Answer:
Ketone
Explanation:
As you are stating here, we have a carbonated chain of three carbons, and the first and last has 3 Hydrogens, then this means that we have CH₃ . The center carbon is a carbon double bonded to oxygen.
In general terms this belongs to the carbonyl group. However, this alone does not represent a functional group, but when it's in a chain with other radycals or chains, it becomes a functional group.
In this case, the molecule you are talking here is the following:
CH₃ - CO - CH₃
This molecule is known as the Acetone, and has the general form of:
R - CO - R'
Which belongs to a ketone as a functional group.
The formula we will use is:
pH = 0.5 (pKa – log C)
where C is the concentration of sulfathiazole in molarity
or mol / L
Since all values are given, we can compute directly for
pH:
pH = 0.5 (7.12 – log 0.002)
<span>pH = 4.91</span>
Answer:
To increase surface area of the platinum electrode which results in superior quality and action of the electrodes as opposed to normal platinum electrodes.
Explanation:
Platinization of Platinum is the process of covering platinum electrode with a layer of platinum black. Platinum black is a finally divided form of platinum, optimized for catalysing the addition of hydrogen to unsaturated organic compound. This increases the surface area of the platinum electrodes and therefore exhibits action superior to that of normal electrodes.
Nitrate ions
and potassium ions
.
<h3>Explanation</h3>
Potassium dichromate undergoes a double-decomposition reaction with barium nitrate to produce barium dichromate and potassium nitrate. The reaction is possible due to the low solubility of barium dichromate that precipitate out of the solution readily after its production.
Start with the balanced chemical equation for this process:

Rewrite the chemical equation as an ionic one; express all soluble salts- those with state symbol (aq)- as their constituting ions while leaving the insoluble (s) intact.

and
are found on both sides of the equation by the same quantity. The two ions thus took no part in the net reaction and act as spectator ions.