The descriptive term applied to the type of diene represented by 2,4-hexadiene is conjugated diene.
Dienes are compounds which contains two double bonds. These dienes can be non conjugated or conjugated.
Conjugated diene are those compound which have two double bonds joined by a single σ bond. Conjugated dienes can also be called 1,3-diene. To know if diene is conjugated or non conjugated, sp³ hybridization is to b checked and the number of double bonds and single sigma bond is checked.
Conjugated dienes are found in many different molecules. 2,4-hexadiene is a conjugated diene with two carbon-carbon double bonds that are separated by one sigma bond.
The stabilization of dienes by conjugation is better than the aromatic stabilization. Conjugated dienes are more stable than non conjugated or cumulative diene because it has higher electron density of molecules delocalized.
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Answer:
1.9 L
Explanation:
Step 1: Given data
- Initial number of moles of air (n₁): 4.0 mol
- Initial volume of the balloon (V₁): 2.5 L
- Final number of moles of air (n₂): 3.0 mol
- Final volume of the balloon (V₂): ?
Step 2: Calculate the final volume of the balloon
According to Avogadro's law, the volume of an ideal gas is directly proportional to the number of moles. We can calculate the final volume of the balloon using the following expression.
V₁ / n₁ = V₂ / n₂
V₂ = V₁ × n₂ / n₁
V₂ = 2.5 L × 3.0 mol / 4.0 mol
V₂ = 1.9 L
Practical work refers to the art of conducting experiments in order to answer certain research questions.
<h3>What is practical work?</h3>
In science, practical work refers to the art of conducting experiments in order to answer certain research questions. This could occur in a laboratory under controlled conditions or in the field.
In the physical sciences, most of the practical work is conducted in the laboratory under controlled conditions. However, some experiments in the biological sciences and most experiments in the social sciences are conducted outside the laboratory.
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Answer is: A) 124 s.
c₀ = 3 mol/L.
c₁ = 0,700 mol/L.
k = 8,8·10⁻³ 1/M·s.
Integrated second order rate law is: 1/c₁ = 1/c₀ + k·t.
k·t = 1/0,700 - 1/3.
0,0088·t = 1,095.
t = 1,095 ÷ 0,0088.
t = 124 s.
c₀ - <span>initial concentration.
c</span>₁ - <span> concentration at a particular time.
k - </span><span>the rate constant.
t - time.</span>
<span>Assuming that there are 36 strontium and 24 phosphate, there
aren’t any equal cations and anoins because in theory only one ionic bond is
formed by a strontium with each phosphate ion. To the point that a cation will
eventually have an excess.</span>
