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uysha [10]
3 years ago
15

Advantages and disadvantages of extracting petroleum products from oil sands

Chemistry
1 answer:
Brut [27]3 years ago
6 0
<span> Advantages
- large resource of potential oil
- biggest resource is in Canada, whereas the largest reserves of crude oil are in Saudi Arabia and Venezuela
</span><span><span>- It requires less processing than oil shale and presents fewer problems than oil shale
- Processing is mainly by hot water.</span>- Exploitation has only started recently so low cost reserves are still available.



Disadvantages
- Most of the resource is uneconomic at present oil prices - so exploitable reserves are less than for crude oil
- Oil sands require mining, which is more complex and costly than drilling and pumping crude oil.
- Processing requires more energy than for crude oil.
- Resources are less than for oil shale - which is available in very large reserves in the US and Australia.</span>
<span>- Refining is more difficult because the product is in the form of bitumen - requiring cracking to produce a saleable product.
- because of the increased processing and refining, more carbon is released per unit of useable energy than for crude oil</span>

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Consider the following intermediate chemical equations.
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Answer: 250 kJ

Explanation: According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to Hess’s law, the chemical equation can be treated as algebraic expressions and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

P_4(s)+6Cl_2\rightarrow 4PCl_3  \Delta H_1=-2439kJ (1)

4PCl_5(g)\rightarrow P_4(s)+10Cl_2(g)  \Delta H_2=3438kJ (2)

Net chemical equation:

PCl_5(g)\rightarrow PCl_3(g)+Cl_2(g)  \Delta H=? (3)

Adding 1 and 2 we get,

4PCl_5(g)\rightarrow 4PCl_3(g)+4Cl_2 \Delta H_3=\Delta H_1+\Delta H_2=-2439+3438=1000kJ   (4)

Now dividing equation (4) by 4, we get

PCl_5(g)\rightarrow PCl_3(g)+Cl_2

\Delta H=\frac{\Delta H_3}{4}=\frac{1000kJ}{4}=250kJ   (4)

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Why can't the reaction, ZnCl2 + H2 → Zn + 2HCI, occur naturally?
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Answer:

It is fairly obvious that zinc metal reacts with aqueous hydrochloric acid! The bubbles are hydrogen gas. ... In fact, electrons are being transferred from the zinc atoms to the hydrogen atoms (which ultimately make a molecule of diatomic hydrogen), changing the charges on both elements.

Explanation:

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To use the gas law constant R=0.0821 the unit for temp should be kelvin and the unit for pressure should be atmospheres True or
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4 years ago
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X) = 3x - 13<br> g(x) = 2x2 - 4x-5<br> {x) = -4-7<br> 6. Find (f- g(x).
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3 years ago
NEED HELP ASAP WITH THESE QUESTIONS GIVING FAIR AMOUNT OF POINTS IF HELPED WITH ALL QUESTIONS Violet light has a wavelength of 4
scoray [572]

Answer:

a) 7.14e19 Hz

b) 2.298e-27 J

c) 2.793e-19 J; 7.117e9 nm

d) 7.5e14 Hz; 4.96e-19 J

e) 6.2947e-18 J; 31.6 nm

f) 2.21e-22 J

g) 7.1e-19 J; 1.1e15 Hz

h) 3.422e-19 J; 581 nm

i) 4.2e14 Hz

j) 1.92e8 m

k) 7.14e16 Hz; Ultraviolet

Explanation:

Frequency: ν       Wavelength: λ       Energy: E       Speed of light: C (3.00e8)       Planck's Constant: h (6.626e-34)

ν -> λ    λ = C/ν

λ -> ν    ν = C/λ

For either of these equations, wavelength must be converted to meters or nanometers, depending on the equation.

For ν -> λ, after doing the equation, convert the wavelength into nanometers by dividing by 1e-9.

For converting λ -> ν, convert the wavelength into meters by multiplying by 1e-9.

For energy: E = hν = hc/λ

Now that the setup is out of the way:

a) Violet light has a wavelength of 4.20 x 10-12 m. What is the frequency?

λ -> ν    ν = C/λ

\frac{3.00e8}{4.20e-12} = 7.14e19 Hz

b) A photon has a frequency (n) of 3.468 x 106 Hz. Calculate its energy

E = hν = hc/λ

(6.626e-34) (3.468e6) = 2.298e-27 J

c) Calculate the energy (E) and wavelength (l) of a photon of light with a frequency of 4.215 x 1014 Hz.

E = hν = hc/λ

(6.626e-34) (4.215e14) = 2.793e-19 J

ν -> λ    λ = C/ν

\frac{3.00e8}{4.215e14} = 7.117 m

\frac{7.117m}{1}*\frac{1nm}{1e-9m} = 7.117e9 nm

d) Calculate the frequency and the energy of blue light that has a wavelength of 400 nm  (h = 6.62 x 10-34 J-s).

λ -> ν    ν = C/λ

\frac{400 nm}{1} *\frac{1e-9m}{1nm} = 4e-7 m

\frac{3.00e8}{4e-7} = 7.5e14 Hz

E = hν = hc/λ

(6.626e-34) (7.5e14) = 4.96e-19 J

e) Calculate the wavelength and energy of light that has a frequency of 9.5 x 1015 Hz.

ν -> λ    λ = C/ν

\frac{3.00e8}{9.5e15} = 3.16e-8 m

\frac{3.16e-8m}{1}*\frac{1nm}{1e-9m} = 31.6 nm

E = hν = hc/λ

(6.626e-34) (9.5e15) = 6.2947e-18 J

f) A photon of light has a wavelength of 0.090 cm. Calculate its energy.

E = hν = hc/λ

Convert the wavelength from cm to meters:

\frac{0.090cm}{1} *\frac{1m}{100cm} = 9e-4 m

\frac{(6.626e-34)(3.00e8)}{9e-4} = 2.21e-22 J

g) Calculate the energy and frequency of red light having a wavelength of 2.80 x 10-5 cm.

E = hν = hc/λ

Convert the wavelength from cm to meters:

\frac{2.80e-5cm}{1} *\frac{1m}{100cm} = 2.8e-7 m

\frac{(6.626e-34)(3.00e8)}{2.8e-7} = 7.1e-19 J

λ -> ν    ν = C/λ

Convert the wavelength from cm to meters:

\frac{2.80e-5cm}{1} *\frac{1m}{100cm} = 2.8e-7 m

\frac{3.00e8}{2.8e-7} = 1.1e15 Hz

h) Calculate the energy (E) and wavelength (l) of a photon of light with a frequency of 5.165 x 1014 Hz.

E = hν = hc/λ

(6.626e-34) (5.165e14) = 3.422e-19 J

ν -> λ    λ = C/ν

\frac{3.00e8}{5.165e14} = 5.81e-7 m

\frac{5.81e-7m}{1}*\frac{1nm}{1e-9m} = 581 nm

i) The wavelength of green light from a traffic signal is centered at 7.20 x 10-5 cm. Calculate the frequency.

λ -> ν    ν = C/λ

Convert the wavelength from cm to meters:

\frac{7.20e-5 cm}{1} *\frac{1m}{100cm} = 7.2e-7 m

\frac{3.00e8}{7.2e-7} = 4.2e14 Hz

j) If it takes 1.56 seconds for radio waves (which travel at the speed of light) to reach the moon from Earth, how far away is the moon?

  All we want to do here is to convert frequency (speed) to wavelength (distance). This problem requires a bit of thought, but it isn't bad once you realize that frquency is speed and wavelength is distance. It becomes just like the other problems after that. Also, I'll leave this distance in meters, but I think you can figure out how to convert it if it wants it in another unit.

  One second is equal to 1 Hertz, so our frequency is 1.56 Hz.

ν -> λ    λ = C/ν

\frac{3.00e8}{1.56} = 1.92e8 m

  The actual distance from the earth to the moon via google is 3.84e7, but sometimes problems like this will mess with the numbers to make sure that you didn't just look up the answer. I'm still pretty sure that this is right, however.

k) Calculate the frequency of light that has a wavelength of 4.20 x 10-9m. Identify the type of electromagnetic radiation.

First, we convert wavelength to frequency, as normal:

λ -> ν    ν = C/λ

\frac{3.00e8}{4.20e-9} = 7.14e16 Hz

Then we identify the electromagnetic wave type. You can look up a conversion chart for these on google, but since our frequency is in the e15 - e17 range, this light is considered ultraviolet.

5 0
4 years ago
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