<h3><u>Answer;</u></h3>
321.8 g CaF2
321.5 g Al2(CO3)3
<h3><u>Explanation;</u></h3>
The equation for the reaction is;
3 CaCO3 + 2 AlF3 → 3 CaF2 + Al2(CO3)3
Number of moles of CaCO3 will be;
=(412.5 g CaCO3) / (100.0875 g CaCO3/mol)
= 4.12139 mol CaCO3
Number of moles of AlF3 will be;
= (521.9 g AlF3) / ( 83.9767 g AlF3/mol)
= 6.21482 mol AlF3
But;
4.12139 moles of CaCO3 would react completely with 4.12139 x (2/3) = 2.74759 moles of AlF3.
Thus; there is more AlF3 present than that, so AlF3 is in excess, and CaCO3 is the limiting reactant.
Therefore;
Mass of CaF2 will be;
(4.12139 mol CaCO3) x (3/3) x (78.0752 g CaF2/mol) = 321.8 g CaF2
Mass of Al2(CO3)3 on the other hand will be;
(4.12139 mol CaCO3) x (1/3) x (233.9903 g Al2(CO3)3/mol) = 321.5 g Al2(CO3)3
Answer: The greatest concentration of C would be produced if the equilibrium constant of the reaction is equal to 
Explanation:
Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as 
For the given chemical reaction:
The expression for is written as:
![K_c=\frac{[C]}{[A][B]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BC%5D%7D%7B%5BA%5D%5BB%5D%7D)
Thus higher is the value of , higher will be the concentration of C. Thus the greatest concentration of C would be produced if the equilibrium constant of the reaction is equal to
Answer:
The valence electrons in the metal atom in the lattice structure can delocalize and move freely. The electrons are "loosely" connected with their parent atom in the metallic bond because most metals have excess electrons in their outer orbitals above the stable configuration.
Explanation:
Says google
Answer:
How to Formulate an Effective Research Hypothesis
State the problem that you are trying to solve. Make sure that the hypothesis clearly defines the topic and the focus of the experiment.
Try to write the hypothesis as an if-then statement. ...
Define the variables.
Explanation:
Answer:soluble
Explanation: bc its less then NaOh
