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Dahasolnce [82]
2 years ago
6

Tag all the carbon atoms with pi bonds in this molecule. If there are none, please check the box.

Chemistry
1 answer:
snow_tiger [21]2 years ago
4 0

Answer:

Pi bonds (π bonds) are covalent chemical bonds where two lobes of an orbital involved in the bond overlap with two lobes of the other orbital involved. These orbitals share a nodal plane that passes through the nuclei involved. Are generally weaker than sigma links, because their negatively charged electronic density is further from the positive charge of the atomic nucleus, which requires more energy.

They are frequent components of multiple bonds, as is the molecule indicated in our exercise.

The characteristics that distinguish pi bonds from other kinds of interactions between atomic species are described below, beginning with the fact that this union does not allow the free rotation movement of atoms, such as carbon. For this reason, if there is rotation of the atoms, the bond is broken.

Explanation:

In order to describe the formation of the pi bond, first we must talk about the hybridization process, as this is involved in some important links.

Hybridization is a process where hybrid electronic orbitals are formed; that is, where orbitals of atomic sub-levels s and p can get mixed. This causes the formation of sp, sp2 and sp3 orbitals, which are called hybrids.

In this sense, the formation of pi bonds occurs thanks to the overlapping of a pair of lobes belonging to an atomic orbital over another pair of lobes that are in an orbital that is part of another atom.

This orbital overlap occurs laterally, so the electronic distribution is mostly concentrated above and below the plane formed by the linked atomic nuclei, and causes the pi bonds to be weaker than the sigma bonds.

When talking about the orbital symmetry of this type of junction, it should be mentioned that it is equal to that of the p-type orbitals as long as it is observed through the axis formed by the bond. In addition, these junctions are mostly made up of p orbitals.

Since pi bonds are always accompanied by one or two more links (one sigma or another pi and one sigma), it is relevant to know that the double bond that is formed between two carbon atoms has less bond energy than that corresponding to two Sometimes the sigma link between them.

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you push a friend sitting on a swing. she has a mass of 50kg and accelerates at a rate of 4 m/s2. find the force you exerted
Hoochie [10]
Mass of the friend sitting on the swing = 50 kg
Rate of acceleration = 4 m/s^2
The above information's are already given in the question. We need to utilize these information's in such a manner that the answer to the question can be determined easily.
We already know that
Force = Mass * acceleration.
This formula has to be used to find the force exerted.
Then
Force exerted = Mass of the friend * Acceleration
                       = (50 * 4) kg m/s^2
                       = 200 Newton
So the force exerted by me is 200 Newton. I hope the explanation and the procedure is clear to you and you are satisfied with it.
6 0
2 years ago
Read 2 more answers
A gas sample occupies 3.50 liters of volume at 20.°c. what volume will this gas occupy at 100.°c (reported to three significant
VLD [36.1K]

Explanation:

According to Charle's law, at constant pressure the volume of an ideal gas is directly proportional to the temperature.

That is,             Volume \propto Temperature

Hence, it is given that V_{1} is 3.50 liters, T_{1} is 20 degree celsius, and T_{2} is 100 degree celsius.

Therefore, calculate V_{2} as follows.

                           \frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}

                           \frac{3.50 liter}{20^{o}C} = \frac{V_{2}}{100^{o}C}

                                V_{2} = 17.5 liter

Thus, we can conclude that volume of gas required at 100 degree celsius is 17.5 liter.

6 0
2 years ago
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Predict what will observe in below mention experiment.
Sloan [31]

Predict what will be observed in each experiment below. Rock candy is formed when excess sugar is dissolved in hot water followed by crystallization. A student wants to make two batches of rock candy. He finds an unopened box of "cane sugar" in the pantry. He starts preparing batch A by dissolving sugar in 500 mL of hot water (70 degree C). He keeps adding sugar until no more sugar dissolves in the hot water. He cools the solution to room temperature. He prepares batch B by dissolving sugar in 500 mL of water at room temperature until no more sugar is dissolved. He lets the solution sit at room temperature

a. It is likely that more rock candy will be formed in batch A.

b. It is likely that less rock candy will be formed in batch A.

c. It is likely that no rock candy will be formed in either batch.

d. I need more information to predict which batch is more likely to form rock candy.

Answer: Option A

Explanation:

More rock candy will be formed in the batch A because it is dissolved in hot water and less rock candy will be formed in batch B because the water is not hot.

Formation of the candies require hot water as the solubility of sugar is more in hot water as compared to normal water.

The sugar will be dissolved in water until the time all the space is filled sugar molecules.

Hence, the correct answer is Option A.

6 0
2 years ago
The small intestine shown below contains lymphatic vessels but no capillaries. Which nutrient will be absorbed by this small int
Bumek [7]

Answer:

fatty acids

Explanation:

8 0
1 year ago
Read 2 more answers
(A) A chemical reaction takes place in a container of cross-sectional area 100 cm2. As a result of the reaction, a piston is pus
balandron [24]

Answer:

(A) The work done by the system is -101.325J

(B) The workdone by the system is -90.75J

Explanation:

(A) Workdone = -PΔV

Given that A = 100cm2 = 0.01m2

distance d = 10cm = 0.1m

ΔV= Area × distance

ΔV= 0.01 ×0.1

ΔV = 0.001m3

P= external pressure = 1atm = 101325Pa

Workdone = -0.001 × 101325

W= - 101.325Pa m3

1Pam3 = 1J

Therefore W = - 101.325J

The work done on the system is -101.325J

(B) Workdone = -PΔV

Given that A = 50cm2 = 0.005m2

distance d = 15cm = 0.15m

ΔV= Area × distance

ΔV= 0.005×0.15

ΔV = 0.00075m3

P=121kPa = 121000Pa

W= - 121000 × 0.00075

W= -90.75Pa m3

1Pam3 = 1J

W = - 90.75J

The woekdone by the system is -90.75J

5 0
2 years ago
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