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When 50.0 mL of 0.900 M Ca(NO3)2 is added to 50.0 mL of 1.80 M NaF,...

When 50.0 mL of 0.900 M Ca(NO3)2 is added to 50.0 mL of 1.80 M NaF, CaF2 precipitates, as shown in the net ionic equation below. The initial temperature of both solutions is 30.00°C. As the reaction proceeds, the temperature rises to 31.50 °C. Assuming that the reaction goes to completion, and that the resulting solution has a mass of 100.0 g and a specific heat of 4.18 J/(g ∙ °C), calculate the ΔH° for this reaction, expressed in kJ/mol Ca+2.

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Expert Solution

Ca(NO3)2(aq) + 2NaF(aq) --------------> CaF2(s) + 2NaNO3(aq)

no of moles of Ca(NO3)2   = molarity * volume in L

                                          = 0.9*0.05   = 0.045 moles

no of moles of NaF            = molarity * volume in L

                                         = 1.8*0.05   = 0.09moles

Ca^2+(aq)   + 2F^-(aq) --------------> CaF2(s)

ΔT   = T2-T1

         = 31.5-30   = 1.50C

m   = 100g

c    = 4.18J/g-0C

q   = mcΔT

        = 100*4.18*1.5

         = 627J

no of moles of Ca^2+   = 0.045moles

ΔH°    = -627J

         = -627J/0.045mole    = -13933J/mole   = -13.933KJ/mole

queen_honey_blossom answered 2 years ago
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