Answer:
The metals in this group are lithium, sodium, potassium, rubidium, cesium, and francium. The gas hydrogen is also put in this group because it shares similar reactivity with the alkali metals.
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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:
Rate = 1.321M/s[H₂CO₃]¹ ; k = 1.32M/s
Explanation: