Answer : The % of (+) limonene isomer = 79%
The % of (-) limonene isomer = 0%
The % of enantiomeric excess = 58%
Explanation : Enantiomeric excess (ee) is the measurement of purity used for chiral substances.
Given,
% of pure limonene enantiomer = The % of (+) limonene isomer = 79%
Therefore, The % of (-) limonene isomer = 0%
Formula used :
Where, ee → enantiomeric excess
Now, put all the values in above formula, we get the value of enantiomeric excess (ee).
= 58%
Answer:
Na₁₁ = 1s² 2s² 2p⁶ 3s¹
Explanation:
Sodium is present in group 1.
It is alkali metal.
It has one valence electron.
The atomic number of sodium is 11.
Its atomic mass is 23 amu.
The longhand notation of electronic configuration of sodium can be written as,
Na₁₁ = 1s² 2s² 2p⁶ 3s¹
The electronic configuration in shorthand notation( noble gas) would be written as,
Na₁₁ = [Ne] 3s¹
Sodium loses its one valence electron to complete the octet and get stable thus form +1 cation.
It react with halogen and form salt. Such as sodium chloride.
2Na + Cl₂ → 2NaCl
I don’t see what you need help with but thanks:)
Answer:
P2N5
Explanation:
you have to plus it 2 times
