25 x 10.5 = 262.5
So the approx. is 260 g which is answer D.
If 1.2 L of solution contains 0.97 mol
then let 1 L of solution contain x mol
⇒ (1.2 L) x = (0.97 mol) (1 L)
x = (0.97 mol · L) ÷ (1.2 L)
x = 0.8083 mol
Thus the molarity of the Barium Chloride solution is 0.808 mol / L OR 0.808 mol/dm³.
Test tubes heat small amounts of liquids while boiling tube boils liquids
Answer:
0.5ppm
Explanation:
Step 1:
Data obtained from the question.
Volume of water = 2500L
Mas of Cu = 1.25 g
Step 2:
Determination of the concentration of Cu in g/L. This is illustrated below:
Volume of water = 2500L
Mas of Cu = 1.25 g
Conc. of Cu In g/L =?
Conc. g/L = Mass /volume
Conc. of Cu in g/L = 1.25/2500
Conc. of Cu in g/L = 5x10^–4 g/L
Step 3:
Conversion of the concentration of Cu in g/L to ppm. This is illustrated below
Recall:
1g/L = 1000mg/L
Therefore, 5x10^–4 g/L = 5x10^–4 x 1000 = 0.5mg/L
Now, we know that 1mg/L is equal to 1ppm.
Therefore, 0.5mg/L is equivalent to 0.5ppm
