If Compound A (256 mg) was hydrolyzed using 12.5 mL of 1.0 M KOH
and the...
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Chemistry
If Compound A (256 mg) was hydrolyzed using 12.5 mL of 1.0 M KOHand the amount of remaining KOH was back titrated and required 44.9mL of 0.25 M HCL to neutralize the excess NaOH, what were thesaponification equivalent and neutralization equivalent, and whatis compound A?
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KOH HCL KCL H2O 1 mole 1 mole no of moles of HCl molarity volume in L 02500449 0011225 moles of HCl no of moles of KOH
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