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
Answer:
B I believe im sorry if it's not right
Answer:
Explanation:
Hello!
In this case, since the study of the bond energy allows us to compute the enthalpies of some reactions, for this combination reaction by which ammonia is yielded, we understand the enthalpy of reaction equals the enthalpy of formation of ammonia, and, in terms of the bonds energy we can write:
Whereas the bonds enthalpy of those bonds that get broken cover the N≡N and the three H-H bonds at the reactants side and the enthalpy of those bonds that are formed cover the six N-H bonds at the products; which means we obtain:
Which differs from the theoretical value that is -46 kJ/mol.
Best regards!
