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10 Nov 2019
A volume of 70.0mL of aqueous potassium hydroxide (KOH) wastitrated against a standard solution of sulfuric acid (H2SO4). Whatwas the molarity of the KOH solution if 12.7mL of 1.50 MH2SO4 was needed? The equation is
2KOH(aq) + H2SO4(aq) ? K2SO4(aq) + 2H2O(l)
Please, show me.
A volume of 70.0mL of aqueous potassium hydroxide (KOH) wastitrated against a standard solution of sulfuric acid (H2SO4). Whatwas the molarity of the KOH solution if 12.7mL of 1.50 MH2SO4 was needed? The equation is
2KOH(aq) + H2SO4(aq) ? K2SO4(aq) + 2H2O(l)
Please, show me.
Bunny GreenfelderLv2
31 Aug 2019
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