What is the percent yield of a reaction?

Using your knowledge of reactions and mechanisms, draw the mechanism for the formation of 2-methylpropene from 2-methyl-2-propan

BARSIC [14]

Answer:

Here's what I get

Explanation:

The reaction is an E1 elimination of an alcohol to form an alkene. It has three steps:

1. Protonation

The alcohol is protonated with aqueous sulfuric acid to convert it into a better leaving group.

2. Loss of the leaving group

A water molecule leaves in a unimolecular process to form a stable 3° carbocation.

3. Loss of an α-hydrogen

A water molecule removes an α-hydrogen, forming 2-methylpropene and regenerating the original hydronium ion.

4 0

3 years ago

Phosphoric acid is a polyprotic acid, with p K values of 2.14, 6.86, and 12.38. Which ionic form predominates at pH 9.3

Vlad1618 [11]

Answer:

HPO₄⁻² predominates at pH 9.3

Explanation:

These are the equilibriums of the phosphoric acid, a tryprotic acid where 3 protons (H⁺) are realesed.

H₃PO₄ + H₂O ⇄ H₂PO₄⁻ + H₃O⁺ pKa 2.14

H₂PO₄⁻ + H₂O ⇄ HPO₄⁻² + H₃O⁺ pKa 6.86

HPO₄⁻² + H₂O ⇄ PO₄⁻³ + H₃O⁺ pKa 12.38

The H₂PO₄⁻ works as amphoterous, it can be a base and acid, according to these equilibriums.

H₂PO₄⁻ + H₂O ⇄ HPO₄⁻² + H₃O⁺

H₂PO₄⁻ + H₂O ⇄ H₃PO₄ + OH⁻

pH 9.3 is located between 6.86 and 12.38 where we have this buffer system HPO₄⁻² / PO₄⁻³, where the HPO₄⁻² is another amphoterous:

HPO₄⁻ + H₂O ⇄ H₂PO₄⁻ + OH⁻

HPO₄⁻² + H₂O ⇄ PO₄⁻³ + H₃O⁺

The media from the two pKa, indicates the pH where the protonated form is in the same quantity as the unpronated form, so:

(6.86 + 12.38) /2 = 9.62

Above this pH, [PO₄⁻³] > [HPO₄⁻²].

In conclussion, at pH 9.3, [HPO₄⁻²] > [PO₄⁻³]

6 0

3 years ago

Explain how the type of an atom is defined?

eimsori [14]

Answer:

the type of atom is defined by how many protons, neutrons, and electrons the atom has

6 0

4 years ago

Read 2 more answers

What best explains the observed trend in atomic radii going down the periodic table

Serggg [28]

going down the the periodic table, the number of shells increases resulting in an obvious increase in the atomic radii of the elements

5 0

3 years ago

An acne preparation contains 600 milligrams of clindamycin in 100 milliliters of solution. How many 150 milligram capsules will

REY [17]

Answer:

8 capsules of clindamycin are required.

Explanation:

Amount of clindamycin in 100 mL solution = 600 mg

Amount of clindamycin in 1 mL = $\frac{600 mg}{100}=6 mg$

Amount of clindamycin in 200 mL solution :

$200\times 6 mg=1200 mg$

Mass of 1 capsule = 150 mg

Number of capsules required to make 200 milliters of solution:

$\frac{1200 mg}{150 mg}=8$

8 capsules of clindamycin are required.

7 0

4 years ago