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# Petrucci Answers _1 - 4B The assumptions include no heat...

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Unformatted text preview: 4B The assumptions include no heat loss to the smmundings or to the calorimeter, a solution density of 1.00 gImL, a speciﬁc heat of 4.18 J g" 'C‘1 , and that the initial and ﬁnal solution volumes are the same. The equation for the reaction that occurs is NaOH(aq)+ HCl(aq) -) NaCl(aq) + 1130(1). Since the two reactants combine in a one to one mole ratio, the limiting reactant is the one present in smaller amount (Le. the one with a smaller molar quantity). amount HCl = 100.0 mL x 1_.02_o____ “mm = 102.0 mmol HCl 1 mL soln amount NaOH = 50.0 mL x W = 99.4 mmol NaOH 1 mL soln Thus, NaOH is the limiting reactant. 1 mmol H10 1 mol H20 —56 1:] q“ = 99.4 mmol NaOHx = -5.5-, kl ....—..—-u—-x ___—- 1 mmol NaOl-I 1000 mmol H20 x1111011120 1.00g 4.131 11:1 X x lmL g'C 10001 Quin-5m = -qneutr = 5'52 k1 = (100-0 + 50'0) mL x x(l‘ — 24521:) I: 5.51 +153, = 33.4'C 0.62'1‘r = 0627:4531 g]; Determine the initial number of moles: 1 molN = 50.0 N x ———3—— = 1.785 moles of N n g 2 28.014 g N2 — 2 V = nRT = (1.785 mol N2)(0.08206 Latm K"mol")(293.15 K) = 17 2 L P 2.50 atm ' AV= 17.2 — 75.0 L = —57.8 L 101.33 .1 11:] ————x =41 V=—.50 tnl-~57.8L w A 2a( )leatmIOOOJ = +14.6 it] work done on system. _ J .. i cm —375 gx4.18E;E(87—26) C=9.5§x104 J=-qh a“ =_9.5§,x104 J = 46ng0'449_ng(87'ﬁ)=1-31§x10‘J—2.08§x102 2; —9.56x 10‘ - 1.816): 104 *1133x10“ T =-—:——é=__:——_= 2. l —2.08§x101 “2.08§x102 SAﬁXIO C or 545 °C The number of signiﬁcant ﬁgures in the ﬁnal answer is limited by the two signiﬁcant ﬁgures for the given temperatures. 0.335 1 . . 1.25 g _ _ 44.8 x 31.1 C—143.2c: =165 mLx x .111. 31.1C—24.8C g 3.0 x( ) lmL 8p x( ) 3 3.23x10‘=1.3x103x(sp.ht.) sp_ht.=§-2_3ﬂ=2.5 13-113-. 13):“) 92.1 g nnlarheateapacity = 2.5 Jg"'C" x = 2.3 x 102 Ind—“C" 1 mol C3H,03 10. The additional water simply acts as a heat transfer medium. The essential relationship is heat lost by iron = heat gained by water (of unknown mass) —[1.23 kgxlooo gJ0.449—'_'-—(25.15—-68.5)“C=Jlrgl-I,Ox45.18-41—(25.15-—18.5)'C 1 kg g C g'C 2.37x10‘ 1 mLH 0 2.37 10‘J=29.7 =—————=79s H 0 ——"’-——=8.0 102 mLH o x "x x 29.1 - g 1 x1.00 31-110 x 1 36. Note that the heat evolved is the negative of the heat absorbed. Imoll£3.,H.§O3 30231:] 120ng —.x .— heat evolved 138.123C71-I‘03 1molC1.I-I¢.,03 _ . =__=_________4.28kJI°C heat capacity A T (29.32 - 23.68) “C ...
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Petrucci Answers _1 - 4B The assumptions include no heat...

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