Answer:
5.01%
Explanation:
Density of vinegar = mass/volume
Mass of 10.00 mL = density x volume
= 1.006 x 10 = 10.06 g
From the equation of reaction:
1 mole pf CH3COOH requires 1 mole of NaOH for neutralization.
mole of NaOH = molarity x volume
= 0.5062 x 0.01658
= 0.008392796 mole
0.008392796 mole of NaOH will therefore require 0.008392796 mole of CH3COOH.
mass of CH3COOH = mole x molar mass
= 0.008392796 x 60.052
= 0.504 g
Percentage by mass of acetic acid in the vinegar = 0.504/10.06 x 100%
= 5.01%
The percent by mass of acetic acid in the vinegar is 5.01%
Answer is: 12 grams of the isotope carbon-12 (¹²C).
The Avogadro’s number is the number of atoms in 12 grams of the isotope carbon-12 (¹²C).
Na is Avogadro number or Avogadro constant (the number of particles, in this example carbon, that are contained in the amount of substance given by one mole).
The Avogadro number has value 6.022·10²³ 1/mol in the International System of Units; Na = 6.022·10²³ 1/mol.
1.50*10^23 molecules/ (6.02*10^23 molecules/mol)= 0.249 mol.
There is 0.249 mol in 1.50*10^23 molecules~
I think the correct answer is C