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4 years ago

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

Physics

2 answers:

den301095 [7]4 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]4 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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An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 1.60 mm. If a

nata0808 [166]

Answer:

Explanation:

(a) For the calculation of the Electric field we use

$E=\frac{V}{d}=\frac{15.0V}{1.6*10^{-3}m} =9375\frac{N}{C}$

(b) The capacitance is calculate by using the expression

$C=\frac{\epsilon_{0}A}{d}=\frac{8.85*10^{-12}C^{2}/(Nm^{2})*(7.6*10^{-4}m^{2})}{1.6*10^{-3}m}=4.2*10^{-12}C$

(c) Finally, the charge on each plate is

$Q=CV=(4.2*10^{-12}C)(15V)=6.3*10^{-11}C$

I hope this is useful for you

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4 0

3 years ago

Read 2 more answers

A proton initially moves left to right along the x-axis at a speed of 2.00 x 103 m/s. It moves into an uniform electric field, w

FinnZ [79.3K]

Answer:

E = 1.04*10⁻¹ N/C

Explanation:

Assuming no other forces acting on the proton than the electric field, as this is uniform, we can calculate the acceleration of the proton, with the following kinematic equation:

$vf^{2} -vo^{2} = 2*a*x$

As the proton is coming at rest after travelling 0.200 m to the right, vf = 0, and x = 0.200 m.

Replacing this values in the equation above, we can solve for a, as follows:

$a = \frac{vo^{2}*mp}{2*x} = \frac{(2.00e3m/s)^{2}}{2*0.2m} = 1e7 m/s2$

According to Newton´s 2nd Law, and applying the definition of an electric field, we can say the following:

F = mp*a = q*E

For a proton, we have the following values:

mp = 1.67*10⁻²⁷ kg

q = e = 1.6*10⁻¹⁹ C

So, we can solve for E (in magnitude) , as follows:

$E = \frac{mp*a}{e} =\frac{1.67e-27kg*1e7m/s2}{1.6e-19C} = 1.04e-1 N/C$

⇒ E = 1.04*10⁻¹ N/C

5 0

3 years ago

Two blocks of weight 3.0N and 7.0N are connected by a massless string and slide down a 30 degree inclined plane. The coefficient

Luba_88 [7]

Answer:

a. a = $6.41 m/s^2$

b. T = -0.81 N

Explanation:

Given,

weight of the lighter block = $w_1\ =\ 3.0\ N$
weight of the heavier block = $w_2\ =\ 7.0\ N$
inclination angle = $\theta\ =\ 30^o$
coefficient of kinetic friction between the lighter block and the surface = $\mu_1\ =\ 0.13$

coefficient of kinetic friction between the heavier block and the surface = $\mu_2\ =\ 0.31$
friction force on the lighter block = $f_1\ =\ \mu_1N_1\ =\ \mu_1 w_1cos\theta$
friction force on the heavier block = $f_2\ =\ \mu_2N_2\ =\ \mu_2w_2cos\theta$

Let 'a' be the acceleration of the blocks and 'T' be the tension in the string.

From the f.b.d. of the lighter block,

$w_1sin\theta\ -\ T\ -\ f_1\ =\ \dfrac{w_1a}{g}\\\Rightarrow T\ =\ w_1sin\theta\ -\ \dfrac{w_1a}{g}\ -\ f_1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (1)$

From the f.b.d. of the heavier block,

$w_2sin\theta\ +\ T\ -\ f_2\ =\ \dfrac{w_2a}{g}\\\Rightarrow T\ =\ \dfrac{w_2a}{g}\ -\ w_2sin\theta\ -\ f_2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (2)$

From eqn (1) and (2), we get,

$w_1sin\theta\ -\ \dfrac{w_1a}{g}\ -\ f_1\ =\ \dfrac{w_2a}{g}\ -\ w_2sin\theta\ -\ f_2\\\Rightarrow w_1gsin\theta \ -\ w_1a\ -\ \mu_1w_1gcos\theta\ =\ w_2a\ -\ w_2gsin\theta\ -\ \mu_2 w_2gcos\theta\\$

$\Rightarrow a\ =\ \dfrac{g(w_1sin\theta\ -\ \mu_1w_1cos\theta\ +\ w_2sin\theta\ +\ \mu_2w_2cos\theta)}{w_1\ +\ w_2}\\\Rightarrow a\ =\ \dfrac{g[sin\theta(w_1\ +\ w_2)\ +\ cos\theta(\mu_2w_2\ -\ \mu_1w_1)]}{w_1\ +\ w_2}\\$

$\Rightarrow a\ =\ \dfrac{9.81\times [sin30^o\times (3.0\ +\ 7.0)\ +\ cos30^o\times (0.31\times 7.0\ -\ 0.13\times 3.0)]}{3.0\ +\ 7.0}\\\Rightarrow a\ =\ 6.4\ m/s.$

part (b)

From the eqn (2), we get, $T\ =\ \dfrac{w_2a}{g}\ -\ w_2sin\theta\ -\ \mu_2w_2cos\theta\\\Rightarrow T\ =\ \dfrac{7.0\times 6.41}{9.81}\ -\ 7.0\times sin30^o\ -\ 0.31\times 7.0\times cos30^o\\\Rightarrow T\ =\ -0.81\ N$