Answer:
80.0 g Na and 20.0 g N2.
Explanation:
This means the limiting reactant determines the maximum mass of the product formed.
<span> Ksp = [Ag+]^2[CO32-]that should be it </span>
Answer:
<h3>601.93 g/mol</h3>
<h3>explanation:</h3>
Problem: The Ba3(PO4)2 (molar mass = 601.93 g/mol) precipitate that formed from a salt mixture has a mass of 0.667 g.
CH3CH2OH = 46.068 (molar mass)
1.50g(1 mole / 46.0680) =
0.03256 moles