Question

1) when the temperature-time data graph was reviewed, it was found that the error had been made determining DeltaT. Instead of 4.76C, Delta T was actually 4.70C. Based on this change only, calculate the correct deltaH neutralization for the reaction oh NH3 and acetic acid. 2) calculate percent error if for correct Delta H (neutzn) if Delta T 4.76C had been used? <-----thew second question is what I want

Answer 1

Answer 8 - Roge no-10 (1) let, Volume of C-13CCOH = 500 ml volume of NH3 = 500ml concentration of CH2C00H = LOM or mor Concentration of NH3 = HOM HQ! Moles of Both CH₂ CCOH G NH3 = lomu 20.000 = 0.05 Holes. Reaction Ht to oH JOHOR) (aa) Cool) Density of H₂O = igimi so, mass of watu = 100.00 some Hel + 50 ml NaOH = 100ml to] WHO = mx CXAT : Given data R = 100 gx 4.184 x 4.700 AT = 4.70cl AR WHO - 1966-48 Jouls AH Neutralisation - HD 1966.48 Toiles o.oso mol - 39329.6 Joulymol As Neutoalization is an exothermic Cheat released) - 39329-65 Coy) - 39.3 kJ mol mol molls : - 39329-65 4Houts

Page No - (2) Given, Дт: •16". QE MXOYOT = 100 of x4.184 I x4,216d = 1991.58uJ At Neuta = 1991-584 T 0.05 MOI = 398.31.685 mol - 398 les mo1 experimental value Percent error = - True Value True value x 100% True value = Actual value we got AT = 4.700c using • 39.8 - 39.3 2 x 1001. 39,3 1 % erren - 1.27% error]