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# image005 - 15.83 In order to solve this problem we must be...

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Unformatted text preview: 15.83 In order to solve this problem, we must be able to add reactions {a}. {b}. and {c} to give the reaction desired. The most logical starting point for solving this problem is to find the reactant or product that appears onlyr once in {a}, (b). or {c}. as this will determine how that equation must be used. Because 01 onlyr appears in {a}, equation {a} must be maltipled by 1i and left in its original direction in order to obtain the correct amount of 01(5)- iCHtis} + stats) —>i C0: is} + ‘2‘ “20(3) H402 H) We now consider how to combine this equation with {b} and [c] to give the desired overall reaction. Because Haﬂig) appears in {c} and not in (b), we mustadd equation to} to tltctraasfonned is} sues to obtain atotal of two 1410(3). To do this, we must reverse [a] and multiplyr it by +: +Cﬁ(g]+4}H3fg] —>-l-CH‘{g} +12-Hiﬂig) HHS-i? kl) Equation {b} must be muitiplied by -} and left in its original direction in order that the COﬁg} will be cancelled ﬁ'om the overall equation: %CH4[g}+i-C01{g]—r§C0{g)+§-Hlig} -}[+206Ir.l) The net result is Alb-SEAS} +-} mfg—a. % Cﬁﬁgj + {- HIO-{g} =}{—3l]2 k1} tints) +iHlisJ—4t CHttsJ+s Hattie] iii-24'? to iCHASH'iCDﬂSJ ﬂicqa) +§H1tgﬂrt+2iiﬁ i=3} CH,[gJ+e}ﬂ,[g]-—r CtIgJ+2 H101?) oH°=—5'Fﬂk.l 631 [a}1‘lteheetoutputofeaehaampleisfoundusingtheheat capacityot‘the calorimeteraudtheobservedchaogointemperature ofthe calorinmteI. For BmdxdTwasﬁﬂTandﬂoeenergvcontentoflﬂﬁgis: {\$.E"C)-[dﬂﬂJ-K"}=SJDDJ-g". amenaaemusmarmsssnc andtiteeoergv mammnoogo: (s.stc)-[sas 1-K") =5lﬂﬂJ~g4. (bjﬁﬂgofﬂrandXoerealoontaios . -I Icalﬂﬁﬂ- a: . . 3“ not {393](53kﬁl g }[4.Iﬂ4lci] 38 Catoneswltﬂe gofABCprod I Calorie 4.134 lei maﬁa g](5.lk.l- g")[ )=ss Calories ...
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