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An aliquot (28.7 mL) of a KOH solution required 31.3 mL of 0.118 M HCl for...

An aliquot (28.7 mL) of a KOH solution required 31.3 mL of 0.118 M HCl for neutralization. What mass (g) of KOH was in the original sample? Question 4 options: A) 0.173 B) 7.28 C) 1.64 D) 0.207 E) 0.414

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= 28.7 m2 IL Sol=> An aliquot (28.7 m2) of a kot solution required 31.3 mL of 0.118 m Hel for neutralization, what mass cym)=> 2) converts moles of Hel into moles of koh:- According to baluneed equation 1 mole Hel neutralized requires 1 mole koH. 1

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