For 1 mole of any substance, there are 6.02214086 × 10^<span>23 particles (atoms, molecules, ions, etc. depending on the substance)
So, for 1 mole of water, there are </span>6.02214086 × 10^23 water particles, and so on
So, in 1 mole of Lithium, there are 6.02214086 × 10^23 Lithium atoms
We are given the starting amount of Al metal to be used. This will be the starting point of the calculations. We do as follows:
74.00 g ( 1 mol / 26.98 g ) ( 1 mol Al2O3 / 2 mol Al ) ( 101.96 g / 1 mol ) = 139.83 g Al2O3 produced
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Answer:
No, it is not feasible because the Gibbs free energy change is positive
Explanation:
∆Hreaction= (-602 KJ/mol) - (-348 KJ/mol) = -254 KJ/mol
∆Sreaction = (42 + 27) J/Kmol - (33 + 44) J/K = -8J/Kmol
From;
∆G = ∆H - T∆S
∆G = 254 × 10^3 J/mol - [340K × (-8 J/Kmol)]
∆G = 2.57 × 10^5 J/mol
Note that when the change in free energy is positive, a reaction is non spontaneous. Only a reaction that has a negative change in free energy is spontaneous.
Answer:
Explanation:
Toe change the retention factor of a TLC analysis, you can change your solvent for a more or less polar one, depending on your analyte. You can use a mix of solvents too.
You can also change the your method to visualize the spots, you can use fluorescent compounds that can only be seen in black light, you can use Iodine, Bromine and so on.
Explanation:
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