化学(山东联盟)》章节测试答案第一章1、重量分析法和滴定分析法通常用于高含量中含量组分的测定，两者都有各自的优点。
以下哪个选项是重量分析法的优点。
答案: 准确度较高2、分析方法可分为两大类：化学分析方法和仪器分析方法;化学分析又可以分为重量分析法和滴定分析法。
答案: 对3、分析化学依据分析的目的、任务可分为：答案: 定性分析、定量分析、结构分析4、组分分析中的组分包括元素、分子、官能团、化合物以及其他复合物组分组成的物相。
答案: 错5、分析化学的研究对象结构分析和成分分析，成分分析可分为定性分析和定量分析。
答案: 对第二章3、用 25ml 移液管移出的溶液体积应记录为：答案: 25.004、置信度越高，则平均值的置信区间越大。
答案: 对5、滴定分析法要求相对误差为±0.1%，若使用灵敏度为0.0001g 的天平称取试样时，至少应称取：答案: 0.2g第三章5、对于速度较慢的反应，可以采用下列哪种方式进行测定( )答案:返滴定法第四章1、下列物质中，能用氢氧化钠标准溶液直接滴定的是( )A.苯酚B.氯化氨C.醋酸钠D.草酸答案:D2、下列物质不能烘干的是( )A.碳酸钠B.重铬酸钾C.邻苯二甲酸氢钾D.硼砂答案:D3、下列对影响酸碱指示剂变色范围的主要因素错误的是( )A.指示剂用量B.温度C.溶剂D.试液的酸碱性答案:D4、下列物质不能在溶液中共存的一组是( )A.Na2CO3+NaHCO3B.NaOH+Na2CO3C.NaOH+NaHCO3D.Na2HPO4+NaH2PO4答案:C5、在NaOH标准滴定溶液滴定盐酸中，若酚酞指示用量过多则会造成终点( )A.推迟到达B.正常到达C.无影响D.提前到达答案:D6、用0.1mol.L-1NaOH滴定0.1mol.L-1的甲酸(pKa =3.74)，适用的指示剂为( )A.甲基橙(3.46)B.百里酚兰(1.65)C.甲基红(5.00)D.酚酞(9.1)答案:D第五章1、在配位滴定中，下列说法何种正确( )：A.酸效应使条件稳定常数增加有利于反应完全;B.配位效应使条件稳定常数增加不利于反应完全;C.所有副反应都不利于滴定反应;D.其余三项都不对答案:D2、在EDTA配位滴定中，下列有关酸效应的叙述正确的是：A.酸效应系数越大，配合物的稳定性越高;B.反应的pH值越大，EDTA的酸效应系数越大;C.EDTA的酸效应系数越大，滴定曲线的突跃范围越大D.酸效应系数越小，配合物的稳定性越高;答案:A3、溶液中存在M、N两种金属离子时，准确滴定M，而N不干扰的条件是( )A.ΔlgK≥0.3B.ΔlgK≤0.5%C.ΔlgK≥5D.ΔlgK≥8答案:D第六章1、高锰酸钾标准溶液直接配制。
南京大学-杨荣武分子生物学课件2
Nirenberg, 1967 "When man becomes capable of instructing his own cells, he must refrain from doing so until he has sufficient wisdom to use this knowledge for the benefit of mankind.... [D]ecisions concerning the application of this knowledge must ultimately be made by society, and only an informed society can make such decisions wisely."
The DNA structure provided a new template for hypotheses regarding biological phenomena (amenability to study)
DNA is easy to work with…
• Readily isolated--plasmid isolation, PCR • Stable--not chemically reactive like RNA (even archaeologically stable!) • Easy to propagate and move from cell to cell • Easy to make specific constructs • Easy to make specific mutations • Very easy to sequence (record-keeping) • Predictable behavior • Sequence lends itself to analysis--genome projects
智慧树答案海洋数据处理与可视化知到课后答案章节测试2022年
第一章1.下列哪个不是本门课程的关键词。
（）答案:海洋观测2.Ocean data view 软件可以通过编程实现功能。
（）答案:错3.NaN表示有效数据。
（）答案:错4.同学们可以自主选题。
（）答案:对5.个人成绩由小组成绩和个人贡献计算得到。
（）答案:对第二章1.在石油开发等工程建设的规划设计和施工都需详细掌握各种海洋气象、水文动力要素的变化规律，但是，不需要准确计算哪些要素多年一遇极值？（）答案:海底地形2.下列哪些不是海洋资料必须具备的属性？（）答案:相同地点3.以下哪个类别的资料不是以描述空间分布特征为划分依据的？（）答案:浮标、平台长期连续观测资料4.以下资料中为确定性资料的是？（）答案:潮位5.Matlab不能读取.nc格式？（）答案:错6.描点法不可以作为数据获取的手段？（）答案:错7.西北太平洋海洋环流与气候试验（简称NPOCE），是中国发起的第一个海洋领域大型国际合作计划？（）答案:对第三章1.测量值与真值之间的差值为（）答案:绝对误差2.以下哪种方法不是判别异常值的方法（）答案:傅里叶准则3.衡量测量值的均值与被测量真值的符合程度的是（）答案:准确度4.在一定的测量条件下，多次重复测量，所得的误差的量值和正负号的变化呈较明显的规律性，则该误差为偶然误差。
（）答案:错5.可以通过剩余误差代数和法将系统误差从数据中分离出来。
（）答案:对第四章1.检验两个总体的方差是否存在显著差异，可以用————。
( )答案:F检验2.对于不清楚分布形式的变量，使用哪种变换公式最合适？( )答案:幂变换3.常用的数据正态化变换有：对数变换、平方根变换、角变换、幂变换。
（）答案:对4.F检验可以用于总体均值的检验，可用于检验某地的气候是否稳定。
（）答案:错5.对于不清楚分布形式的变量，使用角变换是最合适的。
（）答案:错1.不是选配影响因子的注意事项的是（）答案:影响因子之间要相互联系2.下列关于阶段回归挑选法的说法错误的是（）答案:因子被选入之后依旧可以剔除3.如下图所示，表现的相关性如何。
智慧树知到《医学英语词汇学》章节测试答案
智慧树知到《医学英语词汇学》章节测试答案第一章1、The symbol of medicine is called ____.A:the rod of AaronB:the Rod of AsclepiusC:the Rod of JesseD:the Rod of Iron正确答案：the Rod of Asclepius2、The first bone of the spine is known as ____.A:AtlasB:AxisC:SacrumD:Coccyx正确答案：Atlas3、The heel cord, i.e. the largest and strongest tendon in the human body, is called _.A:Patella tendonB:Deltoid tendonC:Achilles tendonD:Trapezius tendon正确答案：Achilles tendon4、A 18-year-old man has an exaggerated sense of self-importance and always believes that he is better than others. This man may be diagnosed with having a(n) _.A:Oedipus complexB:Cain complexC:Electra complexD:Narcissistic personality disorder正确答案：Narcissistic personality disorder5、A senior woman always repeats statements and questions over and over, not realizing that she has asked the question before. She tends to forget conversations, appointments or events. This woman may be diagnosed with _.A:Parkinson’s diseaseB:Alzheimer’s diseaseC:Down’s syndromeD:Crohn’s disease正确答案：Alzheimer’s disease6、A boy always feels tired. Blood test shows that he is not iron deficient, but his MCV（平均红细胞体积）is very low. The name of this condition derives from the Greek Thalassa sea. What may be the possible diagnosis of him?A:Aplastic anemiaB:SyphilisC:Mediterranean anemiaD:Cataract正确答案：Mediterranean anemia7、In the brain, there is a region involved in memory forming, organizing, and storing, which is called the ___ because it has a shape of a seahorse.A:Pineal glandB:Pituitary glandC:ThalamusD:Hippocampus正确答案：Hippocampus8、A newborn baby is with reddish purple marks on the face, which is caused by a localized area of abnormal blood vessels. This condition is called __.A:Butterfly diseaseB:AcneC:Port-wine stainD:Pimple正确答案：Port-wine stain第二章1、Which of the following word parts is a combining form?A:hyper-B:ot/o-C:-iatricsD:penta-正确答案：ot/o-2、Which of the following words is spelled correctly ?A:enterogastritisB:enterogastroitisC:gastroenteritisD:gastrenteritis正确答案：gastroenteritis3、The correct pronunciation of “pt” in the word “gastroptosis” and “ptosis” is __. A:[t] and [t]B:[pt] and [pt]C:[t] and [pt]D:[pt] and [t]正确答案：[pt] and [t]4、Words ending with __are stressed on the next to last syllable.A:-tomyB:-omaC:-ectomyD:-rrhea正确答案：-oma5、The word “rhinoplasty” is pronounced_______.A:[‘rainəu,plæsti]B:[rai’nəuplæsti]C:[rainəuplæs’ti]D:[,rainəu’plæsti]正确答案：[‘rainəu,plæsti]第三章1、fibrous adhesionA:纤维组织B:纤维结构C:纤维固定D:纤维性粘连正确答案：D2、afferent neuronsA:传入神经元B:传出神经元C:传入神经递质D:传出神经递质正确答案：A3、回顾性研究A:a prospective studyB:a retrospective studyC:an extrovert studyD:an introvert study正确答案：A4、小脑下静脉A:inferior veins of cerebrum B:Superior veins of cerebrum C:inferior veins of cerebellum D:superior veins of cerebellum 正确答案：C5、glaucomaA:灰发病B:青光眼C:紫癜病D:发绀正确答案：B第四章1、Cardiopathy means ___.A:heart failureB:heart diseaseC:brain diseaseD:skeletal disease正确答案：B2、The suffix “-centesis” in the word arthrocentesis means __. A:destruction ofB:puncture ofC:pertaining toD:originating in正确答案：B3、The suffix “-malacia” in the word chondromalacia means __. A:flowingB:hardeningC:softeningD:producing正确答案：C4、Downward displacement of the stomach is termed ___.A:gastritisB:gastroptosisC:mammoplasiaD:sternoschisis正确答案：B5、Which of the following physicians specializes in treating patients with diseases of the liver?A:hepatologistB:hematologistC:nephrologistD:rheumatologist正确答案：hepatologist6、___ means surgical repair of an organ.A:phag/o-B:-plasmC:-plastyD:-pathy正确答案：C7、A record of electric wave occurring in the brain is called ___.A:electroencephogramB:electromyogramC:electrocardiogramD:electroencephalogram正确答案：electroencephalogram8、Which of the following suffix es does not mean “pertaining to”?A:-icB:-alC:-arD:-our正确答案：-our9、The suffix “-megaly” in the word hepatosplenomegaly means ___.A:swellingB:softeningC:hardeningD:enlargement正确答案：enlargement10、The instrument for viewing is -scope whereas the instrument for cutting is______. A:-tomeB:-scopyC:-meterD:-graph正确答案：-tome第五章1、Surgical reconstruction or cosmetic alteration of the nose is termed______.A:rhinopexyB:rhinoplastyC:rhinotomyD:rhinoscopy正确答案：rhinoplasty2、A faster than normal respiratory rate of breathing is termed _.A:dyspneaB:apneaC:tachypneaD:bradypnea正确答案：tachypnea3、Hemoptysis is a term describing_______.A:a bloody noseB:bleeding from the gumsC:a clot in a pulmonary arteryD:coughing up blood from the lungs正确答案：coughing up blood from the lungs4、A bluish discoloration of the skin and mucous membranes is termed_______, which is a sign that oxygen in the blood is dangerously diminished (as in carbon monoxide poisoning).A:cyanosisB:cyanideC:albinismD:erythematosus正确答案：cyanosis5、_is an instrument for measuring the vital capacity of the lungs.A:pneumographB:spirographC:spirometerD:spirometry正确答案：spirometer6、A special procedure to examine the blood vessels of the lungs by X-ray is called A:a lung scanB:bronchoscopyC:thoracocentesisD:pulmonary angiography正确答案：pulmonary angiography第六章1、aortaA:capn/oB:aort/oC:spir/oD:ventricul/o正确答案：aort/o 2、heartA:coron/oB:ton/oC:chrom/oD:cardi/o正确答案：cardi/o 3、arteryA:arteri/oB:cephal/oC:cerebr/oD:encephal/o正确答案：4、veinA:phon/oB:dent/oC:astr/oD:ven/o正确答案：5、bloodA:plasm/oB:ser/oC:hem/oD:femor/o正确答案：angi/o 6、atriumA:atri/oB:melan/oC:chrom/oD:erythr/o正确答案：7、blood vessle A:duoden/oB:col/oC:angi/oD:jejun/o正确答案：angi/o 8、valveA:hepat/oB:valvul/oC:ren/oD:adren/o正确答案：valvul/o9、inflammationA:-itisB:-aseC:-lysisD:-ide正确答案：-itis10、incisionA:-tomyB:-stomyC:-ectomyD:-scopy正确答案：-tomy第七章1、medullar oblongataA:脑干B:海马C:延髓D:杏仁体正确答案：延髓2、somatic nervous system A:肠神经系统B:躯体神经系统C:自主神经系统D:网状神经系统正确答案：躯体神经系统3、脑皮层运动区A:cerebellar motor cortexB:trabecular motor cortexC:limbic motor cortexD:cerebral motor cortex正确答案：cerebral motor cortex 4、顶叶A:frontal lobeB:parietal lobeC:occipital lobeD:temporal lobe正确答案：occipital lobe5、基底神经节A:the basal gangliaB:the basal microgliaC:the oligodendrocyteD:the glial cells正确答案：the basal ganglia第八章1、An enzyme that digests starch into disaccharides and it is usually secreted by salivary glands and by the pancreas is ___A:amylaseB:pepsinC:lipaseD:trypsin正确答案：amylase2、The term gastrectomy refers to the excision or removal of _A:liverB:gallbladderC:stomachD:pancreas正确答案：stomach3、Which of the following word parts indicates the first part of a small intestine?A:col/oB:ile/oC:duoden/oD:jejun/o正确答案：duoden/o4、gastroenteritis is an inflammation on ___(parts of the body).A:stomach and liverB:entrance of the stomachC:pylorus and cardiaD:bottom stomach & small intestines正确答案：bottom stomach & small intestines5、Gastrojejunostomy describes a procedure that____A:surgically creates an artificial opening between the stomach and jejunumB:surgically creates an opening between the intestines and the abdominal wallC:surgically creates a passageway between the gallbladder duct to the intestineD:surgically removes gallstones through an opening in the abdomen.正确答案：surgically creates an artificial opening between the stomach and jejunum第九章1、_ is a hormone secreted by the thymus that can stimulate the immunological activity of lymphoid tissue.A:ThymineB:SomatotropinC:AdrenalinD:Thymosin正确答案：Thymosin2、The term “phagocytosis” means_______A:the process by which a cell, such as a white blood cell, is producedB:a cell that can ingest microorganisms, other cells, and foreign particlesC:the process by which a cell, such as a white blood cell, ingests microorganisms, other cells, and foreign particlesD:the death of a phagocyte正确答案：the process by which a cell, such as a white blood cell, ingests microorganisms,other cells, and foreign particles3、The movement of a microorganism or cell in response to a chemical stimulus is termed _.A:chemokineB:chemotaxisC:cytokineD:chemotherapy正确答案：chemotaxis4、The word part “-phylaxis” denotes______.A:protectionB:allergyC:preventionD:diagnosis正确答案：protection5、Grafts used for transplantation from a different species are ____.A:autograftsB:isograftsC:homograftsD:heterografts正确答案：heterografts第十章1、Inflammation of joints is termed as __.A:arthrodesisB:arthroplastyC:arthriticD:arthritis正确答案：arthritis2、A sheet or band of fibrous tissue which lies deep to the skin or forms an investment for muscles and various organs of the body is __.A:faciesB:faceC:facetD:fascia正确答案：fascia3、Pick out the one which aids cartilaginous activity and thus promote the growth of new cartilage.A:chondroblastB:synchondrosisC:perichondriumD:chondroma正确答案：chondroblast4、The medical apparatus used to invest patella is __.A:patellarB:patellapexyC:patellectomyD:patellometer正确答案：patellometer5、Which of the following combining forms means “bone”? A:ren/o-B:myel/o-C:oste/o-D:gastr/o-正确答案：oste/o-6、The stem “chondro- ”in the word chondritis me ans________ . A:tendonB:cartilageC:jointD:rib正确答案：cartilage7、Radius means ___ lower arm bone.A:superiorB:lateralC:medialD:handful正确答案：superior8、The flesh is expressed in the combining form of ___.A:top/o-B:ten/o-C:thym/o-D:sarc/o-正确答案：sarc/o-9、Which of the following has the meaning of the joint?A:athr/o-B:arthr/o-C:articul/o-D:both B and C正确答案：both B and C10、Abnormal condition of being hard is referred to as __.A:sclerosisB:sclirosisC:barometerD:thermometer正确答案：sclerosis第十一章1、The term for scanty or less than normal urine formation is___ A:anuriaB:oliguriaC:polyuriaD:nocturia正确答案：oliguria2、Presence of a kidney “stone” is termed___.A:nephrosisB:hydronephrosisC:nephrolithiasisD:pyelonephrosis正确答案：nephrolithiasis3、The inflammation of the final pathway for urine in both sexes is termed___.A:nephritisB:urethritisC:nephrosisD:pyelonephrosis正确答案：urethritis4、A specialist in diseases of the lower urinary tract, bladder and urethra, is called a __.A:nephrologistB:urologistC:proctologistD:serologist正确答案：urologist5、A procedure that allows a physician to look into the bladder and examine its interior is termed a_.A:cystoscopyB:cystogramC:voiding cystourethrogram D:intravenous pyelogram 正确答案：cystoscopy第十二章1、glandA:chondr/oB:aden/oC:oste/oD:femor/o正确答案：aden/o2、secretionA:crin/oB:col/oC:ile/oD:prot/o正确答案：crin/o3、pituitary glandA:hemat/oB:encephal/o-C:hypophys/oD:choledoch/o正确答案：hypophys/o4、adrenal gland A:adren/oB:pneum/oC:immun/oD:ser/o正确答案：adren/o 5、thyroid gland A:pharyng/oB:thyr/oC:laryng/oD:bronch/o正确答案：thyr/o 6、pinealA:pineal/oB:lingu/oC:glomerul/oD:odont/o正确答案：pineal/o 7、thymus gland A:duoden/oB:thym/oC:nephr/o正确答案：thym/o8、thalamusA:capn/oB:pulm/oC:thalam/oD:scoli/o正确答案：thalam/o9、pancreasA:hepat/oB:gastr/oC:enter/oD:pancreat/o正确答案：pancreat/o10、glucoseA:muc/oB:gen/oC:gluc/oD:lip/o正确答案：gluc/o第十三章1、The word root “odonto-” denotes______.B:tongueC:mouthD:cheek正确答案：teeth2、The word root “cise-” denotes______. A:to exciseB:to cutC:to killD:to measure正确答案：to cut3、The word root “alveolo-” denotes______. A:tooth socketB:gumC:root canalD:lips正确答案：tooth socket4、Canines are also termed __.A:premolarsB:cuspidsC:bicuspidsD:baby teeth正确答案：cuspids5、A tender tooth is __.A:softB:aching all the timeC:a prosthesisD:sensitive and painful when touched正确答案：sensitive and painful when touched第十四章1、an instrument used to determine refractive errors of the eye iscalled _ A:otoscopeB:retinoscopeC:salpingoscopeD:osteoscope正确答案：retinoscope2、Inflammation of the middle eardrum is __A:tympanitisB:sclerosisC:conjunctivitisD:otitis正确答案：tympanitis3、Convulsion of the eyelid is called_____A:cochleoconstrictionB:corneodesisC:dacryotoniaD:blepharospasm正确答案：blepharospasm4、Surgical incision of the eardrum is called____ A:myringotomyB:lacrimationC:stapedectomyD:ophthalmostomy正确答案：myringotomy5、paralysis of the iris is called_____A:iridectomyB:iridodialysisC:iridoplegiaD:iridocyclitis。
英语材基试卷
英文原版教材班“材料科学基础”考试试题试卷一Examination problems of the course of “fundament of materials science”姓名：班级：记分：1. Glossary (2 points for each)1) crystal structure:2) basis (or motif):3) packing fractor:4) slip system:5) critical size:6) homogeneous nucleation:7) coherent precipitate:8) precipitation hardening:9) diffusion coefficient:10) uphill diffusion:2. Determine the indices for the planes in the cubic unit cell shown in Figure 1. (5 points)Fig. 13. Determine the crystal structure for the following: (a) a metal with a0 =4.9489 Å, r = 1.75 Å and one atom per lattice point; (b) a metal with a0 = 0.42906 nm, r = 0.1858 nm and one atom per lattice point. (10 points)4-1. What is the characteristic of brinell hardness test, rockwell hardness test and Vickers hardness test? What are the effects of strain rate and temperature on the mechanical properties of metallic materials? (15 points)4-2. What are the effects of cold-work on metallic materials? How to eliminate those effects? And what is micro-mechanism for the eliminating cold-work effects? (15 points)5-1. Based on the Pb-Sn-Bi ternary diagram as shown in Fig. 2, try to(1)Show the vertical section of 40wt.%Sn; (4 points)(2) Describe the solidification process of the alloy 2# with very low cooling speed (includingphase and microstructure changes); (4 points)(3)Plot the isothermal section at 150o C. (7 points)Fig. 25-2. A 1mm sheet of FCC iron is used to contain N2in a heated exchanger at 1200o C. The concentration of N at one surface is 0.04 atomic percent and the concentration at the second surface is 0.005 atomic percent. At 1000 o C, if same N concentration is demanded at the second surface and the flux of N becomes to half of that at 1200o C, then what is the thickness of sheet?(15 points)6-1. Supposed that a certain liquid metal is undercooled until homogeneous nucleation occurs. (15 points)(1)How to calculate the critical radius of the nucleus required? Please give the deductionprocess.(2)For the Metal Ni, the Freezing Temperature is 1453︒C, the Latent Heat of Fusion is 2756J/cm3, and the Solid-liquid Interfacial Energy is 255⨯10-7 J/cm2. Please calculate the critical radius at 1353︒C. (Assume that the liquid Ni is not solidified.)6-2. Fig.3 is a portion of the Mg-Al phase diagram. (15 points)(1)If the solidification is too rapid, please describe the solidification process of Mg-10wt%Alalloy.(2)Please describe the equilibrium solidification process of Mg-20wt%Al alloy, and calculate theamount of each phase at 300︒C.Fig. 37-1. Figure 4 shows us the Al-Cu binary diagram and some microstructures found in a cooling process for an Al-4%Cu alloy. Please answer following questions according to this figure. (20 points)Fig. 4(1)What are precipitate, matrix and microconstituent? Please point them out in the in the figure and explain.(2)Why is need-like precipitate not good for dispersion strengthening? The typical microstructure shown in the figure is good or not? why?(3)Please tell us how to obtain the ideal microstructure shown in this figure.(4)Can dispersion strengthened materials be used at high temperature? Please give the reasons (comparing with cold working strengthening)7-2. Please answer following questions according to the time-temperature-transformation (TTT) diagram as shown in Fig. 5. (20 points)(1)What steel is this TTT diagram for? And what means P, B, and M in the figure? (2)Why dose the TTT diagram exhibi ts a ‘C’ shape?(3)Point out what microconstituent will be obtained after austenite is cooled according to the curves I, II, III and IV .(4)What is microstructural difference between the curve I and the curve II? (5)How to obtain the steel with the structure of(a) P+B(b) P+M+A (residual) (c) P+B+M+A (residual)(d) Full tempered martensiteIf you can, please draw the relative cooling curve or the flow chart of heat treatment.Fig. 5III III IV英文原版教材班“材料科学基础”考试试题答案Solution s of the course of “fundament of materials science”1. Glossary (2 points for each)1) The arrangement of the atoms in a material into a repeatable lattice.2) A group of atoms associated with a lattice.3) The fraction of space in a unit cell occupied by atoms.4) The combination of the slip plane and the slip direction.5) The minimum size that must be formed by atoms clustering together in the liquid before thesolid particle is stable and begins to grow.6) Formation of a critically sized solid from the liquid by the clustering together of a largenumber of atoms at a high undercooling (without an external interface).7) A precipitate whose crystal structure and atomic arrangement have a continuousrelationship with matrix from which precipitate is formed.8) A strengthening mechanism that relies on a sequence of solid state phase transformationsin a dispersion of ultrafine precipitates of a 2nd phase. This is same as age hardening. It is a form of dispersion strengthening.9) A temperature-dependent coefficient related to the rate at which atom, ion, or otherspecies diffusion. The DC depends on temperature, the composition and microstructure of the host material and also concentration of the diffusion species.10) A diffusion process in which species move from regions of lower concentration to that ofhigher concentration.2. Solution: A(-364), B(-340), C(346).3. Solution: (a)fcc; (b) bcc.4-1. What is the characteristic of brinell hardness test, rockwell hardness test and Vickers hardness test? What are the effects of strain rate and temperature on the mechanical properties of metallic materials? (15 points)4-2. What are the effects of cold-work on metallic materials? How to eliminate those effects? And what is micro-mechanism for the eliminating cold-work effects? (15 points)5-1. Based on the Pb-Sn-Bi ternary diagram as shown in Fig. 2, try to(1)Show the vertical section of 40wt.%Sn; (5 points)(2) Describe the solidification process of the alloy 2# with very low cooling speed (includingphase and microstructure changes); (5 points)(3)Plot the isothermal section at 150o C. (5 points)Fig. 25-2. A 1mm sheet of FCC iron is used to contain N2in a heated exchanger at 1200o C. The concentration of N at one surface is 0.04 atomic percent and the concentration at the second surface is 0.005 atomic percent. At 1000 o C, if same N concentration is demanded at the second surface and the flux of N becomes to half of that at 1200o C, then what is the thickness of sheet?(15 points)6-1. Supposed that a certain liquid metal is undercooled until homogeneous nucleation occurs. (15 points)(3)How to calculate the critical radius of the nucleus required? Please give the deductionprocess.(4)For the Metal Ni, the Freezing Temperature is 1453︒C, the Latent Heat of Fusion is 2756J/cm3, and the Solid-liquid Interfacial Energy is 255⨯10-7 J/cm2. Please calculate the critical radius at 1353︒C. (Assume that the liquid Ni is not solidified.)6-2. Fig.3 is a portion of the Mg-Al phase diagram. (15 points)(3)If the solidification is too rapid, please describe the solidification process of Mg-10wt%Alalloy.(4)Please describe the equilibrium solidification process of Mg-20wt%Al alloy, and calculate theamount of each phase at 300︒C.Fig. 37-1. Figure 4 shows us the Al-Cu binary diagram and some microstructures found in a cooling process for an Al-4%Cu alloy. Please answer following questions according to this figure. (20 points)Fig. 4(1)What are precipitate, matrix and microconstituent? Please point them out in the in the figure and explain.(2)Why is need-like precipitate not good for dispersion strengthening? The typical microstructure shown in the figure is good or not? why?(3)Please tell us how to obtain the ideal microstructure shown in this figure.(4)Can dispersion strengthened materials be used at high temperature? Please give the reasons (comparing with cold working strengthening)7-2. Please answer following questions according to the time-temperature-transformation (TTT) diagram as shown in Fig. 5. (20 points)(1)What steel is this TTT diagram for? And what means P, B, and M in the figure? (2)Why dose the TTT diagram exhibits a ‘C’ shape?(3)Point out what microconstituent will be obtained after austenite is cooled according to the curves I, II, III and IV .(4)What is microstructural difference between the curve I and the curve II? (5)How to obtain the steel with the structure of(a) P+B(b) P+M+A (residual) (c) P+B+M+A (residual)(d) Full tempered martensiteIf you can, please draw the relative cooling curve or the flow chart of heat treatment.Fig. 5III III IV英文原版教材班“材料科学基础”考试试题试卷二Examination problems of the course of “fundament of materials science”姓名：班级：记分：1. You would like to be able to physically separate different materials in a scrap recycling plant. Describe some possible methods that might be used to separate materials such as polymers, aluminum alloys, and steels from one another. (5 points)2. Plot the melting temperature of the elements in the 1A column of the periodic table versus atomic number (i.e., plot melting temperatures of Li through Cs). Discuss this relationship, based on atomic bonding and binding energy. (10 points)3.Above 882℃, titanium has a BCC crystal structure, with a = 0.332 nm. Below this temperature, titanium has a HCP structure, with a = 0.2978 nm and c = 0.4735 nm. Determine the percent volume change when BCC titanium transforms to HCP titanium. Is this a contraction or expansion? (10 points)4. The density of BCC iron is 7.882 g/cm3and the lattice parameter is 0.2866 nm whenhydrogen atoms are introduced at interstitial positions. Calculate (a) the atomic fraction of hydrogen atoms and (b) the number of unit cells required on average to contain one hydrogen atom. (15 points)5. A carburizing process is carried out on a 0.10% C steel by introducing 1.0% C at the surface at 980℃, where the iron is FCC. Calculate the carbon content at 0.01 cm, 0.05 cm, and 0.10 cm beneath the surface after 1 h. (15 points)6. The following data were collected from a standard 0.505-in.-diameter test specimen of acopper alloy (initial length (t o) = 2.0 in.):Load Gage Length Stress Strain(lb) (in.) (psi) (in/in.)0 2.00000 0 0.03,000 2.00167 15,000 0.0008356,000 2.00333 30,000 0.0016657,500 2.00417 37,500 0.0020859,000 2.0090 45,000 0.004510,500 2.040 52,500 0.0212,000 2.26 60,000 0.1312,400 2.50 (max load) 62,000 0.2511,400 3.02 (fracture) 57,000 0.51After fracture, the gage length is 3.014 in. and the diameter is 0.374 in. Plot the data and calculate (a) the 0.2% offset yield strength, (b) the tensile strength, (c) the modulus of elasticity, (d) the %Elongation, (e) the %Reduction in area, (f) the engineering stress at fracture, (g) the true stress at fracture, and (h) the modulus of resilience. (15 points)7. A 1.5-em-diameter metal bar with a 3-cm gage length is subjected to a tensile test. Thefollowing measurements are made.Change in Force (N) Gage length (cm) Diameter (cm)16,240 0.6642 1.202819,066 1.4754 1.088419,273 2.4663 0.9848Determine the strain hardening coefficient for the metal. Is the metal most likely to be FCC, BCC, or HCP? Explain.(15 points)8. Based on Hume-Rothery’s conditions, which of the following systems would be expected todisplay unlimited solid solubility? Explain. (15 points)(a) Au-Ag (b) Al-Cu (c) Al-Au (d)U-W(e) Mo-Ta (f) Nb-W (g) Mg-Zn (h) Mg-Cd英文原版教材班“材料科学基础”考试试题答案Solutions of the course of “fundament of materials science”1.Steels can be magnetically separated from the other materials; steel (or carbon-containing iron alloys) are ferromagnetic and will be attracted by magnets. Density differences could be used—polymers have a density near that of water; the specific gravity of aluminum alloys is around2.7;that of steels is between 7.5 and 8. Electrical conductivity measurements could be used—polymers are insulators, aluminum has a particularly high electrical conductivity.(5 points)2.T (o C)L i–180.7N a– 97.8K – 63.2R b– 38.9As the atomic number increases, the melting temperature decreases, (10 points)3. We can find the volume of each unit cell. Two atoms are present in both BCC and HCP titanium unit cells, so the volumes of the unit cells can be directly compared.V BCC = (0.332 nm)3 = 0.03659 nm3V HCP= (0.2978 nm)2(0.4735 nm)cos30 = 0.03637 nm3△V=x 100 =×100= -0.6%Therefore titanium contracts 0.6% during cooling. (10 points)4. (a) 7.882 g/cm3 =x = 0.0081 H atoms/cellThe total atoms per cell include 2 Fe atoms and 0.0081 H atoms. Thus:(10 points)(b) Since there is 0.0081 H/cell, then the number of cells containing H atoms is:cells = 1/0.0081 = 123.5 or 1 H in 123.5 cells (5 points)5. D = 0.23 exp[-32,900/(1.987)(1253)] = 42 × 10-8 cm2/sC x= 0.87% CC x= 0.43% CC x= 0.18% C(15 points)6. σ=FI (π/4)(0.505)2 = F/0.2ε = (l-2)/2(a) 0.2% offset yield strength = 45,000 psi(b)tensile strength = 62,000 psi(c) E = (30,000 - 0) / (0.001665 - 0) = 18 x 106 psi(d)%Elongation =(e) %Reduction in area =(f) engineering stress at fracture = 57,000 psi(g)true stress at fracture = 11,400 lb / (TC/4)(0.374)2= 103,770 psi (h) From the graph, yielding begins at about 37,500 psi. Thus:(15 points)7.Force(lb) Gage length(in.) Diameter(in.) True stress(psi) True strain(psi)16,240 3.6642 12.028 143 0.20019,066 4.4754 10.884 205 0.40019,273 5.4663 9.848 249 0.600σt=Kεt2or ln143=ln K + n ln0.2ln 249 = ln K + nln 0.6n=0.51A strain hardening coefficient of 0.51 is typical of FCC metals.(15 points)8.The Au–Ag, Mo–Ta, and Mg–Cd systems have the required radius ratio, the same crystal structures, and the same valences. Each of these might be expected to display complete solid solubility. [The Au –Ag and Mo –T a d o have isomorphous phase diagrams. In addition, the Mg–Cd alloys all solidify like isomorphous alloys; however a number of solid state phase transformations complicate the diagram.] (15 points)英文原版教材班“材料科学基础”考试试题试卷三Examination problems of the course of “fundament of materials science”姓名：班级：记分：1. You would like to be able to identify different materials without resorting to chemical analysis or lengthy testing procedures. Describe some possible testing and sorting techniques you might be able to use based on the physical properties of materials. (5 points)2. Plot the melting temperatures of elements in the 4A to 8-10 columns of the periodic table versus atomic number (i.e., plot melting temperatures of Ti through Ni, Zr through Pd, and Hf through Pt). Discuss these relationships, based on atomic bonding and binding energy, (a) as the atomic number increases in each row of the periodic table and (b) as the atomic number increases in each column of the periodic table. (10 points)3. Beryllium has a hexagonal crystal structure, with a o= 0.22858 nm and c o= 0.35842 nm. The atomic radius is 0.1143 nm, the density is 1.848 g/cm3, and the atomic weight is 9.01 g/mol. Determine (a) the number of atoms in each unit cell and (b) the packing factor in the unit cell.(10 points)4. Suppose we introduce one carbon atom for every 100 iron atoms in an interstitial position in BCC iron, giving a lattice parameter of 0.2867 nm. For the Fe-C alloy, find (a) the density and (b) the packing factor. (15 points)5. Iron containing 0.05% C is heated to 912oC in an atmosphere that produces 1.20% C at the surface and is held for 24 h. Calculate the carbon content at 0.05 cm beneath the surface if (a) the iron is BCC and (b) the iron is FCC. Explain the difference. (15 points)6. The following data were collected from a 0.4-in. diameter test specimen of poly vinyl chloride(l0 = 2.0 in):Load(lb) Gage Length(in.) Stress(psi) Strain(in/in.)0 2.00000 0 0.0300 2.00746 2,387 0.00373600 2.01496 4,773 0.00748900 2.02374 7,160 0.011871200 2.032 9,547 0.0161500 2.046 11,933 0.0231660 2.070 (max load) 13,206 0.0351600 2.094 12,729 0.0471420 2.12 (fracture) 11,297 0.06After fracture, the gage length is 2.09 in. and the diameter is 0.393 in. Plot the data and calculate (a) the 0.2% offset yield strength, (b) the tensile strength, (c) the modulus of elasticity, (d) the %Elongation, (e) the %Reduction in area, (f) the engineering stress at fracture, (g) the true stress at fracture, and (h) the modulus of resilience. (15 points)7. A titanium alloy contains a very fine dispersion of tiny Er203 particles. What will be the effectof these particles on the grain growth temperature and the size of the grains at any particular annealing temperature? Explain. (15 points)8. Suppose 1 at% of the following elements is added to copper (forming a separate alloy witheach element) without exceeding the solubility limit. Which one would be expected to give the higher strength alloy? Is any of the alloying elements expected to have unlimited solid solubility in copper?(a) Au (b) Mn (c) Sr (d) Si (e) Co (15 points)英文原版教材班“材料科学基础”考试试题答案Solutions of the course of “fundament of materials science”1.Steels can be magnetically separated from the other materials; steel (or carbon-containing iron alloys) are ferromagnetic and will be attracted by magnets. Density differences could be used—polymers have a density near that of water; the specific gravity of aluminum alloys is around2.7;that of steels is between 7.5 and 8. Electrical conductivity measurements could be used—polymers are insulators, aluminum has a particularly high electrical conductivity.(5 points)2. Ti –1668 Zr – 1852 Hf – 2227V –1900 Nb –2468 Ta – 2996Cr –1875 Mo–2610 W–3410Mn–1244 Tc –2200 Re–3180Fe –1538 Ru –2310 Os–2700Co –1495 Rh –1963 Ir –2447Ni –1453 Pd –1552 Pt –1769For each row, the melting temperature is highest when the outer “d” energy level is partly full. In Cr, there are 5 electrons in the 3d shell; in Mo, there are 5 electrons in the 4d shell; in W there are 4 electrons in the 5d shell. In each column, the melting temperature increases as the atomic number increases—the atom cores contain a larger number of tightly held electrons, making the metals more stable. (10 points)3.V= (0.22858 nm)2(0.35842 nm)cos 30 = 0.01622 nm3 = 16.22 × 10-24 cm3(a)From the density equation:1.848 g/cm3 =x = 2 atoms/cell(b)The packing factor (PF) is:PF == 0.77 (10 points)4. There is one carbon atom per 100 iron atoms, or 1 C/50 unit cells, or 1/50 C per unit cell:(a)(b)(15 points)5. t= (24 h)(3600 s/h) = 86,400 sD BCC = 0.011 exp[-20,900/(1.9871185)] = 1.54 × 10-6 cm2/sD FCC = 0.23 exp[-32,900/(1.987)(1185)] = 1.97×10-7 cm2/sBCC: = erf[0.0685] = 0.077C x= 1.11% CFCC: = erf[0.192] = 0.2139C x = 0.95% CFaster diffusion occurs in the looser packed BCC structure, leading to the higher carbon content at point “x”. (15 points)6. σ=F /(π/4)(0.4)2 = F/0.1257ε = (l-2)/2(a)0.2% offset yield strength = 11,600 psi(b) tensile strength = 12,729 psi(c) E= (7160 - 0) / (0.01187 - 0) = 603,000 psi(d)%Elongation =(e) %Reduction in area =(f) engineering stress at fracture = 11,297 psi(g)true stress at fracture = 1420 lb / (TC/4)(0.393)2= 11,706 psi (h) From the figure, yielding begins near 9550 psi. Thus:(15 points)7. These particles, by helping pin die grain boundaries, will increase the grain growth temperature and decrease the grain size. (15 points)8.The Cu-Sr alloy would be expected to be strongest (largest size difference). The Cu-Au alloy satisfies Hume-Rothery ’s conditions and might be expected to display complete solid solubility—in fact it freezes like an isomorphous series of alloys, but a number of solid state transformations occur at lower temperatures.(15 points)英文原版教材班“材料科学基础”考试试题试卷四Examination problems of the course of “fundament of materials science”姓名：班级：记分：1.Aluminum has a density of2.7 g/cm3. Suppose you would like to produce a compositematerial based on aluminum having a density of 1.5 g/cm3. Design a material that would have this density. Would introducing beads of polyethylene, with a density of 0.95 g/cm3, into the aluminum be a likely possibility? Explain. (5 points)2. (a) Aluminum foil used for storing food weighs about 0.3 g per square inch. How many atomsof aluminum are contained in this sample of foil?(b) Using the densities and atomic weights given in Appendix A, calculate and compare thenumber of atoms per cubic centimeter in (i) lead and (ii) lithium. (10 points)3. The density of potassium, which has the BCC structure and one atom per lattice point, is0.855 g/cm3. The atomic weight of potassium is 39.09 g/mol. Calculate (a) the latticeparameter; and (b) the atomic radius of potassium. (10 points)4. The density of a sample of HCP beryllium is 1.844 g/cm3 and the lattice parameters are a0=0.22858 nm and c0= 0.35842 nm. Calculate (a) the fraction of the lattice points that containvacancies and (b) the total number of vacancies in a cubic centimeter. (15 points)5. A ceramic part made of MgO is sintered successfully at 1700℃in 90 minutes. To minimizethermal stresses during the process, we plan to reduce the temperature to 1500℃. Which will limit the rate at which sintering can be done: diffusion of magnesium ions or diffusion of oxygen ions? What time will be required at the lower temperature? (15 points)6. (a) A thermosetting polymer containing glass beads is required to deflect 0.5 mm when aforce of 500 N is applied. The polymer part is 2 cm wide, 0.5 cm thick, and 10 cm long. If the flexural modulus is 6.9 GPa, determine the minimum distance between the supports. Will the polymer fracture if its flexural strength is 85 MPa? Assume that no plastic deformation occurs.(b) The flexural modulus of alumina is 45 x 106 psi and its flexural strength is 46,000 psi. Abar of alumina 0.3 in. thick, 1.0 in. wide, and 10 in. long is placed on supports 7 in. apart.Determine the amount of deflection at the moment the bar breaks, assuming that no plastic deformation occurs. (15 points)7. Based on the following observations, construct a phase diagram. Element A melts at 850°Cand element B melts at 1200°C. Element B has a maximum solubility of 5% in element A, and element A has a maximum solubility of 15% in element B. The number of degrees of freedom from the phase rule is zero when the temperature is 725°C and there is 35% B present. At room temperature 1% B is soluble in A and 7% A is soluble in B. (15 points)8.Suppose that age hardening is possible in the Al-Mg system (see Figure 10-11). (a)Recommend an artificial age-hardening heat treatment for each of the following alloys, and(b) compare the amount of the precipitate that forms from your treatment of each alloy. (i)Al-4% Mg (ii) Al-6% Mg (iii) Al-12% Mg (c) Testing of the alloys after the heat treatment reveals that little strengthening occurs as a result of the heat treatment. Which of the requirements for age hardening is likely not satisfied? (15 points)英文原版教材班“材料科学基础”考试试题答案Solutions of the course of “fundament of materials science”1. In order to produce an aluminum-matrix composite material with a density of 1.5 g/cm 3, we wouldneed to select a material having a density considerably less than 1.5 g/cm 3. While polyethylene’s density would make it a possibility, the polyethylene has a very low melting point compared to aluminum; this would make it very difficult to introduce the polyethylene into a solid aluminum matrix —processes such as casting or powder metallurgy would destroy the polyethylene .Therefore polyethylene would NOT be a likely possibility.One approach, however, might be to introduce hollow glass beads .Although ceramic glasses have densities comparable to that of aluminum, a hollow bead will have a very low density. The glass also has a high melting temperature and could be introduced into liquid aluminum for processing as a casting. (5 points)2. (a) In a one square inch sample:number ==6.69 × 1021 atoms(b) (i) In lead:= 3.3 × 1022 atoms/cm 3(ii) In lithium:= 4.63 × 1022 atoms/cm 3 (10 points)3. (a) Using Equation 3-5:0.855 g/cm 3 =a o 3 = 1.5189 × 10-22 cm 3 or a o = 5.3355 × 10-8 cm(b) From the relationship between atomic radius and lattice parameter:r == 2.3103 × 10-8cm (10 points)4. V u.c.= (0.22858 nm)2(0.35842 nm)cos30 = 0.01622 nm 3= 1.622 x 10~23 cm 3 (a) From the density equation:x = 1.9984fraction =29984.12 = 0.0008(b) number == 0.986 x 1020 vacancies/cm 3 (15 points)5. Diffusion of oxygen is the slower of the two, due to the larger ionic radius of the oxygen.D 1700= 0.000043 exp[-82,100/(1.987)(1973)] = 3.455 × 10-14 cm 2/sD1500= 0.000043 exp[-82,100/(1.987)(1773)] = 3.255 × 10-15 cm2/st1500 = D1700 t1700/D1500== 955 min = 15.9 h (15 points)6. (a) Solution:The minimum distance L between the supports can be calculated from the flexural modulus.L3 = 4w/z3δ(flexural modulus)/3FL3 = (4)(20 mm)(5 mm)3(0.5 mm)(6.9 GPA)(1000 MPa/GPa) / 500 NL3 = 69,000 mm3 or L = 41 mmThe stress acting on the bar when a deflection of 0.5 mm is obtained isσ= WL/2wh2 = (3)(500 N)(41 mm) / (2)(20 mm)(5 mm)2 = 61.5 MPaThe applied stress is less than the flexural strength of 85 MPa; the polymer is not expected to fracture.(b) Solution:The force required to break the bar isF = 2w/z2(flexural strength)/3LF= (2)(1 in)(0.3 in)2(46,000 psi / (3)(7 in.) = 394 lbThe deflection just prior to fracture is8 = FZ3/4wh3(flexural modulus)8 = (394 lb)(7 in)3/(4)(l in)(0.3 in)3(45 x 106 psi) = 0.0278 in. (15 points)7.(15 points)8. (a) The heat treatments for each alloy might be:Al-4% Mg Al-6% Mg Al-12% MgT Eutectic451°C 451°C 451°CT Solvs210°C 280°C 390°CSolutionTreat at: 210-451°C 280-451°C 390-451°CQuench Quench QuenchAge at: <210°C <280°C <390°C(b) Answers will vary depending on aging temperature selected. If all threeare aged at 200°C, as an example, the tie line goes from about 3.8 to 35% Mg:A1-4% Mg: %β = (4− 3.82)/(35 − 3.82) X 100 = 0.6%Al-6% Mg: %β = (6 − 3.82)/(35 − 3.82) X 100 = 7.1%Al-12% Mg: %β = (12 −3.82)/(35− 3.82) X 100 = 26.8%(c) Most likely, a coherent precipitate is not formed; simple dispersionstrengthening, rather than age hardening, occurs. (15 points)英文原版教材班“材料科学基础”考试试题试卷五Examination problems of the course of “fundament of materials science”姓名：班级：记分：1. You would like to design an aircraft that can be flown by human power nonstop for adistance of 30 km. What types of material properties would you recommend? What materials might be appropriate? (5 points)2. Boron has a much lower coefficient of thermal expansion than aluminum, even though bothare in the 3B column of the periodic table. Explain, based on binding energy, atomic size, and the energy well, why this difference is expected. (10 points)3. Determine the ASTM grain size number if 20 grains/square inch are observed at amagnification of 400. (10 points)4. We currently can successfully perform a carburizing heat treatment at 1200o C in 1 h. In aneffort to reduce the cost of the brick lining in our furnace, we propose to reduce the carburizing temperature to 950℃. What time will be required to give us a similar carburizing treatment? (15 points)5.The data below were obtained from a series of Charpy impact tests performed on foursteels, each having a different manganese content. Plot the data and determine (a) the transition temperature (defined by the mean of the absorbed energies in theductile and brittle regions) and (b) the transition temperature (defined as the temperature that provides 50 J absorbed energy). Plot the transition temperature versus manganese content and discuss the effect of manganese on the toughness of steel. What would be the minimum manganese allowed in the steel if a part is to be used at 0°C?Test temperature°C Impact snergy (J)0.30% Mn 0.39% Mn 1.01% Mn 1.55% Mn-100 2 5 5 15-75 2 5 7 25-50 2 12 20 45-25 10 25 40 700 30 55 75 11025 60 100 110 13550 105 125 130 14075 130 135 135 140100 130 135 135 140(15 points)。
材料分析测试技术A卷
材料分析测试技术A卷一、选择题（每题1分，共15分）1、x射线衍射方法中，最常用的是（）a．劳厄法b.粉末多晶法c.转晶法2、未知x射线定性分析中存有三种索引，未知物质名称可以使用（）a.哈式无机相数值索引b.无机相字母索引c.芬克无机数值索引3、电子束与液态样品相互作用产生的物理信号中能用作测试1nm厚度表层成分分后析的信号是（）a.背散射电子b.俄歇电子c.特征x射线4、测定钢中的奥氏体含量，若采用定量x射线物相分析，常用的方法是（）a.外标法b.内标法c.轻易比较法d.k值法5、以下分析方法中分辨率最低的就是（）a.semb.temc.特征x射线6、表面形貌分析的手段包括（）a.semb.temc.wdsd.dsc7、当x射线将某物质原子的k层电子打出去后，l层电子回迁k层，多余能量将另一个l层电子拿下核外，这整个过程将产生（）a.光电子b.二次电子c.俄歇电子d.背散射电子8、透射电镜的两种主要功能（）a.表面形貌和晶体结构b.内部组织和晶体结构c.表面形貌和成分价键d.内部组织和成分价键9、已知x射线光管是铜靶，应选择的滤波片材料是（）a.cob.nic.fed.zn10、采用复型技术测得材料表面组织结构的式样为（）a.非晶体样品b.金属样品c.粉末样品d.陶瓷样品11、在电子探针分析方法中，把x射线谱仪固定在某一波长，使电子束在样品表面读取获得样品的形貌阴之木元素的成分原产像是，这种分析方法就是（）第1页/共6页a.点分析b.线分析c.面分析12、下列分析测试方法中，能够进行结构分析的测试方法是（）a.xrdb.temc.semd.a+b13、在x射线定量分析中，不需要做标准曲线的分析方法是（）a.外标法b.内标法c.k值法14、热分析技术无法测试的样品就是（）a.固体b.液体c.气体15、以下热分析技术中，（）就是对样品池及滴定法池分别冷却的测试方法a.dtab.dscc.tga二、填空题（每空1分，共20分）1、由x射线管升空出的x射线可以分成两种类型，即为和。
临床科研实验技能知到章节答案智慧树2023年中南大学
临床科研实验技能知到章节测试答案智慧树2023年最新中南大学第一章测试1.验证假设的最好方法是参考答案:实验2.形态学实验中基本的步骤脱水包埋是指对固定后的（）样品进行脱水包埋形成（）或冰冻组织块。
参考答案:组织，蜡块3.学生进入实验室开展实验的基本要求包括参考答案:全部都是4.实验数据记录、处理和分析的要求是参考答案:全部都是5.以下关于分享的实验心得说法不恰当的是参考答案:只需要通过实验课就可以学会做实验6.实验方法应具有科学性、先进性和可行性，并符合伦理。
参考答案:对7.实验室相关制度中包括每月对新进实验人员进行培训，讲解实验室注意事项和仪器操作使用。
参考答案:对8.未经实验室工作人员同意任何人不得把实验室内的仪器或物品拿出室外使用。
参考答案:对9.学生进入实验室开展实验，需要遵守实验室相关规章制度，如不慎发生事故，应及时向实验指导人员和相关负责人员报告并采取相应措施。
参考答案:对10.实验动物饲养过程中的生物危害不包括存在于动物粪便和尿液中的病原体。
参考答案:错第二章测试1.下列行为不属于科研不端行为的是参考答案:投稿时建议审稿员2.学术不端行为的危害包括参考答案:全部都是3.2014年12月，《科学美国人》(Scientific American) 称期刊《诊断病理学》上的一些论文像是工业规模批量生产出来的，经国家自然科学基金委员会调查，涉假论文被移交给纪检监察审计局处理。
此事件给我们的启示是参考答案:违背科研诚信和学术规范的行为终将受到严惩4.下列属于学术规范类型的是参考答案:全部都是5.黄禹锡因宣布攻克体细胞克隆胚胎干细胞的科学难题而轰动世界，却因被揭发伪造多项研究成果而身陷囹圄。
对于此事件的理解下列不正确的是参考答案:为了出名可以美化和篡改自己的研究成果6.科研活动是我们获得有关人类的新知识的过程。
参考答案:错7.遵纪守法，弘扬科学精神是学术规范的基本准则。
参考答案:对8.涉及以人类为对象的试验只能由具备科研资格的人员操作，如果有学生参加研究，应有相关教师负责安排和监管。
智慧树答案光谱分析(山东联盟)知到课后答案章节测试2022年
绪论1.下列方法属于光谱分析法的包括（）答案:核磁共振波谱法;X-射线荧光光谱法2.基于外层电子能级跃迁的光谱分析法有（）答案:紫外-可见分光光度法;分子荧光光谱法;原子吸收分光光度法3.基于振动-转动能级跃迁的光谱分析法有（）答案:拉曼光谱法;红外光谱法4.下列分析法属于发射光谱法的有（）答案:分子磷光光谱法;原子荧光光谱法;化学发光分析法5.下列分析法属于分子光谱法的有（）答案:分子荧光光谱法;拉曼光谱法;紫外-可见分光光度法;红外光谱法6.当前光谱分析方法的主要进展体现在（）答案:高通量分析;一机多能;联用技术;仪器自动化和微型化7.下列部件属于波长选择器的有（）答案:干涉仪;棱镜;光栅;滤光片8.紫外-可见分光光度法的合适光源是（）答案:氘灯9.原子吸收分光光度法的合适光源是（）答案:空心阴极灯10.红外吸收光谱法的合适光源是（）答案:改进型硅碳棒第一章1.下列关于紫外-可见吸收光谱描述不正确的有（）答案:末端吸收也可用于定性或定量分析2.能够快速获得全光光谱信息的紫外-可见分光光度计是（）答案:二极管阵列分光光度计3.紫外-可见吸收光谱属于（）答案:吸收光谱;电子能级跃迁光谱4.下列关于紫外-可见吸收光谱描述正确的有（）答案:由分子轨道上的电子能级跃迁产生;π→π*跃迁产生的吸收带通常具有较大的吸收强度，适用于定量分析。
5.下列关于吸光系数的描述正确的有（）答案:是物质的特性常数;与辐射的波长、溶剂及温度有关;是吸收光度法进行定性和定量分析的重要依据;其物理意义是吸光物质在单位浓度及单位光程长度时的吸光度6.导致Beer定律偏离的可能因素包括（）答案:溶液的酸碱性;非平行光;溶液的浓度;非单色光7.下列关于透光率测量误差的描述正确的有（）答案:是测量中的随机误差;可通过控制待测体系的吸光度值，有效减免透光率测量误差;来自仪器的噪声8.下列关于二极管阵列分光光度计描述正确的有（）答案:检测器为阵列二极管;光源发出的是连续光;单色器为全息光栅9.分光光度计校正和检定的主要指标通常有（）答案:杂散光检查;吸光度校正10.采用紫外-可见分光光度进行单组分样品定量测定时，分析条件的选择通常包括（）答案:测定浓度选择;参比溶液选择;测定波长选择;溶剂选择第二章1.与火焰原子化相比，GF-AAS测定的突出优点在于（）答案:绝对灵敏度高2.下列关于预混合型火焰原子化器的描述不正确的是（）答案:试样利用率高3.AAS的定量分析通常采用标准加入法以（）答案:抑制各种干扰4.下列关于Lorentz变宽效应的相关描述不正确的有（）答案:导致标准曲线弯向浓度轴5.AAS测定条件选择主要考虑（）答案:测量浓度范围;原子化条件;灯电流;分析线6.与紫外-可见分光光度计相比，原子吸收分光光度计在仪器结构上的主要区别包括（）答案:单色器置于吸收系统与检测器之间;光源采用发射待测元素特征谱线的空心阴极灯;吸收在原子化器中进行7.原子吸收线的峰值吸收值与原子浓度呈线性关系的前提条件包括（）答案:锐线光源产生的发射光谱线的半宽度远小于吸收轮廓线;实验温度保持稳定;待测基态原子浓度低且变化不大;原子共振发射线与吸收轮廓线的峰值频率相同8.影响原子吸收线形状的因素主要有（）答案:压力;激发态原子寿命;温度;浓度9.描述原子光谱项的量子数包括（）答案:主量子数;角量子数;磁量子数;自旋量子数10.原子吸收光谱属于（）答案:电子光谱;线状光谱第三章1.基于分子荧光寿命差异进行混合物分别测定的分析方法是（）答案:时间分辨荧光分析法2.在分子荧光测定时可能产生干扰的主要是（）答案:拉曼光3.发生在S1*最低振动能级向基态转换过程中的有（）答案:荧光发射;外部能量转换4.溶液体系中的分子荧光光谱通常具有以下特征（）答案:斯托克斯位移;形状与激发波长无关;与激发光谱呈镜像对称5.影响溶液中分子荧光强度的外部因素主要有（）答案:共存荧光熄灭剂分子;散射光;溶剂;温度6.分子荧光分析法的灵敏度影响因素包括（）答案:待测物质的吸光系数;待测物质的荧光效率;光源强度7.与紫外-可见分光光度法相比，分子荧光分析法主要优势在于（）答案:专属性强;灵敏度高8.下列关于荧光分光光度计基本结构描述正确的是（）答案:光源发射波长范围在UV-Vis光区的连续光;有激发和发射两个单色器;检测器一般采用光电管9.荧光分光光度计的校正包括（）答案:灵敏度校正;波长校正;激发光谱校正;发射光谱校正10.进行荧光定量分析时，需要（）答案:选定合适的激发波长和发射波长;使用标准溶液校正仪器刻度;选定狭缝宽度;使用空白溶液校正仪器零点第四章1.分子中含H基团的倍频吸收出现在（）答案:近红外区2.红外吸收光谱常用（）表示答案:T-σ曲线3.引起中红外吸收光谱的是（）答案:分子轨道上的振动-转动能级跃迁4.下列关于红外吸收光谱的描述不正确的有（）答案:起源于电子能级跃迁，光谱简单5.FT-IR与色散型红外光谱仪最主要的区别在于（）答案:单色器6.下列关于红外光谱测定试样制备描述不正确的有（）答案:可制成水溶液7.下列关于红外吸收光谱的描述正确的有（）答案:包括特征区和指纹区;通常利用相关峰进行官能团指认;吸收峰的强度与跃迁概率有关8.影响红外吸收光谱基频峰数目的因素有（）答案:红外非活性振动;仪器灵敏度低;简并;仪器分辨率低9.FT-IR所采用的检测器有（）答案:光电导型MCT检测器;热电型DTGS检测器10.红外光谱仪的最主要性能指标为（）答案:分辨率;波数准确度与重复性第五章1.核磁共振光谱属于（）答案:分子光谱2.下列最适于进行有机化学结构分析的光谱方法是（）答案:核磁共振光谱3.描述核磁共振光谱图中出峰位置的是（）答案:化学位移4.自旋量子数为1/2的核，其在外磁场中的核磁矩取向有（）答案:2种5.下列关于PFT-NMR仪器描述不正确的是（）答案:采用扫场方式进行测定6.下列组内所有核都具有自旋现象的是（）答案:14N,15N;1H,2H,3H7.影响核磁共振光谱分析灵敏度的因素有（）答案:温度;自旋核的磁旋比;磁场强度;自旋核的自然丰度8.影响自旋核进动频率的因素有（）答案:自旋核磁旋比;外磁场强度9.宏观物质体系产生持续可观测的核磁共振现象必须有（）答案:有效的弛豫过程;磁性原子;频率为兆赫数量级的电磁辐射;外磁场10.核磁共振氢谱提供的谱图信息包括有（）答案:各类质子的相对数量;质子类型;质子周围的化学环境。
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CellTransplantation,Vol.12,pp.33–41,20030963-6897/03$20.00+.00PrintedintheUSA.Allrightsreserved.Copyright2003CognizantComm.Corp.www.cognizantcommunication.com
ArealDensityMeasurementIsaConvenientMethodfortheDeterminationofPorcineIsletEquivalentsWithoutCountingandSizingIndividualIslets
N.Lembert,*†J.Wesche,*P.Petersen,†M.Doser,‡H.D.Becker,†andH.P.T.Ammon**DepartmentofPharmacology,AufderMorgenstelle8,UniversityofTu¨bingen,72076Tu¨bingen,Germany†DepartmentofGeneralSurgery,Hoppe-SeylerStraße3,UniversityofTu¨bingen,72076Tu¨bingen,Germany‡InstituteofTextileandProcessEngineering(ITV),Stuttgart-Denkendorf,73770Denkendorf,Germany
Thedeterminationofisletmassisimportantforthenormalizationofisletexperimentsinthelaboratoryandfortheprecisedosingofisletsfortransplantation.Thecommonmicroscopicalanalysisisbasedonindividualisletsizing,calculationofthefrequencydistribution,andconversionintoisletequivalents(IEQ),whichisthevolumeofasphericalisletwithadiameterof150µm.However,isletsareofirregularform,whichmakesthisdeterminationuserdependent,andtheanalysisisirreproducibleoncetheoriginalsampleisdiscarded.Thisroutinetechniqueofisletquantificationwascomparedwiththeanalysisofarealdensitymeasurements.ItwasassumedthattheentireareaoccupiedbyisletscanbeexpressedinIEQwithoutsizingandcountingindividualislets.Porcineisletswereisolatedbycontinuousdigestion/filtrationandpurifiedbygradientcentrifugation.Purifiedisletswerestainedwithdithizoneandwererepeatedlypicturedunderthemicroscopewithrandomareaselection.Atotalof51pictureswastakenfrom11differentpurificationsandstainedisletsweredetectedbydigitalimageanalysis.Thecorrelationcoefficient(r)betweenbothanalyseswas0.977withanunderestimationofisletyieldbyarealdensitydetection(slope:0.75±0.03).Arealdensityanalysisperpicturetookabout1min,whichisabout10timesfasterthanthetraditionalmethodwithoutincreasingthemethoderror(CV2.1%vs.2.7%).Insummary,arealdensitymeasurementsallowarapidandreproducibleestimationofIEQwithoutcountingindividualislets.Itcanbeperformedinasinglestepanalysiswithoutcomputerprogrammingandisvaluableforonlinedeterminationsofisletyieldprecedingtransplantation.
Keywords:Porcineislets;Isletquantification;Islettransplantation;Areadensity;Bioartificialpancreas
INTRODUCTIONingtoaninternationalguideline(8).Theyaregroupedin50-µm-diameterclassesneglectingisletssmallerthanOneapproachtotreattypeIdiabetesistransplanta-50µm.Althoughisletsareofirregularformwithonly
tionofisletsofLangerhans.AccordingtotheEdmontonaminorityofroundislets(15),theisletvolumeineach
protocolforclinicalislettransplantation,insufficientisletgroupiscalculatedassumingasphericalformandnor-
numberscanbecorrectedbyrepeatedtransplantationsofmalizedagainstthevolumeofanisletequivalent(IEQ),
purifiedislets(11,12).However,suchaprocedureisnotwhichisaperfectlysphericalisletwithadiameterof
applicablewhentransplantingmacroencapsulatedisletsin150µm.Thisprocedureofisletmassdeterminationis
abioartificialpancreas(5,14).Wedevelopedabioartifi-highlyoperatordependentduringsizingandirreproduci-
cialpancreasasavascularprothesiswithacapillaryasbleoncetheoriginalsamplehasbeendiscarded.
isletcompartmenttooptimizethediffusionofnutrientsArealdensitymeasurementsareappliedtoquantify
andinsulin(3,4,7).However,thisdesignrestrictsthethetotalamountofpurifiedisletsasvolume(mm3)or
availablevolumeforisletencapsulation,thusloadingre-intermsofpurifiedisletpertissue(%)butgivenor-
quiresareliableisletmassdetermination.ExaminingthemallynoinformationabouttheyieldintermsofIEQ.
currentlyavailabletechniquesforisletmassdetermina-Weextendarealdensitymeasurementstothedetermina-
tion,wefoundthatsuchaprocedureisnotcommonlytionofisletyieldintermsofIEQ,whichistherelevant
available.numberforisletquantificationinislettransplantation.
Dithizone-stainedhumanorpigisletsareusuallyWeestablishedalinearrelationbetweenthetraditional
manuallycountedandsizedunderthemicroscopeaccord-determinationofIEQbyindividualisletcountingand
AcceptedOctober31,2002.AddresscorrespondencetoDr.NicolasLembert,DepartmentofPharmacology,AufderMorgenstelle8,UniversityofTu¨bingen,72076Tu¨bingen,Germany.Tel:+4970712972471;E-mail:nicolas.lembert@uni-tuebingen.de
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