A volume of 70.0 mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2SO4). What was the molarity of… | Cheap Nursing Papers
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A volume of 70.0 mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2SO4). What was the molarity of…

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A volume of 70.0 mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2SO4). What was the molarity of the KOH solution if 18.2 mL of 1.50 M H2SO4 was needed? The equation is 2KOH+H2SO4K–>2SO4+2H2O

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