BIT Chapters = =
|
|
- Wendy Nichols
- 5 years ago
- Views:
Transcription
1 BIT Chapters K w = [H + ][OH ] = [H + ] = [OH ] = ph = 6.51 The slutin is neither acidic nr basic because the cncentratin f the hydrnium in equals the cncentratin f the hydride in.. The cncentratin f OH is g L 1 The mlar cncentratin f OH is: [OH ] = g OH 1 ml OH 1 L slutin g OH poh = lg[oh ] = lg( M) = 7.56 ph = 14 poh = = 6.44 The water is acidic.. (a) C 6 H 5 OH(aq) + H O(l) C 6 H 5 O (aq) + H O + (aq) (b) K a = (c) C H O 6 5 C H OH 6 5 H O = M [C 6 H 5 OH] = M since the amunt f dissciatin is negligible [H O + ] = lg(ph) = lg(5.07) = M [C 6 H 5 O ] = M since all f the H O + cmes frm the dissciatin f the phenl K a = C H O 6 5 C H OH 6 5 H O pk a = lg K a = lg( ) = = (d) K b = = Ka pk b = lg K b = lg( ) = = (a) pk b = 14 pk a = =. (b) C 7 H SO + H O C 7 H SO H + OH The slutin will be basic K b = C H SO H 7 C H SO 7 OH = = [C 7 H SO ] [C 7 H SO H] [OH ] I C + + E will be large cmpared t M. Therefre we need t slve it either by successive apprimatins r the quadratic equatin. = = [OH ] poh =.1 ph = 11.69
2 BIT Chapters Let Cd stand fr cdeine Cd + H O CdH + + OH K b = CdH = Cd OH = [Cd] [CdH + ] [OH ] I C + + E is large cmpared t , therefre slve fr using either the quadratic equatin r successive apprimatins = = [OH ] poh =.86 ph = 14 ph = CH NH (aq) + H O CH NH + (aq) + OH (aq) K b = CHNH CHNH OH 7. pk b =.6 pk a f its cnjugate acid = 14 pk b = 14.6 = (a) M ascrbic acid slutin =.1 g H C H O 1 ml H C H O (b) H C 6 H 6 O 6 HC 6 H 6 O 6 + H O + K a1 = HC 6 H 6 O 6 C 6 H 6 O 6 + H O + K a = L g H C H O HC H O H C H O H O C H O HC H O H O = 0.14 M H C 6 H 6 O 6 = = Since all f the H O + will cme frm the first equilibrium reactin, it can be calculated as fllws: = = [H C 6 H 6 O 6 ] [HC 6 H 6 O 6 ] [H O + ] I 0.14 C + + E 0.14 HC H O H C H O H O Slve fr =.7 10 = [H O + ] ph = lg[h O + ] = lg(.7 10 ) =.47 [C 6 H 6 O 6 ] = [HC 6 H 6 O 6 ] = [H O + ] in the K a equatin, these cancel
3 BIT Chapters ph = pk a + lg H 4.50 = lg H 0.4 = lg H = H H = (a) C H O + H + HC H O (b) HC H O + OH C H O + H O 11. K a = = 0.10 HO 0.15 [H O + ] = ph = 4.57 C H O HC H O = H O [HC H O ] final = ( )M = 0.11 M [C H O ] final = ( )M = 0.14 M 0.14 HO 0.11 [H O + ] = ph = 4.85 = The change in ph is 0.8 ph units. 1. (a) Prpanic acid, pk a = 4.89 and its salt, sdium prpinate (b) This prblem is best slved using tw equatins and tw unknwns: ph = pk a + lg ml ml H 5.10 = lg H = lg H
4 BIT Chapters = lg 0.11 = lg H H = = H H [ ] = [H] = H H ([H] 0.005) = [H] [H] = 1.88[H] [H] = [H] = M [ ] = (0.095 M) = M (c) Fr 65 ml (0.65 L) f slutin: Fr determining the mass f prpinic acid needed: g C H 6 O = 0.65 L ml C H O g C H O L 1 ml C H O Fr determining the mass f sdium prpinate: g NaC H 5 O = 0.65 L NaC H 5 O Frm the abve calculatin: The cncentratin f prpinic acid is M. The cncentratin f sdium prpinate is M ml NaC H O g NaC H O L 1 ml NaC H O = 1.8 g C H 6 O 5 =.04 g 1. (a) Neutral (b) cidic (c) cidic (d) Basic 14. pk b = 14 pk a = = 9.1 K b = H OH = ll f the acid has been cnverted int the salt at the equivalence pint. (0.115 M)(50.00 ml) = (0.100 M)( ml) = ml ttal vlume is ml [ ] = M [H] = [OH ] =
5 BIT Chapters K b1 = = = M = [OH ] poh = 5.01 ph = Thyml blue r phenlphthalein wuld be gd indicatrs. K w K a HC H O C H O OH 14 1 = [C 6 H 6 O 6 ] [HC 6 H 6 O 6 ] [OH ] I C + + E = = is t large t make a simplifying assumptin, therefre slve fr either using the quadratic equatin r by successive apprimatins. = = [OH ] poh = 1.8 ph = Since half f the acid is neutralized the cncentratin f the acid is equal t the cncentratin f its cnjugate base, the pk a can be determined: ph = pk a + lg H.56 = pk a + lg 1 pk a = ph =.56 K a = 10 pka = = ph = pk a + lg H pk a = 14 pk b pk b = lg K b = lg pk a = = = lg H [ ] = [H] [NH + 4 ] = befre the additin f NaOH. In rder fr the tw cncentratins t be equal, half as much NaOH must be added: (0.100 L NH + 4 )(0.100 M NH slutin) = ml NH 4 ( ml NH + 4 )(0.5) = ml NaOH g NaOH ml NaOH = 0.00 g NaOH 1 ml NaOH 18. K sp = [g + ] [CrO 4 ] [g + ] = ( M) = M [CrO 4 ] = M K sp = ( ) ( )
6 BIT Chapters 17-0 K sp = Mg(OH) (s) Mg + (aq) + OH (aq) K sp = = [Mg + ][OH ] = [][] = = = [OH ] poh = lg[oh ] = lg( ) =.66 ph = = Fe(OH) (s) Fe + (aq) + OH (aq) k sp = = [Fe + ][OH ] ph = poh = = 4.00 [OH ] = = = [Fe + ][ ] M = [Fe + ] ml Fe 1 ml Fe(OH) g Fe(OH) g L 1 = 1 L slutin + 1 ml Fe 1 ml Fe(OH) 1. (e) = g L 1. (a) This is a limiting reagent prblem. (b) 1 ml Pb(NO ml KI needed = ) 1.04 g Pb(NO ) 1. g ml KI 1000 ml KI slutin 1 ml Pb(NO ) ml KI 0.0 ml KI slutin is supplied. KI is in ecess. 1 ml Pb(NO g PbI = ) 1.04 g Pb(NO ) 1. g 1 ml PbI g PbI 1 ml Pb(NO ) 1 ml PbI = 1.45 g PbI = 1.6 ml KI slutin The cncentratin f the spectatr ins: [K + ] = (0.500 M K + )(0.0 ml slutin) 50.0 ml = 0.00 M K + [NO ] = (0.100)()(0.0 ml slutin) 50.0 ml = 0.1 M NO [I ]: Ttal mles f I = (0.500 M I )(0.000 L slutin) = ml I ml I used = 1.04 g Pb(NO ) 1 ml Pb(NO ) ml I + 1. g 1 ml Pb = ml I [I ] = ( ml I ml I )/(0.050 L slutin) = M I [Pb + ]: This will be the amunt f Pb + that is in slutin after the slid PbI reaches equilibrium PbI (s) Pb + (aq) + I (aq)
7 BIT Chapters 17-0 [Pb + ] [I ] I M C + + E K sp = = [Pb + ][I ] = ()( ) << = ()(0.0740) = M = [Pb + ]. This is a simultaneus equilibrium prblem. CuCO (s) Cu + (aq) + CO (aq) K sp = [Cu + ][CO ] = Cu + (aq) + 4NH (aq) Cu(NH ) 4 + (aq) K frm = Cu NH Cu 4 4 NH = CuCO (s) + 4NH (aq) K verall = Cu(NH ) 4 + (aq) + CO (aq) Cu NH CO 4 NH 4 = ll f the Cu(NH ) + 4 cmes frm the CuCO [Cu(NH ) + 1 ml CuCO 4 ] = 1.00 g CuCO 1 = g CuCO 1.00 L slutin s the Cu + is used t frm Cu(NH ) + 4, the [CO ] = [Cu(NH ) + 4 ] = = Cu NH CO 4 NH 4 = = M NH ml NH ml NH = 1.00 L slutin 1 L slutin =0.014 ml NH 4. ssume the slutin is saturated with CO and therefre the cncentratin f H CO is 0.00 M H CO H + + CO K = K a1 K a = ( )( ) = = H CO H CO First, calculate the [H + ] at which thepbco will begin t precipitate: PbCO (s) Pb + (aq) + CO (aq) K sp = = [Pb + ][CO ] [Pb + ] = M [CO ] = ( )/(0.010) = M [H + ] = ph = H CO CO = Nw, determine the [H + ] at which the BaCO will begin t precipitate: BaCO (s) Ba + (aq) + CO (aq) K sp = = [Ba + ][CO ]
8 BIT Chapters 17-0 [Ba + ] = M [CO ] = ( )/(0.010) = M [H + ] = ph = H CO CO.6 10 = The ph range is between.50 and 5.78, abve 5.78, BaCO will begin t precipitate. 5. The less sluble substance is PbS. We need t determine the minimum [H + ] at which NiS will precipitate. + Ni HS (0.100)(0.1) spa + + K = = = 40 (frm Table 18.) H [H ] + (0.10)(0.1) [H ] = = ph = lg[h + ] = t a ph lwer than 1.80, PbS will precipitate and NiS will nt. t larger values f ph, bth PbS and NiS will precipitate. We als need t determine the [H + ] at which PbS will start t precipitate + Pb HS (0.100)(0.1) 7 spa + + K = = = 10 (frm Table 18.) H [H ] + (0.10)(0.1) [H ] = = ph = lg[h + ] =.6. ny acid in water will precipitate the PbS. The ph range is.6 t 1.80 t allw the PbS t precipitate withut the NiS. 6. The less sluble substance is SnS. We need t determine the minimum [H S] at which FeS will precipitate. + spa + Fe H S (0.10) H S K = = = 600 (frm Table 18.) H [1 10 ] (600)(1 10 ) HS = = 6 10 (0.1) The cncentratin f Sn + can nw be determined. + + Sn HS Sn spa + K = = = 1 10 (frm Table 18.) H [1 10 ] + 5 (1 10 )(1 10 ) 9 Sn = = (6 10 ) 7. (a) MS(s) M + (aq) + S (aq) K sp = [M + ][S ] = MS(s) + H O M + (aq) + HS (aq) + OH (aq)
9 BIT Chapters 17-0 K spa = M H H S = ( )(10 1 ) = (b) (c) 0.0 = = M MS wuld be cnsidered an acid insluble sulfide making M a grup in. Grup ins frm insluble sulfides in base. 8. NO(g) NO (g) + N O(g) G = {1 ml G f [NO (g)] + 1 ml G f [N O(g)]} { ml G f [NO(g)]} G = {1 ml (51.84 kj/ml) + 1 ml (10.6 kj/ml)} { ml ( kj/ml)} G = 416 kj G = RTlnK P J = (8.14 J ml 1 K 1 )(98 K)(ln K P ) ln K P = 168 K P = K P = K c (RT) ng = K c [(0.081 L atm ml 1 K 1 )(98)] 1 K c = CH 4 (g) + Cl (g) CH Cl(g) + HCl(g) f f ΔG = { ΔG [CH Cl(g)] + ΔG [HCl(g)]} { ΔG [CH 4 (g)] + ΔG = {1 ml ( 58.6 kj/ml) + 1 ml ( 95.7 kj/ml)} {1 ml ( kj/ml) + 1 ml (0 kj/ml)} ΔG = kj = J f ΔG f [Cl (g)]} G = RTlnK P J = (8.14 J ml 1 K 1 )(47 K)(ln K P ) ln K P = 6.19 K P = K P = K c (RT) ng = K c [(0.081 L atm ml 1 K 1 )(47)] 0 K c = CH Cl HCl Kc.4 10 CH Cl ml 4.56 g g CH4 0.14M.00 L
10 BIT Chapters ml 8.67 g g Cl M.00 L Sme CH 4 is lst t frm CH Cl and sme Cl is lst t frm HCl. [CH 4 ] [Cl ] [CH Cl] [HCl] I C + + E Kc The equilibrium lies s far t the right that yu d nt need t slve the equatin. Cl is the limiting reagent and thus will be cmpletely cnsumued in the reactin. t equilibrium the fllwing cncentratins eist: [Cl ] = 0 ( actually it is slightly greater than zer but etremely small) [CH 4 ] = (.0 L 0.14 M.0 L M)/.0 L = M [CH Cl] = [ HCl] = M 0. First calculate the atmizatin energy fr the frmatin f C H 6, and the atmizatin energy fr the C H and H. The difference is the energy required fr the reactin. Then calculate the amunt f heat required fr 5.0 g f C H 6. atmizatin energy 1 = (6 ml C H B.E.) + (1 ml C C B.E.) = (6 ml 41 kj ml 1 ) + (1 ml 48 kj ml 1 ) = 80 kj atmizatin energy = ( ml C H B.E.) + (1 ml C C B.E.) + (1 ml H H B.E.) = ( ml 41 kj ml 1 ) + (1 ml 960 kj ml 1 ) + (1 ml 46 kj ml 1 ) = 0 kj atmizatin energy 1 atmizatin energy = 80 kj 0 kj = 600 kj 600 kj are absrbed. kj fr 5.0 g = 5.0 g 1 ml CH6 600 kj 6.04 g C H 1 ml C H 6 6 = 576 kj 1. (a) mle e = 0.0 min 60 s 1.00 C 1 ml e 4 1 min 1 s C = ml e mle OH = ml e 1 ml e 1 ml OH [OH ml OH ] = 0.50 L NaCl slutin poh = lg[oh ] = 1.0 ph = 14 poh ph = = = M OH = ml OH
11 BIT Chapters 17-0 (b) mle e 60 s 5.00 C 1 ml e = 10.0 min 4 1 min 1 s C = ml e 1 ml H mle H = ml e ml e = ml H ml H = ml H L atm ml K 7 K 1000 ml 1 atm 1 L = 48 ml. (a) Eternal circuit ( ) electrn flw (+) Fe Salt Bridge Cu Fe + (aq) nde Cu + (aq) Cathde (b) (c) (d) (e) Fe(s) + Cu + (aq) Cu(s) + Fe + (aq) Fe Fe + Cu + Cu E cell = E substance reduced E substance idized Ecell E + E Cu Fe E cell = 0.4 V ( 0.44 V) = 0.78 V e = 50.0 h 600 s 0.10 C 1 ml e 1 h s 96,500 C = ml e 1 ml Fe ml Fe + = ml e ml e = ml Fe + The change in cncentratin f Fe + will be ml Fe = M Fe L slutin The final cncentratin f Fe + will be: 1.00 M M Fe + = 1.94 M Fe + The change in cncentratin f Cu + will be ml Cu = 0.95 M Cu L slutin The final cncentratin f Cu + will be: 1.00 M 0.95 M Cu + = M Cu +
12 BIT Chapters RT Fe cell cell E = E ln nf Cu -1-1 (8.14 J ml K )(98 K) E cell =0.78 V ln -1 (96,500 C ml ) = 0.78 V (.96) E cell = 0.76 V (a) gcl(s) + Ni(s) g(s) + Cl (aq) + Ni + (aq) (b) E cell = E reductin E idatin = 0. V ( 0.57 V) = V 4. E cell 0.80 V 0.4 V 0.46 V + RT Cu cell cell E = E ln nf g -1-1 (8.14 J ml K )(98 K) 0.00 E =0.46 V ln (96,500 C ml ) cell -1 = 0.46 V (.996) V = 0.4 V cell ptential drp f 10% wuld be 0.04 V s the new ptential wuld be 0.78 V (8.14 J ml K )(98 K) = 0.46 V ln (96,500 C ml ) V = V ln e = M This is the cncentratin increase that wuld ccur during the time perid that the cell ptential decreased by 10 %. Each cell has a vlume f 15 ml. ml Cu + = M 0.15 L = ml Cu +
13 BIT Chapters 17-0 t = q/i t = ml Cu 0.10 ml e 96,500 C ml Cu 1 ml e 966 s r.67 hr Fr a 15 ml cell, the cpper electrde wuld lse: 6.55 g ml Cu 0.17 g 1 ml Cu The silver electrde wuld gain: g ml g 1.08 g 1 ml Cu The ttal mass f the cell remains unchanged accrding t the law f cnservatin f mass-energy.
Semester 2 AP Chemistry Unit 12
Cmmn In Effect and Buffers PwerPint The cmmn in effect The shift in equilibrium caused by the additin f a cmpund having an in in cmmn with the disslved substance The presence f the excess ins frm the disslved
More informationCHM 152 Practice Final
CM 152 Practice Final 1. Of the fllwing, the ne that wuld have the greatest entrpy (if cmpared at the same temperature) is, [a] 2 O (s) [b] 2 O (l) [c] 2 O (g) [d] All wuld have the same entrpy at the
More informationUniversity Chemistry Quiz /04/21 1. (10%) Consider the oxidation of ammonia:
University Chemistry Quiz 3 2015/04/21 1. (10%) Cnsider the xidatin f ammnia: 4NH 3 (g) + 3O 2 (g) 2N 2 (g) + 6H 2 O(l) (a) Calculate the ΔG fr the reactin. (b) If this reactin were used in a fuel cell,
More informationChapter 19. Electrochemistry. Dr. Al Saadi. Electrochemistry
Chapter 19 lectrchemistry Part I Dr. Al Saadi 1 lectrchemistry What is electrchemistry? It is a branch f chemistry that studies chemical reactins called redx reactins which invlve electrn transfer. 19.1
More informationFind this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.
Find this material useful? Yu can help ur team t keep this site up and bring yu even mre cntent cnsider dnating via the link n ur site. Still having truble understanding the material? Check ut ur Tutring
More informationFind this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.
Find this material useful? Yu can help ur team t keep this site up and bring yu even mre cntent cnsider dnating via the link n ur site. Still having truble understanding the material? Check ut ur Tutring
More informationFind this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.
Find this material useful? Yu can help ur team t keep this site up and bring yu even mre cntent cnsider dnating via the link n ur site. Still having truble understanding the material? Check ut ur Tutring
More informationCHEM 2400/2480. Lecture 19
Lecture 19 Metal In Indicatr - a cmpund whse clur changes when it binds t a metal in - t be useful, it must bind the metal less strngly than EDTA e.g. titratin f Mg 2+ with EDTA using erichrme black T
More informationCHEM 1032 FALL 2017 Practice Exam 4 1. Which of the following reactions is spontaneous under normal and standard conditions?
1 CHEM 1032 FALL 2017 Practice Exam 4 1. Which f the fllwing reactins is spntaneus under nrmal and standard cnditins? A. 2 NaCl(aq) 2 Na(s) + Cl2(g) B. CaBr2(aq) + 2 H2O(aq) Ca(OH)2(aq) + 2 HBr(aq) C.
More informationChapter 17 Free Energy and Thermodynamics
Chemistry: A Mlecular Apprach, 1 st Ed. Nivald Tr Chapter 17 Free Energy and Thermdynamics Ry Kennedy Massachusetts Bay Cmmunity Cllege Wellesley Hills, MA 2008, Prentice Hall First Law f Thermdynamics
More informationUniversity of Waterloo DEPARTMENT OF CHEMISTRY CHEM 123 Test #2 Wednesday, March 11, 2009
University f Waterl DEPARTMENT OF CHEMISTRY CHEM 13 Test # Wednesday, March 11, 009 This is test versin 001. Fill in vals 001 fr the Card Number (r Test Master) n yur cmputer answer card. Name (Print in
More informationElectrochemistry. Reduction: the gaining of electrons. Reducing agent (reductant): species that donates electrons to reduce another reagent.
Electrchemistry Review: Reductin: the gaining f electrns Oxidatin: the lss f electrns Reducing agent (reductant): species that dnates electrns t reduce anther reagent. Oxidizing agent (xidant): species
More informationCHAPTER 6 / HARVEY A. CHEMICAL EQUILIBRIUM B. THERMODYNAMICS AND EQUILIBRIUM C. MANUPULATING EQUILIBRIUM CONSTANTS
CHPTER 6 / HRVEY. CHEMICL B. THERMODYNMICS ND C. MNUPULTING CONSTNTS D. CONSTNTS FOR CHEMICL RECTIONS 1. Precipitatin Reactins 2. cid-base Reactins 3. Cmplexatin Reactins 4. Oxidatin-Reductin Reactins
More informationIn the half reaction I 2 2 I the iodine is (a) reduced (b) oxidized (c) neither of the above
6.3-110 In the half reactin I 2 2 I the idine is (a) reduced (b) xidized (c) neither f the abve 6.3-120 Vitamin C is an "antixidant". This is because it (a) xidizes readily (b) is an xidizing agent (c)
More informationThree Definitions of Acids/Bases Type Acid Base Problems with it Arrhenius Bronsted-Lowry Lewis. NH 3(aq) + H 2O(l)
CP NT Ch 19: Acid and Bases An Intrductin Prperties f Acids 1. taste 2. Can prduce H + ( ) ins ( ) 3. Change the clr f litmus frm t 4. Reacts with such as Zn and Mg t prduce gas. Ba(s) + H 2SO 4 BaSO 4(aq)
More informationFind this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.
Find this material useful? Yu can help ur team t keep this site up and bring yu even mre cntent cnsider dnating via the link n ur site. Still having truble understanding the material? Check ut ur Tutring
More informationNUPOC STUDY GUIDE ANSWER KEY. Navy Recruiting Command
NUPOC SUDY GUIDE ANSWER KEY Navy Recruiting Cmmand CHEMISRY. ph represents the cncentratin f H ins in a slutin, [H ]. ph is a lg scale base and equal t lg[h ]. A ph f 7 is a neutral slutin. PH < 7 is acidic
More informationUnit 14 Thermochemistry Notes
Name KEY Perid CRHS Academic Chemistry Unit 14 Thermchemistry Ntes Quiz Date Exam Date Lab Dates Ntes, Hmewrk, Exam Reviews and Their KEYS lcated n CRHS Academic Chemistry Website: https://cincchem.pbwrks.cm
More informationCHEMISTRY 16 HOUR EXAM IV KEY April 23, 1998 Dr. Finklea. 1. The anti-cancer drug cis-platin is the complex: cis-[pt(nh ) (Cl) ]. In this complex, the
CHEMISTRY 16 HOUR EXAM IV KEY April 23, 1998 Dr. Finklea Sme useful cnstants: ln(10) = 2.303, R = 8.314 J/ml@K, F = 96,00 cul/ml, 2.303RT/F = 0.0916 V at 2EC. Assume a temperature f 2EC unless tld therwise.
More informationChapter 8 Reduction and oxidation
Chapter 8 Reductin and xidatin Redx reactins and xidatin states Reductin ptentials and Gibbs energy Nernst equatin Disprprtinatin Ptential diagrams Frst-Ebswrth diagrams Ellingham diagrams Oxidatin refers
More informationCHAPTER Read Chapter 17, sections 1,2,3. End of Chapter problems: 25
CHAPTER 17 1. Read Chapter 17, sectins 1,2,3. End f Chapter prblems: 25 2. Suppse yu are playing a game that uses tw dice. If yu cunt the dts n the dice, yu culd have anywhere frm 2 t 12. The ways f prducing
More informationChem 163 Section: Team Number: ALE 24. Voltaic Cells and Standard Cell Potentials. (Reference: 21.2 and 21.3 Silberberg 5 th edition)
Name Chem 163 Sectin: Team Number: ALE 24. Vltaic Cells and Standard Cell Ptentials (Reference: 21.2 and 21.3 Silberberg 5 th editin) What des a vltmeter reading tell us? The Mdel: Standard Reductin and
More informationFind this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.
Find this material useful? Yu can help ur team t keep this site up and bring yu even mre cntent cnsider dnating via the link n ur site. Still having truble understanding the material? Check ut ur Tutring
More informationFind this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.
Find this material useful? Yu can help ur team t eep this site up and bring yu even mre cntent cnsider dnating via the lin n ur site. Still having truble understanding the material? Chec ut ur Tutring
More informationAP CHEMISTRY CHAPTER 6 NOTES THERMOCHEMISTRY
AP CHEMISTRY CHAPTER 6 NOTES THERMOCHEMISTRY Energy- the capacity t d wrk r t prduce heat 1 st Law f Thermdynamics: Law f Cnservatin f Energy- energy can be cnverted frm ne frm t anther but it can be neither
More informationChem 112, Fall 05 (Weis/Garman) Exam 4A, December 14, 2005 (Print Clearly) +2 points
+2 pints Befre yu begin, make sure that yur exam has all 7 pages. There are 14 required prblems (7 pints each) and tw extra credit prblems (5 pints each). Stay fcused, stay calm. Wrk steadily thrugh yur
More informationAcids and Bases Lesson 3
Acids and Bases Lessn 3 The ph f a slutin is defined as the negative lgarithm, t the base ten, f the hydrnium in cncentratin. In a neutral slutin at 25 C, the hydrnium in and the hydrxide in cncentratins
More informationExamples: 1. How much heat is given off by a 50.0 g sample of copper when it cools from 80.0 to 50.0 C?
NOTES: Thermchemistry Part 1 - Heat HEAT- TEMPERATURE - Thermchemistry: the study f energy (in the frm f heat) changes that accmpany physical & chemical changes heat flws frm high t lw (ht cl) endthermic
More informationLecture 16 Thermodynamics II
Lecture 16 Thermdynamics II Calrimetry Hess s Law Enthalpy r Frmatin Cpyright 2013, 2011, 2009, 2008 AP Chem Slutins. All rights reserved. Fur Methds fr Finding H 1) Calculate it using average bnd enthalpies
More informationSCH4U: End of Year Review
SCH4U: End f Year Review Unit 1: Energy and Rates 2 1) When 13.4 g f ammnium chlride disslve int 2.00x10 g f water the temperature changes frm 20.0 C t 15.3 C. Determine the mlar enthalpy f slutin f ammnium
More informationChapter 9 Chemical Reactions NOTES
Chapter 9 Chemical Reactins NOTES Chemical Reactins Chemical reactin: Chemical change 4 Indicatrs f Chemical Change: (1) (2) (3) (4) Cnsist f reactants (starting materials) and prducts (substances frmed)
More informationA Chemical Reaction occurs when the of a substance changes.
Perid: Unit 8 Chemical Reactin- Guided Ntes Chemical Reactins A Chemical Reactin ccurs when the f a substance changes. Chemical Reactin: ne r mre substances are changed int ne r mre new substances by the
More informationThermochemistry. The study of energy changes that occur during chemical : at constant volume ΔU = q V. no at constant pressure ΔH = q P
Thermchemistry The study energy changes that ccur during chemical : at cnstant vlume ΔU = q V n at cnstant pressure = q P nly wrk Fr practical reasns mst measurements are made at cnstant, s thermchemistry
More informationEdexcel IGCSE Chemistry. Topic 1: Principles of chemistry. Chemical formulae, equations and calculations. Notes.
Edexcel IGCSE Chemistry Tpic 1: Principles f chemistry Chemical frmulae, equatins and calculatins Ntes 1.25 write wrd equatins and balanced chemical equatins (including state symbls): fr reactins studied
More informationThe nature acids: taste sour bases: taste bitter, feel slippery
7 Acids and bases The nature acids: taste sur bases: taste bitter, feel slippery Definitin Arrhenius (1859-197) acid and base (1887) Acid Prduces hydrgen ins (H + ) in aqueus slutin Base Prduces hydride
More informationCHAPTER 21 ELECTROCHEMISTRY: CHEMICAL CHANGE AND ELECTRICAL WORK
CHAPTR 1 LCTROCHMISTRY: CHMICAL CHANG AND LCTRICAL WORK 1.1 Oxidatin is the lss f electrns (resulting in a higher xidatin number), while reductin is the gain f electrns (resulting in a lwer xidatin number).
More informationN 2 (g) + 3H 2 (g) 2NH 3 (g) o Three mole ratios can be derived from the balanced equation above: Example: Li(s) + O 2 (g) Li 2 O(s)
Chapter 9 - Stichimetry Sectin 9.1 Intrductin t Stichimetry Types f Stichimetry Prblems Given is in mles and unknwn is in mles. Given is in mles and unknwn is in mass (grams). Given is in mass and unknwn
More informationHow can standard heats of formation be used to calculate the heat of a reaction?
Answer Key ALE 28. ess s Law and Standard Enthalpies Frmatin (Reerence: Chapter 6 - Silberberg 4 th editin) Imprtant!! Fr answers that invlve a calculatin yu must shw yur wrk neatly using dimensinal analysis
More informationThermodynamics and Equilibrium
Thermdynamics and Equilibrium Thermdynamics Thermdynamics is the study f the relatinship between heat and ther frms f energy in a chemical r physical prcess. We intrduced the thermdynamic prperty f enthalpy,
More informationFinals Study Guide. AP Chemistry. Spring Made by Ashley Thomas
Spring 2016 Finals Study Guide AP Chemistry Table f Cntents Sectin 1: Acids and Bases 1.1 Titratins 1.2 Indicatrs 1.3 Buffers 1.4 Slutin Preparatin Sectin 2: Calrimetry 2.1 Specific Heat 2.2 Metals Sectin
More informationIn the spaces provided, explain the meanings of the following terms. You may use an equation or diagram where appropriate.
CEM1405 2007-J-2 June 2007 In the spaces prvided, explain the meanings f the fllwing terms. Yu may use an equatin r diagram where apprpriate. 5 (a) hydrgen bnding An unusually strng diple-diple interactin
More informationTuesday, 5:10PM FORM A March 18,
Name Chemistry 153-080 (3150:153-080) EXAM II Multiple-Chice Prtin Instructins: Tuesday, 5:10PM FORM A March 18, 2003 120 1. Each student is respnsible fr fllwing instructins. Read this page carefully.
More informationChem 116 POGIL Worksheet - Week 3 - Solutions Intermolecular Forces, Liquids, Solids, and Solutions
Chem 116 POGIL Wrksheet - Week 3 - Slutins Intermlecular Frces, Liquids, Slids, and Slutins Key Questins 1. Is the average kinetic energy f mlecules greater r lesser than the energy f intermlecular frces
More informationChapter 4 Thermodynamics and Equilibrium
Chapter Thermdynamics and Equilibrium Refer t the fllwing figures fr Exercises 1-6. Each represents the energies f fur mlecules at a given instant, and the dtted lines represent the allwed energies. Assume
More informationReview Material for Exam #2
Review Material fr Eam # 1. a. Calculate the mlarity f a slutin made with 184.6 mg sample f ptassium dichrmate disslved in enugh water t give 500.0 ml f slutin. 1 g 184.6 mg K Cr O 1 mle K Cr O 7 7 1000
More informationChem 111 Summer 2013 Key III Whelan
Chem 111 Summer 2013 Key III Whelan Questin 1 6 Pints Classify each f the fllwing mlecules as plar r nnplar? a) NO + : c) CH 2 Cl 2 : b) XeF 4 : Questin 2 The hypthetical mlecule PY 3 Z 2 has the general
More informationChapters 29 and 35 Thermochemistry and Chemical Thermodynamics
Chapters 9 and 35 Thermchemistry and Chemical Thermdynamics 1 Cpyright (c) 011 by Michael A. Janusa, PhD. All rights reserved. Thermchemistry Thermchemistry is the study f the energy effects that accmpany
More informationChemistry 132 NT. Electrochemistry. Review
Chemistry 132 NT If yu g flying back thrugh time, and yu see smebdy else flying frward int the future, it s prbably best t avid eye cntact. Jack Handey 1 Chem 132 NT Electrchemistry Mdule 3 Vltaic Cells
More informationChem 75 February 16, 2017 Exam 2 Solutions
1. (6 + 6 pints) Tw quick questins: (a) The Handbk f Chemistry and Physics tells us, crrectly, that CCl 4 bils nrmally at 76.7 C, but its mlar enthalpy f vaprizatin is listed in ne place as 34.6 kj ml
More information2004 AP CHEMISTRY FREE-RESPONSE QUESTIONS
2004 AP CHEMISTRY FREE-RESPONSE QUESTIONS 6. An electrchemical cell is cnstructed with an pen switch, as shwn in the diagram abve. A strip f Sn and a strip f an unknwn metal, X, are used as electrdes.
More informationChem 116 POGIL Worksheet - Week 4 Properties of Solutions
Chem 116 POGIL Wrksheet - Week 4 Prperties f Slutins Key Questins 1. Identify the principal type f slute-slvent interactin that is respnsible fr frming the fllwing slutins: (a) KNO 3 in water; (b) Br 2
More informationHow can standard heats of formation be used to calculate the heat of a reaction?
Name Chem 161, Sectin: Grup Number: ALE 28. Hess s Law and Standard Enthalpies Frmatin (Reerence: Chapter 6 - Silberberg 5 th editin) Imprtant!! Fr answers that invlve a calculatin yu must shw yur wrk
More informationSECTION I (Multiple Choice Questions)
ANDHERI / BORIVALI / DADAR / CHEMBUR / THANE / MULUND/ NERUL / POWAI IIT JEE - 09 CRASH COURSE (MAIN) MARKS:90 TIME: 5 MIN. TOPIC: CHEMICAL & IONIC EQUILIBRIUM DATE:9//8 SECTION I (Multiple Chice Questins)
More information15.0 g Cr = 21.9 g Cr O g Cr 4 mol Cr mol Cr O
WYSE Academic Challenge Sectinal Chemistry Exam 2008 SOLUTION SET 1. Crrect answer: B. Use PV = nrt t get: PV = nrt 2. Crrect answer: A. (2.18 atm)(25.0 L) = n(0.08206 L atm/ml K)(23+273) n = 2.24 ml Assume
More informationCHEM Dr. Babb s Sections Exam #4 Review Sheet
CHEM 116 - Dr. Babb s Sections Exam #4 Review Sheet 158. Explain using the HC 2 H 3 O 2 /NaC 2 H 3 O 2 buffer system how a buffer maintains a relatively constant ph when small quantity of acid (HCl) or
More informationFind this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.
Find this material useful? Yu can help ur team t keep this site up and bring yu even mre cntent cnsider dnating via the link n ur site. Still having truble understanding the material? Check ut ur Tutring
More information4 Fe + 3 O 2 2 Fe 2 O 3
UNIT 7: STOICHIOMETRY NOTES (chapter 9) INTRO TO STOICHIOMETRY Reactin Stichimetry: Stichimetry is simply a way t shw f smething this is. Relatinship between a given and an unknwn: GIVEN UNKNOWN Type 1
More informationUnit 13 Acids and Bases. Name: Period: TEST: Wednesday 4/27/16
Unit 13 Acids and Bases Name: Perid: TEST: Wednesday 4/27/16 1 Unit 13 Acids and Bases Calendar Mnday Tuesday Wednesday Thursday Friday APRIL 7th Slutins Test 8 th Vide: Acids & Bases Ntes #1 Acids and
More informationCHM 112 Dr. Kevin Moore
CHM 112 Dr. Kevin Moore Reaction of an acid with a known concentration of base to determine the exact amount of the acid Requires that the equilibrium of the reaction be significantly to the right Determination
More informationPart One: Heat Changes and Thermochemistry. This aspect of Thermodynamics was dealt with in Chapter 6. (Review)
CHAPTER 18: THERMODYNAMICS AND EQUILIBRIUM Part One: Heat Changes and Thermchemistry This aspect f Thermdynamics was dealt with in Chapter 6. (Review) A. Statement f First Law. (Sectin 18.1) 1. U ttal
More informationCHEM Thermodynamics. Change in Gibbs Free Energy, G. Review. Gibbs Free Energy, G. Review
Review Accrding t the nd law f Thermdynamics, a prcess is spntaneus if S universe = S system + S surrundings > 0 Even thugh S system
More information19 Applications of Standard Electrode Potentials
9 Applicatins f Standard lectrde Ptentials ( Calculating thermdynamic cell ptentials ( Calculating equilibrium cnstants fr redx reactins ( Cnstructing redx titratin curves 9A Calculating Ptentials f lectrchemical
More informationCHAPTER PRACTICE PROBLEMS CHEMISTRY
Chemical Kinetics Name: Batch: Date: Rate f reactin. 4NH 3 (g) + 5O (g) à 4NO (g) + 6 H O (g) If the rate f frmatin f NO is 3.6 0 3 ml L s, calculate (i) the rate f disappearance f NH 3 (ii) rate f frmatin
More informationCHEM 1001 Problem Set #3: Entropy and Free Energy
CHEM 1001 Prblem Set #3: Entry and Free Energy 19.7 (a) Negative; A liquid (mderate entry) cmbines with a slid t frm anther slid. (b)psitive; One mle f high entry gas frms where n gas was resent befre.
More informationStrategy Write the two half-cell reactions and identify the oxidation and reduction reactions. Pt H2 (g) H + (aq)
Slutins manual fr Burrws et.al. Chemistry 3 Third editin 16 Electrchemistry Answers t wrked examples WE 16.1 Drawing a cell diagram (n p. 739 in Chemistry 3 ) Draw a cell diagram fr an electrchemical cell
More informationThermodynamics Partial Outline of Topics
Thermdynamics Partial Outline f Tpics I. The secnd law f thermdynamics addresses the issue f spntaneity and invlves a functin called entrpy (S): If a prcess is spntaneus, then Suniverse > 0 (2 nd Law!)
More informationChapter 17: Thermodynamics: Spontaneous and Nonspontaneous Reactions and Processes
Chapter 17: hermdynamics: Spntaneus and Nnspntaneus Reactins and Prcesses Learning Objectives 17.1: Spntaneus Prcesses Cmparing and Cntrasting the hree Laws f hermdynamics (1 st Law: Chap. 5; 2 nd & 3
More informationWhen a substance heats up (absorbs heat) it is an endothermic reaction with a (+)q
Chemistry Ntes Lecture 15 [st] 3/6/09 IMPORTANT NOTES: -( We finished using the lecture slides frm lecture 14) -In class the challenge prblem was passed ut, it is due Tuesday at :00 P.M. SHARP, :01 is
More informationAP Chemistry Assessment 2
AP Chemistry Assessment 2 DATE OF ADMINISTRATION: January 8 January 12 TOPICS COVERED: Fundatinal Tpics, Reactins, Gases, Thermchemistry, Atmic Structure, Peridicity, and Bnding. MULTIPLE CHOICE KEY AND
More informationTHE ANSWER KEY TO THIS EXAM WILL BE POSTED ON BULLETIN BOARD #4 IN THE HALLWAY EAST OF ROOM 1002 GILMAN AND ON THE CHEM 167 WEBSITE.
PROF. JOHN VERKADE SPRING 2005 THIS EXAM CONSISTS OF 3 QUESTIONS ON 9 PAGES CHEM 67 HOUR EXAM II FEBRUARY 28, 2005 SEAT NO. NAME RECIT. INSTR. RECIT. SECT. GRADING PAGE Page 2 Page 3 Page 4 Page 5 Page
More informationA.P. CHEMISTRY. SOLUTIONS AND ACID BASE CHEMISTRY. p 1
A.P. CHEMISTRY. SOLUTIONS AND ACID BASE CHEMISTRY. p 1 (Nte: questins 1 t 14 are meant t be dne WITHOUT calculatrs!) 1.Which f the fllwing is prbably true fr a slid slute with a highly endthermic heat
More informationUnit 11 Solutions- Guided Notes. What are alloys? What is the difference between heterogeneous and homogeneous mixtures?
Name: Perid: Unit 11 Slutins- Guided Ntes Mixtures: What is a mixture and give examples? What is a pure substance? What are allys? What is the difference between hetergeneus and hmgeneus mixtures? Slutins:
More informationCHEM 103 Calorimetry and Hess s Law
CHEM 103 Calrimetry and Hess s Law Lecture Ntes March 23, 2006 Prf. Sevian Annuncements Exam #2 is next Thursday, March 30 Study guide, practice exam, and practice exam answer key are already psted n the
More information( ) kt. Solution. From kinetic theory (visualized in Figure 1Q9-1), 1 2 rms = 2. = 1368 m/s
.9 Kinetic Mlecular Thery Calculate the effective (rms) speeds f the He and Ne atms in the He-Ne gas laser tube at rm temperature (300 K). Slutin T find the rt mean square velcity (v rms ) f He atms at
More informationExample 15.1 Identifying Brønsted Lowry Acids and Bases and Their Conjugates
Example 15.1 Identifying Brønsted Lowry Acids and Bases and Their Conjugates For Practice 15.1 In each reaction, identify the Brønsted Lowry acid, the Brønsted Lowry base, the conjugate acid, and the conjugate
More informationCHEM 116 Electrochemistry at Non-Standard Conditions, and Intro to Thermodynamics
CHEM 116 Electrchemistry at Nn-Standard Cnditins, and Intr t Thermdynamics Imprtant annuncement: If yu brrwed a clicker frm me this semester, return it t me at the end f next lecture r at the final exam
More information2.303 [ OH ] [ H ] CT
CEE 680 25 Octber 2011 FIRST EXAM Clsed bk, ne page f ntes allwed. Answer all questins. Please state any additinal assumptins yu made, and shw all wrk. Yu are welcme t use a graphical methd f slutin if
More informationALE 21. Gibbs Free Energy. At what temperature does the spontaneity of a reaction change?
Name Chem 163 Sectin: Team Number: ALE 21. Gibbs Free Energy (Reference: 20.3 Silberberg 5 th editin) At what temperature des the spntaneity f a reactin change? The Mdel: The Definitin f Free Energy S
More informationName:. Correct Questions = Wrong Questions =.. Unattempt Questions = Marks =
Name:. Crrect Questins = Wrng Questins =.. Unattempt Questins = Marks = 1. Which metal reacts mst vigrusly with water? (A) Al (B) Ca (C) Fe (D) K 2. Which are strng acids? I. HI II. HNO 3 III. H 2 SO 3
More informationFind this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.
Find this material useful? Yu can help ur team t keep this site up and bring yu even mre cntent cnsider dnating via the link n ur site. Still having truble understanding the material? Check ut ur Tutring
More informationA. Lattice Enthalpies Combining equations for the first ionization energy and first electron affinity:
[15.1B Energy Cycles Lattice Enthalpy] pg. 1 f 5 CURRICULUM Representative equatins (eg M+(g) M+(aq)) can be used fr enthalpy/energy f hydratin, inizatin, atmizatin, electrn affinity, lattice, cvalent
More information188 CHAPTER 6 THERMOCHEMISTRY
188 CHAPTER 6 THERMOCHEMISTRY 4. a. ΔE = q + w = J + 100. J = 77 J b. w = PΔV = 1.90 atm(.80 L 8.0 L) = 10.5 L atm ΔE = q + w = 50. J + 1060 = 1410 J c. w = PΔV = 1.00 atm(9.1 L11. L) = 17.9 L atm 101.
More informationThermochemistry. Thermochemistry
Thermchemistry Petrucci, Harwd and Herring: Chapter 7 CHEM 1000A 3.0 Thermchemistry 1 Thermchemistry The study energy in chemical reactins A sub-discipline thermdynamics Thermdynamics studies the bulk
More informationSession #22: Homework Solutions
Sessin #22: Hmewrk Slutins Prblem #1 (a) In the cntext f amrphus inrganic cmpunds, name tw netwrk frmers, tw netwrk mdifiers, and ne intermediate. (b) Sketch the variatin f mlar vlume with temperature
More information7. A solution has the following concentrations: [Cl - ] = 1.5 x 10-1 M [Br - ] = 5.0 x 10-4 M
Solubility, Ksp Worksheet 1 1. How many milliliters of 0.20 M AlCl 3 solution would be necessary to precipitate all of the Ag + from 45ml of a 0.20 M AgNO 3 solution? AlCl 3(aq) + 3AgNO 3(aq) Al(NO 3)
More informationUNIT III: SOLUBILITY EQUILIBRIUM YEAR END REVIEW (Chemistry 12)
I. Multiple Choice UNIT III: SOLUBILITY EQUILIBRIUM YEAR END REVIEW (Chemistry 12) 1) Which one of the following would form an ionic solution when dissolved in water? A. I 2 C. Ca(NO 3 ) 2 B. CH 3 OH D.
More informationHess Law - Enthalpy of Formation of Solid NH 4 Cl
Hess Law - Enthalpy f Frmatin f Slid NH 4 l NAME: OURSE: PERIOD: Prelab 1. Write and balance net inic equatins fr Reactin 2 and Reactin 3. Reactin 2: Reactin 3: 2. Shw that the alebraic sum f the balanced
More informationx x
Mdeling the Dynamics f Life: Calculus and Prbability fr Life Scientists Frederick R. Adler cfrederick R. Adler, Department f Mathematics and Department f Bilgy, University f Utah, Salt Lake City, Utah
More informationSecondary Topics in Equilibrium
Secondary Topics in Equilibrium Outline 1. Common Ions 2. Buffers 3. Titrations Review 1. Common Ions Include the common ion into the equilibrium expression Calculate the molar solubility in mol L -1 when
More information" 1 = # $H vap. Chapter 3 Problems
Chapter 3 rblems rblem At 1 atmsphere pure Ge melts at 1232 K and bils at 298 K. he triple pint ccurs at =8.4x1-8 atm. Estimate the heat f vaprizatin f Ge. he heat f vaprizatin is estimated frm the Clausius
More informationChem Practice Exam Two (Chapters 19, 20 and 21)
Chem 203 - Practice Exam Two (Chapters 19, 20 and 21) 1. Consider the dissolution of MnS in water (K sp = 3.0 10 14 ). MnS(s) + H 2O(l) Mn 2+ (aq) + HS (aq) + OH (aq) How is the solubility of manganese(ii)
More information**DO NOT ONLY RELY ON THIS STUDY GUIDE!!!**
Tpics lists: UV-Vis Absrbance Spectrscpy Lab & ChemActivity 3-6 (nly thrugh 4) I. UV-Vis Absrbance Spectrscpy Lab Beer s law Relates cncentratin f a chemical species in a slutin and the absrbance f that
More informationName: Date: Class: a. How many barium ions are there per formula unit (compound)? b. How many nitride ions are there per formula unit (compound)?
NOTES Name: Date: Class: Lessn 15 Part 2: Binary II Inic Bnding, Plyatmic Ins Bx 1: 1. Ba 3N 2 is the frmula fr. (name) a. Hw many barium ins are there per frmula unit (cmpund)? b. Hw many nitride ins
More informationEquilibrium HW Holt May 2017
Equilibrium HW Holt May 2017 Answer Key p. 595 (PP 1-3, SR 1-10), p. 604 (SR 1-6); p. 616 (PP 1&2); p. 618 (PP 1&2); p. 620 (PP 1&2, SR 1-7) pp. 622-624 (2-11, 14-16, 27, 29, 32, 33, 34, 37, 39, 40 (review
More informationREVIEW QUESTIONS Chapter 18. H = H (Products) - H (Reactants) H (Products) = (1 x -125) + (3 x -271) = -938 kj
Chemistry 102 ANSWER KEY REVIEW QUESTIONS Chapter 18 1. Calculate the heat reactin ( H ) in kj/ml r the reactin shwn belw, given the H values r each substance: NH (g) + F 2 (g) NF (g) + HF (g) H (kj/ml)
More informationElectrochemistry. Half-Reactions 1. Balancing Oxidation Reduction Reactions in Acidic and Basic Solutions
Electrchemistry Half-Reactins 1. Balancing Oxidatin Reductin Reactins in Acidic and Basic Slutins Vltaic Cells 2. Cnstructin f Vltaic Cells 3. Ntatin fr Vltaic Cells 4. Cell Ptential 5. Standard Cell Ptentials
More informationAdvanced Chemistry Practice Problems
Advanced Chemistry Practice Prblems Thermdynamics: Gibbs Free Energy 1. Questin: Is the reactin spntaneus when ΔG < 0? ΔG > 0? Answer: The reactin is spntaneus when ΔG < 0. 2. Questin: Fr a reactin with
More informationMore Tutorial at
Answer each questin in the space prvided; use back f page if extra space is needed. Answer questins s the grader can READILY understand yur wrk; nly wrk n the exam sheet will be cnsidered. Write answers,
More informationCHEM 1413 Chapter 6 Homework Questions TEXTBOOK HOMEWORK
CHEM 1413 Chapter 6 Hmewrk Questins TEXTBOOK HOMEWORK 6.25 A 27.7-g sample f the radiatr clant ethylene glycl releases 688 J f heat. What was the initial temperature f the sample if the final temperature
More informationElectrochemistry. Learning Objectives. Half-Reactions 1. Balancing Oxidation Reduction Reactions in Acidic and Basic Solutions
Electrchemistry 1 Learning Objectives Electrchemistry Balancing Oxidatin Reductin Reactins in Acidic and Basic Slutins a. Learn the steps fr balancing xidatin reductin reactins using the half-reactin methd.
More information