Answer:
B)−6,942 J
/mol
Explanation:
At constant temperature and pressure, you cand define the change in Gibbs free energy, ΔG, as:
ΔG = ΔH - TΔS
Where ΔH is enthalpy, T absolute temperature and ΔS change in entropy.
Replacing (25°C = 273 + 25 = 298K; 25.45kJ/mol = 25450J/mol):
ΔG = ΔH - TΔS
ΔG = 25450J/mol - 298K×108.7J/molK
ΔG = -6942.6J/mol
Right solution is:
<h3>B)−6,942 J
/mol</h3>
For an aqueous solution of MgBr2, a freezing point depression occurs due to the rules of colligative properties. Since MgBr2 is an ionic compound, it acts a strong electrolyte; thus, dissociating completely in an aqueous solution. For the equation:
ΔTf<span> = (K</span>f)(<span>m)(i)
</span>where:
ΔTf = change in freezing point = (Ti - Tf)
Ti = freezing point of pure water = 0 celsius
Tf = freezing point of water with solute = ?
Kf = freezing point depression constant = 1.86 celsius-kg/mole (for water)
m = molality of solution (mol solute/kg solvent) = ?
i = ions in solution = 3
Computing for molality:
Molar mass of MgBr2 = 184.113 g/mol
m = 10.5g MgBr2 / 184.113/ 0.2 kg water = 0.285 mol/kg
For the problem,
ΔTf = (Kf)(m)(i) = 1.86(0.285)(3) = 1.59 = Ti - Tf = 0 - Tf
Tf = -1.59 celsius
The correct chemical formulae is CsBr
Astatine. Because it has the smaller shell of electrons. I believe
How much it has to drop and how heavy it is. Hope this is what you're looking for:)
