Answer:
B
Explanation:
to find how many moles we have you need to use the molarity equation which for this is
40g * (1mole/200g) = .2 moles
Given we know that we have .2 moles
then all you have to do is divide amount of moles by amount of liters to find the molarity
for this problem it is
.2mol/2L which is .1 M
0.01742919 is the answer because i worked it out
I believe the precipitate that would most likely form between this double replacement reaction would be
Copper (II) Sulphide. CuS.
0.25 L Barium nitrate solution contains 0.02 mol nitrate ions
<h3>Further explanation</h3>
Given
0.04 M Barium nitrate, Ba(NO₃)₂(aq)
Required
The volume of Barium nitrate
Solution
Ionization of Barium nitrate in 1 L solution :
Ba(NO₃)₂ ⇒ Ba²⁺ + 2NO₃²⁻
mol :
0.04 0.04 0.08
There is 0.08 mol of nitrate ions in 1 L solution
So for 0.02 mol nitrate ions :
= 0.02/0.08 x 1 L
= 0.25 L