Gold! it's boiling point is 2807 according to this chart:
http://www.lenntech.com/periodic-chart-elements/boiling-point.htm
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The reaction will produce 12.1 g Ag₂S.
<em>Balanced equation</em> = 2Ag + S ⟶ Ag₂S
<em>Mass of Ag₂S</em> = 10.5 g Ag × (1 mol Ag/107.87 g Ag) × (1 mol Ag₂S/2 mol Ag)
× (247.80 g Ag₂S/1 mol Ag₂S) = 12.1 g Ag₂S
The most reactive metal on the periodic table is Francium, However, Francium is an artificial element and only minimal quantities have been produced, so for all practical purposes, the most reactive metal available is Cesium, in the alkali metal family.
ANSWER In theory, Francium, in practice the Cesium
Answer:
131.5 kJ
Explanation:
Let's consider the following reaction.
CaCO₃(s) → CaO(s) + CO₂(g)
First, we will calculate the standard enthalpy of the reaction (ΔH°).
ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g)
) - 1 mol × ΔH°f(CaCO₃(s)
)
ΔH° = 1 mol × (-634.9 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1207.6 kJ/mol)
ΔH° = 179.2 kJ
Then, we calculate the standard entropy of the reaction (ΔS°).
ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g)
) - 1 mol × S°(CaCO₃(s)
)
ΔS° = 1 mol × (38.1 J/mol.K) + 1 mol × (213.8 J/mol.K) - 1 mol × (91.7 J/mol.K)
ΔS° = 160.2 J/K = 0.1602 kJ/K
Finally, we calculate the standard Gibbs free energy of the reaction at T = 25°C = 298 K.
ΔG° = ΔH° - T × ΔS°
ΔG° = 179.2 kJ - 298 K × 0.1602 kJ/K
ΔG° = 131.5 kJ
