Molar mass
Ca₃(PO₄)₂ = 310 g/mol
1 mole -------------------> 310 g
2.3x10⁻⁴ mole ---------> ?
m = 2.3x10⁻⁴ * 310 / 1
m = 0.0713 g
hope this helps!
Answer:
60 cm³ of water
Explanation:
We'll begin by calculating the volume of the diluted solution. This can be obtained as follow:
Concentration of stock solution (C₁) = 17 M
Volume of stock solution (V₁) = 25 cm³
Concentration of diluted solution (C₂) = 5 M
Volume of diluted solution (V₂) =?
C₁V₁ = C₂V₂
17 × 25 = 5 × V₂
425 = 5 × V₂
Divide both side by 5
V₂ = 425 / 5
V₂ = 85 cm³
Thus, the volume of the diluted solution is 85 cm³
Finally, we shall determine the volume of water needed to dilute the solution. This can be obtained as follow:
Volume of stock solution (V₁) = 25 cm³
Volume of diluted solution (V₂) = 85 cm³
Volume of water =?
Volume of water = V₂ – V₁
Volume of water = 85 – 25
Volume of water = 60 cm³
Therefore, 60 cm³ of water is needed to dilute the solution.
Answer:
.125 M
Explanation:
.15 M/L * .125 L = .01875 moles
now dilute to 150 cc (by adding 25 cc)
.01875M / (150/1000) = .125M
Answer:
The concentration of HI present at equilibrium is 0.471 M.
Explanation:
