1. A 100g coffee sample (specific heat = 0.385 J/gC) at 100C is added to 50.0g of water at 26.5C what is the final T of the mixture? (T in Fahrenheit could be asked) qcoffee + qwater = 0 qcoffee = - qwater 100g x 0.385J/gC x (Xc – 100C) = -50g x 4.184T/gC x (XC – 26.5C) 38.5Xc -3850J = -209Xc +5539J Xc =37.9C 2. 10g ice is heated to 60C… what is the total ene involved in this sample? Specific heat of water 4.184 J/gC and enthalpy of fusion = 6.02 KJ/mol Initial temp = -10C. desity of water is 1.00 g/L -10C – q1 0C – q2water – q3 60C 3. The heat of combustion of benzoic acid in -26.42 KJ/g. The combustion of 1.176g sample of benzoic acid causes a temp increase of 4.96C in bomb calorimeter. What is the heat capacity of the of the calorimeter assembly? qrxn = 1.176g x -26.42 KJ/g = -31.07 KJ = - qcal qcal = c x ∆T so, c = 6.26 KJ/C Aqueous silver ion reacts with aqueous chloride ion to yield a white precipitate of silver chloride. + Ag + cl agcl When 10.0 ml of 1.00 M AgNo3 is added to 10.0 ml of 1.00 M Nacl solution at 25C in a calorimeter a white precipitate of Agcl forms and the temperature of the aqueous mixture increases to 32.6C Assuming the specific heat of the aqueous mixture is 4.18 J/gC the density of the mixture is 1.00 g/ml, calorimeter absorbs negligible heat. Calculate ∆H in kJ in the reaction. Ans. * Balance the equation first! In such cases m = 20.0 ml x 1 g/ml = 20 g ∆T = 32.6C – 25.0C = 7.6C q = 4.18 J/gC x 20.0 g x 7.6C = 6.4 x 102 J moles of Ag+ = 10.0 ml X (1.00 mol Ag+ / 1000 ml ) = 1.00 X 102 mol Ag+ Therefore, moles of AgCl Heat evolved / mole AgCl = 6.4 X102 J / (1.00 X 10-2 mol AgCl) = -64 KJ/ mol Agcl *Either mention heat is evolved or put the negative sign 4) Oxyac welding torches burn acetylene gas C2H2O. calculate ∆H in KJ for combustion reaction of acetylene to yield Co2 and H2O ∆Hf , C2H2 = 227.4 KJ/mol H2O = -241.8 KJ/mol CO2 = -393.5 KJ/mol Ans. *First step balance the equation 2C2H2 (g) + 5O2 (g) 4CO2 (g) +2H2O(g) ∆Hf = [4∆Hf0CO2 + 2∆Hf0 H2O]-[2∆Hf0C2H2+ 5∆Hf0 O2] = [4mol x -393.5 KJ /mol + 2mol x -241.8 KJ/mol]-[2mol + 227.4 KJ/mol] = 5) A reaction inside a cylindrical container with a moveable piston causes volume change from 12.0L to 18.0L while the pressure on outside the container remain constant at 0.975 atm. What is the value in J of the work w done during the reaction? Ans. ∆V = 18.0 L – 12.0 L = 6.0L W = -P = 0.975 atm x 6.0 L = -585L.atm x 101.325 J/ L. atm = 593 J (we say evolved so positive) V = 6.0L X (1000ml/1L) X (1cm3/1ml) = 6000 cm3 h= 6000 cm3 / π x (8.50 cm)2 = 26.4 cm

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