Cl2(g) -------> Cl-(aq) + ClO-(aq)
2e- + Cl2(g) -------> 2Cl-(aq) [reduction]
4OH-(aq) + Cl2(g) -----------> 2ClO-(aq) + 2H2O(l) + 2e- [oxidation]
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2OH-(aq) + Cl2(g) --------> Cl-(aq) + ClO-(aq) + H2O(l)
A, They Lower The Activation Energy.
Answer:- 13.6 L
Solution:- Volume of hydrogen gas at 58.7 Kpa is given as 23.5 L. It asks to calculate the volume of hydrogen gas at STP that is standard temperature and pressure. Since the problem does not talk about the original temperature so we would assume the constant temperature. So, it is Boyle's law.
Standard pressure is 1 atm that is 101.325 Kpa.
Boyle's law equation is:

From given information:-
= 58.7 Kpa
= 23.5 L
= 101.325 Kpa
= ?
Let's plug in the values and solve it for final volume.

On rearranging the equation for 

= 13.6 L
So, the volume of hydrogen gas at STP for the given information is 13.6 L.
Answer:

Explanation:
Hello,
For the given chemical reaction:

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

Finally, we compute the percent yield with the obtained 2.10 g:

Best regards.