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SrSO4 (Strontium Sulfate) has a Ksp of 3.2E-7. In a crystal lattice solid, each SrSO4 formula unit has a volume of 0.90 nm by 0.45 nm by 0.45 nm. At exposed regions, the region is 0.9 nm by 0.45 nm.

a) Calculate the standard free energy change per mole assocatiated with the formation of solid SrSO4. Use the solubility product. Remember G(standard) equals -RTln(K) when all surface effects are neglected.

b) Convert (a) to a standard free energy change per SrSO4 molecule (that is, per SrSO4 formula unit).

c) The surface free energy of BaSO4 is 3.8E-20 J/nm2 while SrSO4 is 1.05E-19 J/nm2. The solubility product is Ksp of SrSO4 is 3.5E-7 while BaSO4 is 1.1E-10. Which one will be easier or more difficult to nucleate.

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