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How many moles of NaOH are present in 13.0 mL of 0.220 M NaOH?
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How many moles of NaOH are present in 13.0 mL of 0.220 M NaOH?

User Clearly
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1 Answer

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v=13.0 mL=0.0130 L
c=0.220 mol/L

n=vc

n=0.0130*0.220=0.00286=2.86*10⁻³ mol
User Iayork
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