Answer:
C₂H₇F₂P
Explanation:
Given parameters:
Composition by mass:
C = 24%
H = 7%
F = 38%
P = 31%
Unknown:
Empirical formula of compound;
Solution :
The empirical formula is the simplest formula of a compound. To solve for this, follow the process below;
C H F P
% composition
by mass 24 7 38 31
Molar mass 12 1 19 31
Number of
moles 24/12 7/1 38/19 31/31
2 7 2 1
Dividing
by the
smallest 2/1 7/1 2/1 1/1
2 7 2 1
Empirical formula C₂H₇F₂P
Answer is: 6.022·10²² molecules of glucose.
c(glucose) = 100 mM.
c(glucose) = 100 · 10⁻³ mol/L.
c(glucose) = 0.1 mol/L; concentration of glucose solution.
V(glucose) = 1 L; volume of glucose solution.
n(glucose) = c(glucose) · V(glucose).
n(glucose) = 0.1 mol/L · 1 L.
n(glucose) = 0.1 mol; amount of substance.
N(glucose) = n(glucose) · Na (Avogadro constant).
N(glucose) = 0.1 mol · 6.022·10²³ 1/mol.
N(glucose) = 6.022·10²².
D
Avogadro's number allows us to measure the amount of atoms or molecules in one mole of a substance.
D has a total of four significant figures.
Displacement = √(3² + 4²)
Displacement = 5 meters north east
Velocity = displacement / time
Velocity = 5 / 35
Velocity = 0.14 m/s northeast