To solve this we use the equation,
M1V1 = M2V2
where M1 is the concentration
of the stock solution, V1 is the volume of the stock solution, M2 is the
concentration of the new solution and V2 is its volume.
2 M x V1 = 0.1 M x .5 L
<span>V1 = 0.025 L or 25 mL of the
2 M KCl solution is needed</span>
Answer:
32 g Cu
Explanation:
1 mol Cu -> 63.5 g
0.5 mol Cu ->x
x=(0.5 mol *63.5 g)/1 mol x= 32 g Cu
1) acid
2) ether
3) ester
4) aldehyde
5) ketone
6) amine
7) alcohol
The reactant in a chemical reaction that limits the amount of product that can be formed. The reaction will stop when all of the limiting reactant is consumed
SrCl2 + Li2(SO4) —> 2LiCl + Sr(SO4)