10.) 0.085 g of an unknown diprotic acid is titrated with 0.095 M NaOH. The first...

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10.) 0.085 g of an unknown diprotic acid is titrated with 0.095 M NaOH. The first equivalence point occurred in the titration at a volume of 9.70 mL of NaOH added; The second equivalence point occurred at a volume of 19.40 mL of NaOH added. How many moles of NaOH were used to reach the second equivalence point in this diprotic acid titration? 10.) 0.085 g of an unknown diprotic acid is titrated with 0.095 M NaOH. The first equivalence point occurred in the titration at a volume of 9.70 mL of NaOH added; The second equivalence point occurred at a volume of 19.40 mL of NaOH added. How many moles of NaOH were used to reach the second equivalence point in this diprotic acid titration?

Answer rating: 100% (QA)

Cuestion 0 085g of unknown diprotic acid i Hitrated with 0 0 95 Florst equival... View the full answer