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Margarita [4]
3 years ago
11

Di- n- pentyl ether can be converted to 1- bromopentane by treatment with HBr through essentially a(n) ________ mechanism.

Chemistry
1 answer:
mamaluj [8]3 years ago
3 0

Answer:

SN1 mechanism

Explanation:

The mechanism of this reaction is shown in the image attached.

The Di- n- pentyl ether is first protonated. The CH3(CH2)4OH is now a good leaving group as shown.

The attack of the bromide ion on the cation formed completes the mechanism to yield 1- bromopentane as shown in the mechanism.

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If the 3.90 m solution from Part A boils at 103.45 âC, what is the actual value of the van't Hoff factor, i? The boiling point o
poizon [28]

Answer:

Van't Hoff factor is 1.72 ≅ 2

Explanation:

Let's apply the colligative property of boiling point elevation:

ΔT = Kb . m . i

ΔT = Boiling T° of solution - T° boiling of pure solvent

103.45 °C - 100°C = 0.512 °C/m . 3.90 m . i

3.45 °C /  0.512 m/°C . 3.90 m = i

1.72 = i  ≅ 2

7 0
3 years ago
If carbon-15 has a half-life of 2.5 seconds, % of the sample would still be
kramer

Answer:

After 5 second 25% C-15 will remain.

Explanation:

Given data:

Half life of C-15 = 2.5 sec

Original amount = 100%

Sample remain after 5 sec = ?

Solution:

Number of half lives = T elapsed / half life

Number of half lives = 5 sec / 2.5 sec

Number of half lives = 2

At time zero = 100%

At first half life = 100%/2 = 50%

At second half life = 50%/2 = 25%

Thus after 5 second 25% C-15 will remain.

8 0
4 years ago
CH4 + 2O2 → CO2 + 2H2O In the chemical reaction, if 10 moles of H2O are produced, moles of CO2 are also produced
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Answer:

The correct answer is 5 moles of CO2 are produced.

Explanation:

The given reaction:  

CH₄ (g) + 2O₂ (g) ⇔ CO₂ (g) + 2H₂O (g)

The given reaction is an illustration of a combustion reaction. Any reaction in which a substance is burnt in excess of oxygen to generate water and carbon dioxide is termed as a combustion reaction. From the given equation, it is clear that the moles of the formation of the products are in the ratio 1: 2, that is, if 10 moles of H₂O is produced, the production of 5 moles of CO₂ will be produced.  

Let us multiply, the given equation with 5 we get,  

5CH₄ + 10O₂ ⇔ 5CO₂ + 10H₂O

Hence, it is clear that with the formation of 10 moles of H₂O, formation of 5 moles of CO₂ will also take place.  

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Answer:

Localized flooding may be caused or exacerbated by drainage obstructions such as landslides, ice, debris, or beaver dams. Slow-rising floods most commonly occur in large rivers with large catchment areas. The increase in flow may be the result of sustained rainfall, rapid snow melt, monsoons, or tropical cyclones.

Explanation:

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