Answer:
pH at equivalence point is 8.52
Explanation:

1 mol of HCOOH reacts with 1 mol of NaOH to produce 1 mol of 
So, moles of NaOH used to reach equivalence point equal to number of moles
produced at equivalence point.
As density of water is 1g/mL, therefore molarity is equal to molality of an aqueous solution.
So, moles of
produced = 
Total volume of solution at equivalence point = (25+29.80) mL = 54.80 mL
So, at equivalence point concentration of
= 
At equivalence point, pH depends upon hydrolysis of
. So, we have to construct an ICE table.

I: 0.1940 0 0
C: -x +x +x
E: 0.1940-x x x
So, ![\frac{[HCOOH][OH^{-}]}{[HCOO^{-}]}=K_{b}(HCOO^{-})=\frac{10^{-14}}{Ka(HCOOH)}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BHCOOH%5D%5BOH%5E%7B-%7D%5D%7D%7B%5BHCOO%5E%7B-%7D%5D%7D%3DK_%7Bb%7D%28HCOO%5E%7B-%7D%29%3D%5Cfrac%7B10%5E%7B-14%7D%7D%7BKa%28HCOOH%29%7D)
species inside third bracket represent equilibrium concentrations
So, 
or,
So, 
So, 
So, ![pH=14-pOH=14+log[OH^{-}]=14+logx=14+log(3.285\times 10^{-6})=8.52](https://tex.z-dn.net/?f=pH%3D14-pOH%3D14%2Blog%5BOH%5E%7B-%7D%5D%3D14%2Blogx%3D14%2Blog%283.285%5Ctimes%2010%5E%7B-6%7D%29%3D8.52)
Answer:
An unbalanced single replacement reaction
The amount of energy required to remove the most loosely bound electron in1 mole of an atom in its neutral gaseous state. Atomic radius is smaller, great nuclear charge (pulling power of protons on less electrons)
I’m not a 100% shure but I would personally say OIL SPILLS.