Answer:
Butanoic acid present in solution
Explanation:
In this case, we have a buffer solution of butanoic acid and sodium butanoate. In other words a reaction like this:
HC₄H₇O₂ + H₂O <------> C₄H₇O₂⁻ + H₃O⁺ Ka = 1.5x10⁻⁵
The low value of Ka means that this is a weak acid. So, after this, the NaOH is added to the solution.
The NaOH is a really strong base, so we might expect that the pH of the solution increase drastically, however this do not occur.
The reason for this is because the first thing to happen in this reaction is an acid base reaction.
The NaOH react with the butanoic acid still present in solution, because is a weak acid, so in solution, this acid is not completely dissociated into it's respective ions. So the butanoic acid reacts with the NaOH and the products:
HC₄H₇O₂ + NaOH <------> Na⁺C₄H₇O₂⁻ + H₂O
So, because of this, the pH increase but not much.
Increase the surface area
Answer:
T2 = 547.2K
Explanation:
Given:
V1 = 135 ml. V2 = 250 ml
T1 = 295.5K. T2 = ?
Using Charles's law and solving for T2,
V1/T1 = V2/T2
T2 = (V2/V1)T1
= (250 ml/135 ml)(295.5K)
= 547.2K (or 273.2°C)
Answer:
I'm not the smartest human being on Earth, so I'm not 100% sure, but I think that the answer is either B or D. Again I don't know for sure, but that's what I think
