Triprotic acid is a class of Arrhenius acids that are capable of donating three protons per molecule when dissociating in aqueous solutions. So the chemical reaction as described in the question, at the third equivalence point, can be show as: H3R + 3NaOH ⇒ Na3R + 3H2O, where R is the counter ion of the triprotic acid. Therefore, the ratio between the reacted acid and base at the third equivalence point is 1:3.
The moles of NaOH is 0.106M*0.0352L = 0.003731 mole. So the moles of H3R is 0.003731mole/3=0.001244mole.
The molar mass of the acid can be calculated: 0.307g/0.001244mole=247 g/mol.
I am going to say it is false.
The atoms undergo chemical changes.
New substances may be a product of the reaction.
Answer:
0.00757 grams
Explanation:
Find the molar mass of the compound: which is 60.05.
The molar mass is basically just the sum of all the atomic masses of each of the elements.
Then multiply the molar mass by the number of moles in the compound, which is 1.26 x 10^-4 moles.
Your answer should be 0.00757 grams.
