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'd say the correct answer is: Noodles rising and falling apart in boiling water.
If one starts with 0.020 g of Mg, 0.0008 moles of H2 would be made if the reaction is complete.
Going by the balanced equation of reaction in the image, 1 mole of Mg will produce 1 mole of H2 in a complete reaction.
If 0.020 g of Mg is started with:
mole of Mg = mass/molar mass
= 0.020/24.3
= 0.0008 moles
Since the mole of Mg to H2 is 1:1, thus, 0.0008 moles of H2 will also be made from the reaction.
More on stoichiometry can be found here: brainly.com/question/9743981