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Every cube would contain 1 milliliter of water, or .01 liter
Answer:- 544.5 mL of water need to be added.
Solution:- It is a dilution problem. The equation used for solving this type of problems is:
where, 
is initial molarity and 
is the molarity after dilution. Similarly, 
is the volume before dilution and 
is the volume after dilution.
Let's plug in the values in the equation:
Volume of water added = 907.5mL - 363mL = 544.5 mL
So, 544.5 mL of water are need to be added to the original solution for dilution.
16) Na (s) + H2O(L) ---> H2 (g) + NaOH (aq)
17) O2 (g) + NH3 (g) --->H2O (L) + HNO3 (aq)
18) K (s) + Cl2 (g) ---> KCl (s)
19) Al (s) + HCl (aq) ---> H2 (g) + AlCl (aq)
20) Na3PO4 (aq) + CaCl2 (aq) ---> NaCl (s) + Ca3(PO4)2 (s)
Answer:
Explanation:
4
N
a
+
O
2
→
2
N
a
2
O
.
By the stoichiometry of this reaction if 5 mol natrium react, then 2.5 mol
N
a
2
O
should result.
Explanation:
The molecular mass of natrium oxide is
61.98
g
⋅
m
o
l
−
1
. If
5
m
o
l
natrium react, then
5
2
m
o
l
×
61.98
g
⋅
m
o
l
−
1
=
154.95
g
natrium oxide should result.
So what have I done here? First, I had a balanced chemical equation (this is the important step; is it balanced?). Then I used the stoichiometry to get the molar quantity of product, and converted this molar quantity to mass. If this is not clear, I am willing to have another go.
