Osmolarity=osmole of the solute/litres of the solution
ionic equation for dissociation of CaCl2 is
CaCl2--->Ca2+ +2Cl-
total osmoles for reaction are 1(Ca2+) + 2(Cl-)= 3 osmoles
therefore
0.50 moles of CaCl2 x 3 osmoles/ 1mole of CaCl2 = 1.5osmoles
osmolarity=1.5 /1.0 L=1.5 osmol/l
Answer: The concentration of KOH for the final solution is 0.275 M
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

where,
n = moles of solute
= volume of solution in ml = 150 ml
moles of solute =
Now put all the given values in the formula of molality, we get

According to the dilution law,

where,
= molarity of stock solution = 1.19 M
= volume of stock solution = 15.0 ml
= molarity of diluted solution = ?
= volume of diluted solution = 65.0 ml
Putting in the values we get:


Therefore, the concentration of KOH for the final solution is 0.275 M
Answer:
5.65 is the pH.
Explanation:
I am assuming that you are asking for confirmation on your answer. The answer is 5.65.
You would do:
[pOH] = -log[OH-]
= -log[4.5*10^-9]
equals about 8.3468
To find pH your would subtract the pOH from 14.
14-8.3468 = 5.65 << Rounded to match the answer choices.
Answer:
7.5 L of the 10% and 22.5 L of the 30% acid solution, she should mix.
Explanation:
Let the volume of 10% acid solution used to make the mixture = x L
So, the volume of 30% acid solution used to make the mixture = y L
Total volume of the mixture = <u>x + y = 30 L .................. (1)
</u>
For 10% acid solution:
C₁ = 10% , V₁ = x L
For 30% acid solution :
C₂ = 30% , V₂ = y L
For the resultant solution of sulfuric acid:
C₃ = 25% , V₃ = 30 L
Using
C₁V₁ + C₂V₂ = C₃V₃
10×x + 30×y = 25×30
So,
<u>x + 3y = 75 .................. (2)
</u>
Solving 1 and 2 we get,
<u>x = 7.5 L
</u>
<u>y = 22.5 L</u>
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Some of false positive drug use is because of this medications: antihistamines, decongestants, antibiotics, antidepressants, analgesics and antipsychotics.