l 1219) + F29) 55 2HF(g) DeltaHrn.Chegg.com 8/1/2016 11:55 PM 39.5/100 Grad Prin
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l 1219) + F29) 55 2HF(g) DeltaHrn.Chegg.com 8/1/2016 11:55 PM 39.5/100 Grad Print Caculator Periodic Table il Caculatr Print -Periodic Table stion 9 of 15 Sapling Learning Map A coffee cup calorimeter with a heat capacity of 5.30 J'C was used to measure the change in enthalpy of a precipitation reaction. A 50.0 mL solution of 0.360 M AgNOs was mixed with 50.0 mL of 0.290 M KI. After mixing, the temperature was observed to increase by 387 °C. Calculate the enthalpy of reaction, n, per mole of formed (Agi). Assume the specific heat of the product solution is 4.10 Mg·) and that the density of both the reactant solutions is 1.00 g/mL Hn per mole of precipitate Calculate the theoretical moles of precipitate formed from AgNOs (left) and KI (right) Number Number moles moles Calculate the heat change expenenced by the calorimeter contents, nnes. Calculate the heat change experienced by the calorimeter, qal Number Calculate the heat change produced by the solution process, quolution Using the mole values calcuated above, calulate Hsoution for one mole of precipitate formed. kJ/ mole Previous ®Gwe Up & View Solon O Check Answer Next Ent about us | careers partnersprvacy poiy terms of use contact us hep 888 a 4 5 8Explanation / Answer
nO of mol of AgNO3 = 50/1000*0.36 = 0.018 mol
nO of mol of KI = 50/1000*0.29 = 0.0145 mol
AgNO3 + KI ----> AgI + KNO3
NO of mol of AgI produced from AgNO3 = 0.018 mol
NO of mol of AgI produced from KI = 0.0145 mol
Heat change experienced by calorimeter contents
= msDT
= 100*4.1*3.87
= 1586.7 joule
Heat change experienced by calorimeter , (qcalorimeter)
= = 5.3*3.87 = 20.51 joule
calculate the heat change produced = 1586.7+20.51 = 1607.21 joule
= 1.607 kj
no of mol of AgI produced actually = 0.0145 mol
DHrxn = 1.607/0.0145 = 110.83 kj/mol
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