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B2QfcϾ >XJ:=>W)egK}zotaf<ﱸy, I+)XnGo=6i4NMkS>ET4C!T`~aWlHـ=;✕EM$o; diAM_qA^9fRkS=G2ВH>@ָV!1 ~?C s?hbƶk:uې68QBi4qk ooc夜I<2u蠒` f _ Variables de estadoMagnitudes que pueden variar a lo largo de un proceso (por ejemplo, en el transcurso de una reaccin qumica) . Ejemplos: Presin. Temperatura. Volumen. Concentracin. :zZ/Zp D/68 ;Funciones de estadoTienen un valor nico para cada estado del sistema. Su variacin solo depende del estado inicial y final y no del camino desarrollado. S son: Presin, temperatura, energa interna, entalpa. NO son: calor, trabajoZ 973 $Primer principio de la TermodinmicaENERGA INTERNA (U) es la energa total del sistema. Es imposible medir. Su variacin s se mide. D U = Q + W Q y W > 0 si se realizan a favor del sistema. U es funcin de estado.8-. 8- . Calor a volumen constante (Qv)0 Es el intercambio de energa en un recipiente cerrado que no cambia de volumen. Si V= constante, es decir, DV = 0 W = 0 Qv = DU~ l &P / Calor a presin constante (Qp)&.La mayora de los procesos qumicos ocurren a presin constante, normalmente la atmosfrica. Si p = cte W =  p D V (expansin) D U = Qp  p D V U2  U1 = Qp  p (V2  V1) Qp = (U2 + p V2)  (U1 + p V1) H2 H1 (entalpa) r> h   ,dP_Calor a presin constante (Qp)&H1= U1 + p V1 H2= U2 + p V2 Qp = H2  H1 = D H H es una funcin de estado. 6$4    Relacin Qv con Qp (gases).J ,  D H = D U + p D V Aplicando la ecuacin de los gases: p V = n R T y si p y T son constantes la ecuacin se cumplir para los estados inicial y final: p D V = D n R T D H = D U + D n R T6H(:k@(    (Relacin Qv con Qp (slidos y lquidos)T) , En reacciones de slidos y lquidos apenas se produce variacin de volumen y ... Qv @ Qp es decir: DU @ DH Q Q,,,,,, ,,QE9vEjemplo: Determinar la variacin de energa interna para el proceso de combustin de 1 mol de propano a 25C y 1 atm, si la variacin de entalpa, en estas condiciones, vale  2219,8 kJ. <  C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l) D H =  2219,8 kJ nreactivos = 1+5 = 6 ; nproductos = 3 D n =  3 Despejando en D U = D H  D n R T =  2219800 J + 3 mol (8,3 J/mol.K) 298 K =  2212379 J D U =  2212 kJ E          D ,F    Entalpa estndar de la reaccinEs el incremento entlpico de una reaccin en la cual, tanto reactivos como productos estn en condiciones estndar (p = 1 atm; T = 298 K = 25 C; conc. = 1 M). Se expresa como DH0 y como se mide en J o kJ depende de cmo se ajuste la reaccin. As, DH0 de la reaccin  2 H2 + O2 2 H2O es el doble del de  H2 + O2 H2O . DH0 = H0productos  H0reactivosI"H   a  '  L # Ecuaciones termoqumicasFExpresan tanto los reactivos como los productos indicando entre parntesis su estado fsico, y a continuacin la variacin energtica expresada como DH (habitualmente como DH0). Ejemplos: CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l); DH0 =  890 kJ H2(g) + O2(g) H2O(g); DH0 =  241 4 kJ r      Ecuaciones termoqumicasCUIDADO!: DH depende del nmero de moles que se forman o producen. Por tanto, si se ajusta poniendo coeficientes dobles, habr que multiplicar DH0 por 2: 2 H2(g) + O2(g) 2 H2O(g) ; DH0 = 2 ( 241 4 kJ) Con frecuencia, suelen usarse coeficientes fraccionarios para ajustar las ecuaciones: H2(g) + O2(g) H2O(g) ; DH0 =  241 4 kJ `QH(   V    4Entalpa estndar de formacin (calor de formacin).Es el incremento entlpico (DH) que se produce en la reaccin de formacin de un mol de un determinado compuesto a partir de los elementos en estado fsico normal (en condiciones estndar). Se expresa como DHf0. Se trata de un  calor molar , es decir, el cociente entre DH0 y el nmero de moles formados de producto. Por tanto, se mide en kJ/mol. Ejemplos: C(s) + O2(g) CO2(g) DHf 0 =  393 13 kJ/mol H2(g) + O2(g) H2O(l) DHf0 =  285 8 kJ/mol1z<K  B : J kL<JClculo de D H0 (calor de reaccin) 6& HAplicando la ley de Hess podemos concluir que : D H0 = S npDHf0(productos) S nrDHf0(reactivos) Recuerda que DHf0 de todos los elementos en estado original es 0. .10C0  1M=Ejemplo: Conocidas las entalpas estndar de formacin del butano (C4H10), agua lquida y CO2, cuyos valores son respectivamente  124 7,  285 8 y  393 5 kJ/mol, calcular la entalpa estndar de combustin del butano.~  <   {nLa reaccin de combustin del butano es: C4H10(g) +13/2O2(g) 4 CO2(g) + 5H2O(l) DH0comb= ? D H0 = S npDHf0(product.)  S nrDHf0(reactivos) = 4 mol( 393 5 kJ/mol) + 5 mol( 285 8 kJ/mol)  1 mol( 124 7 kJ/mol) =  2878 3 kJ Luego la entalpa estndar de combustin ser: D H0combustin =  2878 3 kJ/mol&/K% #)                  Ley de HessDH en una reaccin qumica es constante con independencia de que la reaccin se produzca en una o ms etapas. Recuerda que H es funcin de estado. Por tanto, si una ecuacin qumica se puede expresar como combinacin lineal de otras, podremos igualmente calcular DH de la reaccin global combinando los DH de cada una de las reacciones.NQl'!Ejemplo: Dadas las reacciones (1) H2(g) + O2(g) H2O(g) DH10 =  241 8 kJ (2) H2(g) + O2(g) H2O(l) DH20 =  285 8 kJ calcular la entalpa de vaporizacin del agua en condiciones estndar. n   #  P# La reaccin de vaporizacin es... (3) H2O(l) H2O(g) DH03 = ? (3) puede expresarse como (1)  (2), luego DH03 = DH01  DH02 =  241 8 kJ  ( 285 8 kJ) = 44 kJ DH0vaporizacin = 44 kJ /mol:ZZ8 Z2# 2)  Esquema de la ley de Hess-'Ejercicio A: Conocidas las entalpas estndar de formacin del butano (C4H10), agua lquida y CO2, cuyos valores son respectivamente  124 7,  285 8 y  393 5 kJ/mol, calcular la entalpa estndar de combustin del butano. v  ;   }Si utilizamos la ley de Hess, la reaccin: (4) C4H10(g) +13/2O2(g) 4 CO2(g) + 5H2O(l) DH0comb=? Puede obtenerse a partir de: (1) H2(g) + O2(g) H2O(l) DH10 =  285 8 kJ (2) C(s) + O2(g) CO2(g) DH20 =  393 5 kJ (3) 4 C(s) + 5 H2(g) C4H10(g) DH30 =  124 7 kJ (4) = 4 (2) + 5 (1)  (3) 4 C(s) + 4 O2(g) +5 H2(g) + 5/2 O2(g) + C4H10(g) 4 CO2(g) + 5H2O(l) + 4 C(s) + 5 H2(g) D H04 = 4 mol( 393 5 kJ/mol) + 5 mol( 285 8 kJ/mol)  1 mol( 124 7 kJ/mol) =  2878 3 kJ+              !     G <Ejercicio B: Determinar D Hf0 del eteno (C2H4) a partir de los calores de reaccin de las siguientes reacciones qumicas: (1) H2(g) + O2(g) H2O(l) DH10 =  285 8 kJ (2) C(s) + O2(g) CO2(g) DH20 =  393 13 kJ (3) C2H4(g) + 3O2(g) 2CO2(g) + 2 H2O(l) DH30 =  1422 kJ8         S     \La reaccin de formacin del eteno C2H4(g) a partir de sus constituyentes en estado normal es: (4) 2 C(s) + 2 H2(g) C2H4(g) (4) se puede expresar como 2(2) + 2(1)  (3)_!/$I3Ejercicio C: Determinar DHf0 del eteno (C2H4) a partir de los calores de reaccin de las siguientes reacciones qumicas:z        N:(1) H2(g) + O2(g) H2O(l) DH10 =  285 8 kJ (2) C(s) + O2(g) CO2(g) DH20 =  393 13 kJ (3) C2H4(g) + 3O2(g) 2CO2(g) + 2 H2O(l) DH30 =  1422 kJ (4) 2 C(s) + 2 H2(g) C2H4(g) (4) = 2(2) + 2(1)  (3) luego DH40 = 2DH20 + 2DH10  DH30 = = 2 ( 393 13 kJ) + 2 ( 285 8 kJ)  ( 1422 kJ) = = 64 14 kJ es decir DHf0 (eteno) = 64 14 kJ/mol Se trata, pues, de una reaccin endotrmica.Z      &'$    3`Ejercicio D: Las entalpas de combustin de la glucosa (C6H12O6) y del etanol (C2H5OH) son  2815 kJ/mol y  1372 kJ/mol, respectivamente. Con estos datos determina la energa intercambiada en la fermenta cin de un mol de glucosa, reaccin en la que se produce etanol y CO2. Es exotrmica la reaccin?&1 ,        Las reacciones de combustin son, respectivamente: (1) C6H12O6 + 6 O2 6 CO2 + 6 H2O ; DH1 =  2815 kJ (2) C2H5OH + 3 O2 2 CO2 + 3 H2O ; DH2 =  1372 kJ La reaccin de fermentacin de la glucosa es: (3) C6H12O6 2 C2H5OH +2 CO2 DH3 = ? (3) puede expresarse como (1)  2 (2), luego DH3 = DH1  2DH2 =  2815 kJ  2 ( 1372 kJ) =  71 kJ y la reaccin es exotrmica.5         7    #      Energa de enlace. Es la energa necesaria para romper un enlace de un mol de sustancia en estado gaseoso En el caso de molculas diatmicas es igual que la energa de disociacin: A B(g) A(g) + B(g) DHdis = Eenlace= Ee Ejemplo: H2(g) 2 H(g) DH = 436 kJ Es positiva (es necesario aportar energa al sistema) Es difcil de medir. Se suele calcular aplicando la ley de Hess. lVT6Y  ( Clculo de DH0 a partir de las Energa de enlace (disociacin).HA "Aplicando la ley de Hess en cualquier caso se obtiene la siguiente frmula: DH0 = S ni Ee(enl. rotos)  S nj Ee(enl. formados) en donde ni representa el nmero de enlaces rotos y formados de cada tipo. M8ML ^L H;lEjemplo: Sabiendo que las energa de los siguientes enlaces (kJ/mol): C=C : 611; C C : 347; C H : 413 y H H : 436, calcular el valor de DH0 de la reaccin de hidrogenacin del eteno. J"  "  " , 4         H Reaccin: CH2=CH2(g) + H2(g) CH3 CH3(g) En el proceso se rompe un enlace C=C y otro H H y se forman 2 enlaces C H nuevos (el etano tiene 6 mientras que el eteno tena slo 4) y un enlace C C. D H0 = S Ee(enl. rotos)  S Ee(enl. formados) = D H0 = 1Ee(C=C) + 1 Ee(H H)  1Ee(C C)  2 Ee(C H) D H0 = 1 mol 611 kJ/mol + 1mol 436 kJ/mol  (1 mol 347 kJ/mol + 2 mol 413 kJ/mol) =  126 kJ L             X oEjercicio E: Calcula el calor de combustin de propano a partir de los datos de energa de enlace de la tabla.&p  d07 6 bC3H8 + 5 O2 3 CO2 + 4 H2O Enlaces rotos: 8 C H, 2 C C y 5 O=O Enlaces formados: 6 C=O y 8 O H DH0 = S Ee(e. rotos)  S Ee(e. form.) DH0 = 8 Ee(C H) + 2 Ee(C C) + 5 Ee(O=O)  [6 Ee(C=O) + 8 Ee(O H)] DH0 = 8413 kJ + 2347 kJ +5499 kJ  (6745 kJ + 8460 kJ) =  1657 kJ DH0comb(C3H8) =  1657 kJ/mol ^   *            B      # Entropa (S)PEs una medida del desorden del sistema que s puede medirse y tabularse. DS = Sfinal  Sinicial Existen tablas de S0 (entropa molar estndar) de diferentes sustancias. En una reaccin qumica: DS0 = S np S0productos  S nr S0reactivos La entropa es una funcin de estado.\IZZJZF:FF&ZI \  )D8Ejemplo: Calcula DS0 para las siguientes reacciones qumicas: a) N2(g) + O2(g) 2 NO(g); b) 3 H2(g) + N2(g) 2 NH3(g). Datos: S0 (Jmol 1K 1): H2(g) = 130,6; O2(g) =205; N2(g) = 191,5; NO(g) = 210,7; NH3(g) =192,3  )     DS0 = S np S0productos  S nr S0reactivos a) DS0 = 2 mol 210,7 J mol 1 K 1  (191,5 JK 1 + 205 JK 1 ) = 24,9 JK 1 b) DS0 = 2192,3 JK 1  (3 mol 130,6 J mol 1K 1 + 191,5 JK 1 ) =  198,7 JK 1      $&Segundo principio de la Termodinmica. En cualquier proceso espontneo la entropa total del universo tiende a aumentar siempre . DSuniverso = DSsistema + DSentorno 0 A veces el sistema pierde entropa (se ordena) espontneamente. En dichos casos el entorno se desordena.ji^n% $Tercer principio de la Termodinmica La entropa de cualquier sustancia a 0 K es igual a 0 (mximo orden). Equivale a decir que no se puede bajar de dicha temperatura. CUIDADO! Las S de los elementos en condiciones estndar no son 0 sino que es positiva.2 N'" *En procesos reversibles y a temperatura constante se puede calcular DS de un sistema como: Q DS =  T y si el proceso qumico se produce a presin constante: DHsistema  DHsistema DSsistema =    ; DSentorno=     T T S0 (entropa molar estndar) se mide en Jmol 1K 1. DSreaccin se mide en JK 1. \ZG<;Zd<TZ&&  % (!<Energa libre de Gibbs (G) (energa libre o entalpa libre).=% En procesos a T constante se define como: G = H  T S DG = D H  T DS En condiciones estndar: DG0 = DH0  T DS0 Si DG < 0 la reaccin ES espontnea Si DG. > 0 la reaccin NO es espontnea Si DG. = 0 el sistema est en equilibrio6+Z#Z.ZZzZ<    -)#B6`Incremento de energa libre de una reaccin (DG)$1-0 G es una funcin de estado. Al igual que el incremento entlpico el incremento de energa libre de una reaccin puede obtenerse a partir de DGf0 de reactivos y productos: DG0 = S npDGf0(productos) S nrDGf0(reactivos) /  &7  +%9Energa libre y Espontaneidad de las reacciones qumicasReac. no espontnea 2"*Espontaneidad en las reacciones qumicas.:No siempre las reacciones exotrmicas son espontneas. Hay reacciones endotrmicas espontneas: Evaporacin de lquidos. Disolucin de sales... Ejemplos de reacciones endotrmicas espontneas: NH4Cl(s) NH4+(aq) + Cl (aq) D H0 = 14 7 kJ H2O(l) H2O(g) DH0 = 44 0 kJ :`0`01      )#1Espontaneidad de las reacciones qumicas (cont).+Una reaccin es espontnea cuando DG (DH  T x DS) es negativo. Segn sean positivos o negativos los valores de DH y DS (T siempre es positiva) se cumplir que: D H < 0 y D S > 0 DG < 0 Espontnea D H > 0 y D S < 0 DG > 0 No espontnea D H < 0 y D S < 0 DG < 0 a T bajas DG > 0 a T altas D H > 0 y D S > 0 DG < 0 a T altas DG > 0 a T bajast#>(    x#   @  , ,&1Espontaneidad de las reacciones qumicas (cont).+.)Ejemplo: Ser o no espontnea la siguiente reaccin 2H2O2(l) 2H2O (l) + O2(g) en condiciones estndar? Datos: DH0f (kJ/mol) H2O(l) =  285,8; H2O2(l) =  187,8 ; S0 (Jmol 1 K1) H2O(l) = 69,9; H2O2(l) = 109,6; O2(g) =205,0. .                  PD H0 = S npDHf0(productos) S nrDHf0(reactivos) = = 2 DHf0(H2O) + DHf0(O2)  2 DHf0(H2O2) = 2 mol( 285,8 kJ/mol)  2 mol( 187,8 kJ/mol) =  196,0 kJ DS0 = S np S0productos  S nr S0reactivos = 2 S0(H2O) + S0(O2)  2 S0(H2O2) = 2 mol(69,9 J/molK) + 1 mol(205, J/molK)  2mol(109,6 J/molK) = 126,0 J / K = 0,126 kJ / K DG0 = D H0  T D S0 =  196,0 kJ  298 K 0,126 kJ/ K = DG0 =  233,5 kJ luego ser espontnea                 0           O   $&/FPJ  0` ̙33` ` ff3333f` 333MMM` f` f` 3>?" dd@,?nKd@   d @nA`  d n?" dd@   @@``PR    @ ` ` p>>  b(  8 1 1  S BC1DEF &L1?Z B 0Z @1  S BB C DEF&L1?rA r @\   S B+ C DEF&L1?v*  v @   S BCIDEF&L1??H\ @ I    `xaxa1 ?  e1Haga clic para editar el estilo del ttulo patrn2 2  Zxaxa1 ?<$D  0  iHaga clic para editar el estilo del texto del patrn Segundo nivel Tercer nivel Cuarto nivel Quinto nivel5    j   <41? j*&(2  <1?P@ 0 0Z  Bfd޽h @ ? 3eX{IZ(a==___PPT10=.+.đDz' = @B D5' = @BA?%,( < +O%,( < +Dl' =%(D' =%(D' =A@BBBB0B%(E5' =1B B`BPB,54*3>B ppt_c='`B@BPB<*5E' =4 B`BPB`B?<*%(/%,( < +D' =1:Bvisible*o3>+B#style.visibility<*5%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*5D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*5D' =A@BBBB0B%(E5' =1B B`BPB,54*3>B ppt_c='`B@BPB<*5CE' =4 B`BPB`B?<*%( /%,( < +D' =1:Bvisible*o3>+B#style.visibility<*5C%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*5CD' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*5CD' =A@BBBB0B%(E5' =1B B`BPB,54*3>B ppt_c='`B@BPB<*CPE' =4 B`BPB`B?<*%( /%,( < +D' =1:Bvisible*o3>+B#style.visibility<*CP%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*CPD' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*CPD' =A@BBBB0B%(E5' =1B B`BPB,54*3>B ppt_c='`B@BPB<*P]E' =4 B`BPB`B?<*%(/%,( < +D' =1:Bvisible*o3>+B#style.visibility<*P]%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*P]D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*P]D' =A@BBBB0B%(E5' =1B B`BPB,54*3>B ppt_c='`B@BPB<*]jE' =4 B`BPB`B?<*%(/%,( < +D' =1:Bvisible*o3>+B#style.visibility<*]j%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*]jD' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*]j+++0 + +D' =4@BBBB%(E5' =1B B`BPB,54*3>B ppt_c='`B@BPB<*E' =4 B`BPB`B?<*%(/%,( < +D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<* +D' =4@BBBB%(E5' =1B B`BPB,54*3>B ppt_c='`B@BPB<*E' =4 B`BPB`B?<*%(/%,( < +D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<* +D' =4@BBBB%(E5' =1B B`BPB,54*3>B ppt_c='`B@BPB<*E' =4 B`BPB`B?<*%(/%,( < +D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<* +D' =4@BBBB%(E5' =1B B`BPB,54*3>B ppt_c='`B@BPB<*E' =4 B`BPB`B?<*%(/%,( < +D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<* +D' =4@BBBB%(E5' =1B B`BPB,54*3>B ppt_c='`B@BPB<*E' =4 B`BPB`B?<*%(/%,( < +D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<* Diagazus 0 [SPd(  dW  ZNxaxa1 ?p @ Y iHaga clic para editar el estilo del texto del patrn Segundo nivel Tercer nivel Cuarto nivel Quinto nivel5    jp  01 ?  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3eX{IZ(a___PPT10z.+OD' = @B D' = @BA?%,( < +O%,( < +D' =%(D' =%(DR' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*<%(D' =-6B%strips(downRight)*<3<*<D' =%(D' =%(DR' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*<#%(D' =-6B%strips(downRight)*<3<*<#+8+0+<0 +  0N0 @PD(  @l @ C 4    @ C <$ 0  H @ 0޽h ? 3eX{IZ(a  ___PPT10 .+WD6 ' = @B D ' = @BA?%,( < +O%,( < +D' =%(D' =%(DT' =A@BBB B0B%(D' =1:Bvisible*o3>+B#style.visibility<*@%(D' =-6B'blinds(horizontal)*<3<*@D' =%(D' =%(DT' =A@BBB B0B%(D' =1:Bvisible*o3>+B#style.visibility<*@%(D' =-6B'blinds(horizontal)*<3<*@D' =%(D' =%(DT' =A@BBB B0B%(D' =1:Bvisible*o3>+B#style.visibility<*@$%(D' =-6B'blinds(horizontal)*<3<*@$D' =%(D' =%(DT' =A@BBB B0B%(D' =1:Bvisible*o3>+B#style.visibility<*@$Q%(D' =-6B'blinds(horizontal)*<3<*@$Q+8+0+@0 +6"  0N0 PDD(  Dl D C      D C 0 <$  0  H D 0޽h ? 3eX{IZ(a  ___PPT10.+TD' = @B DE' = @BA?%,( < +O%,( < +D' =%(DA' =%(D' =A@BBB$B0B%(D' =1:Bvisible*o3>+B#style.visibility<*D%(Dc' =++4 8?CBa(6*min(max(#ppt_w*#ppt_h,.3),1)-7.4)/-.7*#ppt_wBCB#ppt_wB*Y3>B ppt_w<*DDc' =++4 8?CBa(6*min(max(#ppt_w*#ppt_h,.3),1)-7.4)/-.7*#ppt_hBCB#ppt_hB*Y3>B ppt_h<*DD' =+4 8?RCB?BCB#ppt_xB*Y3>B ppt_x<*DDk' =+34 8?CBi1+(6*min(max(#ppt_w*#ppt_h,.3),1)-7.4)/-.7*#ppt_h/2BCB#ppt_yB*Y3>B ppt_y<*DD' =%(DA' =%(D' =A@BBB$B0B%(D' =1:Bvisible*o3>+B#style.visibility<*D>%(Dc' =++4 8?CBa(6*min(max(#ppt_w*#ppt_h,.3),1)-7.4)/-.7*#ppt_wBCB#ppt_wB*Y3>B ppt_w<*D>Dc' =++4 8?CBa(6*min(max(#ppt_w*#ppt_h,.3),1)-7.4)/-.7*#ppt_hBCB#ppt_hB*Y3>B ppt_h<*D>D' =+4 8?RCB?BCB#ppt_xB*Y3>B ppt_x<*D>Dk' =+34 8?CBi1+(6*min(max(#ppt_w*#ppt_h,.3),1)-7.4)/-.7*#ppt_h/2BCB#ppt_yB*Y3>B ppt_y<*D>D' =%(DA' =%(D' =A@BBB$B0B%(D' =1:Bvisible*o3>+B#style.visibility<*D>\%(Dc' =++4 8?CBa(6*min(max(#ppt_w*#ppt_h,.3),1)-7.4)/-.7*#ppt_wBCB#ppt_wB*Y3>B ppt_w<*D>\Dc' =++4 8?CBa(6*min(max(#ppt_w*#ppt_h,.3),1)-7.4)/-.7*#ppt_hBCB#ppt_hB*Y3>B ppt_h<*D>\D' =+4 8?RCB?BCB#ppt_xB*Y3>B ppt_x<*D>\Dk' =+34 8?CBi1+(6*min(max(#ppt_w*#ppt_h,.3),1)-7.4)/-.7*#ppt_h/2BCB#ppt_yB*Y3>B ppt_y<*D>\D' =%(DA' =%(D' =A@BBB$B0B%(D' =1:Bvisible*o3>+B#style.visibility<*D\%(Dc' =++4 8?CBa(6*min(max(#ppt_w*#ppt_h,.3),1)-7.4)/-.7*#ppt_wBCB#ppt_wB*Y3>B ppt_w<*D\Dc' =++4 8?CBa(6*min(max(#ppt_w*#ppt_h,.3),1)-7.4)/-.7*#ppt_hBCB#ppt_hB*Y3>B ppt_h<*D\D' =+4 8?RCB?BCB#ppt_xB*Y3>B ppt_x<*D\Dk' =+34 8?CBi1+(6*min(max(#ppt_w*#ppt_h,.3),1)-7.4)/-.7*#ppt_h/2BCB#ppt_yB*Y3>B ppt_y<*D\+8+0+D0 +  0 =(  x  c $     c $P<$ 0    B1?p ,$D  0 H      H"p[1))?P  ZMUY IMPORTANTE0 2H  0޽h ? 3eX{IZ(a___PPT10.+}*D' = @B D' = @BA?%,( < +O%,( < +D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*0%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*0D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*0D' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*1a%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*1aD' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*1aDn' =A@BBBB0B%()))D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-m6Bbox(in)*<3<*D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*b%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*bD' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*b+p+0+0 ++0+0 +  0 zr (    c $    e  c $|0<$ 0    <1? @ ,$D  0H  0޽h ? 3eX{IZ(a___PPT10.+DV' = @B D' = @BA?%,( < +O%,( < +D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*)%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*)D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*)D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*)]%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*)]D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*)]D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*]%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*]D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*]D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*D<' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*8%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*8D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*8D' =%(Da' =4@BBBB%()))D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-m6Bbox(in)*<3<*+8+0+0 +>  0N0 `H/( l@ Hl H C  T  T  H C T<$  0 T  H H"T1))?P  ZMUY IMPORTANTE0 2H H 0޽h ? 3eX{IZ(a/'___PPT10.+-D' T= @B Db' = @BA?%,( < +O%,( < +D' =%(D+' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*Hn%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*HnD' =+4 8?dCB1+#ppt_h/2BCB#ppt_yB*Y3>B ppt_y<*HnD' =%(D+' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*Hn%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*HnD' =+4 8?dCB1+#ppt_h/2BCB#ppt_yB*Y3>B ppt_y<*HnD' =%(D+' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*HQ%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*HQD' =+4 8?dCB1+#ppt_h/2BCB#ppt_yB*Y3>B ppt_y<*HQ+8+0+H0 +  0N0 bZh( x hz h C T  T e h C 5T`0<$D  0 T  h <1?@0p,$D 0H h 0޽h ? 3eX{IZ(a___PPT10.+D3' 7T= @B D' = @BA?%,( < +O%,( < +D' =%(D+' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*h#%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*h#D' =+4 8?dCB1+#ppt_h/2BCB#ppt_yB*Y3>B ppt_y<*h#D' =%(D+' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*h#C%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*h#CD' =+4 8?dCB1+#ppt_h/2BCB#ppt_yB*Y3>B ppt_y<*h#CD' =%(D+' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*hCn%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*hCnD' =+4 8?dCB1+#ppt_h/2BCB#ppt_yB*Y3>B ppt_y<*hCnD' =%(D+' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*hn%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*hnD' =+4 8?dCB1+#ppt_h/2BCB#ppt_yB*Y3>B ppt_y<*hnD' =%(D+' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*h%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*hD' =+4 8?dCB1+#ppt_h/2BCB#ppt_yB*Y3>B ppt_y<*hDn' =%(D' =%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*h%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*hD' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*h+8+0+h0 +   0N0 $  l (  ll l C d  N pB l HLD1?PQpdB l <D1?` `dB l <D1? 0 dB  l <D1?@ @@ jB  l BD1?` jB l BD1?` @ jB l BD1?@  l <X1?  t"DH10 =  241 8 kJ6 l <N1?` P  v"DH20 =  285 8 kJ8 2 l <,N1?   DH30 = 44 kJ^ 2 l <1?  3H 2 l <N1?0  H2(g) + O2(g)f 2  l <N1?    jH2O(g)H 2 l <N1?P @  @ 2   l HPN1?c   hH2O(l)FH l 0޽h ? 3eX{IZ(ad5  0 ( $ z  C AT   e  C AT  <$  0 T dB @ <D1?  l  p  p,$D  0lB B <D1? plB B <D1? lB B <D1? pH  0޽h ? 3eX{IZ(a10___PPT100.+= D~0' ߑ|= @B D90' = @BA?%,( < +O%,( < +D' =%(DD' =%(D' =A@BBBB0B%(E5' =1B B`BPB,54*3>B ppt_c='`B@BPB<*+E' =4 B`BPB`B?<*%(/%,( < +D' =1:Bvisible*o3>+B#style.visibility<*+%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*+D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*+D' =%(DD' =%(D' =A@BBBB0B%(E5' =1B B`BPB,54*3>B ppt_c='`B@BPB<*+dE' =4 B`BPB`B?<*%( /%,( < +D' =1:Bvisible*o3>+B#style.visibility<*+d%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*+dD' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*+dD' =%(DD' =%(D' =A@BBBB0B%(E5' =1B B`BPB,54*3>B ppt_c='`B@BPB<*dE' =4 B`BPB`B?<*%(/%,( < +D' =1:Bvisible*o3>+B#style.visibility<*d%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*dD' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*dD' =%(DD' =%(D' =A@BBBB0B%(E5' =1B B`BPB,54*3>B ppt_c='`B@BPB<*)E' =4 B`BPB`B?<*%(/%,( < +D' =1:Bvisible*o3>+B#style.visibility<*)%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*)D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*)D' =%(DD' =%(D' =A@BBBB0B%(E5' =1B B`BPB,54*3>B ppt_c='`B@BPB<*)HE' =4 B`BPB`B?<*%(/%,( < +D' =1:Bvisible*o3>+B#style.visibility<*)H%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*)HD' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*)HD' =%(D}' =%(D' =A@BBBB0B%(E5' =1B B`BPB,54*3>B ppt_c='`B@BPB<*HE' =4 B`BPB`B?<*%(#/%,( < +D' =1:Bvisible*o3>+B#style.visibility<*H%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*HD' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*HD1' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-m6Bbox(in)*<3<*D_' =%(D' =%(D' =A@BBBB0B%(E' =4 B`BPB`B?<*%(,/%,( < +D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*+8+0+0 +a  0N0 p|R(  Pz P C ,[TP  T e P C ~T@<$  0 T H P 0޽h ? 3eX{IZ(a/'___PPT10.+.ɷND' ]T= @B Db' = @BA?%,( < +O%,( < +D' =%(D+' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*P_%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*P_D' =+4 8?dCB1+#ppt_h/2BCB#ppt_yB*Y3>B ppt_y<*P_D' =%(D+' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*P_%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*P_D' =+4 8?dCB1+#ppt_h/2BCB#ppt_yB*Y3>B ppt_y<*P_D' =%(D+' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*P%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*PD' =+4 8?dCB1+#ppt_h/2BCB#ppt_yB*Y3>B ppt_y<*P+8+0+P0 +  0N0 zT( G.x7@ Tz T C $   e T C T<$  0  ,]emu T <1? P`e,$D  0H T 0޽h ? 3eX{IZ(a___PPT10.+[9D~' T= @B D9' = @BA?%,( < +O%,( < +D' =%(D+' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*T%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*TD' =+4 8?dCB1+#ppt_h/2BCB#ppt_yB*Y3>B ppt_y<*TD' =%(D+' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*T%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*TD' =+4 8?dCB1+#ppt_h/2BCB#ppt_yB*Y3>B ppt_y<*TD' =%(D+' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*TE%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*TED' =+4 8?dCB1+#ppt_h/2BCB#ppt_yB*Y3>B ppt_y<*TEDD' =%(D+' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*TEn%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*TEnD' =+4 8?dCB1+#ppt_h/2BCB#ppt_yB*Y3>B ppt_y<*TEnD' =%(Da' =4@BBBB%()))D' =1:Bvisible*o3>+B#style.visibility<*T%(D' =-m6Bbox(in)*<3<*TD' =%(D+' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*Tn%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*TnD' =+4 8?dCB1+#ppt_h/2BCB#ppt_yB*Y3>B ppt_y<*Tn+8+0+T0 +  0N0 h`(   X S    4 X <1?P ,$D  0 X C P @<$  0  H  0޽h ? 3eX{IZ(a ___PPT10.+6D' = @B D>' = @BA?%,( < +O%,( < +Dg' =%(D' =%(D' =A@BBBB0B%(E' =4 B`BPB`B?<*%(/%,( < +D' =1:Bvisible*o3>+B#style.visibility<*X%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*XD' =+4 8?dCB1+#ppt_h/2BCB#ppt_yB*Y3>B ppt_y<*XDg' =%(D' =%(D' =A@BBBB0B%(E' =4 B`BPB`B?<*%( /%,( < +D' =1:Bvisible*o3>+B#style.visibility<*X%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*XD' =+4 8?dCB1+#ppt_h/2BCB#ppt_yB*Y3>B ppt_y<*XD(' =%(D' =%(D' =A@BBBB0B%(E' =4 B`BPB`B?<*%(/%,( < +D' =1:Bvisible*o3>+B#style.visibility<*Xi%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*XiD' =+4 8?dCB1+#ppt_h/2BCB#ppt_yB*Y3>B ppt_y<*XiD' =%(Da' =4@BBBB%()))D' =1:Bvisible*o3>+B#style.visibility<*X%(D' =-m6Bbox(in)*<3<*XDg' =%(D' =%(D' =A@BBBB0B%(E' =4 B`BPB`B?<*%(/%,( < +D' =1:Bvisible*o3>+B#style.visibility<*Xi%(D' =+4 8?dCB1+#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*XiD' =+4 8?dCB1+#ppt_h/2BCB#ppt_yB*Y3>B ppt_y<*Xi+8+0+X0 +a  0 t(  tl t C Ij   l t C T  H t 0޽h ? 3eX{IZ(ay___PPT10Y+D=' = @B +;  0N0 >6(  l  C  TPP  T   C LTp<$  0 T *X`hpd  <1?a H  0޽h ? 3eX{IZ(a___PPT10e.+렓\D=' = @B +  0N0 nf`X(  X X S kWQ1  T e X S l <$  0 T  X <1?9 ,$D  0H X 0޽h ? 3eX{IZ(a___PPT10.+@Do' = @B D*' = @BA?%,( < +O%,( < +D' =%(D' =%(D@' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*X*%(D' =-o6Bwipe(up)*<3<*X*D' =%(D' =%(D@' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*X*%(D' =-o6Bwipe(up)*<3<*X*D' =%(D' =%(D@' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*X%(D' =-o6Bwipe(up)*<3<*XD' =%(D' =%(D@' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*X&%(D' =-o6Bwipe(up)*<3<*X&D' =%(D' =%(D@' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*X&%(D' =-o6Bwipe(up)*<3<*X&D' =%(D1' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*X%(D' =-m6Bbox(in)*<3<*X+8+0+X0 +:;  0N0 0-{( L z  C T00  T e  C 0T` <$  0 T  x   #"B6    6@? x  i432 '   @`  6T? x  nCl H '   @`  6Y?   i460 '   @`  6W?   lO H '   @`  6W?   i745 '   @`  6W?   iC=O '   @`  6"W?  i315 '   @`  6`+W?  lC O '   @`  65W? i413 '   @`  6|>W? lC H '   @`  6GW? i243 '   @`  6OW? nCl C '   @`  6YW? i499 '   @`  6bW? iO=O '   @`  6kW? i812 '   @`  6uW? 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LBU ,$D  0H x 0޽h ? 3eX{IZ(a___PPT10i.+-D ' Y= @B D' = @BA?%,( < +O%,( < +D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*H-%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*H-D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*H-D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*H-%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*H-D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*H-D' =%(D' =%(D1' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*x%(D' =-m6Bbox(in)*<3<*xD ' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*H%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*HD' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*HD' =%(D1' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*x%(D' =-m6Bbox(in)*<3<*x+8+0+H0 +   0N0 D(  l  C M  M   C M<$  0 M H  0޽h ? 3eX{IZ(a  ___PPT10x .+'D ' = @B D ' = @BA?%,( < +O%,( < +D' =%(D' =%(DN' =A@BB BB0B%(D' =1:Bvisible*o3>+B#style.visibility<*]%(D' =-}6B!slide(fromLeft)*<3<*]D' =%(D' =%(DN' =A@BB BB0B%(D' =1:Bvisible*o3>+B#style.visibility<*]%(D' =-}6B!slide(fromLeft)*<3<*]D' =%(D' =%(DN' =A@BB BB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-}6B!slide(fromLeft)*<3<*+8+0+0 +   0N0 |D(  |l  C $Y  Y   C %Y<$ 0 Y H | 0޽h ? 3eX{IZ(av n ___PPT10N .+NwnD ' = @B D ' = @BA?%,( < +O%,( < +D' =%(D' =%(D@' =A@BB BB0B%(D' =1:Bvisible*o3>+B#style.visibility<*H%(D' =-o6Bdissolve*<3<*HD' =%(D' =%(D@' =A@BB BB0B%(D' =1:Bvisible*o3>+B#style.visibility<*H%(D' =-o6Bdissolve*<3<*HD' =%(D' =%(D@' =A@BB BB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-o6Bdissolve*<3<*+8+0+0 +  0N0 @8(    C P2Y` <$  0 Y H  0޽h ? 3eX{IZ(a4,___PPT10 .0  0N0 ( w l  C :Y  Y   C ;Y<$  0 Y d  <1?P pd  <1?P pH  0޽h ? 3eX{IZ(a4,___PPT10 .l  0 4,.(  @r @ S GY  Y r @ S HY Y  @ <1? p,$D 0H @ 0޽h ? 3eX{IZ(a___PPT10+D' = @B DW' = @BA?%,( < +O%,( < +D' =%(%(D' =%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*@%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*@D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*@+  0 t@(  tl  C QY  Y   B4SY1?   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