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:
We need the picture to answer the question
Explanation:
but i did do this test and number 4 was biotic and number 5 was abotic hope this helped
Answer:
The answer to your question is: 79.6 L
Explanation:
Data:
n = 60 moles
Pressure = P = 20.0 atm
Temperature = T = 50°C
Volume = V = ?
Convert temperature to Kelvin
°K = 273 + 50
°K = 323
Formula (ideal gas law)
PV = nRT solve for V
V = nRT / P
V = (60)(0.08206)(323) / 20 Substitution
V = 1590.32 /20
V = 79.6 L
