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Svetllana [295]
4 years ago
15

For the following dehydrohalogenation (E2) reaction, draw the Zaitsev product(s) resulting from elimination involving C3–C4 (i.e

., the carbon atoms depicted with stereobonds). Show the product stereochemistry clearly. If there is more than one organic product, both products may be drawn in the same box. Ignore elimination involving C3 and any carbon atom other than C4.

Chemistry
1 answer:
Bond [772]4 years ago
8 0

Answer:

2-methyl-butene

Explanation:

For the E2 mechanism, we have an <u>anti-elimination</u>. The Br leaves the molecule and the base removes the hydrogen in the anti position to form the double that's why only one structure is produced. (See figure 1)

Since we have 2 hydrogens on the right carbon, we cannot indicate a <u>specific stereoisomer</u>, in other words, it is not possible to assign a <u>Z or E</u> configuration for this alkene.

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At construction sites, workers often reduce erosion by ____.
Naddik [55]
<h3><u>Answer;</u></h3>

<em>All the above</em>

Workers at construction sites often reduce erosion by;

  • <em>Moving excess sediment back to its original location </em>
  • <em>Planting trees </em>
  • <em>Spraying water on bare soil</em>
<h3><u>Explanation;</u></h3>
  • Soil erosion is a naturally occurring process which involves the wearing away of the topsoil by natural forces such as wind, water or other forces associated with farming.
  • <em><u>Construction of roads and buildings results to large amounts of soil erosion around the world. It is therefore important to put measures that would help reduce soil erosion at construction sites</u></em>. These measures uses principals of soil control such as implementing sediment control, limiting soil exposure, reducing the runoff velocity, and modifying topography among others.
5 0
3 years ago
50 examples word equation with balanced chemical equarion​
gulaghasi [49]
This question may only be ansewered by frequent mattrrs
5 0
2 years ago
Any two objects that have Mass will also have which type of force between them
saul85 [17]
Gravitational <span>force between them</span>
5 0
3 years ago
If you have a 1500 g aluminum pot, how much heat energy is needed to raise its temperature by 100°C?
Nataly [62]

The heat energy required to raise the temperature of 1500 g of aluminium pot by 100°C is 135 kJ.

The heat energy required to raise the temperature of 1500 g of copper pot by 100 °C is 57.75 kJ.

Explanation:

The heat energy required to raise the temperature of any body can be obtained from the specific heat formula. As this formula states that the heat energy required to raise the temperature of the body is directly proportional to the product of mass of the body, specific heat capacity of the material and temperature change experienced by the material.

So in this problem, the mass of the aluminium is given as m = 1500 g, the specific heat of the aluminium is 0.900 J/g °C. Then as it is stated that the temperature is raised by 100 °C, so the pots are heat to increase by 100 °C from its initial temperature. This means the difference in temperature will be 100°C (ΔT = 100°C).

Then, the heat energy required to raise the temperature will be

q = m*c*del T = 1500 * 0.900 * 100 = 135000 = 135 kJ

Thus, the heat energy required to raise the temperature of 1500 g of aluminium pot by 100 °C is 135 kJ.

Similarly, the mass of copper pot is given as 1500 g, the specific heat capacity of copper is 0.385 and the difference in temperature is 100  °C.

Then, the heat energy required to raise its temperature will be

q = m*c*del T = 1500 * 0.385 * 100 = 57750 = 57.75 kJ

And the heat energy required to raise the temperature of 1500 g of copper pot by 100°C is 57.75 kJ.

So, the heat energy required to raise the temperature of 1500 g of aluminium pot by 100°C is 135 kJ. And the heat energy required to raise the temperature of 1500 g of copper pot by 100 °C is 57.75 kJ.

8 0
3 years ago
under certain water​ conditions, the free chlorine​ (hypochlorous acid,​ hocl) in a swimming pool decomposes according to the la
ivolga24 [154]

Answer:

1.7 ppm

Explanation:

Original amount N' = 2.6 ppm

time to testing t = 24 hr

final amount N = 2.1 ppm

Using exponential inhibited decay, we have

N = N'e^(-kt)

Where

N is the new reading

N' is the original reading

t is the decay time

k is the decay constant

Substituting, we have

2.1 = 2.6 x e^(-k x 24)

2.1 = 2.6 x e^(-24k)

0.808 = e^(-24k)

We take the natural log of both sides of the equation

Ln 0.808 = Ln (e^(-24k))

-0.213 = - 24k

K = 0.213/24 = 0.00886

After 48 hrs, the reading of free chlorine will be

N = 2.6 x e^(-0.00886 x 48)

N = 2.6 x e^(-0.425)

N = 2.6 x 0.654

N = 1.7 ppm

5 0
3 years ago
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