The mass of hydrated salt - 2.123 g
mass of anhydrous salt - 1.861 g
mass that has been reduced is the mass of water that has been heated and lost from the compound thereby making the salt anhydrous.
therefore mass of water lost - 2.123 - 1.861 = 0.262 g
number of moles of water lost - 0.262 g / 18 g/mol = 0.0146 mol
number of moles of salt - 1.861 g / 380.6 g/mol = 0.00490 mol
molar ratio of moles of water to moles of salt
molar ratio = 0.146 mol / 0.00490 mol = 2.98 rounded off to 3
for every 1 mol of salt there are 3 moles of water
therefore empirical formula - Cu₃(PO₄)₂.3H₂O
Answer:
-0.050 kJ/mol.K
Explanation:
- A certain reaction is thermodynamically favored at temperatures below 400. K, that is, ΔG° < 0 below 400. K
- The reaction is not favored at temperatures above 400. K, that is. ΔG° > 0 above 400. K
All in all, ΔG° = 0 at 400. K.
We can find ΔS° using the following expression.
ΔG° = ΔH° - T.ΔS°
0 = -20 kJ/mol - 400. K .ΔS°
ΔS° = -0.050 kJ/mol.K