1. Decreasing the temperature of the water.
2. One new atom with a large nucleus forms.
3.The reaction is an example of oxidation.
Answer:

Explanation:
When percentage composition is given, and asked for the empirical formula, it is simplest to assume 100 g of material. Thus,
Mass C = 40.92 g. Moles C = 40.92 g x 1 mole/12 g = 3.41 moles C
Mass H = 4.58 g. Moles H = 4.58 g x 1 mole/1.0 g = 4.58 moles H
Mass O = 54.50 g. Moles O = 54.50 g x 1 mole/16 g = 3.41 moles O
Now, we want to get the moles into whole numbers, so we begin by dividing all by the smallest, i.e. divide all values by 3.41.
Moles C = 3.41/3.41 = 1
Moles H = 4.58/3.41 = 1.34
Moles O = 3.41/3.41 = 1
Now, in order to get 1.34 to be a whole number we multiply it (and all others) by 3
Moles C = 1x3 = 3
Moles H = 1.34x3 = 4
Moles O = 1x3 = 3
Empirical Formula 
POH value was calculated by the negative logarithm of hydroxide ion concentration.
To know the hydrogen ion concentration, we need to know the pH value, that can be found out if pOH is known
pH + pOH = 14
pH = 14 - pOH
pH = 10.65
once the pH is known we have to find the antilog.
[H⁺] = antilog (-pH)
antilog can be found by
[H⁺] = 10^(-10.65)
[H⁺] = 2.2 x 10⁻¹¹ M
PH is defined as the negative log of Hydrogen ion concentration. Mathematically we can write this as:
![pH=-log[H^{+}]=-log[H_{3}O]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%7B%2B%7D%5D%3D-log%5BH_%7B3%7DO%5D%20%20)
We are given the concentration of

. Using the value in formula, we get:
Therefore, the pH of the solution will be 3.745