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(8} {3 soluts] The farmsbon cocstunt, K. of CoilNHy n; Co' (ag) + 6 NThy{ay) = CoiNH,)E s 45 % 16%%, We dlwilse Blake the nppe tracwoms of Do 3 1 L0 L of nenuear 0988 M ic NI%: (ag). uws e vali e remnaice constans at 1! sl an el L e P » 0 =7 Lo(NPaly o OO = [(*] = ——00 e él- ] e (0. ey 5 poinits) The solubility produce of Cof€ 0.19 M soluticn of Bu( )y, Ihe ter Arature is 25.0°C \S ey Feloll, =2 20 e i fal('“l;,, ey ’U..“'\L\'I 4 aalyen Please answer thase questions an your Page 14 We nave o 5535 M squeous schation of CyllgOfag) (U 1L is the only soluse). The demsity of chiv solution is 0.977 g/mb. Calculare the molality, the percers mans, and the mole b solutise, The wolar wasees o C, 1, aod Hoare 12,01, 16.00, wod 101 gfeaul, (b pemts) The teaction Alag) — B (s¢] 13 & et crder resclion wita respact to Alan). Uhe cunceatration wf Alag) Lells from 0.477 A 2 1.322 M cver the span at 44.4 5. What will the crmeratration of Aag) he after soothee 222 87 Tl st ace i 26 0°C. a) Fve £ e B Fe t4] afle onoer 2225 C4l= Cry e e ~t0.coyes M 6e.6) = evn)e —_ ks 1 [l ks = ( - 1) o BUIBIVR 3 ¢ - —Zalws fnl —— ) S.cietipe ~ 4912 ¢ | | _ es Gams Lsraus ~ Aaens - L% \C PP A — P g —— s § £ L - Nt W)y 43 16 ¢ 10 ¥, Celoulats s solubility {in g/L) in Kalarty - Vo atsol n el @ Na ot moka /1000 realehity = fdass of szlvert ograms ¢ Na, o Bole tretlon - No. of males of 3olute £ Tete! ro. of moles of 2ol and i of (3lly@) in this aquecus Male Fracaan (X) — oy i the sofafion ). » ¥ 150 (4 pelats Tae reaction 2 Ajag » Blag) | Clagi s = first order reaction vt respectts Alg) Vobume of soltion = 103 UdmL = (IL) When te corcentration of Alacliz 0500 M ata tempcrature of 250-C, the rate ot rcactor is R LO0IA0 MACWHer wie st Thes oncnlialise ol S L 0,100 M and vas ine s e = 0105041 Tttt alines [z 75000 e e alee saf et Leann 18 LOOTO0 M. Winsl s T ac Livalion enmigy fr s TR s 1] W knows that. far a glven msetior as & a 2 A ) —» Blaag) + Cla) Calculate the mclarity now: Rale ul reaciivn= K AL Moles of C2ITROTT ALTL= 259 - 25- 273 - 230K, | A| = 0.5 M, Relie ol reaslion = 0,001 M Mol (M) — e ey Mty (2 0001 M1 - KOS M 0.875males K1 - G.0u2s™. —_— A.T2= 7590 ~T5-273 - MG K, | Al SLIA w4 B.9M 0.007 M/ - K20.L M| K2 = nais Feom Arheaniis equatian v ke s {Ky/Ky) = RafR (171 — Mass of C2H50H (40.3% of 100g) = 40.3¢ 220 - L/MD) s roazion 5 61508 foimal We heve an aqueous solutiocn with & mess percentage o 42.2% tor C2HSOH {C2H5CH is the orly solute). The density of this solution is 0.952 g / mL. Calculate the molerity, mo_ality, end mole traction of C2HSCH in this esguecus rzles of xolule - TULL solution. i ) Wheas of C2D) i Wi baoe Ar ajuesons sx:lu.-mwilh " ll‘::ll‘!l fr e L x-:f 0250 fen €21 60 (CINE is The orily BUH M: = Malar o m.R!mM‘ CIH50H colution). | ne density of Tl solution Is U956 ¢/mL Calzulate the molartty, molality, and ¥, mass percentage of CiH6U I this squeous solution. T aunr X i wr = -l e e =¢ U.NTSmoles bt L 3 22 ML ¢ e ke = Mass of waber = Total o of solntton Mase of CLHEOR Mass of water by M ,J"_xjfl_'v & - A = — —= - -mgles - e R poee, Fe wter mass shaald reves be ronverted o dingrame: . g mwrua-.u.nx( ] fRnp Ay 2L ey Deeermilee the maalny: 3 z' - g v Malalily (] %mu..un,;y..| i N L A Tmaks § .Lm\-\:_;l;lq - ULOITER 1 - N w14 filn 2 FeTTRAT gt —_ — - reass iy el T d Tkl il = o P i g wals 2 Tokid mokess = mless of C2ZHEOH i e sl udion —3)— —vﬂw*l'\% Vo] S sreraat LT Woles of C2H.OH T Risse sy SRR i == Y 15 <l e h —10 = @Fws & iy ‘71 = l/\(?) S . —Lnlr) —:/Ta.f)J o.Clukog —T —’—R OOty 1ud | / | N / L”q,t La.p¢ , / | pressure o Ayl? Aig! )2 Big) - 3 Cig), w2 bagin with orly Alg). A 2quillbrium, the partlal 1420 atm. | he eculllerium constant for this reaction 15 1 . What vall be the Tenuilinmum: Waat was the Inltia | pressure of fla)? 1he tenperatire In 2500 e el s jom B ‘ ¥ snone | 3 yae ebm - i) lorthe reaction Mg < Ligd+ & Lial. we start off with Just pure /g (there 1o no Ul or Qg when v:e reach ecullibaum, the partial prascurs of Llg ks 115 3tm. (he 2quillbrim conetant for Uis e lion & 5008 . Whal is e parlisl pressore of Ag) sl sguiitiuen? Whal wae he ivilial . - == Bl A =— Mg «2c@ . PR B - = — 2 o L s R ) Lx I L fi Paubin) ‘_ yath o 4 | & 2 WS * 2 RS .JHJ‘“ S0 - iy B L G 3 | === . =L | - [ 3 b — T | = Lo 3 = Lol [y Y 1§ 304 A atoig g K r- = 1 - I ] | ——y

In a 10.0 Lcteel zontziner, there are only 88 8 g of CQ2[g) and 522 g of a unknan agas. The temperatiure [ 23.0-C and the tatal precairs Is £.27 atm. \What 15 the malar mass ¢f thls urknown gas? nstant teripe=atre of 25 O $arr avd chen of 4 (5 polnks) A~ {eq) i & weak base with a dissociation constant, Ky, of 4.4 x 1075, We have 1.000 {5 poines) A 5 & wesh scid. We cix 2000 L of & solution .33 M in TTA witl U 200 B L of & buffer solution 0.444 M in HA and 0.777 M in NaA. To this huffer solution, we add 0.050 ' tufer : 0,222 1 0 J3OH. The pH o7t reanlting scusos (=#tk a wlse of 2.0 mal of HCI {the volume does not changa). What was the pH of the solution before the sddition of Y/Eat 15 the value of the hesaciaticn comstual, Ky, for the weak tags 4 (2417 The tempers 3 P3{g) + Claliy = 2 Cllgla) U, muker mnsss of COZ MML = 44.02 gl g > xan s g . . e it the snd of the meac-ion® The wi'oree =NNR50 P the HCI? Wi 3 t h i ar ; : - § , s 25.0°0!, seal o e et the snid ¢ R X608 brefsrvimotek: is 25.0°C ki by the PH of the buffer aolution efter the addition of the 11C1? The letprrature . : e el oo ook o 20D 1 [T T " 8 D E s — 1 o o s Calculzre toe 1otal moles of qaics oa folloves: . ko= LT i © \ ot el &) \ am CAy = l\a,u%}\z,uo\_) =666 mal / Thre totad puanber ol moles of U musivre s calculsied s sbowo below: = — S ST =y pli = log ka =k . P g Extf = e RoT " i Ny = Cxv= (9 25) 050 = i mo) PV Phimag =pha * ""'*;?-[[h]] =3B ¢ g2 - q e "CRT (37100 = A OORIIERTRIG 1 6! < Nk Joaw I = 1 WAy + OF Aot Hy ) - Ta) RT o . R . fa) * LT gy " Mty = pla "“}\ |l Y 2.00 @ x 10.0L N 24 ol bl i TR0 Ry = pka 7 oy (] BB Baw = (YR 00821L uim K™ -mol” <258 K i Y M . J e 5 Ga C-oam 3.\ 33 = +\ 1y 3\ -327 Caluulaie b ol s ol urkeown gas es Tolkses: ¢ dom 0w =Wl 1 oy " oA roum 331 =pla oy (P Hhoses) 27mel @ e ol =q.9t e v3se - — fe , ) P . VNN M - n == pla= 4469 Tk tatal anmber of moles is 3.27 mal. The total mumber of mnles is sum of number ¢ moles of CO. aud N, The number of mokes of N, is [BUAMALL+ (522N ta=p T L P sy ke, o R A NMZ = 33502 gl Number of moles = i1 paints) The reaction 2 Afaq) — Tacg; + Clag) is w secand ovdder reacrion with respect to Afard. When the concentration of Alag) is D300 M snd the remperatinve s 23 of rcaction is .OULLY) N Thrrelors, 99,512 g/mal it 1t requited A aver Within a fixed volume of 20.0 L, at a co the rare When we decrease the conecntration of Afag) to 0.100 M and we £ 95°C i piti) T 2Rgimal cstant emperature of 25°C, we have a mixture of Fa(g) and go Clyfg). The partial pressure of Fulg) is 389 atm wnd thet of Cly 1 277 stin. The folowing rescion inerease the remperaturs to T304C the rare of reactian iz 100300 M2 What is the activaton 178uel talas place: energy for rhiz woaction? fo&T T\ [y % I mumber of moles of N, is 198 mol. 15e minber of moks of CO, is: . = -RT . S ?(x) 0] = 2 ClFy(a) wer L €3 T 298 ¥ =\ Y h} 3F{g) + Clalg) = 2 CIFy(g) ey i} . X = ™2 ] \ ] % oy Numnber of moles €O, = Tatal mumber of moles - Number of moles N, . ; .k, (812 b VLt ) \ 3 ‘ -227mol- What mass of CIF(g) is produced? What is the tata’ pressure a- the end of the reaction? The vo >9& = ’ .,)1' u.oatae M T e na wtee ' ¥ (3ag)¥ (220 “ | S 5“—‘—‘.-— 3.27mol-1.98mol -2 — : i 2011 and temperature remain constant at 20.0 L and 25°C thronghout. | . . —_— = T \ W) \ me ) =129wol £ Y - T’ {a sunm) 5 | 3 _ - - F P —— N ‘The mass of €O, is! pv =nRT, Numbcrof remaining excess reactant Cly it B - mols = (1.99’ - ]_1082) mol - — .08 ME Mass = Nismber of mn!cs » Malar mass limiting re. =0.2885 mol oo 4V =125l ddgnol : FInal partizl Pressure cf excess reactant = S B i B =56.7¢ mol ratio _ nR% 18] 19 polE) ror e remenon ¢ suwng) — o) + Cles), the aquilibrivrn conetant i 3.33. 1f the eon mass x M v cemcrarion of Blan) is 0.222 M ard that of Cleq) is 0.333 M. wher roncentration of Aag) would be The mass of €O, 36.7 . (1. R88Rew ] » L NR2 L2 L 20SK required to have o value ol -6.66 kJ for ACG? The lemperature is 25°C throughout ‘ . _ DARRRund N0 ERE < _ S i . o . (0 A5 KT = -L53EN28E15) 143 43) =- 2902 | 2L b} {4 poizta) We bove 266 m7 of on sguecus solution of KoOH( We add an exeess of Mg{NO3), 50 08 = = 1086 atm tn poradnce the precipitaze Mg{OH)y(8). The mass of Mz(OH)y(=) produced ia 0.44 g What wms the AG=AC - 5 . 4 uriginal N ) uion? The + it 25°C vhreughout, At 25.0°C, we dissolve 4 44 g of an seid, HA, in enough water to produce 25 0 mL of solution. We titrate | a & n).\i ur.‘,]"i :" I\u?ﬂ(f!«l' w‘le :‘f-l . J'“’I‘-K‘“\‘:‘U‘( ' E: Aok Q=-2 the solution with a [.333 M NaOH solution and we require 17.7 ml to reach the equivalece point. T'he i o ) o i - =6 H sl (e squivalence point, is 11,44 i e = -L5E . (¥R I 1 sk | IAL the enuvcione paink, pH = Tnekne pOM =256, (W] <10 = R T (a) (3 points) What 18 the malar mass of HA? Also, (A7]= ™ 0 00589 me BAEEAg15) o o ; & Ao rid)ae L = SoqzaL = 0038 ™ oo _ 2] . ma g \ a0 & - S — o =lBILC] . raq_ {b) (4 points) What 1s the value of K, for HA{ug)? oy =5 Hétﬂ\ + CHany 2\ {% aTHRY =[on] <lc‘w=0.§&2.”_5 J oW s {b) ;;‘ \l-:\.m:‘;; What is che pH of o 1.00 M solutionn of NaA(aq) (where A~ (o] is the conjugate boszof s - - Kb=["“]t;°“j . (0.c0235) " ~ ,l".id eig N ! 088 = x W (0038 -0.002335) It — ¢, -G = = Kesese® | 23T el - s RODwWL O o - _ TN : el n:clao\t b Y %= 0.005899 e Ko= ‘_Q_{‘&_ Sloxw™ e Q" 1 ) - 250mL ) * e T o 1 -0 Yo B P, | - 20N M= Yocosan ey = FI3_gine 1w Sibbs . h ) . I O] * S A5 0 ”»_[’\1 4 —_,l% ) X§[_DH]:(./<D°1HS O 10 L )= A = P - pOH=123 pH=(.8%

(3 points) The density of a roct mean square speed (or average ipr@vlj o Naigl sample is D O'JG /L, axd the pressure Is 1.00 aoc. whe Nyfg) malecules in this zample? What is the HAlaq) is a weak acic with a cissociztion constant, Ka, of 7.7 x 10 12. What is ey seweatan eoenan of 53 A, TT A / . 0011 o 0 “ =X ® % k5 ( e G011 % x ] " SXpIENUN KA = IP H:] e w A A u.:;':‘; X 15 very small hence (0,011-x)=0 011 Q=170 2 =002 10" s=H|=0w~1w " Explanation: e ) WL AT 000 atm wnd il e e do St istendd wr constune vol: 33 ool of butsue ut constuon volune « pressie ol instend of dolug i at woastaat po L) aus W otur the contbastion of AH Q = AU W= p 13 pairts] The entbalpy of formation of liquid water i -285.8 £J mol ', The crthalpy of con- densation of water supour, HyOig). s 44.0 kJ el . What s the value of AH for the neacticon below? The temperatire '8 25°C 2 Hyleg} ~ Oule) — 2 H,0(g) rormation of bquld water: Halg)+ § Cafy) — B0l AH=-285 8 I /el ———— Condensation of water vapors: Hiig] — Eel {11 AH= — 11K uuil— = ————— = () Reversa U [2) rews on 1001 N ) Al 43K rol 1 2H:(g) + Oa2(g) — 2H:0(g) AH = 2x (—211.8 KJ /mol) AH —483.6 KJ/mol pomts] In ¢ bomb celomreter st constant volune, we do tae combastion of 1. N.B. combustion is the resction of 8 suhstance wita Oalg) to produce CO specific heit of water is 4 184 J K- from 21 25-C to 26.40°C x 1l wH ul @ suluiur can be caluleled awpH = —lug H' | 3 AHf(products) - I AHf(reactants) TN - ety Wy B - 3.333 mol * -2877 kJ/mol got Cally and Ll The Luwh calorimeter Lus o Lol u\p.u iy uf 2,50 LI K- wad coutains 3.125 kg of waer, The . The teraperoture of the calorimeter and the C. Afzcrm\rd;e. v do the combustion of LO0D mol of tE1s Colly; (1), vnder o constant oressure of L0 st a: 250°C. Celealate the values of €. W, AlLL e with e e Now, 17 AW -9589 k3 Hea. cdeand =0 0 51 AT-2G43 21.25- 515 Mass of wazer 3.12.x20'g Muse NG tricolar s = © 4008 Bl = ol Tollyz = ma zimal Lo CeHpL ) — ) 1 BHRO Temal heat relensec o constant voume heat acsorked by caanimatar 1 heat used te rise temgeatare of waner = 2HEINK = 5104 = 2125 IOLIgA.1 840K = ROZIL25] dor roses TEAA R LAINATML PR S ¢ subetanee, =-3238M I --38390¢ AU, and the entropy change of the surroundings when we co the combustion of 1000 moles wnler ponsrant nrEsanTE. A8 a0 eatrzlzzszd at sorgtont mrazaL s s 2zual o &l | x number of rolzz 3= ¢ 265 = 0Tl = 1055K) SFIENSK < 136 L -8 333900 M=% 'n Ye 0 =238 3R N- |- SALAE TR - | T4Y ML ve = TRIFETRTE S OGF IS = 1TV RS sints) For the reaction A(s) = the pH el 2 0011 M solulion of & [20)? The lemperalure is 250 T sglan oAl N ) iy o jraw ! athalfthe aqulval.\c. palnt, nalf of =& j2cta has F e ity oewugal > [» and [Ha] - [a ] K= N s will farm 2 buffer soluten and H 12 gven as w2 A il swctebore] % ] oo o- The log beem is 2eac, a8 bog 2= 3, it . - A, | iedeu o, pH=pKa -y uy' U7 6-10g L9526, * Afg). the equilibrum censtant is 277 at 25.0°C ard 3.77 av 0°C. Makirg the approximarion that the enthalpy change and entropy change of this reaction doss not vary with temperasure, find the temperature at which the equilibrium constant will be agnal to 1O " (R H] Y peants; What is the melar waze of T4 - it L T S R T e A PRE T & “Alag 4 Naniag) — P oand KO0 sk W - 933 m ra su3t 26 <) 3 ssae® M 'y ] - iy {zise ) - 2+5:9 ha f W disselve 156 g of or unknewer acld, HA In cncugh weter to procuce The pH ot thlz solut'on [e 1.33. We Hirate thk colutior whha 2277 220 mL of soludon 4 solut'on o NaDk and we need 14.4 mL of this NaDH soluton te reach the equvalence palrt. The temperaturs k£ 2.0 € throughout. FI R peoinkd Whlis e ealar nass of HA? 1) 12 pointe) What 'z the dizzo W prnls] Whinl is Lhe gl al he souivalere e poin? 1.8t Mal = . i s number of moles Ka — 107" x concentration of acid — 12 11%mel Ka— 10 x e Ka — 1.0x107% zlatlon zonstant, Ka, of =Aag)? 25.0mL ~ D27T7M number of males = 14.4mL number of moles = 1.125mol Molar mass = 72 Malar mass — O.UIME% pKa = —log (1.0x 10‘5) pKa="5.0 cl b Ao esplvelinat peinl, s owa ke beor wanalhed od o v Na Wit of Nah = awdh ef A = Trural remel o4 42 T gas Tond villiwe s SRt {250 ) 112G wl, - oriay meyL = O H A ik iy "'".""-' krat ‘l foeaak add) 1r(ae) A0 ey - X, 5 ™ * 5 W s = Lo 1Lew) i 1 A 5 | . 2 | Boowe W = 9 ¥he x L i oo 3 ' Sl | AT & a' = e M")“ RhEl o At M= e YN M 3 od S g (PR€) 7 4ed3 e o - s Low) g 3 L f».w‘ =g oM e \ V-+-9% = 127 = [P @37 rrase log [0.125mal] 25.0mL ired 155 A uulnl< ol pH = pKa+ log{conecentration of A-) pH=5.0 pll — 7.00

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n=1.1=0 gH- pha |o3€‘3 e -log () phhas - Jog (k) a- KoK (10x107%) Xe-bt F'—b? e ehE T o (m z(wa- 19@]\) \= \A[A') [5‘) K= bt (ole= -1 AR | 5(33 1 P— -a( 1\ N=3, 1202 A1 by NG T R e Ea.-RWn (Y7, = 1= Pemaaq0f * 7—Ifi~ ‘ﬂ (‘:] K{ t- ‘ﬂiﬂ th“_‘“_z_ K= ‘ﬂ Z C"Kv Krh/ 'iO"‘ E=hvy 1#23:me-2 -0y T "') [ﬂ w K .I;: woomy T Ve B b acr 0 E--Q %ot V ¢ [A] SR 1ol i S0 w €= Ve~ Nee-J 8¢ (AA" o ER)em m‘:fifim*' el NG = (e - BY- e WOl B (E-8) Duorw rmo: 2O - W= BU-G ™M AC . ink 335 [ At: L e o U:;N:;Jm’g? MU G CAT-mghT . L AN (B%w i, 14 - - E Ca DH= QU W N K= -Al L ' M Hoo ail * \T‘,: 8L= QW-» RT MC A = mCAT + cat C\l"ﬂ p"'.!- goaes n 2\ AL? WE WV Curan " - % I Vq:.w'( -\o-—Jb" VSR ’1“4 = p‘V' T ',“:", “4 T .'Ff --1-‘) Ny o w=-1) pS= Q. Co:0=08 G=n-Tp ol T CV= @Al fw~o AV oo - react ps ﬂ;‘I’ ye PRT nafV s Qo= -Gems b5 uiwme>0 Cy - = AN+ - v U TE TR 0 ey T, TN DO Qi g — RT = -aX. Saded