Answer:
The molar solubility of YF₃ is 4.23 × 10⁻⁶ M.
Explanation:
In order to calculate the molar solubility of YF₃ we will use an ICE chart. We identify 3 stages: Initial, Change and Equilibrium and we complete each row with the concentration of change of concentration. Let's consider the solubilization of YF₃.
YF₃(s) ⇄ Y³⁺(aq) + 3 F⁻(aq)
I 0 0
C +S +3S
E S 3S
The solubility product (Ksp) is:
Ksp = [Y³⁺].[F⁻]³= S . (3S)³ = 27 S⁴
![S=\sqrt[4]{Ksp/27} =\sqrt[4]{8.62 \times 10^{-21} /27}=4.23 \times 10^{-6}M](https://tex.z-dn.net/?f=S%3D%5Csqrt%5B4%5D%7BKsp%2F27%7D%20%3D%5Csqrt%5B4%5D%7B8.62%20%5Ctimes%2010%5E%7B-21%7D%20%20%2F27%7D%3D4.23%20%5Ctimes%2010%5E%7B-6%7DM)
Answer:
Temporary hardness is a type of water hardness caused by the presence of dissolved bicarbonate minerals (calcium bicarbonate and magnesium bicarbonate). ... However, unlike the permanent hardness caused by sulfate and chloride compounds, this "temporary" hardness can be reduced by boiling the water.
Answer:
Cesium has one valence electron that can be removed from itself resulting to its stable monatomic ion of Cs+ .
Explanation:
Molar mass of A l 2 (S O 4 ) 3 = 342.15 g/mol
7.2g/ 342.15 g/mol
you can get number of moles.
then multiply that by12
The actual formula for volume for a cube is the length multiplied by the width and then multiplied by the height. Since all three measurements are the same, the formula results in the measurement of one side cubed. For the example, 5^3 is 125 cm^3. Multiply the volume by the known density, which is the mass per volume.
