Answer:
if it gains or loses neutrons
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
.
Answer: The unknown solution had the lower concentration
Explanation: concentration will always move from higher to lower region. If the concentration of the unknown solution has increased, it therefore means that the initial concentration of the unknown solution was low
Answer: Bromine is the limiting reactant
Explanation:
First of all let's generate a balanced equation for the reaction
2Li + Br2 —> 2LiBr
Molar Mass of Li = 7g/mol
Molar Mass of Br2 = 2x80 = 160g/mol
From the question given, were told that 25g of Li and 25g Br2 were present at the take-off of the reaction. Converting these Masses to mole, we have:
Number of mole of Li = 25/7 = 3.6moles
Number of mole of Br2 = 25/160 = 0.156mol.
To know which is the limiting reactant, we have to compare the ratio of the number of mole of experimental Li and Br2 to that of theoretical Li and Br2
For the experimental yield:
Li : Br2 = 3.6/ 0.156 = 23 : 1
For the theoretical yield:
Li : Br = 2 : 1
From the above, we see clear that Br2 is the limiting reactant because according to the equation( which gives the theoretical yield), for every 2moles of Li, 1mole of Br2 is used up. But this is not so from the experiment conducted as 23moles required 1mole of Br2.
