Molecular is every element present in the compound eg C2H6, empirical is the smallest whole number ratio of elements in a compound so that would be CH3 as you divide by the highest common factor. Some compounds only have 1 formula if they are simple or have no common factors. Eg methane, CH4 is its molecular and empirical because its the simplest whole number ratio and includes every element in the molecule
Answer:
Radicals
Explanation:
A radical refers to a chemical specie that contains unpaired electrons in their dot electron diagrams.
Radicals contain an odd number of electrons. They are commonly called odd electron species.
Radicals participate in a number of important reactions. A typical example is the halogenation of alkanes in the presence of light.
Examples of radicals include; Br. , Cl. , F. etc
<span>P*V/T=constant
so P*V= constant*T
if T doesn't change then
P*V= constant
so 150kPa*0.8L=75kPa*xL
xL=150kPa*0.8L/75kPa=1.6L
hope it help</span>
The molecular formula of the compound is C12H15O3 hence the molar mass of the compound is 207 g/mol.
We need to obtain the number of moles of carbon, hydrogen and oxygen in the compound;
Carbon = 24.91 g/44g/mol × 1 mole of carbon = 0.566 moles
Mass of carbon = 0.566 moles × 12 g/mol = 6.792 g
Number of moles of hydrogen = 6.522 g/18 g/mol × 2 moles = 0.725 moles
Mass of hydrogen = 0.725 moles × 1 g/mol = 0.725 g
Mass of oxygen = 10 - (6.792 g + 0.725 g) = 2.483 g
Number of moles of oxygen = 2.483 g/16 g/mol = 0.155 moles
Now we must divide through by the lowest number of moles;
C - 0.566/0.155 H - 0.725/0.155 O - 0.155/0.155
C - 4 H - 5 O - 1
The simplest formula is C4H5O Recall that the molar mass of the compound lies between 150.0 and 220.0 g/mol
4(12) + 5(1) + 16 = 69
Hence; n = 3 and the molecular formula of the compound is C12H15O3
The molar mass of the compound is; 12(12) + 15(1) + 3(16) = 207 g/mol
Learn more: brainly.com/question/15180604
D. A mixture
If the water is evaporating while the salt remains, it means the two are not chemically bonded and therefore are not a compound.
