- 0.05 L needs to be added to the original 0.12 L solution in order to dilute it from 0.13 M to 0.23 M.
The dilution problem uses the equation :
The initial molarity (concentration) 0.13 M
The initial volume = 0.12 L
The desired molarity (concentration) = 0.23 M
The volume of the desired solution = ( 0.12 + x L )
Substituting values in above equation;
(0.13 M ) (0.12 L) = (0.23 M ) (0.12 L + x L)
0.0156 M L = 0.0276 M L + 0.23 x M L
- 0.012 M L = 0.23 x M L
x = - 0.05
Therefore, - 0.05 L needs to be added to the original 0.12 L solution in order to dilute it from 0.13 M to 0.23 M.
Learn more about molarity here:
brainly.com/question/26873446
#SPJ4
Answer:
The molarity of the chromium(III) acetate solution 0.3982 M.
0.3982 M is the concentration of the chromium(III) cation.
1.1946 M is the concentration of the acetate anion.
Explanation:
Mass of chromium(III) acetate = 22.8 g
Molar mass of chromium(III) acetate = 229 g/mol
Total volume of the solution of chromium(III) acetate = V = 250 mL = 0.250 L
1 mL = 0.001 L
The molarity of the chromium(III) acetate solution :
The molarity of the chromium(III) acetate solution 0.3982 M.
According to reaction, 1 mole of chromium(III) acetate gives 1 mole of chromium(III) ions and 3 moles of acetate ions.
So
0.3982 M is the concentration of the chromium(III) cation.
1.1946 M is the concentration of the acetate anion.
Answer:
A
Explanation:
Because it is the greatest
The answer would be D. Good luck my dude
The solution for this problem would be:
H + ions from pH = 4.0 x 10^-4 moles.
But H2SO4 gives off 2 moles of H+ ions / mole.
Moles of H2SO4 = 2.0 x 10^-4
Mass = 2 x 10^-4 moles x 98 = 1.96 x 10^-3 grams
1.96 x 10^-3 grams / 10^3 = 1.96 x 10^-6 Kg.