There are 2071.79 grams of glucose in 11.5 moles.
Answer:
43.0 kJ
Explanation:
The free energy (ΔG) measures the total energy that is presented in a thermodynamic system that is available to produce useful work, especially at thermal machines. In a reaction, the value of the variation of it indicates if the process is spontaneous or nonspontaneous because the free energy intends to decrease, so, if ΔG < 0, the reaction is spontaneous.
The standard value is measured at 25°C, 298 K, and the value of free energy varies with the temperature. It can be calculated by the standard-free energy of formation (G°f), and will be:
ΔG = ∑n*G°f products - ∑n*G°f reactants, where n is the coefficient of the substance in the balanced reaction.
By the balanced reaction given:
2NOCl(g) --> 2NO(g) + Cl2(g)
At ALEKS Data tab:
G°f, NOCl(g) = 66.1 kJ/mol
G°f, NO(g) = 87.6 kJ/mol
G°f, Cl2(g) = 0 kJ/mol
ΔG = 2*87.6 - 2*66.1
ΔG = 43.0 kJ
The solution is as follows:
K = [Partial pressure of isoborneol]/[Partial pressure of borneol] = 0.106
The molar mass of isoborneol/borneol is 154.25 g/mol
Mol isoborneol = 15 g/154.25 = 0.0972 mol
Mol borneol = 7.5 g/154.25 = 0.0486 mol
Use the ICE approach
borneol → isoborneol
I 0.0972 0.0486
C -x +x
E 0.0972 - x 0.0486 + x
Total moles = 0.1458
Using Raoult's Law,
Partial Pressure = Mole fraction*Total Pressure
[Partial pressure of isoborneol] = [(0.0972-x)/0.1458]*P
[Partial pressure of borneol] = [(0.0486+x/0.1458)]*P
0.106 = [(0.0972-x)/0.1458]*P/ [(0.0486+x/0.1458)]*P
Solving for x,
x = 0.0832
Thus,
<em>Mol fraction of borneol = (0.0486+0.0832)/0.1458 = 0.904</em>
<em>Mol fraction of isoborneol = (0.0972-0.0832)/0.1458 = 0.096</em>
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