Answer:
THE EMPIRICAL FORMULA FOR THE UNKNOWN COMPOUND IS C7H9O
Explanation:
The empirical formula for the unknown compound can be obtained by following the processes below:
1 . Write out the percentage composition of the individual elements in the compound
C = 75.68 %
H = 8.80 %
O = 15.52 %
2. Divide the percentage composition by the atomic masses of the elements
C = 75 .68 / 12 = 6.3066
H = 8.80 / 1 = 8.8000
O = 15.52 / 16 = 0.9700
3. Divide the individual results by the lowest values
C = 6.3066 / 0.9700 = 6.5016
H = 8.8000 / 0.9700 = 9.0722
O = 0.9700 / 0.9700 = 1
4. Round up the values to the whole number
C = 7
H = 9
O = 1
5 Write out the empirical formula for the compound
C7H90
In conclusion, the empirical formula for the unknown compound is therefore C7H9O
3s it is what it says sorry if this is wrong but I just did a qu like this and that's what it was
An example is just when they have a a group o numbers, for any reason, and they want to find the average of then all
an example being um idk...numbers resulting from an experiment trying to find different densitys or something and they wanted to find the average density
Answer:
The answer is "Ammonium nitrate is the reactant of this chemical reaction".
Explanation:
The Ammonium nitrate (NH4NO3) is an ammonium and nitrate ionic crystalline solid. In water and hygroscopic as a strong, it's also highly soluble, although it does not form hydrates.
It uses as a nutrient for generating medications as well as leaven fertilizers and explosive materials. Nitrate in ammonium is sodium bicarbonate ammonium chloride. It is incendiary, an oxidizing agent, and a fertilizer.
