CaCl2 and KCl are both salts which dissociate in water
when dissolved. Assuming that the dissolution of the two salts are 100 percent,
the half reactions are:
<span>CaCl2 ---> Ca2+ + 2 Cl-</span>
KCl ---> K+ + Cl-
Therefore the total Cl- ion concentration would be coming
from both salts. First, we calculate the Cl- from each salt by using stoichiometric
ratio:
Cl- from CaCl2 = (0.2 moles CaCl2/ L) (0.25 L) (2 moles
Cl / 1 mole CaCl2)
Cl- from CaCl2 = 0.1 moles
Cl- from KCl = (0.4 moles KCl/ L) (0.25 L) (1 mole Cl / 1
mole KCl)
Cl- from KCl = 0.1 moles
Therefore the final concentration of Cl- in the solution
mixture is:
Cl- = (0.1 moles + 0.1 moles) / (0.25 L + 0.25 L)
Cl- = 0.2 moles / 0.5 moles
<span>Cl- = 0.4 moles (ANSWER)</span>
Answer:
See explanation
Explanation:
If the spot in TLC is below the solvent front, it will be observed that the spot, instead of being separated by the solvent as expected, will just dissolve away in the solvent and zero actual separation of the mixture is achieved.
If the solute is dissolved away instead of being separated by the solvent, then the experiment fails because no actual separation of the mixture is achieved.
Hence, in TLC, the spot must be applied above the solvent front so that the capillary movement of the solvent through the plate can lead to the eventual separation of the components of the mixture since the various components of the mixture will travel at different speeds through the plate.
Also, if the solvent is above the spot, the solvent may evaporate selectively from the points above the spot while separation is ongoing.
Answer:
A.) False B.)False C.) True D.) False E.) False
Explanation:
Im guessing and i think the answer is yes.
Your control group would be the batteries since you CONTROL what brand you're using, for which one lasts the longest...aren't you suppose to figure that out when you do the experiment?
