<u>Given:</u>
Moles of Al = 0.4
Moles of O2 = 0.4
<u>To determine:</u>
Moles of Al2O3 produced
<u>Explanation:</u>
4Al + 3O2 → 2Al2O3
Based on the reaction stoichiometry:
4 moles of Al produces 2 moles of Al2O3
Therefore, 0.4 moles of Al will produce:
0.4 moles Al * 2 moles Al2O3/4 moles Al = 0.2 moles Al2O3
Similarly;
3 moles O2 produces 2 moles Al2O3
0.4 moles of O2 will yield: 0.4 *2/3 = 0.267 moles
Thus Al will be the limiting reactant.
Ans: Maximum moles of Al2O3 = 0.2 moles
Charge = Number of Protons - Number of Electrons = 4
So, Your Answer would be: 4
Hope this helps!
The equivalence point of a titration is equal to its stoichiometric equivalents of analyte and titrant.
Depending on the concentration of titrant we could be adding little excess of it and this may result in persistence of color of solution. After continuous stirring for a while the excess titrant may react with dissolved CO₂ in air and thus decolorizing the solution.
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Answer:
The activation energy is 164.02 kJ/mol
Explanation:
Log (k2/k1) = Ea/2.303R × [1/T1 - 1/T2]
k1 = 8.9×10^-4 s^-1
k2 = 9.83×10^-3 s^-1
R = 8.314 J/mol.K
T1 = 540 K
T2 = 578 K
Log (9.83×10^-3/8.9×10^-4) = Ea/2.303×8.314 × [1/540 - 1/578]
1.043 = 6.359×10^-6Ea
Ea = 1.043/6.359×10^-6 = 164020 J/mol = 164020/1000 = 164.02 kJ/mol
Answer:
Explanation:
From the information given:
Step 1:
Determine the partial pressure of each gas at total Volume (V) = 4.0 L
So, using:





![Total pressure= P [N_2] + P[Ar] \ \\ \\ . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = (0.525 + 1.7)Bar \\ \\ . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2.225 \ Bar](https://tex.z-dn.net/?f=Total%20pressure%3D%20P%20%5BN_2%5D%20%2B%20P%5BAr%5D%20%5C%20%5C%5C%20%5C%5C%20.%20%5C%20%5C%20%20%5C%20%5C%20%20%5C%20%5C%20%20%5C%20%5C%20%5C%20%5C%20%20%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%3D%20%280.525%20%2B%201.7%29Bar%20%5C%5C%20%5C%5C%20.%20%5C%20%5C%20%20%5C%20%5C%20%20%5C%20%5C%20%20%5C%20%5C%20%5C%20%5C%20%20%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%3D%202.225%20%5C%20Bar)
Now, to determine the final pressure using different temperature; to also achieve this, we need to determine the initial moles of each gas.
According to Ideal gas Law.

For moles N₂:



For moles of Ar:





Finally;
The final pressure of the mixture is:

P = 2.217 atm
P ≅ 2.24 bar