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
Chemical reaction to biomass, like kiwi, into synthetic materials, like plastic, impact society because kiwi is biodegradable but plastic is not.
The series of chemical reactions used to change natural resources into synthetic products is termed chemical synthesis. Synthetic materials are formed when humans mix substances together for the purpose of creating new materials with desirable properties. Synthetic materials are produced by chemically changing the initial substances to create some material with different characteristics. Some of the best examples of synthetic materials are plastics and medicines. A synthetic substance may or may not be chemically identical to a naturally-occurring substance.
To know more about synthetic materials check the link below:
brainly.com/question/25834689
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Answer:
Cu(NO3)2(aq)+Pb(s) ⇌ Pb(NO3)2(aq)+Cu(s)
Explanation:
If we look at the both reactions closely, we will quickly discover that the reaction CuSO4(aq)+Pb(s) ⇌ PbSO4(s)+Cu(s) involves PbSO4.
The compound PbSO4 is insoluble in water and sinks to the bottom of the reaction vessel. When this occurs, the concentration of Pb^2+ becomes low. This will bring about a low voltage in the cell.
On the other hand, Pb(NO3)2 is soluble in water hence the cell voltage in this case is higher than the former.
Answer:
Explanation:
The half-cell reduction potentials are
Ag⁺(aq) + e⁻ ⇌ Ag(s) E° = 0.7996 V
Fe²⁺(aq) + 2e⁻ ⇌ Fe(s) E° = -0.447 V
To create a spontaneous voltaic cell, we reverse the half-reaction with the more negative half-cell potential.
The anode is the electrode at which oxidation occurs.
The equation for the oxidation half-reaction is
