Answer:
[C] carbon solid
Explanation:
Pure solids and liquids are never included in the equilibrium constant expression because they do not affect the reactant amount at equilibrium in the reaction, thus since your equation has [C] as solid it will not be part of the equlibrium equation.
Answer:
Al 72.61%
Mg 27.39%
Explanation:
To obtain the mass percentages, we need to place the individual masses over the total mass and multiply by 100%.
If we observe clearly, we can see that the parameters given are the moles. We need to convert the moles to mass.
To do this ,we need to multiply the moles by the atomic masses. The atomic mass of aluminum is 27 while that of magnesium is 24.
Now, the mass of aluminum is thus = 27 * 0.0898 = 2.4246g
The mass of magnesium is 0.0381 * 24 = 0.9144g
We can now calculate the mass percentage.
The total mass is 0.9144 + 2.4246 = 3.339g
% mass of Al = 2.4246/3.339 * 100 = 72.61%
% mass of Mg = 0.9144/3.39 * 100 = 27.39%
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Answer:
Explanation:
Hello!
In this case, since the dissolution of copper (I) chloride is:
And its equilibrium expression is:
![Ksp=[Cu^+][Cl^-]](https://tex.z-dn.net/?f=Ksp%3D%5BCu%5E%2B%5D%5BCl%5E-%5D)
We can represent the molar solubility via the reaction extent as 
, however, since there is 0.050 M KCl we immediately add such amount to the chloride ion concentration since KCl is readily ionized; therefore we write:
Thus, solving for , we obtain:
By using the quadratic equation, we obtain:
Clearly, the solution is 
because no negative results are
allowed. Therefore, the molar solubility is:
Best regards!
Answer:
A titration
Explanation:
A common example of a titration is when we have an acid of unknown concentration, so we add a known volume of a base of known concentration. This process lets us determine the concentration of the acid.
By definition, a titration is a quantitative analysis, as we determine how much of an analyte is there in a sample. However, <u>there are quantitative analyzes which are not titrations</u>. This is why the most appropiate answer is<em> a titration</em>.
