Answer:
A. ΔG° = 132.5 kJ
B. ΔG° = 13.69 kJ
C. ΔG° = -58.59 kJ
Explanation:
Let's consider the following reaction.
CaCO₃(s) → CaO(s) + CO₂(g)
We can calculate the standard enthalpy of the reaction (ΔH°) using the following expression.
ΔH° = ∑np . ΔH°f(p) - ∑nr . ΔH°f(r)
where,
n: moles
ΔH°f: standard enthalpy of formation
ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g)) - 1 mol × ΔH°f(CaCO₃(s))
ΔH° = 1 mol × (-635.1 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1206.9 kJ/mol)
ΔH° = 178.3 kJ
We can calculate the standard entropy of the reaction (ΔS°) using the following expression.
ΔS° = ∑np . S°p - ∑nr . S°r
where,
S: standard entropy
ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g)) - 1 mol × S°(CaCO₃(s))
ΔS° = 1 mol × (39.75 J/K.mol) + 1 mol × (213.74 J/K.mol) - 1 mol × (92.9 J/K.mol)
ΔS° = 160.6 J/K. = 0.1606 kJ/K.
We can calculate the standard Gibbs free energy of the reaction (ΔG°) using the following expression.
ΔG° = ΔH° - T.ΔS°
where,
T: absolute temperature
<h3>A. 285 K</h3>
ΔG° = ΔH° - T.ΔS°
ΔG° = 178.3 kJ - 285K × 0.1606 kJ/K = 132.5 kJ
<h3>B. 1025 K</h3>
ΔG° = ΔH° - T.ΔS°
ΔG° = 178.3 kJ - 1025K × 0.1606 kJ/K = 13.69 kJ
<h3>C. 1475 K</h3>
ΔG° = ΔH° - T.ΔS°
ΔG° = 178.3 kJ - 1475K × 0.1606 kJ/K = -58.59 kJ
Answer: Option (b) is the correct answer.
Explanation:
When there are more number of hydroxide ions in a solution then there will be high concentration of 
or hydroxide ions. As a result, more will be the strength of base in that particular solution.
A base is strong when it readily dissociate into its ions in the solution. When a base is strong, then it does not matter at what concentration it is dissolved in the solution because despite of its low concentration it will remain a strong base.
Thus, we can conclude that out of the given options, the statement even at low concentrations, a strong base is strong best relates the strength and concentration of a base.
Answer:
Iodine
Explanation:
It's in the same group as chlorine.
Answer: 0.075
Explanation:
(concentration in molarity)(volume in liter) = answer
0.15 mol/L *0.500L = 0.075 mol
Answer:
It’s b even check quizzes
Explanation:
