Answer:
The answer to your question is 1.36 x 10²³ atoms
Explanation:
Data
number of atoms = ?
mass of the sample = 34.2 g
Molecule = Cl₂O₅
Process
1.- Calculate the molar mass of Cl₂O₅
Cl₂O₅ = (35.5 x 2) + (16 x 5) = 71 + 80 = 151 g
2.- Calculate the atoms of Cl₂O₅
151 g of Cl₂O₅ ---------------- 6 .023 x 10²³ atoms
34.2 g of Cl₂O₅ ------------ x
x = (34.2 x 6.023 x 10²³) / 151
x = 1.36 x 10²³ atoms
Answer:
3.18 mol
Explanation:
n(CO2) = mass/ Mr.
= 25.5 / 16
= 1.59 mol
As per the equation above,
n(LiOH) : n(CO2)
2 : 1
∴ 3.18 : 1.59
Answer: -
The first step involves protonation of the carbonyl oxygen.
After protonation, the Alcohol oxygen now attacks the carbon of the carbonyl.
Thus a six membered ring is formed with 5 carbon atoms and 1 oxygen atom. The 1st position carbon atom has 2 OH groups.
One of these gets again protonated.
This leaves as water. With the loss of the H+, there results a carbonyl at 1 position.
Thus 5-hydroxypentanoic acid forms a lactone or 2-oxanone in presence of acid.
You can't usually just use a single spectrum line to confirm the identity of an element because there are cases that the emission line id not clearly defined. When the emission line is very weak compared to surrounding noise, in which case the more datapoints you have to build up confidence for the existence of a particular emission spectra, the better.
