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2 years ago

What product is obtained from the aldol condensation of cyclohexanone?

Physics

1 answer:

Evgesh-ka [11]2 years ago

6 0

Answer:

First product is FCH-OH chemically known as 2-[2-furyl(hydroxyl)methyl]-Second product is FCH i.e (2E)-2-[2-furyl-methylene]-cyclohexanone

Explanation:

Please see the attached image for complete chemical reaction of aldol condensation of cyclohexanone

Aldol Condensation is a form of electrophilic substitution reaction in which the alpha carbon in enols or enolate anions is substituted by an electrophile to form carbon-carbon bond. Cyclohexanone also known as the first ketone consists of two alpha-carbons and four potential substitutions i.e alpha-hydrogens but none of the hydrogen on the ring is substituted. Ketones such as cyclohexanone are much more acidic than their parent hydrocarbon.

First product is FCH-OH chemically known as 2-[2-furyl(hydroxyl)methyl]-cyclohexanone that further undergoes dehydration resulting into FCH i.e (2E)-2-[2-furyl-methylene]-cyclohexanone

Based on the explanations above, the compound formed is shown in the image.

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What is cumulative force?

Nadusha1986 [10]

Increasing over time

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2 years ago

Work done can be describe as?

Neko [114]

Explanation:

In physics, work is the energy transferred to or from an object via the application of force along a displacement. In its simplest form, it is often represented as the product of force and displacement. A force is said to do positive work if (when applied) it has a component in the direction of the displacement of the point of application. A force does negative work if it has a component opposite to the direction of the displacement at the point of application of the force.

Quick Facts: Common symbols, SI unit ...

Work

A baseball pitcher does positive work on the ball by applying a force to it over the distance it moves while in his grip.

Common symbols

SI unit

joule (J)

Other units

Foot-pound, Erg

In SI base units

1 kg⋅m2⋅s−2

Derivations from

other quantities

W = F ⋅ s

W = τ θ

Dimension

M L2 T−2

For example, when a ball is held above the ground and then dropped, the work done by the gravitational force on the ball as it falls is equal to the weight of the ball (a force) multiplied by the distance to the ground (a displacement). When the force F is constant and the angle between the force and the displacement s is θ, then the work done is given by:

{\displaystyle W=Fs\cos {\theta }}{\displaystyle W=Fs\cos {\theta }}

Work is a scalar quantity, so it has only magnitude and no direction. Work transfers energy from one place to another, or one form to another. The SI unit of work is the joule (J), the same unit as for energy.

4 0

2 years ago

A uniform rod of mass 3.30×10−2 kg and length 0.450 m rotates in a horizontal plane about a fixed axis through its center and pe

Alex73 [517]

(a) 2.75 rev/min

The moment of inertia of the rod rotating about its center is:

$I_R=\frac{1}{12}ML^2$

where

$M=3.30\cdot 10^{-2} kg$ is its mass

L = 0.450 m is its length

Substituting,

$I_R=\frac{1}{12}(3.30\cdot 10^{-2})(0.450)^2=5.57\cdot 10^{-4} kg m^2$

The moment of inertia of the two rings at the beginning is

$I_r = 2mr^2$

where

m = 0.200 kg is the mass of each ring

$r=5.20\cdot 10^{-2} m$ is their distance from the center of the rod

Substituting,

$I_r=2(0.200)(5.20\cdot 10^{-2})^2=1.08\cdot 10^{-3} kg m^2$

So the total moment of inertia at the beginning is

$I_1=I_R+I_r = 5.57\cdot 10^{-4}+1.08\cdot 10^{-3}=1.64\cdot 10^{-3}kg m^2$

The initial angular velocity of the system is

$\omega_1 = 35.0 rev/min$

The angular momentum must be conserved, so we can write:

$L=I_1 \omega_1 = I_2 \omega_2$ (1)

where $I_2$ is the moment of inertia when the rings reach the end of the rod; in this case, the distance of the ring from the center is

$r=\frac{0.450 m}{2}=0.225 m$

so the moment of inertia of the rings is

$I_r=2(0.200)(0.225)^2=0.0203 kg m^2$

and the total moment of inertia is

$I_2 = I_R + I_r =5.57\cdot 10^{-4} + 0.0203 = 0.0209 kg m^2$

Substituting into (1), we find the final angular speed:

$\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(1.64\cdot 10^{-3})(35.0)}{0.0209}=2.75 rev/min$

(b) 103.0 rev/min

When the rings leave the rod, the total moment of inertia is just equal to the moment of inertia of the rod, so:

$I_2 = I_R = 5.57\cdot 10^{-4}kg m^2$

So using again equation of conservation of the angular momentum:

$L=I_1 \omega_1 = I_2 \omega_2$

We find the new final angular speed:

$\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(1.64\cdot 10^{-3})(35.0)}{5.57\cdot 10^{-4}}=103.0 rev/min$

7 0

2 years ago

Mike walks 200 km in 6 hours.he then walks another 100km in 4 hours .what is his average speed?

kolbaska11 [484]

Average speed is defined as the ratio of total distance covered in total given time

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

here we know that total distance that man moved is

$d_1 = 200 km$

$d_2 = 100 km$

so total distance is

$d = d_1 + d_2$

$d = 200 km + 100 km$

$d = 300 km$

now here total time of the motion is

$t_1 = 6 hours$

$t_2 = 4 hours$

total time will be given as

$t = t_1 + t_2$

$t = 6 + 4 = 10 hours$

now by above formula

$v_{avg} = \frac{300}{10}$

$v_{avg} = 30 km/h$

so his average speed is 30 km/h

8 0

3 years ago

The Event Horizon Telescope needs a 22 micro-arcsecond resolution to view the event horizon regions around black holes. If the a

likoan [24]

Answer:

14869817.395 m

Explanation:

$\theta$ =22 microarcsecond

λ = Wavelength = 1.3 mm

Converting to radians we get

$22\times 10^{-6}\frac{\pi}{180\times 3600}\ radians$

From Rayleigh Criterion

$\theta=1.22\frac{\lambda}{D}\\\Rightarrow D=1.22\frac{\lambda}{\theta}\\\Rightarrow D=1.22\frac{1.3\times 10^{-3}}{22\times 10^{-6}\frac{\pi}{180\times 3600}}\\\Rightarrow D=14869817.395\ m$

Diameter of the effective primary objective is 14869817.395 m

It is not possible to build one telescope with a diameter of 14869817.395 m. But, we need this type of telescope. So, astronomers use an array of radio telescopes to achieve a virtual diameter in order to observe objects that are the size of supermassive black hole's event horizon.

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2 years ago