Find the mass of the empirical formula.
You must be given a sample of some kind to calculate the weight or know how many moles are present. Then you figure out what one mol would be. The key step is multiplying the empirical formula numbers by what it takes to make 1 mol.
It would be clearer if we were working from some choices.
Answer:
The diameter of the oil molecule is 
.
Explanation:
Mass of the oil drop = 
Density of the oil drop = 
Volume of the oil drop: v
Thickness of the oil drop is 1 molecule thick.So, let the thickness of the drop or diameter of the molecule be x.
Radius of the oil drop on the water surface,r = 41.8 cm = 0.418 m
1 cm = 0.01 m
Surface of the sphere is given as: a = 
Volume of the oil drop = v = Area × thickness
The thickness of the oil drop is and so is the diameter of the molecule.
Answer:
The correct answer is 5.447 × 10⁻⁵ vacancies per atom.
Explanation:
Based on the given question, the at 750 degree C the number of vacancies or Nv is 2.8 × 10²⁴ m⁻³. The density of the metal is 5.60 g/cm³ or 5.60 × 10⁶ g/m³. The atomic weight of the metal given is 65.6 gram per mole. In order to determine the fraction of vacancies, the formula to be used is,
Fv = Nv/N------ (i)
Here Nv is the number of vacancies and N is the number of atomic sites per unit volume. To find N, the formula to be used is,
N = NA×P/A, here NA is the Avogadro's number, which is equivalent to 6.022 × 10²³ atoms per mol, P is the density and A is the atomic weight. Now putting the values we get,
N = 6.022 × 10²³ atoms/mol × 5.60 × 10⁶ g/m³ / 65.6 g/mol
N = 5.14073 × 10²⁸ atoms/m³
Now putting the values of Nv and N in the equation (i) we get,
Fv = 2.8 × 10²⁴ m⁻³ / 5.14073 × 10²⁸ atoms/m^3
Fv = 5.44669 × 10⁻⁵ vacancies per atom or 5.447 × 10⁻⁵ vacancies/atom.
Solution :
For the reaction :
we have
![$Ka = \frac{[\text{Tris}^- \times H_3O]}{\text{Tris}^+}$](https://tex.z-dn.net/?f=%24Ka%20%3D%20%5Cfrac%7B%5B%5Ctext%7BTris%7D%5E-%20%5Ctimes%20H_3O%5D%7D%7B%5Ctext%7BTris%7D%5E%2B%7D%24)
Clearing 
, we have 
So to reach 
, one must have the 
concentration of the :
![$\text{[OH}^-]=10^{-pOH} = 6.31 \times 10^{-7} \text{ moles of base}$](https://tex.z-dn.net/?f=%24%5Ctext%7B%5BOH%7D%5E-%5D%3D10%5E%7B-pOH%7D%20%3D%206.31%20%5Ctimes%2010%5E%7B-7%7D%20%5Ctext%7B%20moles%20of%20base%7D%24)
So we can add enough of 1 M NaOH in order to neutralize the acid that is calculated above and also adding the calculated base.
Volume NaOH 
Tris mass 
Now to prepare the said solution we must mix:
gauge to 1000 mL with water.
