Answer:
0.00125 moles H₃X
Solution and Explanation:
In this question we are required to calculate the number of moles of triprotic acid neutralized in the titration.
Volume of NaOH used = final burette reading - initial burette reading
= 39.18 ml - 3.19 ml
= 35.99 ml or 0.03599 L
Step 1: Moles of NaOH used
Number of moles = Molarity × Volume
Molarity of NaOH = 0.1041 M
Moles of NaOH = 0.1041 M × 0.03599 L
= 0.00375 mole
Step 2: Balanced equation for the reaction between triprotic acid and NaOH
The balanced equation is;
H₃X(aq) + 3NaOH(aq) → Na₃X(aq) + 3H₂O(l)
Step 3: Moles of the triprotic acid (H₃X used
From the balanced equation;
1 mole of the triprotic acid reacts with 3 moles of NaOH
Therefore; the mole ratio of H₃X to NaOH is 1 : 3.
Therefore;
Moles of Triprotic acid = 0.00375 mole ÷ 3
= 0.00125 moles
Hence, moles of triprotic acid neutralized during the titration is 0.00125 moles.
Atoms with six protons and six neutrons — carbon. Carbon is a pattern maker. It can link to itself, forming long, resilient chains called polymers. It can also bond with up to four other atoms because of its electron arrangement.
At 1.0 mol/min my teacher told me when I asked her yuh
The answer is: Dividing the number of molecules in the sample by Avogadro's number.
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.
For example:
N(Ba) = 2.62·10²³; number of atoms of barium.
n(Ba) = N(Ba) ÷ Na.
n(Ba) = 1.3·10²⁴ ÷ 6.022·10²³ 1/mol.
n(Ba) = 2.158 mol; amount of substance of barium.
The correct option is B. F
Hope this helps
-AaronWiseIsBae
