Answer : The [α] for the solution is, -118.8
Explanation :
Enantiomeric excess : It is defined as the difference between the percentage major enantiomer and the percentage minor enantiomer.
Mathematically,

Given:
% major enantiomer = 86 %
% minor enantiomer = 14 %
Putting values in above equation, we get:


Now we have to calculate the [α] for the solution.
![[\alpha]=\text{Enantiomer excess}\times [\alpha]_{Pure}](https://tex.z-dn.net/?f=%5B%5Calpha%5D%3D%5Ctext%7BEnantiomer%20excess%7D%5Ctimes%20%5B%5Calpha%5D_%7BPure%7D)
![[\alpha]=0.72\times -165](https://tex.z-dn.net/?f=%5B%5Calpha%5D%3D0.72%5Ctimes%20-165)
![[\alpha]=-118.8](https://tex.z-dn.net/?f=%5B%5Calpha%5D%3D-118.8)
Thus, the [α] for the solution is, -118.8
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hope this helps
Answer:

Explanation:
Hello!
In this case, since the pH of the given metal is 10.15, we can compute the pOH as shown below:

Now, we compute the concentration of hydroxyl ions in solution:
![[OH^-]=10^{-pOH}=10^{-3.95}=1.41x10^{-4}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E%7B-pOH%7D%3D10%5E%7B-3.95%7D%3D1.41x10%5E%7B-4%7DM)
Now, since this hydroxide has the form MOH, we infer the concentration of OH- equals the concentration of M^+ at equilibrium, assuming the following ionization reaction:

Whose equilibrium expression is:
![Ksp=[M^+][OH^-]](https://tex.z-dn.net/?f=Ksp%3D%5BM%5E%2B%5D%5BOH%5E-%5D)
Therefore, the Ksp for the saturated solution turns out:

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Answer: alright now listen fe203(s)=567.66666
Explanation: