The first dissociation for H2X:
H2X +H2O ↔ HX + H3O
initial 0.15 0 0
change -X +X +X
at equlibrium 0.15-X X X
because Ka1 is small we can assume neglect x in H2X concentration
Ka1 = [HX][H3O]/[H2X]
4.5x10^-6 =( X )(X) / (0.15)
X = √(4.5x10^-6*0.15)
∴X = 8.2 x 10-4 m
∴[HX] & [H3O] = 8.2x10^-4
the second dissociation of H2X
HX + H2O↔ X^2 + H3O
8.2x10^-4 Y 8.2x10^-4
Ka2 for Hx = 1.2x10^-11
Ka2 = [X2][H3O]/[HX]
1.2x10^-11= y (8.2x10^-4)*(8.2x10^-4)
∴y = 1.78x10^-5
∴[X^2] = 1.78x10^-5 m
Answer:-ΔG=-101.5KJ
Explanation:We have to calculate ΔG for the reaction so using the formula given in the equation we can calculate the \Delta G for the reaction.
We need to convert the unit ofΔS in terms of KJ/Kelvin as its value is given in terms of J/Kelvin
Also we need to convert the temperature in Kelvin as it is given in degree celsius.
After calculating forΔG we found that the value ofΔG is negative and its value is -101.74KJ
For a reaction to be spontaneous the value of \Delta G \ must be negative .
As the ΔG for the given reaction is is negative so the reaction will be spontaneous in nature.
In this reaction since the entropy of reaction is positive and hence when we increase the temperature term then the overall term TΔS would become more positive and hence the value of ΔG would be less negative .
Hence the value of ΔG would become more positive with the increase in temperature.
So we found the value of ΔG to be -101.74KJ
Molar mass (CaCl2) = 40.1 +2*35.5 = 111.1 g/mol
Molar mass (AlCl3) = 27.0 +3*35.5= 133.5 g/ mol
3CaCl2+Al2O3 -------->3CaO +2AlCl3
mole from reaction 3 mol 2 mol
mass from reaction 3mol* 111.1g/mol 2 mol*133.5g/mol
333.3 g 267.0 g
mass from problem 45.7 g x g
Proportion:
333.3 g CaCl2 ------- 267.0 g AlCl3
45.7 g CaCl2 -------- x g AlCl3
x=45.7*267.0/333.3= 36.6 g AlCl3
Answer:
5.625 moles of oxygen, O₂.
Explanation:
The balanced equation for the reaction is given below:
4Al + 3O₂ —> 2Al₂O₃
From the balanced equation above,
4 moles of Al reacted with 3 moles of O₂.
Finally, we shall determine the number of mole of O₂ required to react with 7.5 moles of aluminum, Al. This can be obtained as illustrated below:
From the balanced equation above,
4 moles of Al reacted with 3 moles of O₂.
Therefore, 7.5 moles of Al will react with = (7.5 × 3)/4 = 5.625 moles of O₂.
Thus, 5.625 moles of O₂ is needed for the reaction.
