2007. Ж. А., А.. З, Ж. Е. Й, З. О. Л Йapple М. К.З. ъ ИОЗ

Save this PDF as:
 WORD  PNG  TXT  JPG

Size: px
Start display at page:

Download "2007. Ж. А., А.. З, Ж. Е. Й, З. О. Л Йapple М. К.З. ъ ИОЗ"

Transcription

1 ЩВИЩ ЕҐЩЙДЕЩ ВЙЗВА ЛЕЖЕҐЩЙДВГ ЩЛЗВВРЕЕ, 007, 41,, Н Д : Рappleapple apple apple apple applefl appleapple apple apple, applefl fl 007. Ж. А., А.. З, Ж. Е. Й, З. О. Л Йapple М. К.З. ИОЗ К apple Вapple apple applefl apple apple apple, apple, apple appleapple appleapple apple apple appleapple apple, apple apple applefl appleapple apple - apple, applefl fl. З appleapple apple appleapple applefl apple 10% - apple 60%. Чapple- apple ( 1 ) apple apple fl,, [1]. А, - fl apple fl apple- apple []. Йfl 1 fl- flfl apple apple- fl- apple, applefl apple apple apple flfl 3 д 94.3 /, apple applefl 60. З apple appleapple apple (РЙД) apple apple apple appleapple apple apple [3]. appleapple fl apple apple - apple 113 ДК apple ( 1 ) 60% [4]. А РЙД apple- apple apple apple apple appleapple 0шЙ applefl 1 / apple apple (N h > N b ); appleapple applefl H : ВЗ + З В H +З В. (1) И apple (1) apple, applefl H apple apple- appleapple [OH ] [H ] [5]. Ж- applefl ( 1 ) fl apple [6]. Кapple apple apple - apple apple - appleapple apple apple apple apple apple apple apple appleapple, apple H 100%- apple ( 1 ) [7]: Cl + H ( ) + Cl + H +, K ч /( ), [8] H + + H H. () Е apple applefl ( 1 ) fl- apple: 1 Cl + ВЗ + З В = ( 1 ) +З В + Йl. (3) Дapplefl OH apple- H apple apple apple OH apple Д [8, 9], - apple apple apple apple OH apple apple- apple. Вapplefl apple apple apple - flfl appleapple apple apple-. Зfl appleapple apple apple- apple apple - apple. Иfl apple appleapple, apple applefl ( 1 ), fl D c, KN b appleapple applefl fl З- D 0 KN b apple ( 1 ), appleapple appleapple, fl apple appleapple applefl -, applefl fl. Зapple- apple, apple N b = 1 /, D 0 ч D c ч / apple- D c 176

2 Рappleapple apple 177 fl apple, apple applefl ( 1 ), appleapple 1, applefl fl ( 1 ) appleapple apple apple appleapple- 10 9, apple 10 6 appleapple [10]. В - applefl fl fl ( 1 ) apple apple fl 100% [11]. Чapple- apple ( 1 ) apple- fl (applefl apple apple apple, 3 /): ( 1 ) + ( 1 ) ( 1 д) + ( 3 д), , (4) ( 1 д) + З В ( 1 ) + З В, , (5) ( 1 ) + ( 1 ) ( 3 д) + ( 3 д), , (6) ( 1 ) +Cl ( 3 д) + Cl, , (7) ( 1 ) +H O ( 3 д) + H O, (8) А apple (4) applefl apple apple apple- fl apple 157 /. Рappleapple apple applefl- - apple, apple apple fl- flfl apple- apple ( 1 ). Д fl apple appleapple РЙД apple [], apple [4], - [1], [13] apple-. Йapple ( 1 ) applefl (4)(8) apple apple- apple. Ч apple applefl appleapple apple apple, apple fl apple apple apple apple apple ( 1 ). - fl apple РЙД apple applefl ( 1 ) apple. Дapple, apple fl, appleapple, apple- apple, apple appleapple H РЙД. А fl apple apple - РЙД apple apple applefl appleapple H, - applefl fl. Е - apple- РЙД fl apple-- apple apple apple apple [14]. apple apple appleapple РЙД, apple applefl ( 1 ), applefl. Кapplefl apple- РЙД apple apple. 1. Аapple apple apple РЙД 135, - 80 apple 50 apple 540 apple fl apple 0.. А apple appleapple applefl apple apple 1-5. Иfl apple. А apple applefl apple,. З applefl 135 apple fl apple- fl applefl apple appleapple apple apple. Вapplefl - apple appleapple apple apple. Рapple apple appleapple apple apple apple, appleapple appleapple. Р apple- apple apple 15 apple 50 apple. РЙД apple- apple fl - fl apple, apple - apple 4. В appleappleapple applefl apple fl V T = = apple apple apple apple fl, apple apple fl applefl apple apple, ( 1 ), apple, appleapple fl. А N c applefl apple 337, N apple- fl apple apple I, appleappleapple ( 1 ) 176, N = AI. Й apple (5), apple N w ( apple apple - РЙД) fl N, N д apple fl apple- fl Аapplefl fl РЙД - appleapple ( 1 ) apple, N - apple (4), (6)(8) apple apple appleapple-. А fl appleapple N д flflfl apple : N д N = N w Й, apple apple-, appleappleapple ( 1 д) 76 I д C I = N w fl apple- fl apple apple, fl РЙД. Дapple applefl - apple apple N c apple - fl fl fl = = 337. Еapple fl apple apple (.3 с 0.01) 10 19, 5

3 178 apple. У appleapple H Д Cl Кapple И. 1. Й apple- РЙД: 1, fl apple, 3 apple, 4 apple. applefl [15]. Д- apple ( 1 ) apple apple apple fl apple apple- apple ( 133 К) applefl apple, 1 : 10, ( 1 ) applefl (4)(8) applefl appleappleapple - apple appleapple apple apple appleapple appleapple. А apple apple ( 1 ) - applefl apple Y 0 apple appleapple H. А - appleapple appleflfl apple fl Y 0 75 [16] 94% [11]. Оapple fl apple ( 1 ) apple Y 0 =. fl - apple applefl apple apple - applefl apple apple appleapple appleapple. Кapple apple apple, apple appleapple appleapple N w apple, applefl - N apple C = w I д. А apple apple fl- I, fl apple apple- fl Y apple с10%, apple - applefl apple apple с0.1 с0.05 apple.. fl apple apple. appleapple apple applefl appleapple, apple- fl - fl ( 1 ) apple. И appleapple РЙД apple- apple applefl fl: CI w = kt, I д c = N c kt,

4 Рappleapple apple 179 Кapple N w > N < , - apple N д N, - apple ( 1 ) fl: N Y N kt = N ч w, c applefl applefl apple fl- fl 1 3 д N o ч w c /kt. Ж applefl applefl applefl РЙД: V з r = / c1 kt GRT 1 applefl appleappleapple з c1 ( V + V T ) T = RT 1 G Кapple з r з appleapple 1 apple- apple appleapple apple apple- apple apple apple. Й, fl ( 1 ) appleapple apple- apple appleapple apple, c, w Y. Й apple apple РЙД applefl apple flfl apple G c = (1 U)G, G o = UG c, G = YUG c, Q(N b N b1 ) = = UG, Q(N h N h1 ) = UG, G w = w (G o + G c ). У appleapple apple apple apple- 14 Ж appleapple ДВЗ 38%- apple apple - apple. Кapple fl РЙД apple - fl fl apple appleapple apple Q = = 1. /. М apple- 10 3, apple apple 13.5 apple 1 1.9, fl apple applefl- apple ~1. apple, appleflfl fl apple applefl apple ~ apple. В appleapple applefl, fl, apple- V = 380 3, V + V T appleapple- apple apple apple w w w =, c w U = , 1 ( w ) c1 = apple Кapple apple apple c 1 apple- 5.4 / apple apple applefl. fl appleapple- apple appleapple H applefl 1 РЙД flfl. З apple- apple appleapple fl. А appleapple РЙД - apple apple, apple, - flapple appleapple appleapple H, apple fl apple. Йapple fl applefl appleapple apple- apple. А appleapple fl, - apple apple, apple appleapple - fl 10. Ч apple apple- apple apple applefl РЙД з r ч apple apple.. А, - apple apple fl. fl apple apple apple apple fl apple applefl apple з r (apple. ). Кapple з r apple fl fl, fl - apple apple apple apple applefl [17]. Йfl - apple apple fl з r, apple apple - appleapple H apple fl H, apple apple apple appleflfl H apple [17]. Кapple apple apple apple apple apple fl РЙД apple fl ( 1 ) applefl (4)(8). В apple- ( 1 ) РЙД appleflfl apple- apple fl apple И 1 apple applefl з,.. И 1 з (apple. 3). Зfl applefl fl apple apple- Y = 5 (1 + И 1 з ) 1. З -, appleapple apple - apple Д appleapple И 1 з , appleapple apple- apple apple - appleapple И 1 (з з r ) apple ( 1 ) apple appleapple- apple applefl apple fl., apple РЙД, apple apple 5*

5 180 apple. U 1.0 () G, / () з r, () t, шc 1.0 () Q, / Y c1 з T И. 3. В apple ( 1 ) И c1 з r, apple Q = 1. /, N b = 6.5 /, N h = 7.5 /, t = = 16шC. И apple G, /: a 0, 40, 57, 75. ( 1 ) apple appleapple -. Й appleapple apple- apple H apple - fl (apple. ), apple - fl fl appleapple, apple - apple H appleapple, apple apple (). Кapple fl apple appleapple - appleapple H apple- apple apple %, ( 1 ) flfl. Н apple - appleapple H apple - apple (apple. ). А fl - applefl apple 4 6% applefl ( 1 ) Кapple - appleapple РЙД apple ДВЗ appleapple H apple () И.. Ч apple: apple apple apple applefl РЙД з r ч c, Q = 1. /, N b = 6.5 /, N h = 7.5 /, t = 16шC; apple applefl РЙД apple Q = 1. /, N b = 6.5 /, N h = 7.5 /, t = 16шC. И apple G, /: 40; 57, 75; appleapple appleapple apple apple- apple Q = 1. /, N b = 6.5 /, N h = 7.5 /, G = = 57 /, c1 = 3.3 apple.., з r ч 10 c; apple appleapple apple apple apple N b = 6.5 /, N h = 7.5 /, G = 57 /, t = N А, / = 16шC, c1 = 0 apple.., з r ч c; - apple ДВЗ apple- apple apple apple apple Q = 1. /, G = = 57 /, t = 10шC, c1 = 0 apple.., з r ч c.

6 Рappleapple apple 181 apple. А 38%- apple apple - fl 14 Ж ДВЗ fl fl apple. apple, appleapple H, apple - applefl fl applefl apple apple. Й apple appleapple apple - apple (apple. ), apple apple apple appleapple- fl apple apple-. А apple apple- N h fl ( 1 ) apple - apple apple. И apple 1 : 4 apple РЙД, apple apple applefl РЙД apple apple apple fl apple apple РЙД fl - apple ( 1 ). И- - appleapple РЙД. Иapple РЙД - ( 1 ) fl apple - apple- apple, apple [18]. Кapple apple apple 57 / - fl apple 150 А. Ч- apple, apple apple apple РЙД applefl [18]. В apple apple- РЙД, apple - appleapple H appleapple - apple. К 60 appleapple apple apple apple apple apple appleapple. Йapple- РЙД - ( 1 ) fl apple apple- fl, fl apple- apple. З fl apple apple- fl apple appleapple H apple-, РЙД applefl [18]. Лfl РЙД applefl- fl apple ~5 1, fl apple apple-- РЙД, apple apple apple fl (apple appleapple H, - apple, applefl applefl ) - apple apple РЙД applefl. А - fl apple apple appleapple applefl - apple appleapple H,, -, apple apple. П fl apple apple apple apple- РЙД flflfl apple apple apple. И apple apple Щapple apple - appleapple (apple ) - apple apple Жapple - apple (apple 30). Вfl О apple ; Й apple ; D c apple - appleapple apple apple, /; D o apple - appleapple apple apple, /; G flapple apple apple РЙД, /; G c flapple apple apple РЙД, /; G o flapple apple apple РЙД, /; G flapple apple apple РЙД, /; G w flapple apple apple РЙД, /; I apple apple, apple- appleapple apple fl 1 ; I д apple apple, apple- appleapple apple fl 1 д; Д apple apple, /( ); k flfl П; N w applefl apple apple-, 3 ; N c applefl apple apple, 3 ; N b, N b1 flapplefl applefl - appleapple apple apple РЙД, /; N h, N h1 flapplefl applefl apple - apple appleapple apple apple РЙД, /; N o applefl applefl apple - flfl 1 3 д, 3 ; N applefl apple fl 1, 3 ; N д applefl apple fl 1 д, 3 ; И 1 РЙД, apple.. И apple, apple..; И 1 apple РЙД, apple..; И apple apple, apple..;

7 18 apple. И w apple apple, apple..; Q apple apple- apple apple apple, /; R applefl fl flfl; 1 appleapple РЙД, Д; appleapple apple, Д; t appleapple appleapple apple apple, шй; U apple РЙД; V fl РЙД, apple- applefl, 3 ; V T appleappleapple apple- fl apple, 3 ; N Y = apple ( 1 ); N o Y 0 Y apple appleapple apple apple; w apple apple ; з r applefl applefl applefl, ; з T applefl appleappleapple apple apple,. ЙКЕЙВД Е ЩИО НИ 1. Khan A.U. Singlet molecular oxygen. A new kind of oxygen // The Journal of hysical Chemistry V McDermott W.E., chelkin N.R., Benard D.J., Bousek R.R. An electronic transition chemical laser// Appl.hys.Lett. 1978, V Browne R.J., Ogryzlo E.A. Chemiluminescence from the reaction of chlorine with aqueous hydrogen peroxide // roc.chem.soc Оfl А.З., Ж.А., З А.. apple. Йapple appleapple ( 1 ) apple 13.3 К // Дfl apple Й Balej J, Spalek O. Calculation of equilibrium composition in more concentrated systems H KOH (or NaOH)H O // Collection Czechoslov.Chem.Commun. 1979, V Richardson R.J., Kelley J.D., Wiswall C.E. ( 1 ) generation mechanisms in the chemically pumped iodine laser // J. Appl. hys V Storch D.M., Dymek C.J., Davis L.. MNDO study of the mechanism of ( 1 ) formation by reaction of Cl with basic H // J. Am. Chem. Soc V Оfl А.З., Ж.А., З А.., Н З.Е. Д apple Cl appleapple H OH KOH H OKOH // Ъapple Й Gershenzon M., Davidovits., Jayne J.T. et al. Rate constant for the reaction of Cl (aq) with OH // J. hys. Chem. A. 00. V Rogers M.A. Lifetime of ( 1 ) in liquid water as determined by time resolved infrared luminescence measurements // J.Am. Chem. Soc V П З.Р., Ж.А., Е А.Е. apple. apple apple- apple // apple МЕОЗ Й Blauer J.A., Munjee S.A., Truesdell K.A. et al. Aerosol generators for singlet oxygen production// J. Appl. hys V Harpole G.M., English W.D., Berd J.G., Miller D.J. Rotating disk oxygen generator // AIAA aper resented on 3 rd lasmadynamics and Lasers conference. July 68, 199. Nashville. USA. 14. Vetrovec J. Singlet oxygen generator with filamentguided jets // roceedings of SIE V В Л. Мfl. Ж.: Жapple, Yang T.T., Copeland D.A., Bauer A.H. et al. Chemical oxygen-iodine laser performance modeling // AIAA aper resented on 8 th lasmadynamics and laser conference. June apple К.А. Р apple. Ж.: Л- fl, Hager G.D., Nikolaev V.D., Svistun M.I., Zagidullin M.V. Lasing performance of a chemical oxygen iodine laser (COIL) with advanced ejector nozzle banks // Appl. hys. A V

ЗВАГ ЙКВЙВП ДВЗ ЩЗ ИЕИВАОЗЕБ Е Е АЩҐЩЗЕБ ВВ О Е РИОМЕ ВАЛ ИН НИРЩЗЩАЙДВРВ ЖЩЙ ВИВЪ ЩЗЕБ, КИЕЖВИ Щ

ЗВАГ ЙКВЙВП ДВЗ ЩЗ ИЕИВАОЗЕБ Е Е АЩҐЩЗЕБ ВВ О Е РИОМЕ ВАЛ ИН НИРЩЗЩАЙДВРВ ЖЩЙ ВИВЪ ЩЗЕБ, КИЕЖВИ Щ ВДО ОДО ЩЖЕЕ ЗОНД, 2008, 423, 1,. 110113 Н Д 553.411+546.59 РЩВЛЕЖЕБ ЗВАГ ЙКВЙВП ДВЗ ЩЗ ИЕИВАОЗЕБ Е Е АЩҐЩЗЕБ ВВ О Е РИОМЕ ВАЛ ИН НИРЩЗЩАЙДВРВ ЖЩЙ ВИВЪ ЩЗЕБ, КИЕЖВИ Щ 2008. О О. Е. Л, Ж. О. Ж,. К. К, А.

More information

ЗВАВЩ ЙЩЖЩГЙ АВ ЕЙКЩИРЕИВАОЗЗЛ КВА ВИВА Е РЩЗВЖО ҐЩХНГҐО Л ИЩК ЕЕГ

ЗВАВЩ ЙЩЖЩГЙ АВ ЕЙКЩИРЕИВАОЗЗЛ КВА ВИВА Е РЩЗВЖО ҐЩХНГҐО Л ИЩК ЕЕГ ЖВЩДНБИЗОБ ПЕВВРЕБ, 2006, 40, 2,. 378382 Н Д 577.21 Дapple ъfl ЗВАВЩ ЙЩЖЩГЙ АВ ЕЙКЩИРЕИВАОЗЗЛ КВА ВИВА Е РЩЗВЖО ҐЩХНГҐО Л ИЩК ЕЕГ 2006. Й. О. Д, В. И. Пapple, А. А. Рapple*,. О. Дappleapple Е flapple ъ

More information

Norpeth. 9 weights 5 variations of numerals opentype features

Norpeth. 9 weights 5 variations of numerals opentype features Norpeth 9 weights 5 variations of numerals opentype features Norpeth 26 pt modern humanist sans serif typeface. The proportions of each character have a strong lateral dynamic that makes it ideal for on-screen

More information

Programming the Microchip Pic 16f84a Microcontroller As a Signal Generator Frequencies in Railway Automation

Programming the Microchip Pic 16f84a Microcontroller As a Signal Generator Frequencies in Railway Automation 988 Programming the Microchip Pic 16f84a Microcontroller As a Signal Generator Frequencies in Railway Automation High School of Transport "Todor Kableshkov" 1574 Sofia, 158 Geo Milev str. Ivan Velev Abstract

More information

Problem A. Nanoassembly

Problem A. Nanoassembly Problem A. Nanoassembly 2.5 seconds One of the problems of creating elements of nanostructures is the colossal time necessary for the construction of nano-parts from separate atoms. Transporting each of

More information

ттгт ци тг цсс пястсгс цсгс TELOS

ттгт ци тг цсс пястсгс цсгс TELOS пепистги ягтгр сг хети епистг тг епистггс упцист гисс пхг еусгс и диеияисгс ттгт ци тг цсс пястсгс цсгс TELOS цв цв applefi евfl г, ж 1994 пепистги ягтгр сг хети епистг тг епистггс упцист гисс пхг еусгс

More information

INFO1 a File-Based Management Information System

INFO1 a File-Based Management Information System БЪЛГАРСКА АКАДЕМИЯ НА НАУКИТЕ. BULGARIAN ACADEMY OF SCIENCES КИБЕРНЕТИКА И ИНФОРМАЦИОННИ ТЕХНОЛОГИИ, 1 CYBERNETICS AND INFORMATION TECHNOLOGIES, 1 София. 2002. Sofia INFO1 a File-Based Management Information

More information

The European Ombudsman

The European Ombudsman Overview The European Ombudsman Е в р о п е й с к и о м б у д с м а н E l D e f e n s o r d e l P u e b l o E u r o p e o E v r o p s k ý v e ř e j n ý o c h r á n c e p r á v D e n E u r o p æ i s k e

More information

User Manual. June 2008 Revision 1.7. D- 2 02 Customer Display

User Manual. June 2008 Revision 1.7. D- 2 02 Customer Display WW User Manual June 2008 Revision 1.7 D- 2 02 Customer Display Copyright 2008 August All Rights Reserved Manual Version 1.7 The information contained in this document is subject to change without notice.

More information

Digital Typography. This reading describes different types of writing systems and the development of computer-based font files for representing them.

Digital Typography. This reading describes different types of writing systems and the development of computer-based font files for representing them. D R A F T - FOR DISCUSSION ONLY - D R A F T Digital Typography and computer fonts Introduction Follow-up Classes Other readings This reading describes different types of writing systems and the development

More information

А ýэ СаЬЬа оча. А а Ьаусап. сар

А ýэ СаЬЬа оча. А а Ьаусап. сар ч к тдв тап дт ФЁ Тч з ха а а п п А а Ьаусап п Ё Т о А е о п е па опа Ё й О о о а На еп ч о а п а ар С М о а Еар Ва е ако М а А агьаусап г Ъч пс А СаЬЬа оча Аупч а есе г А а Ьаусап сё а Ь у сар о чес Э

More information

4 theoretical problems 2 practical problems

4 theoretical problems 2 practical problems 1 st 4 theoretical problems 2 practical problems FIRST INTERNATIONAL CHEMISTRY OLYMPIAD PRAGUE 1968 CZECHOSLOVAKIA THEORETICAL PROBLEMS PROBLEM 1 A mixture of hydrogen and chlorine kept in a closed flask

More information

Reaction Quotient. Trial 1 A (g) + B (g) C (g) Trial 2 A (g) + B (g) C (g) Initial 1.000 M 1.000 M 1.000 M Initial 2.000 M 0.500 M 0.

Reaction Quotient. Trial 1 A (g) + B (g) C (g) Trial 2 A (g) + B (g) C (g) Initial 1.000 M 1.000 M 1.000 M Initial 2.000 M 0.500 M 0. Reaction Quotient How do you predict which direction a reaction will proceed to reach equilibrium? Why? When a reaction reaches equilibrium there must be some non-negligible amount of every species in

More information

EFFICIENCY OF FEED USE OF WET FATTENING PIGS U.

EFFICIENCY OF FEED USE OF WET FATTENING PIGS U. 1 UDC 636.4.033.083 EFFICIENCY OF FEED USE OF WET FATTENING PIGS U. Zasukha doctor of agricultural sciences, professor S. Grishchenko Candidate of Agricultural Sciences N. Gryshchenko graduate student

More information

Test Review # 9. Chemistry R: Form TR9.13A

Test Review # 9. Chemistry R: Form TR9.13A Chemistry R: Form TR9.13A TEST 9 REVIEW Name Date Period Test Review # 9 Collision theory. In order for a reaction to occur, particles of the reactant must collide. Not all collisions cause reactions.

More information

IС A A RT 2013. Proceedings Volume 2. 5th International Conference on Agents and Artificial Intelligence. Barcelona, Spain 15-18 February, 2013

IС A A RT 2013. Proceedings Volume 2. 5th International Conference on Agents and Artificial Intelligence. Barcelona, Spain 15-18 February, 2013 «'.''«ИЧИЧГШ ИШ М Ш * /////>. л ъ и г ш я ш и ъ в т ъ т ', : 4 р * т Ъ ъ ^ Х 'Ш У Л *а * 1 ЛЧй==:й?й!^'ййй IС A A RT 2013. *»ф«ч>»д* 'И И в Я в З Г З г И Ж /а 1 * icw-ia & «:*>if E M e i i i i y. x '-

More information

User Manual. January 2011 Revision 2.0. Galéo 200 Point of - Sale Hardware System

User Manual. January 2011 Revision 2.0. Galéo 200 Point of - Sale Hardware System User Manual January 2011 Revision 2.0 Galéo 200 Point of - Sale Hardware System Copyright 2011 All Rights Reserved Manual Version 2.0 The information contained in this document is subject to change without

More information

Science 1194 SAS Curriculum Pathways Chemical Equations: Journal

Science 1194 SAS Curriculum Pathways Chemical Equations: Journal Chemical Equations: Journal NAME: ray CLASS: chem90 DATE: 10/27/2013 FOCUS QUESTION: How and why are chemical equations balanced? TAB 1: Equations Read the questions below. Then complete this Journal by

More information

Chemistry: Chemical Equations

Chemistry: Chemical Equations Chemistry: Chemical Equations Write a balanced chemical equation for each word equation. Include the phase of each substance in the equation. Classify the reaction as synthesis, decomposition, single replacement,

More information

The 5 Types of Chemical Reactions (Chapter 11) By C B 6 th period

The 5 Types of Chemical Reactions (Chapter 11) By C B 6 th period The 5 Types of Chemical Reactions (Chapter 11) By C B 6 th period 1) Combination Reactions Is also referred to as a synthesis reaction It is a chemical change in which two or more substances react to form

More information

UNDERGRADUATE STUDY SKILLS GUIDE 2014-15

UNDERGRADUATE STUDY SKILLS GUIDE 2014-15 SCHOOL OF SLAVONIC AND EAST EUROPEAN STUDIES UNDERGRADUATE STUDY SKILLS GUIDE 2014-15 ECONOMICS AND BUSINESS HISTORY LANGUAGES AND CULTURE POLITICS AND SOCIOLOGY 1 1. AN INTRODUCTION TO STUDY SKILLS 5

More information

Chem 31 Fall 2002. Chapter 3. Stoichiometry: Calculations with Chemical Formulas and Equations. Writing and Balancing Chemical Equations

Chem 31 Fall 2002. Chapter 3. Stoichiometry: Calculations with Chemical Formulas and Equations. Writing and Balancing Chemical Equations Chem 31 Fall 2002 Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations Writing and Balancing Chemical Equations 1. Write Equation in Words -you cannot write an equation unless you

More information

CLASS TEST GRADE 11. PHYSICAL SCIENCES: CHEMISTRY Test 6: Chemical change

CLASS TEST GRADE 11. PHYSICAL SCIENCES: CHEMISTRY Test 6: Chemical change CLASS TEST GRADE PHYSICAL SCIENCES: CHEMISTRY Test 6: Chemical change MARKS: 45 TIME: hour INSTRUCTIONS AND INFORMATION. Answer ALL the questions. 2. You may use non-programmable calculators. 3. You may

More information

Physical Chemistry. Lecture 4 Introduction to chemical kinetics

Physical Chemistry. Lecture 4 Introduction to chemical kinetics Physical Chemistry Lecture 4 Introduction to chemical kinetics Thermodynamics and kinetics Thermodynamics Observe relative stability of states Energy differences Static comparisons of states Kinetics Observe

More information

APPENDIX B: EXERCISES

APPENDIX B: EXERCISES BUILDING CHEMISTRY LABORATORY SESSIONS APPENDIX B: EXERCISES Molecular mass, the mole, and mass percent Relative atomic and molecular mass Relative atomic mass (A r ) is a constant that expresses the ratio

More information

Unit 2: Quantities in Chemistry

Unit 2: Quantities in Chemistry Mass, Moles, & Molar Mass Relative quantities of isotopes in a natural occurring element (%) E.g. Carbon has 2 isotopes C-12 and C-13. Of Carbon s two isotopes, there is 98.9% C-12 and 11.1% C-13. Find

More information

MOLECULAR MASS AND FORMULA MASS

MOLECULAR MASS AND FORMULA MASS 1 MOLECULAR MASS AND FORMULA MASS Molecular mass = sum of the atomic weights of all atoms in the molecule. Formula mass = sum of the atomic weights of all atoms in the formula unit. 2 MOLECULAR MASS AND

More information

Chapter 3. Chemical Reactions and Reaction Stoichiometry. Lecture Presentation. James F. Kirby Quinnipiac University Hamden, CT

Chapter 3. Chemical Reactions and Reaction Stoichiometry. Lecture Presentation. James F. Kirby Quinnipiac University Hamden, CT Lecture Presentation Chapter 3 Chemical Reactions and Reaction James F. Kirby Quinnipiac University Hamden, CT The study of the mass relationships in chemistry Based on the Law of Conservation of Mass

More information

Chemical Equations & Stoichiometry

Chemical Equations & Stoichiometry Chemical Equations & Stoichiometry Chapter Goals Balance equations for simple chemical reactions. Perform stoichiometry calculations using balanced chemical equations. Understand the meaning of the term

More information

Victims Compensation Claim Status of All Pending Claims and Claims Decided Within the Last Three Years

Victims Compensation Claim Status of All Pending Claims and Claims Decided Within the Last Three Years Claim#:021914-174 Initials: J.T. Last4SSN: 6996 DOB: 5/3/1970 Crime Date: 4/30/2013 Status: Claim is currently under review. Decision expected within 7 days Claim#:041715-334 Initials: M.S. Last4SSN: 2957

More information

Hydrogen Bonds in Water-Methanol Mixture

Hydrogen Bonds in Water-Methanol Mixture Bulg. J. Phys. 34 (2007) 103 107 Hydrogen Bonds in Water-Methanol Mixture G.M. Georgiev, K. Vasilev, K. Gyamchev Faculty of Physics, University of Sofia 5J.Bourchier Blvd., 1164 Sofia, Bulgaria Received

More information

Reaction Mechanisms. 1. Introduction. What is a reaction mechanism?

Reaction Mechanisms. 1. Introduction. What is a reaction mechanism? Reaction Mechanisms In chemistry, we sometimes find that looking at an overall reaction alone fails to tell us accurate information about the dynamics, an in particular the kinetics, of a reaction. Thus,

More information

7-5.5. Translate chemical symbols and the chemical formulas of common substances to show the component parts of the substances including:

7-5.5. Translate chemical symbols and the chemical formulas of common substances to show the component parts of the substances including: 7-5.5 Translate chemical symbols and the chemical formulas of common substances to show the component parts of the substances including: NaCl [salt], H 2 O [water], C 6 H 12 O 6 [simple sugar], O 2 [oxygen

More information

CHEM 105 HOUR EXAM III 28-OCT-99. = -163 kj/mole determine H f 0 for Ni(CO) 4 (g) = -260 kj/mole determine H f 0 for Cr(CO) 6 (g)

CHEM 105 HOUR EXAM III 28-OCT-99. = -163 kj/mole determine H f 0 for Ni(CO) 4 (g) = -260 kj/mole determine H f 0 for Cr(CO) 6 (g) CHEM 15 HOUR EXAM III 28-OCT-99 NAME (please print) 1. a. given: Ni (s) + 4 CO (g) = Ni(CO) 4 (g) H Rxn = -163 k/mole determine H f for Ni(CO) 4 (g) b. given: Cr (s) + 6 CO (g) = Cr(CO) 6 (g) H Rxn = -26

More information

IB Chemistry 1 Mole. One atom of C-12 has a mass of 12 amu. One mole of C-12 has a mass of 12 g. Grams we can use more easily.

IB Chemistry 1 Mole. One atom of C-12 has a mass of 12 amu. One mole of C-12 has a mass of 12 g. Grams we can use more easily. The Mole Atomic mass units and atoms are not convenient units to work with. The concept of the mole was invented. This was the number of atoms of carbon-12 that were needed to make 12 g of carbon. 1 mole

More information

Chemical Reactions. Chemistry 100. Bettelheim, Brown, Campbell & Farrell. Introduction to General, Organic and Biochemistry Chapter 4

Chemical Reactions. Chemistry 100. Bettelheim, Brown, Campbell & Farrell. Introduction to General, Organic and Biochemistry Chapter 4 Chemistry 100 Bettelheim, Brown, Campbell & Farrell Ninth Edition Introduction to General, Organic and Biochemistry Chapter 4 Chemical Reactions Chemical Reactions In a chemical reaction, one set of chemical

More information

Liquid phase. Balance equation Moles A Stoic. coefficient. Aqueous phase

Liquid phase. Balance equation Moles A Stoic. coefficient. Aqueous phase STOICHIOMETRY Objective The purpose of this exercise is to give you some practice on some Stoichiometry calculations. Discussion The molecular mass of a compound is the sum of the atomic masses of all

More information

rate = k [NO] 2 [H 2 ] CHEMICAL KINETICS Review Exam 3 1. How FAST {Speed like miles per hour and 2. By what MECHANISM Does a Reaction Take Place?

rate = k [NO] 2 [H 2 ] CHEMICAL KINETICS Review Exam 3 1. How FAST {Speed like miles per hour and 2. By what MECHANISM Does a Reaction Take Place? Review Chap 14: CHEMICAL KINETICS Review Exam Chapters 14 15 16 CHEMICAL KINETICS DEALS WITH 1. How FAST {Speed like miles per hour and. By what MECHANISM Does a Reaction Take Place? Given the following

More information

Oxidation-Reduction Reactions

Oxidation-Reduction Reactions Oxidation-Reduction Reactions What is an Oxidation-Reduction, or Redox, reaction? Oxidation-reduction reactions, or redox reactions, are technically defined as any chemical reaction in which the oxidation

More information

EXPERIMENT 12: Empirical Formula of a Compound

EXPERIMENT 12: Empirical Formula of a Compound EXPERIMENT 12: Empirical Formula of a Compound INTRODUCTION Chemical formulas indicate the composition of compounds. A formula that gives only the simplest ratio of the relative number of atoms in a compound

More information

INTI COLLEGE MALAYSIA A? LEVEL PROGRAMME CHM 111: CHEMISTRY MOCK EXAMINATION: DECEMBER 2000 SESSION. 37 74 20 40 60 80 m/e

INTI COLLEGE MALAYSIA A? LEVEL PROGRAMME CHM 111: CHEMISTRY MOCK EXAMINATION: DECEMBER 2000 SESSION. 37 74 20 40 60 80 m/e CHM111(M)/Page 1 of 5 INTI COLLEGE MALAYSIA A? LEVEL PROGRAMME CHM 111: CHEMISTRY MOCK EXAMINATION: DECEMBER 2000 SESSION SECTION A Answer ALL EIGHT questions. (52 marks) 1. The following is the mass spectrum

More information

Survival Organic Chemistry Part I: Molecular Models

Survival Organic Chemistry Part I: Molecular Models Survival Organic Chemistry Part I: Molecular Models The goal in this laboratory experience is to get you so you can easily and quickly move between empirical formulas, molecular formulas, condensed formulas,

More information

AP CHEMISTRY 2009 SCORING GUIDELINES (Form B)

AP CHEMISTRY 2009 SCORING GUIDELINES (Form B) AP CHEMISTRY 2009 SCORING GUIDELINES (Form B) Question 3 (10 points) 2 H 2 O 2 (aq) 2 H 2 O(l) + O 2 (g) The mass of an aqueous solution of H 2 O 2 is 6.951 g. The H 2 O 2 in the solution decomposes completely

More information

Molecular Formula: Example

Molecular Formula: Example Molecular Formula: Example A compound is found to contain 85.63% C and 14.37% H by mass. In another experiment its molar mass is found to be 56.1 g/mol. What is its molecular formula? 1 CHAPTER 3 Chemical

More information

Notes on Unit 4 Acids and Bases

Notes on Unit 4 Acids and Bases Ionization of Water DEMONSTRATION OF CONDUCTIVITY OF TAP WATER AND DISTILLED WATER Pure distilled water still has a small conductivity. Why? There are a few ions present. Almost all the pure water is H

More information

(3)

(3) 1. Organic compounds are often identified by using more than one analytical technique. Some of these techniques were used to identify the compounds in the following reactions. C 3 H 7 Br C 3 H 8 O C 3

More information

List of Part Numbers Manufacturer's No. / Hella No.

List of Part Numbers Manufacturer's No. / Hella No. List of Part Numbers Manufacturer's No. / No. List of part numbers.02-.09 Manufacturer's no. / no..-.53 Note: The original numbers listed in the manufacturer number - number comparison serve exclusively

More information

Chem 1B Dr. White 1. Chapter 14 Acids and Bases. 14.1 Nature of Acids and Bases. A. Acids. B. Bases

Chem 1B Dr. White 1. Chapter 14 Acids and Bases. 14.1 Nature of Acids and Bases. A. Acids. B. Bases Chem 1B Dr. White 1 Chapter 14 Acids and Bases 14.1 Nature of Acids and Bases A. Acids B. Bases Chem 1B Dr. White 2 C. Arrhenius Definition 1. acid 2. base 3. Acid-base reaction involving Arrhenius acids

More information

Solutions Review Questions

Solutions Review Questions Name: Thursday, March 06, 2008 Solutions Review Questions 1. Compared to pure water, an aqueous solution of calcium chloride has a 1. higher boiling point and higher freezing point 3. lower boiling point

More information

Ьа ЮВ 20 р сь Р щ БЗ сч ТЭ С

Ьа ЮВ 20 р сь Р щ БЗ сч ТЭ С ЯИ чл Р щ Ьа ЮВ 20 р сь Р щ БЗ сч ТЭ С ЯИ чл, нз Я Р щ ( я ЛМ : Ьа ЮВ 20 р 8 Йе 30 ЛМ ( Во ),31 ЛМ ( ЛМ ), Мч :B-Con Plaza) тт лх, Йф Р щ ( я ЛМ : Ьа ЮВ 20 р 10 Йе 18 ЛМ ( Во ),19 ЛМ ( ЛМ ), Мч : ЛМ эб

More information

Unit 6 The Mole Concept

Unit 6 The Mole Concept Chemistry Form 3 Page 62 Ms. R. Buttigieg Unit 6 The Mole Concept See Chemistry for You Chapter 28 pg. 352-363 See GCSE Chemistry Chapter 5 pg. 70-79 6.1 Relative atomic mass. The relative atomic mass

More information

Chemical Reactions 2 The Chemical Equation

Chemical Reactions 2 The Chemical Equation Chemical Reactions 2 The Chemical Equation INFORMATION Chemical equations are symbolic devices used to represent actual chemical reactions. The left side of the equation, called the reactants, is separated

More information

The Gas Laws. Our Atmosphere. Pressure = Units of Pressure. Barometer. Chapter 10

The Gas Laws. Our Atmosphere. Pressure = Units of Pressure. Barometer. Chapter 10 Our Atmosphere The Gas Laws 99% N 2 and O 2 78% N 2 80 70 Nitrogen Chapter 10 21% O 2 1% CO 2 and the Noble Gases 60 50 40 Oxygen 30 20 10 0 Gas Carbon dioxide and Noble Gases Pressure Pressure = Force

More information

Chapter 5, Calculations and the Chemical Equation

Chapter 5, Calculations and the Chemical Equation 1. How many iron atoms are present in one mole of iron? Ans. 6.02 1023 atoms 2. How many grams of sulfur are found in 0.150 mol of sulfur? [Use atomic weight: S, 32.06 amu] Ans. 4.81 g 3. How many moles

More information

Chemistry Themed. Types of Reactions

Chemistry Themed. Types of Reactions Chemistry Themed Types of Reactions 1 2 Chemistry in the Community-2015-2016 Types of Reactions Date In-Class Assignment Homework T 10/20 TEST on Reactivity of Metals and Redox None W 10/21 Late Start

More information

3. Which of the following describes a conjugate acid-base pair for the following equilibrium? CN - (aq) + CH 3 NH 3 + (aq) H 2 CO 3 (aq) + H 2 O (l)

3. Which of the following describes a conjugate acid-base pair for the following equilibrium? CN - (aq) + CH 3 NH 3 + (aq) H 2 CO 3 (aq) + H 2 O (l) Acids, Bases & Redox 1 Practice Problems for Assignment 8 1. A substance which produces OH ions in solution is a definition for which of the following? (a) an Arrhenius acid (b) an Arrhenius base (c) a

More information

Writing and Balancing Chemical Equations

Writing and Balancing Chemical Equations Name Writing and Balancing Chemical Equations Period When a substance undergoes a chemical reaction, chemical bonds are broken and new bonds are formed. This results in one or more new substances, often

More information

Oxidation-Reduction Reactions

Oxidation-Reduction Reactions CHAPTER 19 REVIEW Oxidation-Reduction Reactions SECTION 1 SHORT ANSWER Answer the following questions in the space provided. 1. All the following equations involve redox reactions except (a) CaO H 2 O

More information

Chemistry B11 Chapter 4 Chemical reactions

Chemistry B11 Chapter 4 Chemical reactions Chemistry B11 Chapter 4 Chemical reactions Chemical reactions are classified into five groups: A + B AB Synthesis reactions (Combination) H + O H O AB A + B Decomposition reactions (Analysis) NaCl Na +Cl

More information

Chapter 17. How are acids different from bases? Acid Physical properties. Base. Explaining the difference in properties of acids and bases

Chapter 17. How are acids different from bases? Acid Physical properties. Base. Explaining the difference in properties of acids and bases Chapter 17 Acids and Bases How are acids different from bases? Acid Physical properties Base Physical properties Tastes sour Tastes bitter Feels slippery or slimy Chemical properties Chemical properties

More information

Chapter 4 Notes - Types of Chemical Reactions and Solution Chemistry

Chapter 4 Notes - Types of Chemical Reactions and Solution Chemistry AP Chemistry A. Allan Chapter 4 Notes - Types of Chemical Reactions and Solution Chemistry 4.1 Water, the Common Solvent A. Structure of water 1. Oxygen's electronegativity is high (3.5) and hydrogen's

More information

IMPORTANT INFORMATION: S for liquid water is 4.184 J/g degree C

IMPORTANT INFORMATION: S for liquid water is 4.184 J/g degree C FORM A is EXAM II, VERSION 2 (v2) Name 1. DO NOT TURN THIS PAGE UNTIL DIRECTED TO DO SO. 2. These tests are machine graded; therefore, be sure to use a No. 1 or 2 pencil for marking the answer sheets.

More information

EQUILIBRIUM. Consider the reversible system initially consisting of reactants only.

EQUILIBRIUM. Consider the reversible system initially consisting of reactants only. EQUILIBRIUM When non reversible chemical reactions proceed to completion, the concentration of the reactants gradually decrease, until there is NO limiting reactant remaining. Most chemical reactions,

More information

Formulas, Equations and Moles

Formulas, Equations and Moles Chapter 3 Formulas, Equations and Moles Interpreting Chemical Equations You can interpret a balanced chemical equation in many ways. On a microscopic level, two molecules of H 2 react with one molecule

More information

1. Oblast rozvoj spolků a SU UK 1.1. Zvyšování kvalifikace Školení Zapojení do projektů Poradenství 1.2. Financování 1.2.1.

1. Oblast rozvoj spolků a SU UK 1.1. Zvyšování kvalifikace Školení Zapojení do projektů Poradenství 1.2. Financování 1.2.1. 1. O b l a s t r o z v o j s p o l k a S U U K 1. 1. Z v y š o v á n í k v a l i f i k a c e Š k o l e n í o S t u d e n t s k á u n i e U n i v e r z i t y K a r l o v y ( d á l e j e n S U U K ) z í

More information

IB Chemistry. DP Chemistry Review

IB Chemistry. DP Chemistry Review DP Chemistry Review Topic 1: Quantitative chemistry 1.1 The mole concept and Avogadro s constant Assessment statement Apply the mole concept to substances. Determine the number of particles and the amount

More information

Formulae, stoichiometry and the mole concept

Formulae, stoichiometry and the mole concept 3 Formulae, stoichiometry and the mole concept Content 3.1 Symbols, Formulae and Chemical equations 3.2 Concept of Relative Mass 3.3 Mole Concept and Stoichiometry Learning Outcomes Candidates should be

More information

Lingjie Zhao Singlet Oxygen 1. Singlet Oxygen. Lingjie Zhao. Free Radical and Radiation Biology Graduate Program. Department of Radiology B-180 ML

Lingjie Zhao Singlet Oxygen 1. Singlet Oxygen. Lingjie Zhao. Free Radical and Radiation Biology Graduate Program. Department of Radiology B-180 ML Lingjie Zhao Singlet Oxygen 1 Singlet Oxygen By Lingjie Zhao Free Radical and Radiation Biology Graduate Program Department of Radiology B-180 ML The University of Iowa Iowa City, IA 52242-1181 For 77:222,

More information

Chapter 6 Oxidation-Reduction Reactions. Section 6.1 2. Which one of the statements below is true concerning an oxidation-reduction reaction?

Chapter 6 Oxidation-Reduction Reactions. Section 6.1 2. Which one of the statements below is true concerning an oxidation-reduction reaction? Chapter 6 Oxidation-Reduction Reactions 1. Oxidation is defined as a. gain of a proton b. loss of a proton c. gain of an electron! d. loss of an electron e. capture of an electron by a neutron 2. Which

More information

Chapter 20. Thermodynamics p. 811 842. Spontaneity. What have we learned about spontaneity during this course?

Chapter 20. Thermodynamics p. 811 842. Spontaneity. What have we learned about spontaneity during this course? Chapter 20 p. 811 842 Spontaneous process: Ex. Nonspontaneous process: Ex. Spontaneity What have we learned about spontaneity during this course? 1) Q vs. K? 2) So.. Spontaneous process occurs when a system

More information

YIELD YIELD REACTANTS PRODUCTS

YIELD YIELD REACTANTS PRODUCTS Balancing Chemical Equations A Chemical Equation: is a representation of a chemical reaction in terms of chemical formulas Example: 1. Word Description of a Chemical Reaction When methane gas (CH 4 ) burns

More information

How to Quickly Solve Spectrometry Problems

How to Quickly Solve Spectrometry Problems How to Quickly Solve Spectrometry Problems You should be looking for: Mass Spectrometry (MS) Chemical Formula DBE Infrared Spectroscopy (IR) Important Functional Groups o Alcohol O-H o Carboxylic Acid

More information

Project of public e-learning earning portal

Project of public e-learning earning portal Masaryk University Faculty of Informatics Project of public e-learning earning portal Diploma thesis Bc. Jan Bleha Brno, 2013 Keywords E-learning, business plan, business model, machine learning, cloud

More information

Titration Curve of a Weak Acid

Titration Curve of a Weak Acid Titration Curve of a Weak Acid Amina Khalifa El-Ashmawy, Ph.D. Collin College Department of Chemistry Introduction: Titration is an analytical process whereby two reactant solutions are carefully reacted

More information

@ Oxford Fajar Sdn. Bhd. (008974-T) 2012. Matter. 1.1 Atoms and Molecules 1.2 Mole Concept 1.3 Stoichiometry

@ Oxford Fajar Sdn. Bhd. (008974-T) 2012. Matter. 1.1 Atoms and Molecules 1.2 Mole Concept 1.3 Stoichiometry 1 Matter 1.1 Atoms and Molecules 1.2 Mole Concept 1.3 Stoichiometry 2 Chemistry for Matriculation Semester 1 1.1 Atoms and Molecules LEARNING OUTCOMES Describe proton, electron and neutron in terms of

More information

Chapter 7: Chemical Equations. Name: Date: Period:

Chapter 7: Chemical Equations. Name: Date: Period: Chapter 7: Chemical Equations Name: Date: Period: 7-1 What is a chemical reaction? Read pages 232-237 a) Explain what a chemical reaction is. b) Distinguish between evidence that suggests a chemical reaction

More information

Periodic Table, Valency and Formula

Periodic Table, Valency and Formula Periodic Table, Valency and Formula Origins of the Periodic Table Mendelѐѐv in 1869 proposed that a relationship existed between the chemical properties of elements and their atomic masses. He noticed

More information

Honors Chemistry: Unit 6 Test Stoichiometry PRACTICE TEST ANSWER KEY Page 1. A chemical equation. (C-4.4)

Honors Chemistry: Unit 6 Test Stoichiometry PRACTICE TEST ANSWER KEY Page 1. A chemical equation. (C-4.4) Honors Chemistry: Unit 6 Test Stoichiometry PRACTICE TEST ANSWER KEY Page 1 1. 2. 3. 4. 5. 6. Question What is a symbolic representation of a chemical reaction? What 3 things (values) is a mole of a chemical

More information

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. A.P. Chemistry Practice Test - Ch. 13: Equilibrium Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) At equilibrium,. A) the rates of the forward

More information

Lecture 22 The Acid-Base Character of Oxides and Hydroxides in Aqueous Solution

Lecture 22 The Acid-Base Character of Oxides and Hydroxides in Aqueous Solution 2P32 Principles of Inorganic Chemistry Dr. M. Pilkington Lecture 22 The Acid-Base Character of Oxides and Hydroxides in Aqueous Solution Oxides; acidic, basic, amphoteric Classification of oxides - oxide

More information

Chem 1101. 2 cases from last 2 lectures. Highlights of last lecture. Electrochemistry. Electrochemistry. A/Prof Sébastien Perrier

Chem 1101. 2 cases from last 2 lectures. Highlights of last lecture. Electrochemistry. Electrochemistry. A/Prof Sébastien Perrier Chem 1101 A/Prof Sébastien Perrier Room: 351 Phone: 9351-3366 Email: s.perrier@chem.usyd.edu.au Unless otherwise stated, all images in this file have been reproduced from: Blackman, Bottle, Schmid, Mocerino

More information

Unit 8: Chemical Reactions and Equations

Unit 8: Chemical Reactions and Equations 1 Chemical Reactions Unit 8: Chemical Reactions and Equations What are chemical reactions and how do they occur? How are chemical reactions classified? How are products of chemical reactions predicted?

More information

Russian Introductory Course

Russian Introductory Course Russian Introductory Course Natasha Bershadski Learn another language the way you learnt your own Succeed with the and learn another language the way you learnt your own Developed over 50 years, the amazing

More information

Work hard. Be nice. Name: Period: Date: UNIT 1: Introduction to Matter Lesson 4: A Fine Line Between Compounds and Mixtures

Work hard. Be nice. Name: Period: Date: UNIT 1: Introduction to Matter Lesson 4: A Fine Line Between Compounds and Mixtures Name: Period: Date: UNIT 1: Introduction to Matter Lesson 4: A Fine Line Between Compounds and Mixtures Do Now: PRE-READING OPEN-NOTES QUIZ! By the end of today, you will have an answer to: How do pure

More information

NET IONIC EQUATIONS. A balanced chemical equation can describe all chemical reactions, an example of such an equation is:

NET IONIC EQUATIONS. A balanced chemical equation can describe all chemical reactions, an example of such an equation is: NET IONIC EQUATIONS A balanced chemical equation can describe all chemical reactions, an example of such an equation is: NaCl + AgNO 3 AgCl + NaNO 3 In this case, the simple formulas of the various reactants

More information

Nomenclature: How to Name Chemicals

Nomenclature: How to Name Chemicals Nomenclature: How to Name Chemicals Introduction Many of the chemicals we use at home have common names. Baking soda is used as a rising agent in cookies. Bleach is used to whiten our clothes. Ammonia

More information

COMPLIANCE OF MANAGEMENT ACCOUNTING WHEN USING INFORMATION TECHNOLOGIES

COMPLIANCE OF MANAGEMENT ACCOUNTING WHEN USING INFORMATION TECHNOLOGIES Margaryta I. Skrypnyk, Mykola M. Matiukha COMPLIANCE OF MANAGEMENT ACCOUNTING WHEN USING INFORMATION TECHNOLOGIES The article studies the correspondence of management accounting structure when using of

More information

Tutorial 3: Ionization of Water

Tutorial 3: Ionization of Water Tutorial 3: Ionization of Water DEMONSTRATION OF CONDUCTIVITY OF TAP WATER AND DISTILLED WATER - Pure distilled water still has a small conductivity. Why? - There are a few ions present. - Almost all the

More information

The component present in larger proportion is known as solvent.

The component present in larger proportion is known as solvent. 40 Engineering Chemistry and Environmental Studies 2 SOLUTIONS 2. DEFINITION OF SOLUTION, SOLVENT AND SOLUTE When a small amount of sugar (solute) is mixed with water, sugar uniformally dissolves in water

More information

Acid Base Concepts. Arrhenius concept. Hydronium Ion. Page 1

Acid Base Concepts. Arrhenius concept. Hydronium Ion. Page 1 Acid Base Concepts Page 1 The Swedish chemist, Svante Arrhenius, framed the first successful concept of acids and bases. He defined acids and bases in terms of their effect on water. According to Arrhenius,

More information

Chapter 13. Chemical Equilibrium

Chapter 13. Chemical Equilibrium Chapter 13 Chemical Equilibrium Chapter 13 Preview Chemical Equilibrium The Equilibrium condition and constant Chemical equilibrium, reactions, constant expression Equilibrium involving Pressure Chemical

More information

Friday 23 May 2014 Morning

Friday 23 May 2014 Morning Friday 23 May 2014 Morning AS GCE CHEMISTRY A F321/01 Atoms, Bonds and Groups *1168266380* Candidates answer on the Question Paper OCR supplied materials: Data Sheet for Chemistry A (inserted) Other materials

More information

The Mole Concept. A. Atomic Masses and Avogadro s Hypothesis

The Mole Concept. A. Atomic Masses and Avogadro s Hypothesis The Mole Concept A. Atomic Masses and Avogadro s Hypothesis 1. We have learned that compounds are made up of two or more different elements and that elements are composed of atoms. Therefore, compounds

More information

Review - After School Matter Name: Review - After School Matter Tuesday, April 29, 2008

Review - After School Matter Name: Review - After School Matter Tuesday, April 29, 2008 Name: Review - After School Matter Tuesday, April 29, 2008 1. Figure 1 The graph represents the relationship between temperature and time as heat was added uniformly to a substance starting at a solid

More information

Final. Mark Scheme. Chemistry CHEM5. (Specification 2420) Unit 5: Energetics, Redox and Inorganic Chemistry

Final. Mark Scheme. Chemistry CHEM5. (Specification 2420) Unit 5: Energetics, Redox and Inorganic Chemistry Version.2 General Certificate of Education (A-level) January 202 Chemistry CHEM5 (Specification 2420) Unit 5: Energetics, Redox and Inorganic Chemistry Final Mark Scheme Mark schemes are prepared by the

More information

Chapter 11. Homework #3. Electrochemistry. Chapter 4 73. a) Oxidation ½ Reaction Fe + HCl HFeCl 4 Fe + 4HCl HFeCl 4 Fe + 4HCl HFeCl 4 + 3H +

Chapter 11. Homework #3. Electrochemistry. Chapter 4 73. a) Oxidation ½ Reaction Fe + HCl HFeCl 4 Fe + 4HCl HFeCl 4 Fe + 4HCl HFeCl 4 + 3H + hapter 4 73. a) Oxidation ½ Reaction Fe + HlHFel 4 Fe + 4HlHFel 4 Fe + 4HlHFel 4 + 3H + Homework #3 hapter 11 Electrochemistry Fe + 4HlHFel 4 + 3H + + 3e - Reduction ½ Reaction H H + H H + + e- H Balanced

More information

Write chemical equations for the following reactions: Reactants Products Unbalanced. Water H 2 O. and

Write chemical equations for the following reactions: Reactants Products Unbalanced. Water H 2 O. and Chemical Equations 17.2 Chemical symbols provide us with a shorth method of writing the name of an element. Chemical formulas do the same for compounds. But what about chemical reactions? To write out,

More information

Balancing REDOX Reactions: Learn and Practice - KEY

Balancing REDOX Reactions: Learn and Practice - KEY Balancing REDOX Reactions: Learn and Practice - KEY Are these reactions are REDOX reactions? If yes, then balance the reaction using the halfreaction method. 1. 2Au 3+ (aq) + 6I (aq) à 2Au (s) + 3I 2 (s)

More information

Equilibrium Notes Ch 14:

Equilibrium Notes Ch 14: Equilibrium Notes Ch 14: Homework: E q u i l i b r i u m P a g e 1 Read Chapter 14 Work out sample/practice exercises in the sections, Bonus Chapter 14: 23, 27, 29, 31, 39, 41, 45, 51, 57, 63, 77, 83,

More information