Theoretical yield of Al₂O₃: 1.50 mol.
<h3>Explanation</h3>
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How many moles of aluminum oxide formula units will be produced <em>if</em> aluminum is the limiting reactant?
Aluminum reacts to aluminum oxide at a two-to-one ratio.
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As a result, 3.00 moles of aluminum will give rise to 1.50 moles of aluminum oxide.
How many moles of aluminum oxide formula units will be produced <em>if</em> oxygen is the limiting reactant?
Oxygen reacts to produce aluminum oxide at a three-to-two ratio.

As a result, 2.55 moles of oxygen will give rise to 1.70 moles of aluminum oxide.
How many moles of aluminum oxide formula units will be produced?
Aluminum is the limiting reactant. Only 1.50 moles of aluminum oxide formula units will be produced. 1.70 moles isn't feasible since aluminum would run out by the time 1.50 moles was produced.
One of the best buffer choice for pH = 8.0 is Tris with Ka value of 6.3 x 10^-9.
To support this answer, we first calculate for the pKa value as the negative logarithm of the Ka value:
pKa = -log Ka
For Tris, which is an abbreviation for 2-Amino-2-hydroxymethyl-propane-1,3 -diol and has a Ka value of 6.3 x 10^-9, the pKa is
pKa = -log Ka
= -log (6.3x10^-9)
= 8.2
We know that buffers work best when pH is equal to pKa:
pKa = 8.2 = pH
Therefore Tris would be a best buffer at pH = 8.0.
Answer:
20cm^2
Explanation:
Here, Density= Mass/ Volume
=100/5
= 20 cm^2
Glucose is the starting molecule for glycolysis.