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3 years ago

What is the relationship between dipole moment and boiling point?

1 answer:

3 0

The dipole moment of a molecule and its boiling point are directly varied. If the dipole moment is large, the attraction between the positive and negative charges of a molecule is strong thus, requiring stronger forces to break them. Hence, they will have higher boiling points.

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Arsenic diffusion in Si: Arsenic is diffused into a thick slice of silicon with no previous arsenic in it at 1100ºC. If the surf

Rainbow [258]

Answer:

Diffusion time is 7.42 h

Solution:

As per the question:

Temperature, T = $1100^{\circ}C$

Surface concentration of arsenic, $C_{S} = 5.0\times 10^{18}\ atoms/cm^{3}$

Surface concentration below Silicon surface, $C_{x} = 1.5\times 10^{16}\ atoms/cm^{3}$

D = $3.0\times 10^{- 14}\ cm^{2}/s$

x = $1.2\mu m = 1.2\times 10^{- 4}\ cm$

Initial concentration at t = 0, $C_{o} = 0$

Now, by using Flick's second eqn:

$\frac{C_{S} - C_{x}}{C_{x} - C_{o}} = erf(\frac{x}{\sqrt{Dt}})$

Thus by putting appropriate values:

$\frac{5.0\times 10^{18} - 1.5\times 10^{16}}{5.0\times 10^{18}} = erf(\frac{1.2\times 10^{- 4}}{2\sqrt{3.0\times 10^{- 14}t}})$

$0.997 = erf(\frac{364.4}{\sqrt{t}})$ (1)

Now,

$erf(z) = 0.997$

Now, from error function values tabulation:

For z = 2.0, erf(z) = 0.998

For z = 2.2, erf(z) = 0.995

Now,

With the help of linear interpolation method:

$\frac{z - 2}{2.2 - 2.0} = \frac{0.997 - 0.995}{0.998 - 0.995}$

z = 2.12

Now, using eqn (1) and above value:

$\frac{364.4}{\sqrt{t}} = 2.12$

$t = (\frac{364.4}{2.12})^{2}$ = 26700 s

t = $\frac{26700}{3600} = 7.42\ h$

7 0

3 years ago

small plastic container, called the coolant reservoir, catches the radiator fluid that overflowswhen the automobile engine becom

Ilia_Sergeevich [38]

Answer:

0.53 quart

Explanation:

The volume expansion of the coolant is gotten from ΔV = VγΔθ where ΔV = change in volume of the coolant, V = initial volume of coolant = 15 quart, γ = coefficient of volume expansion of coolant = 410 × 10⁻⁶ /°C and Δθ = temperature change = θ₂ - θ₁ where θ₁ = initial temperature of coolant = 6 °C and θ₂ = final temperature of coolant = 92 °C. So, Δθ = θ₂ - θ₁ = 92 °C - 6 °C = 86 °C

Since, ΔV = VγΔθ

substituting the values of the variables into the equation, we have

ΔV = VγΔθ

ΔV = 15 × 410 × 10⁻⁶ /°C × 86 °C

ΔV = 528900 × 10⁻⁶ quart

ΔV = 0.528900 quart

ΔV ≅ 0.53 quart

Since the change in volume of the coolant equals the spill over volume, thus the overflow from the radiator will spill into the reservoir when the coolant reaches its operating temperature of 92 °C is 0.53 quart.

4 0

3 years ago

A vehicle travels from a 30m marker to a 100m marker. What is the change in distance?

sesenic [268]

The change in distance is 30 because if you subtract both number you'll get 30

3 0

4 years ago

Read 2 more answers

You make a u turn in your car, what provides the centripetal force on the car and on you

grin007 [14]

Newton's first and third laws

8 0