What is the molarity of a solution prepared from 85. 0 g cacl2 in 300. 0 g of solution? the density of the solution is 1. 05 g/m
1 answer:
You might be interested in
Answer: There is one way to write it but i’ll also provide an unbalanced equation and a balanced one.
Explanation:
Unbalanced : Ba (aq) + Cl2 (aq)—-> BaCl (aq)
Balanced : 2Ba (aq) + Cl2 (aq)—> 2BaCl(aq)
Answer: The volume of required is 25.0 ml
Explanation:
According to the neutralization law,
where,
= basicity
= 1
= molarity of
solution = 2.00 M
= volume of
solution = 50.0 ml
= acidity of
= 1
= molarity of
solution = 4.00 M
= volume of
solution = ?
Putting in the values we get:
Therefore, volume of required is 25.0 ml