Title: If 1.020 g of KHC8H4O4 (204.23 g/mol) is completely neutralized with 0.200 M KOH Post by: OldSpice on Dec 8, 2014 If 1.020 g of KHC8H4O4 (204.23 g/mol) is completely neutralized with 0.200 M KOH, what volume of potassium hydroxide is required?
KHC8H4O4(aq) + KOH(aq) ? K2C8H4O4(aq) + 2 H2O(l) A) 12.5 mL B) 20.0 mL C) 25.0 mL D) 40.0 mL E) 50.0 mL Title: Re: If 1.020 g of KHC8H4O4 (204.23 g/mol) is completely neutralized with 0.200 M KOH Post by: Electric on Dec 11, 2014 Content hidden
Title: Re: If 1.020 g of KHC8H4O4 (204.23 g/mol) is completely neutralized with 0.200 M KOH Post by: OldSpice on Dec 14, 2014 Perfect!
..Thank you so much for your support Title: Re: If 1.020 g of KHC8H4O4 (204.23 g/mol) is completely neutralized with 0.200 M KOH Post by: Electric on Dec 14, 2014 My pleasure
Title: BFSF: If 1.020 g of KHC8H4O4 (204.23 g/mol) is completely neutralized with 0.200 M KOH Post by: Reuben Sequeira on May 22, 2023 Help! The answer is missing an explanation...
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