What volume (in milliliters) of 0.140 M HClO₃ is required to neutralize 60.0 mL of 0.160 M LiOH? _____mL
Added by Marc R.
Step 1
Given: Volume of LiOH = 60.0 mL = 60.0 mL * (1 L / 1000 mL) = 0.060 L Molarity of LiOH = 0.160 M Moles of LiOH = Molarity * Volume Moles of LiOH = 0.160 mol/L * 0.060 L Moles of LiOH = 0.0096 moles Show more…
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