B, Frost Wedging. Have a good rest of your day!
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.
Answer:
4. +117,1 kJ/mol
Explanation:
ΔG of a reaction is:
ΔGr = ΔHr - TΔSr <em>(1)</em>
For the reaction:
2 HgO(s) → 2 Hg(l) + O₂(g)
ΔHr: 2ΔHf Hg(l) + ΔHf O₂(g) - 2ΔHf HgO(s)
As ΔHf of Hg(l) and ΔHf O₂(g) are 0:
ΔHr: - 2ΔHf HgO(s) = <u><em>181,66 kJ/mol</em></u>
<u><em /></u>
In the same way ΔSr is:
ΔSr= 2ΔS° Hg(l) + ΔS° O₂(g) - 2ΔS° HgO(s)
ΔSr= 2* 76,02J/Kmol + 205,14 J/Kmol - 2*70,19 J/Kmol
ΔSr= 216,8 J/Kmol = <em><u>0,216 kJ/Kmol</u></em>
Thus, ΔGr at 298K is:
ΔGr = 181,66 kJ/mol - 298K*0,216kJ/Kmol
ΔGr = +117,3 kJ/mol ≈ <em>4. +117,1 kJ/mol</em>
<em></em>
I hope it helps!
The answer is A warm air rises cool air sinks
P = 11.133 atm (purple)
T = -236.733 °C(yellow)
n = 0.174 mol(red)
<h3>Further explanation </h3>
Some of the laws regarding gas, can apply to ideal gas (volume expansion does not occur when the gas is heated),:
- Boyle's law at constant T, P = 1 / V
- Charles's law, at constant P, V = T
- Avogadro's law, at constant P and T, V = n
So that the three laws can be combined into a single gas equation, the ideal gas equation
In general, the gas equation can be written
where
P = pressure, atm
V = volume, liter
n = number of moles
R = gas constant = 0.08206 L.atm / mol K
T = temperature, Kelvin
To choose the formula used, we refer to the data provided
Because the data provided are temperature, pressure, volume and moles, than we use the formula PV = nRT
T= 10 +273.15 = 373.15 K
V=5.5 L
n=2 mol
V=8.3 L
P=1.8 atm
n=5 mol
T = 12 + 273.15 = 285.15 K
V=3.4 L
P=1.2 atm
