Asked by Madi

How many grams of KOH are needed to neutralize 12.9 mL of 0.16 M HCl in stomach acid?
Answered by R_scott
.0129 L * 0.16 M = 1.9E-3 moles

multiply 1.9E-3 by molar mass of KOH
Answered by anonymous
blub?
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