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.
There are many ways to test and identify metal. The easiest way is observing its color. Also how reflective it is. Other ways would be boiling point, melting point, density, or conductivity of the metal.
Hope This Helps and God Bless!
Answer:
The reaction between magnesium and sulphuric acid is given as :
Mg(s) + H2SO4(aq) --> MgSO4(aq) + H2(g)
Explanation:
pls mark as brainliest