We are told that KOH is being used to completely neutral H₂SO₄ according to the following reaction:
KOH + H₂SO₄ → H₂O + KHSO₄
If KOH can completely neutralize H₂SO₄, then there must be an equal amount of moles of each as they are in a 1:1 ratio:
0.025 L x 0.150 mol/L = .00375 mol KOH
0.00375 mol KOH x 1 mole H₂SO₄/1 mole KOH = 0.00375 mol H₂SO₄
We are told we have 15 mL of H₂SO₄ initially, so now we can find the original concentration:
0.00375 mol / 0.015 L = 0.25 mol/L
The concentration of H₂SO₄ being neutralized is 0.25 M.
I don't know why I am answering this question but assuming C-13 has a natural abundance of 1.07%:
(1.6000x10^4)(0.0107) = 171.2 = 171 atoms of C-13
<span>Catalysts decrease the activation energy and the more collisions result in a </span>reaction<span>, so the </span>rate<span> of </span><span>reaction increases.</span><span />
