For this lab you will utilize the following

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Unformatted text preview: s covered in class, the isotopic fractionation between two species is often a function of temperature, which is governed by an experimentally calibrated fractionation factor. For this lab you will utilize the following fractionation factors: 1) Quartz ­water fractionation: a. 103lnαquartz ­water = 3.38x106/T2 – 3.4 2) Chlorite ­water fractionation: a. 103lnαchlorite ­water = 1.56x106/T2 – 4.74 3) Muscovite ­water fractionation: a. 103lnαmuscovite ­water = 2.38x106/T2 – 3.89 4) Quartz ­cassiterite: a. 103lnαquartz ­cassiterite = 3.08x106/T2  ­0 Remember: 103lnαx ­y = Δx ­y = δx ­δy For the data presented in the attached spreadsheet, answer the following: 1) What are the temperatures of formation of each species (e.g., chlorite, muscovite)? In samples with both muscovite ­quartz and cassiterite ­quartz provide temperatures for both mineral pairs (10 marks). Note...
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