N = 3.2 moles, T = 50 + 273 = 323 K, P = 101.325 kPa, R = 8.314 L.kPa/K.mol
PV = nRT
V = nRT / P substituting.
V = (3.2 mole)(8.314 L.kPa/K.mol )(323 K) / (<span>101.325 kPa)
That is the answer, but it is not among the options you provided. Check your options properly.</span>
Answer:
Option C. Energy Profile D
Explanation:
Data obtained from the question include:
Enthalpy change ΔH = 89.4 KJ/mol.
Enthalpy change (ΔH) is simply defined as the difference between the heat of product (Hp) and the heat of reactant (Hr). Mathematically, it is expressed as:
Enthalpy change (ΔH) = Heat of product (Hp) – Heat of reactant (Hr)
ΔH = Hp – Hr
Note: If the enthalpy change (ΔH) is positive, it means that the product has a higher heat content than the reactant.
If the enthalpy change (ΔH) is negative, it means that the reactant has a higher heat content than the product.
Now, considering the question given, the enthalpy change (ΔH) is 89.4 KJ/mol and it is a positive number indicating that the heat content of the product is higher than the heat content of the reactant.
Therefore, Energy Profile D satisfy the enthalpy change (ΔH) for the formation of CS2 as it indicates that the heat content of product is higher than the heat content of the reactant.
Iodine electron configuration is:
1S^2 2S^2 2P^6 3S^2 3P^6 4S^2 3d^10 4P^6 5S^2 4d^10 5P^5
when Krypton is the noble gas in the row above iodine in the periodic table,
we can change 1S^2 2S^2 2P^6 3S^2 3P^6 4S^2 3d^10 4P^6 by the symbol
[Kr] of Krypton.
So we can write the electron configuration of Iodine:
[Kr] 5S^2 4d^10 5P^5
