The number of moles of b2o3 that will be formed is determined as 4 moles.
<h3>
Limiting reagent</h3>
The limiting reagent is the reactant that will be completely used up.
4 b + 3O₂ → 2b₂O₃
from the equation above;
4 b ------------> 2 b₂O₃
2b ------------> b₂O₃
2 : 1
3O₂ -------------> 2b₂O₃
3 : 2
b is the limiting reagent, thus, the amount of b2o3 to be formed is calculated as;
4 b ------------> 2 moles of b2o3
8 moles -------> ?
= (8 x 2)/4
= 4 moles
Thus, the number of moles of b2o3 that will be formed is determined as 4 moles.
Learn more about limiting reactants here: brainly.com/question/14222359
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Divide this problem in to two steps: 1. Convert millimoles to moles. 2. Convert moles to grams.
1.
2.65 mlmol x 1mol/1000mlmol=.00265mol
2.
.00265mol x 32.065g/1 mol=.0850g S
Let me know if you need more explanation on why I used the conversions that I did.
A amplitude = 0.50 m and wave-length = 1.0 m
B amplitude = 0.40 m and wave-length = 2.0 m
C amplitude = 0.60 m and wave-length = 2.0 m
Option C. Aluminum atoms lose electrons to oxygen atoms.