Given :
A compound has a molar mass of 129 g/mol .
Empirical formula of compound is C₂H₅N .
To Find :
The molecular formula of the compound.
Solution :
Empirical mass of compound :
Now, n-factor is :
Multiplying each atom in the formula by 3 , we get :
Molecular Formula, C₆H₁₅N₃
Answer: CoBr3 < K2SO4 < NH4 Cl
Justification:
1) The depression of the freezing point of a solution is a colligative property, which means that it depends on the number of particles of solute dissolved.
2) The formula for the depression of freezing point is:
ΔTf = i * Kf * m
Where i is the van't Hoof factor which accounts for the dissociation of the solute.
Kf is the freezing molal constant and only depends on the solvent
m is the molality (molal concentration).
3) Since, you are assuming equal concentrations and complete dissociation of the given solutes, the solute with more ions in the molecular formula will result in the solution with higher depression of the freezing point (lower freezing point).
4) These are the dissociations of the given solutes:
a) NH4 Cl (s) --> NH4(+)(aq) + Cl(-) (aq) => 1 mol --> 2 moles
b) Co Br3 (s) --> Co(3+) (aq) + 3Br(-)(aq) => 1 mol --> 4 moles
c) K2SO4 (s) --> 2K(+) (aq) + SO4 (2-) (aq) => 1 mol --> 3 moles
5) So, the rank of solutions by their freezing points is:
CoBr3 < K2SO4 < NH4 Cl
Nitrogen has 7 protons, 7 neutrons, and 7 electrons.
Seven protons, seven neutrons, and seven electrons make up nitrogen-14.
Utilize the atomic number and mass number of an atom to determine the number of subatomic particles it contains: Atomic number Equals proton count. Electron count equals atomic number. Atomic number - mass number equals the number of neutrons.
Seven protons, seven neutrons, and seven electrons make up the atom of nitrogen. The nucleus is the collection of protons and neutrons that make up the center of an atom. The 7 electrons, which are much smaller than the nucleus, orbit it in what is known as orbits. Since nitrogen-14 is a neutral atom, the number of protons in its nucleus must match the number of electrons around it.
Learn more about atomic numbers at brainly.com/question/2942556
#SPJ4.
Answer:
One extraction: 50%
Two extractions: 75%
Three extractions: 87.5%
Four extractions: 93.75%
Explanation:
The following equation relates the fraction q of the compound left in volume V₁ of phase 1 that is extracted n times with volume V₂.
qⁿ = (V₁/(V₁ + KV₂))ⁿ
We also know that V₂ = 1/2(V₁) and K = 2, so these expressions can be substituted into the above equation:
qⁿ = (V₁/(V₁ + 2(1/2V₁))ⁿ = (V₁/(V₁ + V₁))ⁿ = (V₁/(2V₁))ⁿ = (1/2)ⁿ
When n = 1, q = 1/2, so the fraction removed from phase 1 is also 1/2, or 50%.
When n = 2, q = (1/2)² = 1/4, so the fraction removed from phase 1 is (1 - 1/4) = 3/4 or 75%.
When n = 3, q = (1/2)³ = 1/8, so the fraction removed from phase 1 is (1 - 1/8) = 7/8 or 87.5%.
When n = 4, q = (1/2)⁴ = 1/16, so the fraction removed from phase 1 is (1 - 1/16) = 15/16 or 93.75%.
