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3 years ago

What relation does the boiling point of an amine have to a similar hydrocarbon?

Physics

2 answers:

den301095 [7]3 years ago

5 0

Answer:

Amine have higher boiling points than hydrocarbons.

Explanation:

Primary, secondary and tertiary amines have higher boiling points than hydrocarbons because they can engage in intermolecular hydrogen bonding.

Amines has three classes

1. Primary amines

2. Secondary amines

3. Tertiary amines

All this classes of amines have higher boiling point than hydrocarbons due to C-N bond in them

This is because amines can engage in hydrogen bonding with water, amines of low molar mass are quite soluble in water.

Amines are having higher boiling points than hydrocarbons, as C-N bond in amines is more polar than a C-C bond in hydrocarbons. Due to the polar nature of amines, it forms intermolecular H-bonds and exists as associated molecules.

GaryK [48]3 years ago

4 0

Answer:

Amones have higher boiling points than hydrocarbons.

Explanation:

Amines are derivatives of ammonia in which one, two, or even all three of its hydrogen atoms are replaced by hydrocarbon groups. Amines ate soluble in water via hydrogen bond between its molecule (=N-H group) and water molecules. Hydrocarbons on the other hand are insoluble in water (in which this solubility increases as the carbon chain increases)

Amines are having higher boiling points than hydrocarbons , as C-N bond in amines is more polar than a C-C bond in hydrocarbons. Due to the polar nature of amines it forms intermolecular H-bonds and exists as associated molecules.

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Answer:

Option b. it has the same position and the same acceleration as A

Explanation:

Let's analyze every statement:

a. it has the same position and the same velocity as A

In the instant where B passes A, they Do have the same position. Velocity however, cannot be the same because if they were, ball B would never pass ball A. So, this is false.

b. it has the same position and the same acceleration as A

As we said in the previous option, the position is the same. The acceleration is gravity for both balls, so this is true.

c. it has the same velocity and the same acceleration as A

Acceleration is the same but velocities are not, so this is false.

d. it has the same displacement and the same velocity as A

The distance they have traveled is the same, so the displacement is the same, but the velocity is not, so this is false.

e. it has the same position, displacement and velocity as A

The position and displacement is the same but not velocity, so this is false.

Only option b is true.

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3 years ago

The potential difference between two points, A and B, in an electric field is 2.00 volts. The energy required to move a charge o

Pavlova-9 [17]

Answer:

$W_A_B=-1.6\times 10^{-18} J$

Explanation:

Let A and B be two points located in a uniform electric field, A being a distance d from B in the direction of the field. The work that an external force must do to bring a unit positive charge q from the reference point to the point considered against the electric force at constant speed, mathematically is expressed by:

$V_B_A=\frac{W_A_B}{q}$

Therefore, isolating $W_A_B$ and replacing the data provided:

$W_A_B=V_B_A *q=-2*(8\times 10^{-19}) =-1.6\times 10^{-18}J$

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4 years ago

The free-fall acceleration at the surface of planet 1 is 22 m/s^2. The radius and the mass of planet 2 are twice those of planet

algol13

Answer:

g₂ = 11 m/s²

Explanation:

The value of free-fall acceleration on the surface of a planet is given by the following formula:

$g = \frac{Gm}{r^2}$

where,

g = free-fall acceleration

G = Universal Gravitational Constant

m = mass of the planet

r = radius of planet

FOR PLANET 1:

$g_1 = \frac{Gm_1}{r_1^2}\\\\\frac{Gm_1}{r_1^2} = 22 m/s^2$ --------------------- equation (1)

FOR PLANET 2:

$g_2 = \frac{Gm_2}{r_2^2}\\\\g_2 = \frac{G(2m_1)}{(2r_1)^2}\\\\g_2 = \frac{1}{2}\frac{Gm_1}{r_1^2}\\\\$

using equation (1):

$g_2 = \frac{g_1}{2}\\\\g_2 = \frac{22\ m/s^2}{2}$

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