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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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When the frequency of an ac circuit is decreased, the current in the circuit increases. which combination of elements is most li

zaharov [31]

Answer:

The combination of elements most likely to comprise the circuit are resistor, inductor and capacitor

Explanation:

The impedance of an LCR circuit shown as

Z = √R² + (X↓l - X↓c)²

Z = √R² + (2π∨L - 1/2π∨c)²

Variation of Z with respect to υ is shown in the figure.

As υ increases, Z decreases and so the current increases.

At υ = υ↓r

Z is minimum, current is maximum. Beyond

υ = υ↓r

Z increases and so current decreases.

so the combination of circuit elements that is most suitable to comprise

the circuit is R, L and C.

To learn more about these circuits

brainly.com/question/13140756

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5 0

2 years ago

An airplane flies eastward and always accelerates at a constant rate. At one position along its path it has a velocity of 34.3 m

Tomtit [17]

Explanation:

We'll need two equations.

v² = v₀² + 2a(x - x₀)

where v is the final velocity, v₀ is the initial velocity, a is the acceleration, x is the final position, and x₀ is the initial position.

x = x₀ + ½ (v + v₀)t

where t is time.

Given:

v = 47.5 m/s

v₀ = 34.3 m/s

x - x₀ = 40100 m

Find: a and t

(47.5)² = (34.3)² + 2a(40100)

a = 0.0135 m/s²

40100 = ½ (47.5 + 34.3)t

t = 980 s

7 0

3 years ago

Do help me pleaseeeee

Anton [14]

The mass needed at peg 1 is a 5g mass.

The 15g should hang at peg 5.

The reason is force x distance clockwise is equal to force x distance anti-clockwise

8 0

3 years ago

Resonances of the ear canal lead to increased sensitivity of hearing, as we’ve seen. Dogs have a much longer ear canal—5.2 cm—th

krek1111 [17]

Answer:

B. 1700 Hz, 5100 Hz

Explanation:

Parameters given:

Length of ear canal = 5.2cm = 0.052 m

Speed of sound in warm air = 350 m/s

The ear canal is analogous to a tube that has one open end and one closed end. The frequency of standing wave modes in such a tube is given as:

f(m) = m * (v/4L)

Where m is an odd integer;

v = velocity

L = length of the tube

Hence, the two lowest frequencies at which a dog will have increased sensitivity are f(1) and f(3).

f(1) = 1 * [350/(4*0.052)]

f(1) = 1682.69 Hz

Approximately, f(1) = 1700 Hz

f(3) = 3 * [350/(4*0.052)]

f(3) = 5048 Hz

Approximately, f(3) = 5100 Hz

7 0

3 years ago

What fraction of 5 MeV α particles will be scattered through angles greater than 8.5° from a gold foil (Z = 79, density = 19.3 g

aalyn [17]

Answer:

$f(8) = \pi (5.90x10^{28} atoms/m^3) *(10^{-8} m) (\frac{2*79*(1.6x10^{-19}C)^2}{8 \pi (8.85 x10^{-12} C^2/Nm^2)*(5x10^6 Nm) *1.6x10^{-19}C})* cot^2 (8/2)$

And after do all the operations we got:

$f(8) = 1.96 x10^{-4] atoms/m^2$

And that would be the final answer for this case.

Explanation:

For this case we can use the fomrula for the fraction of incident particles scattered by an angle $\theta$ , given by:

$f(\theta) = \pi nt (\frac{Z_1 e Z_2 e}{8 \pi e_o K})^2 cos^2 (\theta/2)$

Where:

$Z_1 e$ represent the charge of the projectile (Z1=2)

$Z_2 e$ is the target charge (z2=79)

$K= 5x10^6 eV$ represent the kinetic energy of incident particle

n represent the density of target particles (we need to find it first)

$t= 10^{-8] m$ represent the thicknss of the foil

The first step would be calculate the density of target particles with the following formula:

$n =\frac{\rho N_A N_M}{M_g}$

Where:

$\rho = 19.3 g/m^3= 19300 Kg/m^3$

$N_A = 6.022 x10^{23} molecules/mol$ the Avogadro's number

$N_M = 1$ represent the atoms per molecule

$M_g = 197 g/mol = 0.197 Kg/mol$ represent the molecular weigth

If we replace we got:

$n = \frac{19300 kg/m^3 *6.022x10^{23} molecu/mol * 1 atom/mole}{0.197 Kg/mol}= 5.90 x106{28] atoms/m^3$

Now we can calculate the fraction of 5 MeV alpha particles that would be scatteres with angle higher than 8 degrees in a piece of thickness $t=10^{-8}m$

And using the first formula we got:

$f(8) = \pi (5.90x10^{28} atoms/m^3) *(10^{-8} m) (\frac{2*79*(1.6x10^{-19}C)^2}{8 \pi (8.85 x10^{-12} C^2/Nm^2)*(5x10^6 Nm) *1.6x10^{-19}C})* cot^2 (8/2)$

And after do all the operations we got:

$f(8) = 1.96 x10^{-4] atoms/m^2$

And that would be the final answer for this case.

4 0

3 years ago