How many mL of 1.000M NaOH solution would it take to neutralize 10.0mL of vinegar that...

Question

Question

Answer 1

Answer #1

A) at neutralization point,

mol of Acid = mole of Base

10×0.50 = 1.00 × V

V = 5.00 mL (d) is correct

1) Moles of a cid m es of NA 0H J0- OX0.10 CB M- 10.201 False Fabbe 2

Answer 2

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