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<span>3O2(g) <--> 2O3(g);
Keq = 1 = [O3]^2/[O2]^3
So [O2]^3 = [O3]^2
Thus A) is correct</span>
1. This can be due to the dissolving of the solid in liquid and form a solution.
Dissolving is a physical property because dissolving doesn't form new substances and the chemical composition of the solid is not changed.
The color building up over the time can be due to the rate of dissolving of the solid and amount of particles have been dissolved.
Example:
- Dissolving of CuSO₄ solid in water.
- This develops a blue color.
2. This can be due to the chemical reaction between the solid and liquid.
Chemical reaction is a chemical property because from reacting substances new substances can be formed which the chemical formula is different from initial substances.
The color building up over the time can be due to the rate of the reaction and the amount of reactants.
Example:
- The reaction between calcium metal with water.
- The color of Ca(OH)₂ is white color.
- Reaction is
Ca(s) + 2H₂O(l) → Ca(OH)₂(aq) + H₂(g)
Hydrogen peroxide in water and adding kool- aid powder to water so the liquid turns red
The idea here is that you need to figure out how many moles of magnesium chloride,
MgCl
2
, you need to have in the target solution, then use this value to determine what volume of the stock solution would contain this many moles.
As you know, molarity is defined as the number of moles of solute, which in your case is magnesium chloride, divided by liters of solution.
c
=
n
V
So, how many moles of magnesium chloride must be present in the target solution?
c
=
n
V
⇒
n
=
c
⋅
V
n
=
0.158 M
⋅
250.0
⋅
10
−
3
L
=
0.0395 moles MgCl
2
Now determine what volume of the target solution would contain this many moles of magnesium chloride
c
=
n
V
⇒
V
=
n
c
V
=
0.0395
moles
3.15
moles
L
=
0.01254 L
Rounded to three sig figs and expressed in mililiters, the volume will be
V
=
12.5 mL
So, to prepare your target solution, use a
12.5-mL
sample of the stock solution and add enough water to make the volume of the total solution equal to
250.0 mL
.
This is equivalent to diluting the
12.5-mL
sample of the stock solution by a dilution factor of
20
.
