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:
Reverse the reaction
Explanation:
Reactions:
Overall:
As can be seen, in the overall reaction we have in the reactants like in the second reaction and
in the products. The
is in the first reaction but as a reactant so we need to reverse that reaction:
Combining:
Answer:
a) 1,6%
b) 64,775 g/mol
c) 3,6×10⁻² M
d) 2,3×10⁻³ g/mL
Explanation:
a) The mass fractium of helium is obtained converting the moles of the four gases to grams with molar weight and then caculating of the total of grams how many are of helium, thus:
Total grams: 1g+10g+20,85g+30,825g= 62,675 g
Mass fraction of helium: × 100 = <em>1,6%</em>
<em />
<em>The mass fraction of Helium is 1,6%</em>
<em />
<em>b)</em><em> </em>Because the mole fraction of all gases is the same the average molecular weight of the mixture is:
= 64,775 g/mol
c) The molar concentration is possible to know ussing ideal gas law, thus:
= M
Where:
P is pressure: 150 kPa
R is gas constant: 8,3145
T is temperature: 500 K
And M is molar concentration. Replacing:
M = 3,6×10⁻² M
d) The mass density is possible to know converting the moles of molarity to grams with average molecular weight and liters to mililiters, thus:
3,6×10⁻² ×
×
=
2,3×10⁻³ g/mL
I hope it helps!