.0002345 I believe this is correct
As,
CuCO₃ ⇆ Cu²⁺ + CO₃²⁻
So,
Kc = [Cu²⁺] [CO₃²⁻] / CuCO₃
Or,
Kc (CuCO₃) = [Cu²⁺] [CO₃²⁻]
Or,
Ksp = [Cu²⁺] [CO₃²⁻]
As,
Ksp = 1.4 × 10⁻¹⁰
So,
1.4 × 10⁻¹⁰ = [x] [x]
Or,
x² = 1.4 × 10⁻¹⁰
Or,
x = 1.18 × 10⁻⁵ mol/L
To cahnge ito g/L,
x = 1.18 × 10⁻⁵ mol/L × 123.526 g/mol
x = 1.45 × 10⁻³ g/L
Answer:
7650
Explanation:
formula- multiply the volume value by 1e+6
Solute of solution = 17.8 g
Solvn = 198 g
% = 17.8 / 198
w% = 0.089 x 100 = 8.9% by mass
hope this helps!
Answer: pH of resulting solution will be 13
Explanation:
pH is the measure of acidity or alkalinity of a solution.
Moles of 
ion = 
Moles of 
ion = 
For neutralization:
1 mole of ion will react with 1 mole of ion
0.01 mol of ion will react with =
of ion
Thus (0.012-0.01)= 0.002 moles of are left in 20 ml or 0.02 L of solution.
![[OH^-]=\frac{0.002}{0.02L}=0.1M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%5Cfrac%7B0.002%7D%7B0.02L%7D%3D0.1M)
![pOH=-log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-log%5BOH%5E-%5D)
![pOH=-log[0.1]=1](https://tex.z-dn.net/?f=pOH%3D-log%5B0.1%5D%3D1)
Thus the pH of resulting solution will be 13
