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
A fact can be disapproved with further evidence different details. So basically evidence can change a fact!! Hope this helps
Answer:
Ag is the oxidizing agent
Explanation:
oxidizing agent in the following equation?
Al (s) + 3 Ag+ (aq) = Al+3 (aq) + 3 Ag (s)
Left side
Al = 1
Ag = 3
Right Side
Al = 1
Ag = 3
So it's balanced already good.
Define
oxidizing agent = An oxidizing agent is the substance that gains electrons and is reduced in a chemical reaction.
Al is the reducing agent.
Ag is the oxidizing agent
I believe the answer is C, n = 3, l = 3, m = 3. The magnetic quantum number, or
<span>ml</span>, can only take values that range from <span>−l</span> to <span>+l</span>, as you can see in the table above.
For option C), the angular momentum quantum number of equal to ++2<span>, which means that <span>ml</span> can have a maximum value of </span>+2<span>. Since it is given as having a value of </span>+3**, this set of quantum numbers is not a valid one.
The other three sets are valid and can correctly describe an electron.
