Crude oil is a mixture of many different components. The extraction of crude oil from the Earth is important, but its refinement into different substances is a key piece to obtaining as many uses as possible from the crude oil. Using the diagram, justify the source of data used to develop the technology for refining the crude oil.


Answer 1

Crude oil is a mixture of nitrogen, oxygen, sulphur, and hydrogen components

What is Crude oil?

A combination of hydrocarbons known as crude oil is one that is found in naturally occurring subsurface reservoirs in the liquid phase and continues to be liquid at atmospheric pressure after passing through surface separation equipment.

Refineries transform crude oil into useful products including gasoline, diesel, and aviation fuels for transportation. Gasoline: A fuel used in both personal and commercial vehicles that are made for internal combustion engines.

In addition to some nitrogen, sulphur, and oxygen, crude oil is a combination of very flammable liquid hydrocarbons (compounds mostly made of hydrogen and carbon).

More about the Crude oil link is given below.


Answer 2

Answer: Different fuel components boil at different temperatures, allowing them to be separated.


Related Questions

A sinusoidal wave is travelling on a string under tension T = 8.0(N), having a mass per unit length of 1 = 0.0128(kg/m). It’s displacement function is D(x,t) = Acos(kx - t). It’s amplitude is 0.001m and its wavelength is 0.8m. It reaches the end of this string, and continues on to a string with 2 = 0.0512(kg/m) and the same tension as the first string. Give the values of A, k, and , for the original wave, as well as k and  the reflected wave and the transmitted wave.
Two kids are playing on a newly installed slide, which is 3 m long. John, whose mass is 30 kg, slides down into William (20 kg), who is sitting at the very bottom end, and whom he holds onto when he arrives. Laughing, John & William leave the slide horizontally and land in the muddy ground near the foot of the slide. (A) If John starts out 1.8 m above William, and the slide is essentially frictionless, how fast are they going when they leave the slide? (B) Thanks to the mud he acquired, John will now experience an average frictional force of 105 N as he slides down. How much slower is he going when he reaches the bottom than when friction was absent?
You have a grindstone (a disk) that is 105.00 kg, has a 0.297-m radius, and is turning at 71.150 rpm, and you press a steel axe against it with a radial force of 46.650 N. Assuming the kinetic coefficient of friction between steel and stone is 0.451. How many turns will the stone make before coming to rest?
A typical atomic polarizability is 1 × 10-40 (C·m)/(N/C). If the q in p = qs is equal to the proton charge e, what charge separation s could you produce in a typical atom by applying
A tuning fork vibrates at 15,660 oscillations every minute. What is the period (in seconds) of one back and forth vibration of the tuning fork?

The volume control on a stereo is designed so that three clicks of the dial increase the output by 10 dB. How many clicks are required to increase the power output of the loudspeakers by a factor of 100?



300 clicks...


Output on 3 clicks = 10 dB

Increasing 10 by a factor of 100 equals 1000 dB so,

Its simple math, clicks will also increase in the same ratio and it shall take 300 clicks to increase the volume by a factor of 100.

What is the average kinetic energy of hydrogen atoms on the 5500°C surface of the sun?


Answer: The average kinetic energy of hydrogen atoms is 1.19562* 10^(-19)J


To calculate the average kinetic energy of the atom, we use the equation:



K = average kinetic energy

k = Boltzmann constant = 1.3807* 10^(-23)J/K

T = temperature = 5500^oC=[5500+273]K=5773K

Putting values in above equation, we get:

K=(3)/(2)* 1.3807* 10^(-23)J/K* 5773K\n\nK=1.19562* 10^(-19)J

Hence, the average kinetic energy of hydrogen atoms is 1.19562* 10^(-19)J

A can of soup has a mass of 0.35 kg. The can is moved from a shelf that is 1.2 m off the ground to a shelf that is 0.40m off the ground. How does the gravitational potential energy of the can change?



2.744 difference


Use Pe=mgh

So when the soup is at a height of 1.2m, its Pe is (.35kg)(9.8m/s^(2))(1.2m)=4.116

when the soup is at a height of .40m, its Pe is (.35kg)(9.8m/s^(2))(.40m)=1.372

So youre looking at a 2.744 difference in pe

Ezra (m = 20.0 kg) has a tire swing and wants to swing as high as possible. He thinks that his best option is to run as fast as he can and jump onto the tire at full speed. The tire has a mass of 10.0 kg and hangs 3.50 m straight down from a tree branch. Ezra stands back 10.0 m and accelerates to a speed of 3.62 m/s before jumping onto the tire swing. (a) How fast are Ezra and the tire moving immediately after he jumps onto the swing? m/s (b) How high does the tire travel above its initial height?



a) v=5.6725\,m.s^(-1)

b) h= 1.6420\,m



  • mass of the body, M=20\,kg
  • mass of the tyre,m=10\,kg
  • length of hanging of tyre, l=3.5m
  • distance run by the body, d=10m
  • acceleration of the body, a=3.62m.s^(-2)


Using the equation of motion :



v=final velocity of the body

u=initial velocity of the body

here, since the body starts from rest state:


putting the values in eq. (1)

v^2=0^2+2* 3.62 * 10


Now, the momentum of the body just before the jump onto the tyre will be:


p=20* 8.5088


Now using the conservation on momentum, the momentum just before climbing on the tyre will be equal to the momentum just after climbing on it.

(M+m)* v'=p

(20+10)* v'=170.1764



Now, from the case of a swinging pendulum we know that the kinetic energy which is maximum at the vertical position of the pendulum gets completely converted into the potential energy at the maximum height.


(1)/(2) (M+m).v'^2=(M+m).g.h

(1)/(2) (20+10)* 5.6725^2=(20+10)* 9.8* h

h\approx 1.6420\,m

above the normal hanging position.

The concentration of Biochemical Oxygen Demand (BOD) in a river just downstream of a wastewater treatment plant’s effluent pipe is 75 mg/L. If the BOD is destroyed through a first-order reaction with a rate constant equal to 0.05/day, what is the BOD concentration 50 km downstream? The velocity of the river is 15 km/day.



The BOD concentration 50 km downstream when the velocity of the river is 15 km/day is 63.5 mg/L


Let the initial concentration of the BOD = C₀

Concentration of BOD at any time or point = C

dC/dt = - KC

∫ dC/C = -k ∫ dt

Integrating the left hand side from C₀ to C and the right hand side from 0 to t

In (C/C₀) = -kt + b (b = constant of integration)

At t = 0, C = C₀

In 1 = 0 + b

b = 0

In (C/C₀) = - kt

(C/C₀) = e⁻ᵏᵗ

C = C₀ e⁻ᵏᵗ

C₀ = 75 mg/L

k = 0.05 /day

C = 75 e⁻⁰•⁰⁵ᵗ

So, we need the BOD concentration 50 km downstream when the velocity of the river is 15 km/day

We calculate how many days it takes the river to reach 50 km downstream

Velocity = (displacement/time)

15 = 50/t

t = 50/15 = 3.3333 days

So, we need the C that corresponds to t = 3.3333 days

C = 75 e⁻⁰•⁰⁵ᵗ

0.05 t = 0.05 × 3.333 = 0.167

C = 75 e⁻⁰•¹⁶⁷

C = 63.5 mg/L

Final answer:

The BOD concentration 50 km downstream from the wastewater treatment plant is approximately 15.865 mg/L.


To calculate the BOD concentration 50 km downstream, we need to consider the rate of dilution due to the flow of the river and the first-order reaction that destroys BOD. The concentration of BOD downstream can be calculated using the equation C2 = C1 * exp(-k * d/v), where C1 is the initial concentration, k is the rate constant, d is the distance, and v is the velocity of the river.

Plugging in the given values, we have C2 = 75 * exp(-0.05 * 50/15), which gives us a BOD concentration of approximately 15.865 mg/L 50 km downstream from the wastewater treatment plant.

Learn more about downstream here:


The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the


Complete question:

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.


The exit velocity is 629.41 m/s



initial temperature, T₁ = 1200K

initial pressure, P₁ = 150 kPa

final pressure, P₂ = 80 kPa

specific heat at 300 K, Cp = 1004 J/kgK

k = 1.4

Calculate final temperature;

T_2 = T_1((P_2)/(P_1))^{(k-1 )/(k)

k = 1.4

T_2 = T_1((P_2)/(P_1))^{(k-1 )/(k)}\n\nT_2 = 1200((80)/(150))^{(1.4-1 )/(1.4)}\n\nT_2 = 1002.714K

Work done is given as;

W = (1)/(2) *m*(v_i^2 - v_e^2)

inlet velocity is negligible;

v_e = \sqrt{(2W)/(m) } = √(2*C_p(T_1-T_2)) \n\nv_e = √(2*1004(1200-1002.714))\n\nv_e = √(396150.288) \n\nv_e = 629.41  \ m/s

Therefore, the exit velocity is 629.41 m/s