Answer:
212.304 grams
Explanation:
similar to your other question, use the same formula
q=mCpΔT
23617=m(4.182)(46.6-20)
23617=111.2412m
m=212.304g
Answer:

Explanation:
Given:
A solution contains one or more of the following ions such as Ag,
and 
Here the Lithium bromide is added to the solution and no precipitate forms
Solution:
Since with LiBr no precipitation takes place therefore Ag+ is absent
Here on adding
to it precipitation takes place.
Precipitate is as follows,

Thus,
is present
When
is added again precipitation takes place.
Therefore the reaction is as follows,

Therefore,
are present in the solution
Balanced chemical equation is
3CaCl2 +2Na3PO4-->6NaCl +Ca3(PO4)2
moles of CaCl2 =89.3g/[ (35.5x2) +40]=0.805moles
from the equation above the ratio of CaCl2 to Ca3(PO4)2 is 3:1 therefore the moles of Ca3(PO4)2 is 0.809/3=0.268moles
mass is therefore 0.268 x310.18(R.F.M of Ca3(PO4)2 ) =83.23grams