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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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What is the gauge pressure of the water right at the point p, where the needle meets the wider chamber of the syringe? neglect t

Helen [10]

Missing details: figure of the problem is attached.

We can solve the exercise by using Poiseuille's law. It says that, for a fluid in laminar flow inside a closed pipe,

$\Delta P = \frac{8 \mu L Q}{\pi r^4}$

where:

$\Delta P$ is the pressure difference between the two ends

$\mu$ is viscosity of the fluid

L is the length of the pipe

$Q=Av$ is the volumetric flow rate, with $A=\pi r^2$ being the section of the tube and $v$ the velocity of the fluid

r is the radius of the pipe.

We can apply this law to the needle, and then calculating the pressure difference between point P and the end of the needle. For our problem, we have:

$\mu=0.001 Pa/s$ is the dynamic water viscosity at $20^{\circ}$

L=4.0 cm=0.04 m

$Q=Av=\pi r^2 v= \pi (1 \cdot 10^{-3}m)^2 \cdot 10 m/s =3.14 \cdot 10^{-5} m^3/s$

and r=1 mm=0.001 m

Using these data in the formula, we get:

$\Delta P = 3200 Pa$

However, this is the pressure difference between point P and the end of the needle. But the end of the needle is at atmosphere pressure, and therefore the gauge pressure (which has zero-reference against atmosphere pressure) at point P is exactly 3200 Pa.

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3 years ago