What volume of a 0.3300-m solution of sodium hydroxide would be required to titrate 15.00 ml of 0.1500 m oxalic acid? c2 o4 h2(aq) + 2naoh(aq) ⟶ na2 c2 o4(aq) + 2h2 o(l)?

13.64 ml

Explanation:

1) Titration will be done when all the moles of the oxalic acid have reacted with the volume of sodium hydroxide.

2) The balanced chemical equation is:

C₂O₄H₂ (aq) + 2 NaOH(aq) ⟶ Na₂C₂O₄(aq) + 2H₂O(l)

3) Mole ratio:

1 mol C₂O₄H₂ (aq) ; 2 mole NaOH(aq)

4) Calculate the number of moles of C₂O₄H₂ (aq)

M = n/V ⇒ n = MV = 0.01500 l × 0.1500M / 1000 = 0.00225 mol

5) Calculate the number of moles of NaOH needed using proportion with the mole ratio:

1 mol C₂O₄H₂ (aq) / 2 mole NaOH(aq) = 0.00225 mol C₂O₄H₂ (aq) / x

⇒ x = 0.0045 mol NaOH

6) Use molarity to find the volume that contains 0.0045 mol of NaOH.

M = n / V ⇒ V = n / M = 0.0045 / 0.3300 = 0.1363 liter = 13.64 ml

metals are good conductor of electricity. it has free electrons on their surface. according to electron pool theory, each atom present in a metal crystal losses all of its valence electron and as a result a pool of free is form. these pool of electrons are free to move and   conduct electricity easily   while making metal as good conductor.

the correct answer is a. the ability of electrons to flow throughout the metal.

if it's not useful, feel free to report for mistake in answer.