Lab 10: Enthalpy of Neutralization(7) Identify the limiting reagent and briefly explain why it is limiting. ds toad sdd sialla ()Jed ofhioea sh ba Im & SO1 aew ouurbimanswerwans(8) Find AHneutzn for the reaction.ardt gaieu mite bee momod asminolso ard yd bodioeda 1sod lo inuorms srlt sdaluole )(A) S nobesuC ni batsluol Instenoo 1odsminolanswer5wansbadioeds lesdl Intoi ont sisiuoin ()O8.0 lo.im 0.001 bas HOM M0080 lo Jm 0.001 lo noibesnsdi yd boenslon tead Indot orl: bnit (a) (4) Calculate the calorimeter constant, using the AT of the cold water.(9.08E 2)- (7.97E2)%3D3.81otai bne aniuqung lo noieeunaibol somce ovitaiolrs ne bea29.13 J Banswer e A.Cat lis dgesSTobooo1g sl gniau oamiroles or lo instenos aisinols ar benimebatzad lo Im 00.08 ot yatew blos to Im 00.0 bobba inobute odT olobom aidt3. The student then determined AHneutzn for the reaction of sodium hydroxide and acetic acid, using theprocedure described in this module. The student added 100.0 mL of 0.8500M NaOH to 100.0 mL of0.8404M acetic acid. Prior to and following the mixing of the acid and base solutions, the followingtemperature-time data were collected.sisw bloo sdt bns 19i ord urfi tol TA orlt onimsioG )temperature, °Cmixturetime,temperature, °CNaOHtime,minHOẠCmin0.17.526.23.280.517.7323.031.017.54823.101.517.7523.062.017.551022.992.517.771122.933.017.571222.893.517.781322.854.017.581422.784.517.801522.735.0mixing1622.70tew forl1722.641Owens1822.59nw blo(1) Plot the temperature-time data and determine the mean temperature of the unmixed reagents.19/ew tord orit yd teol Jaod odl sinlupla )answer(2) Determine AT from the graph.tevr blo d ved bon 1esd arlt oteluain (E

The balanced equation for the reaction of NaOH and acetic acid, CH3COOH is: NaOH(aq) + CH3COOH(aq) -----> CH3COONa(aq) + H2O(l) mole ratio between the two reactants is 1/1.#1: There is a total of 10 temperature measurement values before mixing. Hence we have to add them all and divide by 10 to find the mean temperature. Mean temperature of the unmixed reagents i.e before mixing: = 17.52+17.54+17.55+17.57+17.58+17.73+17.75+17.77+17.78+17.8010= 17.66 Co (Answer) Hence initial temperature, Ti = 17.66 oC # Temperature-time plot:#2: The linear regression equation is: Y = - 0.0504X + 23.495 Final temperature is is the temperature when "X" axis value is 0 => Y = - 0.0504*0 + 23.495 => Y = 23.495 or 23.50 oC So the final temperature is, Tf = 23.50 oC Initial temperature, Ti = 17.66 oC Change in temperature, ∆T = Tf - Ti = 23.50 oC - 17.66 oC = 5.84 oC (Answer) #7: moles of reactant NaOH= C*V= 0.8500 molL*100.0 mL*1 L1000 mL= 0.08500 molmoles of reactant acetic acid= C*V= 0.8404 molL*100.0 mL*1 L1000 mL= 0.08404 mol Since the mole ratio of the reactants in the balanced equation is 1/1, the reactant with smaller mole is the limiting reactant. Hence acetic acid (CH3COOH) is the limiting reactant (Answer) Note: As per the company policies, we can answer the first 3 parts only. Please repost the rest of the parts as a separate question to get the solution.