A chemist needs 1.36 mol of strontium bromide, SrBr2, for a reaction. What mass of strontium chloride should the chemist use?
2 answers:
Answer: 336.5 grams
Explanation:
According to avogadro's law, 1 mole of every substance weighs equal to the molecular mass and contains avogadro's number of particles.
convert given masses into moles.
Number of Moles =
Given number of moles = 1.36
Molar mass of = 247.43 g/mol
Putting in the values we get
Thus mass of strontium bromide used should be 336.5 grams.
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The question is incomplete, complete question is;
A solution of is added dropwise to a solution that contains
of
and
and
.
What concentration of is need to initiate precipitation? Neglect any volume changes during the addition.
value
value
What concentration of is need to initiate precipitation of the first ion.
Answer:
Cadmium carbonate will precipitate out first.
Concentration of is need to initiate precipitation of the cadmium (II) ion is
.
Explanation:
1)
The expression of an solubility product of iron(II) carbonate :
2)
The expression of an solubility product of cadmium(II) carbonate :
On comparing the concentrations of carbonate ions for both metallic ions, we can see that concentration to precipitate out the cadmium (II) carbonate from the solution is less than concentration to precipitate out the iron (II) carbonate from the solution.
So, cadmium carbonate will precipitate out first.
And the concentration of carbonate ions to start the precipitation of cadmium carbonate we will need concentration of carbonate ions greater than the concentration.