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4 years ago

What are the most common petrochemical products and their uses?

1 answer:

7 0

Petrochemicals are separated into 3 groups: Olefins, Aromatics, and Synthesis Gas.

Olefins: Ethelyne (Plastic Products), Propelyne (Plastic Products), and Butadiene (Synthetic Rubber)
Aromatics: Benzene (Dyes), Toluene (Polyurethane), and Xylenes (Synthetic Fibers)
Synthesis Gas: Ammonia (Fertilizer Urea) and Methanol (Chemical intermediate)

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A horse was 6,200 meters north of a barn. For 22.5 minutes, the horse ran toward the barn at 152 meters per minute. In that time

lutik1710 [3]

3420‬m

Explanation:

Given parameters:

Length of path = 6200m

time taken for the horse to run = 22.5min

speed = 152meters/minute

Unknown:

Distance run by the horse = ?

Solution:

Speed is the rate of change of distance with time.

Speed = $\frac{distance}{time}$

Distance = speed x time

Distance = 152 x 22.5 = 3420‬m

learn more:

Speed brainly.com/question/5424148

#learnwithBrainly

6 0

4 years ago

Which statement correctly describes an atom of the element helium

Assoli18 [71]

Is this a multiple choice? Anyways I will just give you a written answer and hopefully it helps.

A: Helium is made up of two electrons held by electromagnetic force to two protons that are inside of a nucleus along with one or two neutrons.

5 0

4 years ago

Draw structures for all constitutional isomers with the molecular

ycow [4]

Answer:

(a)

Cyclobutane.
Methylcyclopropane.

(b)

1-Butene.
2-Butene (two spatial isomers.)
2-Methyl-1-propene.

Refer to the diagram attached for the structures.

Explanation:

<h3>Number of rings and double bonds in each molecule</h3>

Calculate the degree of unsaturation to find the number of rings and multiple bonds (including double bonds) in those isomer molecules. The presence of each ring and each double bond would add one to the degree of unsaturation of that molecule.

Let $C$ , $N$ , $X$ , and $H$ denote the number of carbon, nitrogen, halogen, and hydrogen atoms in each isomer molecule, respectively.

The degree of unsaturation of one such molecule would be:

$\displaystyle \frac{2\, C + 2 + N - X - H}{2}$ .

For $\rm C_4 H_8$ :

$C = 4$ (four carbon atoms in each molecule.)
$N = 0$ (no nitrogen atom.)
$X = 0$ (no halogen atom.)
$H = 8$ (eight hydrogen atoms in each molecule.)

Therefore, the degree of unsaturation of a molecule with the chemical formula $\rm C_4 H_8$ would be:

$\begin{aligned}&\text{degree of unsaturation} \\ &= \frac{2\, C + 2 + N - X - H}{2} \\ &= \frac{2\times 4 + 2 + 0 - 8}{2} = 1\end{aligned}$ .

In other words, the degree of unsaturation is $1$ for all isomers with this particular chemical formula. Therefore, each isomer would contain either a ring or a double bond, but not both.

If there is no double bond in one such molecule, there must be exactly one ring per molecule.

The minimum number of carbon atoms in a ring is three. With four carbon atoms in each molecule, there would be either a three-member ring (with one methyl group attached) or a four-member ring.

Four-membered ring with single bonds only: cyclobutane.
Three-membered ring with single bonds only and one methyl group attached: methylcyclopropane.

If there is one double bond in one such isomer molecule, there would not be a ring. Because of the degree of unsaturation, there would be no more than one double bond in each of these molecules.

With four carbon atoms, there are two possible backbones to consider: backbones with three carbon atoms each, and backbones with four carbon atoms each.

The backbone of those molecules might contain three carbon atoms. There would be one double bond in the backbone and one methyl group attached to the carbon atom at the center of the backbone. The corresponding isomer molecule would be 2-methyl-1-propene.

Alternatively, the backbone of those molecules might contain four carbon atoms. There are two possible locations for the double bond:

Between the first and the second carbon atoms: 1-Butene.
Between the second and third carbon atoms: 2-Butene.

Notice, that the name 2-Butene refers to two distinct spatial isomers. Unlike carbon-carbon single bonds, groups on the two sides of a carbon-carbon double bond are unable to rotate relative to one another. Therefore, the first and fourth carbon atoms would either be:

on the same side of the $120^\circ$ double bond: (2Z)-2-Butene, or
on opposite sides of that double bond: (2E)-2-Butene.

6 0

3 years ago

1. All atoms of the same element have the same

zalisa [80]

Answer:

All atoms of the same element have the same number of protons

Explanation:

All atoms of the same element have the same number of protons. The atoms are formed by protons, neutrons (both located in the nucleus) and electrons located on the outside. The number Z characterizes each element (corresponds to the number of protons located in the nucleus, which coincides with the number of electrons). The number A corresponds to the sum of protons and neutrons of an element (it varies in isotopes, since the number of neutrons varies).

8 0

3 years ago

5. The maximum depth dmax that a diver can snorkel is set by the density of the water and the fact that human lungs can function

Andreyy89

Answer:

The $d_{max}$ value is 17.2 cm.