Answers
Answer:
enantiomeric excess = 68%
Explanation:
Enantiomeric excess is a value used to determine the purity of chiral molecules. It is possible to determine enantiomeric excess (ee) using:
ee = R - S / R + S * 100
Where R is the mass (In this case percentage) of the R enantiomer and S of the S enantiomer.
Replacing with values of the problem:
ee = 84% - 16% / 84% + 16% * 100
ee = 68%
Final answer:
The enantiomeric excess of the mixture, defined as the difference between the concentrations of the R and S enantiomers, is 68.0%.
Explanation:
The enantiomeric excess (ee) is defined as the absolute difference between the mole percentage of the major enantiomer and the minor enantiomer in a mixture. In a sample that contains 84.0 % of the R enantiomer and 16.0 % of the S enantiomer, the enantiomeric excess is calculated as follows:
- Calculation: The enantiomeric excess is 84.0% (R) - 16.0% (S) = 68.0%
Therefore, the enantiomeric excess of the mixture is 68.0%.
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Related Questions
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Answer:
a) H₂SO₄ + Ba(OH)₂ ⇄ BaSO₄(s) + 2 H₂O(l)
b) H₂SO₄, H⁺, HSO₄⁻, SO₄²⁻. H₂O, H⁺, OH⁻.
c) H⁺, HSO₄⁻, SO₄²⁻
d) As the titration takes place, reaction [1] proceeds to the right. The conductivity of the solution decreases because the amount of H⁺, HSO₄⁻, SO₄²⁻ decreases. The formed solid is barium sulfate BaSO₄. Since BaSO₄ is very insoluble, the main responsible for conductivity are still H⁺, HSO₄⁻ and SO₄²⁻,
e) At the equivalence point equivalent amounts of H₂SO₄ and Ba(OH)₂ react. The conducting species are Ba²⁺, SO₄²⁻, H⁺ and OH⁻.
f) After the equivalence point there is an excess of Ba(OH)₂. The ions Ba²⁺ and OH⁻ are responsible for the increase in the conductivity, being the major conducting species.
Explanation:
a) Write an equation (including states of matter) for the reaction between H₂SO₄ and Ba(OH)₂.
The balanced equation is:
H₂SO₄ + Ba(OH)₂ ⇄ BaSO₄(s) + 2 H₂O(l) [1]
b) At the very start of the titration, before any titrant has been added to the beaker, what is present in the solution?
In the beginning there is H₂SO₄ and the ions that come from its dissociation reactions: H⁺, HSO₄⁻, SO₄²⁻. There is also H₂O and a very small amount of H⁺ and OH⁻ coming from its ionization.
H₂SO₄(aq) ⇄ H⁺(aq) + HSO₄⁻(aq)
HSO₄⁻(aq) ⇄ H⁺(aq) + SO₄²⁻(aq)
H₂O(l) ⇄ H⁺(aq) + OH⁻(aq)
c) What is the conducting species in this initial solution?
The main responsible for conductivity are the ions coming from H₂SO₄: H⁺, HSO₄⁻, SO₄²⁻.
d) Describe what happens as titrant is added to the beaker. Why does the conductivity of the solution decrease? What is the identity of the solid formed? What is the conducting species present in the beaker?
As the titration takes place, reaction [1] proceeds to the right. The conductivity of the solution decreases because the amount of H⁺, HSO₄⁻, SO₄²⁻ decreases. The formed solid is barium sulfate BaSO₄. Since BaSO₄ is very insoluble, the main responsible for conductivity are still H⁺, HSO₄⁻ and SO₄²⁻,
e) What happens when the conductivity value reaches its minimum value (which is designated as the equivalence point for this type of titration)? What is the conducting species in the beaker?
At the equivalence point equivalent amounts of H₂SO₄ and Ba(OH)₂ react. Only BaSO₄ and H₂O are present, and since they are weak electrolytes, there is a small amount of ions to conduct electricity. The conducting species are Ba²⁺ and SO₄²⁻ coming from BaSO₄ and H⁺ and OH⁻ coming from H₂O.
f) Describe what happens at additional titrant is added past the equivalence point. Why does the conductivity of the solution increase? What is the conducting species present in the beaker?
After the equivalence point there is an excess of Ba(OH)₂. The ions Ba²⁺ and OH⁻ are responsible for the increase in the conductivity, being the major conducting species.
Final answer:
The chemical reaction between H2SO4 and Ba(OH)2 forms BaSO4 and water, reducing conductivity by reducing the number of free ions. Beyond the equivalence point, the conductivity increases due to the dissociated ions from the excess Ba(OH)2 in the solution.
Explanation:
Chemical Reaction and Metric Titration
Firstly, the equation representing the reaction between sulfuric acid (H2SO4) and barium hydroxide (Ba(OH)2) is:
Ba(OH)2 (aq) + H2SO4 (aq) → BaSO4 (s) + 2H2O (l)
In the beginning, the solution only contains H2SO4 with its dissociated ions serving as the conducting species. As titrant (Ba(OH)2) is added, they react to form BaSO4, a solid precipitate reducing the number of free ions in the solution, thus decreasing conductivity. At the equivalence point, all H2SO4 has reacted, and conductivity reaches its minimum as there are lesser free ions for conduction. If additional titrant is added past the equivalence point, conductivity increases due to excess Ba(OH)2's dissociated ions that increase ion concentration in solution.
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Answers
Answer:
0.2 moles of CO₂ are produced
Explanation:
Given data:
Moles of CO₂ produced = ?
Moles of Na₂CO₃ react = 0.2 mol
Solution:
Chemical equation:
Na₂CO₃ + 2HCl → 2NaCl + CO₂ + H₂O
Now we will compare the moles of CO₂ with Na₂CO₃ .
Na₂CO₃ : CO₂
1 : 1
0.2 : 0.2
Thus, 0.2 moles of CO₂ are produced.
Answers
5.83
B.
3.67
C.
2.08
D.
1.66
Answers
Answer:
B) pH = 3.67
Explanation: