Anyone know the answer to this one?

C) 0.24 M

Explanation:

The given chemical reaction is presented as follows;

H₂SO₄(aq) + 2 KOH(aq) → K₂SO₄(aq) + 2 H₂O(l)

The titration experiment results are;

Volume of H₂SO₄(aq) used = 12.0 mL

Volume of KOH (aq) used = 36.0 mL

Concentration of KOH (aq) = 0.16 M

The number of moles of KOH present, n = 0.16 M × 36/1000 = 0.00576 moles

From the given reaction, 1 mole of H₂SO₄ reacts with 2 moles of KOH to give 1 mole of K₂SO₄ and 2 moles of H₂O

Therefore, 0.00576 moles of KOH reacts with (1/2) × 0.00576 moles = 0.00288 moles of H₂SO₄

Therefore, for the reaction;

The number of moles of H₂SO₄ in 12.0 mL of H₂SO₄ = 0.00288 moles

The concentration of the H₂SO₄ = 0.00288 M/(12.0 mL) = 0.24 M

The concentration of H₂SO₄ in the reaction = 0.24 M.

what is the question here. i think you messed up.

a). in this case, dh> 0 and sd> 0. if t becomes larger, the term "-tds" will become more negative. at some point, it is possible that -dts will be more negative than the dh is positive, and the reaction will become spontaneous. this is probably the correct answer, but you should look at all of the other possibilities and make sure of that.