Answer: There are 0.5 grams of barium sulfate are present in 250 of 2.0 M
solution.
Explanation:
Given: Molarity of solution = 2.0 M
Volume of solution = 250 mL
Convert mL int L as follows.

Molarity is the number of moles of solute present in liter of solution. Hence, molarity of the given
solution is as follows.

Thus, we can conclude that there are 0.5 grams of barium sulfate are present in 250 of 2.0 M
solution.
Since we already have the balanced equation, we know that the ratio between
is
respectively.
So then we can set up a proportion to find the number of moles produced when 2.90 moles of Na react completely:

Then we cross multiply and solve for x:


Therefore, we know that when 2.90 moles of Na react completely, there are 1.45 moles of
that are produced.
Answer:
a) = 0.704%
b) = 1.30%
c) = 2.60%
Explanation:
Given that:
= 
For Part A; where Concentration of A = 0.270 M
Percentage Ionization(∝) 



percentage% (∝) = 
= 0.704%
For Part B; where Concentration of B =
M



percentage% (∝) = 0.0130 × 100%
= 1.30%
For Part C; where Concentration of C= 



percentage% (∝) = 0.02608 × 100%
= 2.60%
Answer:
The answer is its equal to the volume of its container.
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I hope this helps! :)