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How many moles of NaOH are present in 27.5 mL of 0.270 M NaOH?
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How many moles of NaOH are present in 27.5 mL of 0.270 M NaOH?

User Tautologe
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The solution for the question above is:

C = 0.270
V = 0.0275L
n = ?

Use the molar formula which is: C = n/V
Re-arrange it to: n = CV
n = (0.270)*(0.0275)
n = 0.007425 mols
(more precise) n = 7.425 x 10^-3 mols

7.425 x 10^-3 mols is the answer.
User Vladimir Dyuzhev
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