Let us say that R is the major enantiomer, while
S is the minor enantiomer, therefore the formula for enantiomeric excess (ee)
is:
ee = (R – S) * 100%
Let us further say that the fraction of R is x (R
= x), and therefore fraction of S is 1 – x (S = 1 – x), therefore:
75 = (x – (1 – x)) * 100
75 = 100 x – 100 + 100 x
200 x = 175
x = 0.875
Summary of answers:
R = major enantiomer = 0.875 or 87.5%
<span>S = minor enantiomer = (1 – 0.875) = 0.125 or
12.5%</span>
Answer:
Zn(s) + Cu²⁺(aq) ⇒ Zn²⁺(aq) + Cu(s)
Explanation:
Let's consider the molecular single displacement equation between Zn and Cu(NO₃)₂
Zn(s) + Cu(NO₃)₂(aq) ⇒ Zn(NO₃)₂(aq) + Cu(s)
The complete ionic equation includes all the ions and insoluble species.
Zn(s) + Cu²⁺(aq) + 2 NO₃⁻(aq) ⇒ Zn²⁺(aq) + 2 NO₃⁻(aq) + Cu(s)
The net ionic equation includes only the ions that participate in the reaction and insoluble species.
Zn(s) + Cu²⁺(aq) ⇒ Zn²⁺(aq) + Cu(s)
Explanation:
hydrogen + oxygen = water
glucose, ADP (adenosine diphosphate) and NAD.
