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Quiz%206%20Fall%202010

Course: CH 301 301, Fall 2011
School: University of Texas
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001 Version Quiz 6 sparks (50990) O H A H IMForces 001 1.0 points What would be the most signicant type of intermolecular forces in a liquid sample of uoroform (CHF3 )? Consider the two water molecules This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 1 B O H H Which statement is correct? 1. The hydrogen...

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001 Version Quiz 6 sparks (50990) O H A H IMForces 001 1.0 points What would be the most signicant type of intermolecular forces in a liquid sample of uoroform (CHF3 )? Consider the two water molecules This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 1 B O H H Which statement is correct? 1. The hydrogen bond B is stronger than the covalent bond A. 1. dipole-dipole correct 2. The hydrogen bond A is stronger than the covalent bond B. 2. covalent 3. hydrogen bonding 3. The covalent bond A is stronger than the hydrogen bond B. correct 4. dispersion 4. The covalent bond B is stronger than the hydrogen bond A. CIC T05 18 002 1.0 points Explanation: + O H A H Explanation: London forces, dispersion forces, van der Waals or induced dipoles all describe the same intermolecular force. London forces are induced, short-lived, and very weak. Molecules and atoms can experience London forces because they have electron clouds. London forces result from the distortion of the electron cloud of an atom or molecule by the presence of nearby atoms or molecules. Permanent dipole-dipole interactions are stronger than London forces and occur between polar covalent molecules due to charge separation. H-bonds are a special case of very strong dipole-dipole interactions. They only occur when H is bonded to small, highly electronegative atoms F, O or N only. Ion-ion interactions are the strongest due to extreme charge separation and occur between ionic molecules. They can be thought of as both inter- and intramolecular bonding. CHF3 is a polar molecule that does not contain H bonds; therefore, dipole-dipole forces will be the most signicant type of intermolecular forces present. 5. ionic B O H + H A is a covalent bond while B is a hydrogen bond. Covalent bonds are much stronger than H-bonds. Mlib 04 2065 003 1.0 points Which of the following can be expected to exhibit the strongest hydrogen bonding in the liquid state? 1. CH3 CH3 (ethane) 2. CH3 OH (methyl alcohol) correct 3. CH4 4. CH3 COCH3 (acetone) 5. CH3 OCH3 (dimethyl ether) Explanation: Version 001 Quiz 6 sparks (50990) H-bonds are a special case of very strong dipole-dipole They interactions. only occur when H is bonded to small, highly electronegative atoms F, O or N only. The structure is H H C O H H Methyl alcohol is the only molecule given that has H bonded to N, O, or F (O in this case). Mlib 05 0023 004 1.0 points Freezing water is what kind of process? 1. endothermic 2. Neither of these 2 melted to a liquid. Msci 15 0009 006 1.0 points Heat ow is considered positive when heat ows (into, out of) a system; work is considered positive when work is done (by, on) a system. 1. into; on correct 2. out of; on 3. out of; by 4. into; by Explanation: Heat ow q is considered positive when heat ows into a system. Work w is considered positive when work is done on a system. 3. exothermic correct Explanation: Freezing requires the removal of heat from the water. This is exothermic. Sparks phase change 001 005 1.0 points What happens when energy is added to a sample of ice at 0 C? 1. The temperature of the sample increases while the ice melts. 2. None of the other answers is correct. 3. The sample remains solid and its temperature decreases. 4. The temperature of the sample remains constant while the ice melts. correct 5. The temperature of the sample decreases while the ice melts. Explanation: Energy is being added, so the process is endothermic. Energy is required to melt a solid. As energy is added the temperature remains constant until all of the solid has DAL 0301 01a 007 1.0 points WITHDRAWN ChemPrin3e T06 10 008 1.0 points A CD player and its battery together do 500 kJ of work, and the battery also releases 250 kJ of energy as heat and the CD player releases 50 kJ as heat due to friction from spinning. What is the change in internal energy of the system, with the system regarded as the battery and CD player together? 1. 800 kJ correct 2. +200 kJ 3. 750 kJ 4. 700 kJ 5. 200 kJ Explanation: Heat from the CD player is 50 kJ. Heat from the battery is 500 kJ. Work from both together on the surround- Version 001 Quiz 6 sparks (50990) ings is 250 kJ. This question is testing your ability to see what the system is, and then look at ONLY the energy ow for the system. Here the system is the battery and the CD player together. U = q + w = [50 kJ + (250 kJ)] + (500 kJ) = 800 kJ 3
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University of Texas - CH 301 - 301
Version 001 quiz 7 sparks (50990)This print-out should have 7 questions.Multiple-choice questions may continue onthe next column or page nd all choicesbefore answering.Msci 15 0505001 10.0 pointsFor a reaction in which gases are neither produced no
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9/22/2011Review Question! The dot structure for a Si atom would have_ pairs of electrons and _ lone electronsA. 1; 2B. 2; 0C. 0; 2D. 0; 4Thought Questions!1. Lattice Energy will generally be larger for:A. small ionsB. large ions2. Lattice Ener
University of Texas - CH301 - 301
9/8/2011Check Your Understanding!1. Write an electron configuration for Rh (#45).a. [Kr]5s24d7b. [Kr]5s25d7c. [Xe]5s24d7d. [Kr]4d9e. [Kr]5s14d82. How many unpaired electrons does Rh have?a. 1b. 2c. 3 d. 4e. 5Think about it!An element has the
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UPB Colombia - AUTOMATIC - 29
SCD - Seminarul 6Circuite basculante monostabile.Problema 1VCC = 10VRC11KCa100pFRB10KRC21KCT1Acest circiut are doua stari:1.R110K10nF0 = 100UBE = 0,8VUCEsat = 0VT2R210KO stare stabila, in care T2 este insaturatie si T1 blocat, ca
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Seminarul 2Comutatoare electronice. Comutarea tranzistorului bipolarComutator electronic = dispozitiv electronic care permite inchiderea saudeschiderea unui circiut, a unei cai de curent.Prezinta deci doua stari :- inchis (on)- permite conductia cur
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Seminarul 4Circuite logice cu elemente discreteFamilii de circuite logice:IoIiUiDL logica cu diodeDCTL logica cu transiztoare cuplatedirectRTL logica rezistenta tranzistorDTL logica dioda tranzistorTTL logica tranzistor tranzistorECL logica cu
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SCD - seminarul 5Circuite funcionaleInversorul idealuoNeinversor cucaracteristic cuhisterezisNeinversorul idealuouoV0HuiuiVPLuiVPLVPL VPLPentru circuitele ideale, n fiecare "jumtate" avem o singur valoare (1 sau 0).Problema 1VDDR1I11
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UPB Colombia - AUTOMATIC - 29
Circuite elementare de prelucrare a impulsurilorImpulsul electricForm de und care caracterizeaz variaia rapid a unei mrimi fizice(tensiune, curent, sarcin) pe o anumit poriune a axei timpului. Dac acestea serepet spunem c avem de a face cu o serie de
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SCD - seminarul 5Circuite funcionalePentru circuitele ideale, n fiecare "jumtate" avem o singur valoare (1 sau 0).Inversorul idealNeinversor cucaracteristic cuhisterezisNeinversorul idealVDDuoR1uo I112viVPLR2=R1uoC2ui C1=10C2VPL3V0H
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SCD - Seminarul 6Circuite basculante monostabile.Problema 1VCC = 10VRC11KCa100pFRB10KRC21KCT1Acest circiut are doua stari:1.R110K10nF0 = 100UBE = 0,8VUCEsat = 0VT2R210KO stare stabila, in care T2 este insaturatie si T1 blocat, ca
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CAP 4. MEDIUL DE PROGRAMARE LOGIC ISAGRAFISaGRAF este un mediu de programare dezvoltat de CJ Intenational destinat aplicaiilor pentruautomate programabile. Acest mediu permite programarea, simularea, testarea aplicaiilor folosindlimbajele standardului
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Sisteme cuSistememicroprocesoaremicroprocesoareSeria 2007-2008Prof. Dr. Ing. Traian C. IONESCU1Date de contactProfesor: Traian IONESCUEmail:tcion@rdsmail.roAsistenti: Radu PIETRARUAndrei HOHANBogdan CARSTOIULaborator:ED 308, ED 312Website:
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Capitolul 3CapitolulMultiprogramare siMultitaskingMultiprogramare si MultitaskingMijloace de sprijin pentru multiprogramare : a) salvarea si restaurarea starii programelor pentrua permite intreruperea si reluarea corecta aexecutiei, cerute de exec
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Chestiuni privindsistemele de memorie1Ce Este Memoria Flash?Este memorie nevolatila. Utilizeaza tehnologia NOR, care ii permiteutilizatorului sa programeze si sa steargainformatia electric. Programarea are loc cand electronii sunt plasatipe poart
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Protectia MemorieiSegmentare1Selector de segment153Index DTAtribute210TI RPLDimensiune segmentPartea vizibilaAdresa bazaPartea invizibila(Segment CacheRegister )DT Tabel descriptori (Descriptor Table )TI Indicator Tabel ( Table Indicator
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Circuite s uport (pe rife rice )pe mtru microproce s oare8254, 8279 s i alte le1Ceas programabil 8254 schema blocTamponD7 D 0RD /WR/A1magdateLogicaR/WCLK 0Contor0GATE0OUT0Contor1CLK1GATE1OUT1A0CS/Reg.cuvantcda.Contor2CLK2GA
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Microprocesoruldidactic8086celmaiintuitivexemplu1Registrede uz generalRegistre desegmentInstruction Pointer (IP)EUGenerare adresesicomanda magistralaMagistralamultiplexataOperanziSir instructiuniALUIndicatori conditie(Flags)8086Schemab
UPB Colombia - SYSTEM THE - 13
UPB Colombia - SYSTEM THE - 13
UPB Colombia - SYSTEM THE - 13
UPB Colombia - SYSTEM THE - 13
UPB Colombia - SYSTEM THE - 13
UPB Colombia - SYSTEM THE - 13
UPB Colombia - SYSTEM THE - 13
UPB Colombia - SYSTEM THE - 13
UPB Colombia - SYSTEM THE - 13
UPB Colombia - SYSTEM THE - 13
UPB Colombia - SYSTEM THE - 13
UPB Colombia - SYSTEM THE - 13
Subiecte Teoria Sistemelor II1. Se consider sistemul A =a1 203, b=11, cT = [1 1].(a) Calculati evolutia fortat pentru u(t) = t.a(b) Deteminati un estimator de stare.(c) Construiti un compensator stabilizator.2. Dac perechea (C, A) este observ
UPB Colombia - SYSTEM THE - 14
UPB Colombia - SYSTEM THE - 14
Problema 1 Fie u(t) un semnal continuu cu derivata u(t) continu. Care dintre urmtoarele poate oaanorm pentru u?a1. sup |u(t)|t2. sup |u(t)| + sup |u(t)|ttSolutie1. Se constat, prin vericare direct, cu denitia, c nu avem de a face cu o norm, sch
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Problema 1 Fie diagrama Venn din gur. Artati c functiile de la u1 la u9 , denite mai jos, sunt corectaaapozitionate diagram. Toate functiile sunt nule pentru t < 0.nau1 (t) =1/ t, dac t 1a0,dac t > 1au2 (t) =1/t1/4 , dac t 1a0,dac t > 1a
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Teoria Sistemelor IICristian OARAFacultatea de Automatica si CalculatoareUniversitatea Politehnica BucurestiFax: + 40 21 3234 234Email: oara@riccati.pub.roURL: http:/riccati.pub.roTeoria Sistemelor II Capitolul 7CAPITOLUL 7: Loopshaping1. Tehnici
UPB Colombia - SYSTEM THE - 14
Teoria Sistemelor IICristian OARAFacultatea de Automatica si CalculatoareUniversitatea Politehnica BucurestiFax: + 40 1 3234 234Email: oara@riccati.pub.roURL: http:/riccati.pub.roTeoria Sistemelor II Capitolul 1CAPITOLUL 1: SISTEME DINAMICE PE SPA
UPB Colombia - SYSTEM THE - 14
Teoria Sistemelor IICristian OARAFacultatea de Automatica si CalculatoareUniversitatea Politehnica BucurestiFax: + 40 1 3234 234Email: oara@riccati.pub.roURL: http:/riccati.pub.roTeoria Sistemelor II Capitolul 2Exista anumite proprietati fundament
UPB Colombia - SYSTEM THE - 14
Teoria Sistemelor IICristian OARAFacultatea de Automatica si CalculatoareUniversitatea Politehnica BucurestiFax: + 40 21 3234 234Email: oara@riccati.pub.roURL: http:/riccati.pub.roTeoria Sistemelor II Capitolul 3CAPITOLUL 3: SINTEZA ELEMENTARA1.
UPB Colombia - SYSTEM THE - 14
Teoria Sistemelor IICristian OARAFacultatea de Automatica si CalculatoareUniversitatea Politehnica BucurestiFax: + 40 21 3234 234Email: oara@riccati.pub.roURL: http:/riccati.pub.roTeoria Sistemelor II Capitolul 5CAPITOLUL 5: NORME PENTRU SEMNALE S
UPB Colombia - SYSTEM THE - 14
Teoria Sistemelor IICristian OARAFacultatea de Automatica si CalculatoareUniversitatea Politehnica BucurestiFax: + 40 21 3234 234Email: oara@riccati.pub.roURL: http:/riccati.pub.roTeoria Sistemelor II Capitolul 6CAPITOLUL 6: CERINTE DE PROIECTARE
UPB Colombia - AUTOMATIC - 22
Proiect SCPCPreincalzitoarele de aer cu camera de ardere exterioara functioneaza pe acelasiprincipiu cu preincalzitoarele clasice, deosebirea constand in aceea ca acestea au camerade ardere construita separat de camera gratarelor sau preincalzitorul pr
UPB Colombia - AUTOMATIC - 22
Universitatea Politehnica BucuretiFacultatea de Automatic i CalculatoareProiect SCPCTema proiectului :Proiectarea unui sistem numeric de conducere pentru prenclzitoareleaparinnd unui furnal de pe platforma SIDEX Galai.Cuprins :1. Prezentarea instal
UPB Colombia - AUTOMATIC - 22
PROIECT SCRConsiderente initiale:- camera functioneaza independent de cei 2 roboti, iar dupa decodificareatipului de piesa si generarea iesirilor I1 si I2 acestea se mentin pana la o nouacodificare;- I1 si I2 sunt date de intrare atat pentru robotul
UPB Colombia - AUTOMATIC - 22
SISTEME DE CONDUCERE A PROCESELOR CONTINUETEMA PROIECT: Automatizarea unui punct termic dinreteaua de termoficareOBIECTIVE:1. Aspecte tehnologice (prezentarea instalatiei)2. Solutii de automatizare3. Proiectare SRA pentru parametri din proces4. Pro
UPB Colombia - AUTOMATIC - 18
> A=[0 1 ; 0 0 ]A=0100> B=[0;1]B=01> C=[1 0]C=10> D=[0]D=0> sys1=ss(A,B,C,D)a=x1x2x100x1x2u101y1x11x210b=c=x20d=u11y10Continuous-time model.> E=[0 1 ; -2 -3 ]E=0-21-3> F=[0;1]F=01> G=[0 1]G=01> H=[0
UPB Colombia - AUTOMATIC - 18
Liacu Bogdan331AATEMA 21.Sa se precizeze:a) Sa se precizeze cum se pot genera matrice cu valori proprii oarecare (a se vedearand)-pentru a genera o matrice oarecare de orice tip(adica patratica, sau oarecare),folosim functia rand:A=rand(m,n); pt m
UPB Colombia - AUTOMATIC - 18
Flori Catalin331 ACTEMA 3 CNA1.Se considera matricea: 3 3 2 A = 1 5 2 .1 30Sa se calculeze de mana fara a folosi functii Matlab :a ) F = A1b)G = e A ln 2c) H = cos(A)8Comparati rezultatele obtinute cu cele date prin aplicarea functiilor Matl
UPB Colombia - AUTOMATIC - 18
CNA-TEMA 31.Se considera matricea: 3 3 2 A = 1 5 2 .1 30Sa se calculeze de mana fara a folosi functii Matlab :a ) F = A1b)G = e A ln 2c) H = cos(A)8Comparati rezultatele obtinute cu cele date prin aplicarea functiilor Matlabcorespunzatoarea)
UPB Colombia - AUTOMATIC - 18
CNA-TEMA 31.Se considera matricea: 3 3 2 A = 1 5 2 .1 30Sa se calculeze de mana fara a folosi functii Matlab :a ) F = A1b)G = e A ln 2c) H = cos(A)8Comparati rezultatele obtinute cu cele date prin aplicarea functiilor Matlabcorespunzatoarea)
UPB Colombia - AUTOMATIC - 18
Flori Catalin331 ACTema 4 CNA1.Calculul conversiei ss2tf pentru SISOfunction [r,p]=ss2tfSISO(A,B,C,D)p=poly(A);r=poly(A-B*C);r=r+(D-1)*p;2.Calculul conversiei ss2tf pentru SIMOfunction [r,p]=ss2tfSIMO(A,B,C,D)p=poly(A);s=length(C);for i=1:sr(
UPB Colombia - AUTOMATIC - 18
Tema 4 CNA1.Calculul conversiei ss2tf pentru SISOfunction [r,p]=ss2tfSISO(A,B,C,D)p=poly(A);r=poly(A-B*C);r=r+(D-1)*p;2.Calculul conversiei ss2tf pentru SIMOfunction [r,p]=ss2tfSIMO(A,B,C,D)p=poly(A);s=length(C);for i=1:sr(i,:)=poly(A-B(:,1)*C(
UPB Colombia - AUTOMATIC - 18
Tema 4 CNA1.Calculul conversiei ss2tf pentru SISOfunction [r,p]=ss2tfSISO(A,B,C,D)p=poly(A);r=poly(A-B*C);r=r+(D-1)*p;2.Calculul conversiei ss2tf pentru SIMOfunction [r,p]=ss2tfSIMO(A,B,C,D)p=poly(A);s=length(C);for i=1:sr(i,:)=poly(A-B(:,1)*C(
UPB Colombia - AUTOMATIC - 18
Flori Catalin331 ACTEMA 5 CNAPentru toate rutinele MATLAB mentionate in L5, sa se completeze tabelul :Numele rutineinyquistApelare1.NYQUIST(SYS)2.NYQUIST(SYS,cfw_WMIN,WMAX)3.NYQUIST(SYS1,SYS2,.,W)bodenicholsFunctionalitate1.Deseneaza diagr