What volume of the stock solution (Part A) would contain the number of moles present in the diluted solution (Part B)?
1 answer:
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solution:
The quoted atomic mass on the Periodic Table is the WEIGHTED average of the individual isotopic masses. The higher the isotopic percentage, the MORE that isotope will contribute to the isotopic mass. For this reason, most masses that are quoted on the Table are non-integral.
By way of example we could look to the hydrogen atom. The VAST majority of hydrogen atoms (in this universe) are the protium isotope. i.e. 1H, whose nuclei contain JUST the defining proton. There is a smaller percentage (>1%) of hydrogen atoms WITH one NEUTRON in their nuclei to give the deuterium isotope. i.e. 2H, and because this is relatively cheap, and easily incorporated into a molecule, deuterium labelling is routinely used in analysis.
And there is even a smaller percentage of hydrogen atoms with TWO NEUTRONS in their nuclei, to give the tritium isotope. i.e. 3H. The weighted average of the isotopic percentages gives
Answer:
C. will precipitate out first
the percentage of remaining = 12.86%
Explanation:
Given that:
A solution contains:
From the list of options , Let find the dissociation of
where;
Solubility product constant Ksp of is
Thus;
replacing the known values in order to determine the unknown ; we have :
The dissociation of
The solubility product constant of is
The dissociation of is :
Thus;
Thus; the phosphate anion needed for precipitation is smaller i.e in
than in
Therefore:
will precipitate out first
To determine the concentration of when the second cation starts to precipitate ; we have :
This implies that when the second cation starts to precipitate ; the concentration of in the solution is 0.00566
Therefore;
the percentage of remaining = concentration remaining/initial concentration × 100%
the percentage of remaining = 0.00566/0.0440 × 100%
the percentage of remaining = 0.1286 × 100%
the percentage of remaining = 12.86%