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Azulene has dipole moment 0.8 D because

It exists as aromatic compound in which both the rings are aromatic
Charge separation permits conformational stability
The two rings are of different size
The molecules obey (4n + 2) Huckel rule

The correct answer is: It exists as aromatic compound in which both the rings are aromatic

Locate the image of the point object O in the situation shown in figure. The point C denotes the centre of curvature of the separating surface

physics-General

General

Maths-

An ellipse has the point $left parenthesis 1 comma blank minus 1 right parenthesis blank$ and $left parenthesis 2 comma blank minus 1 right parenthesis blank$ as its foci and $x plus y equals 5$ as one of its tangent then the coordinate of the point where this line touches the ellipse are

So, the point of contact of tangent to ellipse is $S o comma space t h e space p o i n t space o f space c o n t a c t space o f space tan g e n t space t o space e l l i p s e space i s space open parentheses 34 over 9 comma 11 over 9 close parentheses.$ .

An ellipse has the point $left parenthesis 1 comma blank minus 1 right parenthesis blank$ and $left parenthesis 2 comma blank minus 1 right parenthesis blank$ as its foci and $x plus y equals 5$ as one of its tangent then the coordinate of the point where this line touches the ellipse are

Maths-General

General

maths-

The arrangement of the ellipses in ascending order of their eccentricities when $open floor S B S to the power of 1 end exponent close$ is given, Where S & $S to the power of 1 end exponent$ are foci& $B$ is one end of the minor axis A) $open floor S B S to the power of 1 end exponent close equals 2 0 to the power of 0 end exponent$ B) $open floor S B S to the power of 1 end exponent close equals 6 0 to the power of 0 end exponent$ C) $open floor S B S to the power of 1 end exponent close equals 3 0 to the power of 0 end exponent$ D) $open floor S B S to the power of 1 end exponent close equals 9 0 to the power of 0 end exponent$

maths-General

General

maths-

$T h e p$ oints o $n t h e e$ llipse 3 $x to the power of 2 end exponent plus y to the power of 2 end exponent equals 37 comma$ where the normals to it are perpendicular to $6 x plus 5 y minus 2 equals 0$ are

maths-General

General

maths-

$E subscript 1 end subscript equals fraction numerator x to the power of 2 end exponent over denominator 25 end fraction plus fraction numerator y to the power of 2 end exponent over denominator 9 end fraction minus 1 equals 0 comma$ $C equals 0$ is a greatest circle incribed in $E subscript 1 end subscript$ and Greatest Ellipse $E subscript 2 end subscript$ is incribed in c = 0 If the eccentricities of $E subscript 1 end subscript comma E subscript 2 end subscript$ are equal then the minor axis of $E subscript 2 end subscript equals 0$ is

maths-General

General

physics-

Two thin lenses of focal lengths 1 f and 2 f are in contact and coaxial. The combination is equivalent to a single lens of power

physics-General

General

chemistry-

Methanol and acetone can be separated by:

chemistry-General

General

chemistry-

Decreasing order of solubility of following compounds is:

chemistry-General

General

chemistry-

Which of the following statement is correct about tropolone?

chemistry-General

General

chemistry-

In which case first has higher solubility than second?
I) Phenol, Benzene
II) Nitrobenzene, Phenol
III) o–Hydroxybenzaldehyde, p–Hydroxy benzaldehyde
IV) CH₃CHO, CH₃–O–CH₃
V) o–Nitrophenol, p–Nitrophenol
VI)

chemistry-General

General

maths-

If $alpha$ and $beta$ are the eccentric angles of the extremities of a focal chord of an ellipse, then the eccentricity of the ellipse is

maths-General

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Azulene has dipole moment 0.8 D because

It exists as aromatic compound in which both the rings are aromatic Charge separation permits conformational stability The two rings are of different size The molecules obey (4n + 2) Huckel rule

The correct answer is: It exists as aromatic compound in which both the rings are aromatic

Related Questions to study

can be differentiated by

can be differentiated by

The order in which the reagent must be used to separate the compound I - IV is

The order in which the reagent must be used to separate the compound I - IV is

can be differentiated by:

can be differentiated by:

Find the size of image formed in the situation shown in figure.

Find the size of image formed in the situation shown in figure.

Locate the image of the point object O in the situation shown in figure. The point C denotes the centre of curvature of the separating surface

Locate the image of the point object O in the situation shown in figure. The point C denotes the centre of curvature of the separating surface

An ellipse has the point and as its foci and as one of its tangent then the coordinate of the point where this line touches the ellipse are

An ellipse has the point and as its foci and as one of its tangent then the coordinate of the point where this line touches the ellipse are

The arrangement of the ellipses in ascending order of their eccentricities when is given, Where S & are foci& is one end of the minor axis A) B) C) D)

The arrangement of the ellipses in ascending order of their eccentricities when is given, Where S & are foci& is one end of the minor axis A) B) C) D)

oints ollipse 3where the normals to it are perpendicular to are

oints ollipse 3where the normals to it are perpendicular to are

is a greatest circle incribed in and Greatest Ellipse is incribed in c = 0 If the eccentricities of are equal then the minor axis of is

is a greatest circle incribed in and Greatest Ellipse is incribed in c = 0 If the eccentricities of are equal then the minor axis of is

Two thin lenses of focal lengths 1 f and 2 f are in contact and coaxial. The combination is equivalent to a single lens of power

Two thin lenses of focal lengths 1 f and 2 f are in contact and coaxial. The combination is equivalent to a single lens of power

Methanol and acetone can be separated by:

Methanol and acetone can be separated by:

Decreasing order of solubility of following compounds is:

Decreasing order of solubility of following compounds is:

Which of the following statement is correct about tropolone?

Which of the following statement is correct about tropolone?

In which case first has higher solubility than second?I) Phenol, BenzeneII) Nitrobenzene, PhenolIII) o–Hydroxybenzaldehyde, p–Hydroxy benzaldehydeIV) CH3CHO, CH3–O–CH3V) o–Nitrophenol, p–NitrophenolVI)

In which case first has higher solubility than second?I) Phenol, BenzeneII) Nitrobenzene, PhenolIII) o–Hydroxybenzaldehyde, p–Hydroxy benzaldehydeIV) CH3CHO, CH3–O–CH3V) o–Nitrophenol, p–NitrophenolVI)

If and are the eccentric angles of the extremities of a focal chord of an ellipse, then the eccentricity of the ellipse is

If and are the eccentric angles of the extremities of a focal chord of an ellipse, then the eccentricity of the ellipse is

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It exists as aromatic compound in which both the rings are aromatic
Charge separation permits conformational stability
The two rings are of different size
The molecules obey (4n + 2) Huckel rule

An ellipse has the point $left parenthesis 1 comma blank minus 1 right parenthesis blank$ and $left parenthesis 2 comma blank minus 1 right parenthesis blank$ as its foci and $x plus y equals 5$ as one of its tangent then the coordinate of the point where this line touches the ellipse are

An ellipse has the point $left parenthesis 1 comma blank minus 1 right parenthesis blank$ and $left parenthesis 2 comma blank minus 1 right parenthesis blank$ as its foci and $x plus y equals 5$ as one of its tangent then the coordinate of the point where this line touches the ellipse are

$T h e p$ oints o $n t h e e$ llipse 3 $x to the power of 2 end exponent plus y to the power of 2 end exponent equals 37 comma$ where the normals to it are perpendicular to $6 x plus 5 y minus 2 equals 0$ are

$T h e p$ oints o $n t h e e$ llipse 3 $x to the power of 2 end exponent plus y to the power of 2 end exponent equals 37 comma$ where the normals to it are perpendicular to $6 x plus 5 y minus 2 equals 0$ are

In which case first has higher solubility than second?
I) Phenol, Benzene
II) Nitrobenzene, Phenol
III) o–Hydroxybenzaldehyde, p–Hydroxy benzaldehyde
IV) CH₃CHO, CH₃–O–CH₃
V) o–Nitrophenol, p–Nitrophenol
VI)

In which case first has higher solubility than second?
I) Phenol, Benzene
II) Nitrobenzene, Phenol
III) o–Hydroxybenzaldehyde, p–Hydroxy benzaldehyde
IV) CH₃CHO, CH₃–O–CH₃
V) o–Nitrophenol, p–Nitrophenol
VI)

If $alpha$ and $beta$ are the eccentric angles of the extremities of a focal chord of an ellipse, then the eccentricity of the ellipse is

If $alpha$ and $beta$ are the eccentric angles of the extremities of a focal chord of an ellipse, then the eccentricity of the ellipse is