All categories

3 years ago

WEEKEND ASSIGNMENT

Chemistry

1 answer:

Schach [20]3 years ago

6 0

Answer:

1. (a) on land and offshore

2. (c) Boiling, evaporation and condensation

3.(d) Haematite and asphalt.

4. (a) octane

5.(d) petroleum gases, liquids and solids.

Explanation:

Cracking refers to the the breakdown of heavier fractions of crude oil to give the lighter fractions of crude oil having greater economic value. For instance, diesel oil may be cracked to yield gasoline which is composed of lighter hydrocarbons and has greater economic value. Cracking is of importance in the petroleum industry because it is used to produce lighter hydrocarbons from heavier ones.

Reforming is the conversion of straight chain hydrocarbons to branched chain hydrocarbons having a lesser tendency towards knocking of engines. It is very important in the petroleum industry because it leads to the production of better quality fuels with higher octane rating.

b) In Nigeria, oil producing communities are usually faced with the problem of damage to arable land due to oil exploration activities. Many oil producing communities in Nigeria are facing severe damage to the ecosystem due to oil drilling in those areas. Also, oil rich communities are at risk of crud oil spillage which contaminates water bodies and leads to loss of employment for the local people and death of aquatic animals.

You might be interested in

The main building block of a silicate mineral is _____.

galben [10]

<span>The main building block of all silicate mineral is tetrahedron</span>

8 0

4 years ago

Read 2 more answers

liquid nitrogen is generally held at a temperature of 77 k. a) liquid nitrogen is generally held at a temperature of °ra. b) liq

Ivahew [28]

The temperature of liquid nitrogen (LN), an inert cryogenic fluid, is 196 °C (320 °F). To drop the temperature of the concrete as much as is possible without causing it to freeze, LN is directly injected by lances into the batch water storage tank, the aggregate, or the mixer.

<h3>What is Cryopreservation ?</h3>

Cryopreservation is the process of preserving live cells, tissues, and other biological samples by placing them in a deep freezer at extremely low temperatures. The sample is typically stored at 196°C or lower. All of a cell's biological processes come to an end at such low temperatures, and the cell dies.

Cells and tissues intended for transplantation are frequently stored via cryopreservation. A extremely stable ultra-low temperature environment is provided by liquid nitrogen. It is therefore employed for long-term storage.

Learn more about Cryopreservation here:

brainly.com/question/24314907

#SPJ4

7 0

2 years ago

If you remove one bulb from a string of lights, all the remaining lights will go out. Are the lights in a series or parallel cir

Aleksandr [31]

Answer:

series bro. i think bro

6 0

3 years ago

Read 2 more answers

What Celsius temperature, T2, is required to change the volume of the gas sample in Part A (T1 = 23 ∘C , V1= 1.69×103 L ) to a v

andrezito [222]

Answer:

319.15^{o}C[/tex]

Explanation:

When all other variables are constant, we are allowed to use the formula

$\frac{T_{2} }{V_{2} } = \frac{T_{1} }{V_{1} } \\Which can be rewritten as T_{2} = \frac{T_{1} V_{2} }{V_{1} }if you make T2 the subject of the formula. This formula is true only if temperature is in Kelvin not degrees Celsius so T1 must be converted to KelvinNow to calculate T2[tex]T_{2}= \frac{296.15K*3.38.10^{3}L }{1.69.10^{3}L }= 592.3K$ = $319.15^{o} C$

3 0

3 years ago

HELP!! Suppose that the pressure of 1.00 L of gas is 380 mm Hg when the temperature is 200. K. At what

Korolek [52]

Numa máquina térmica uma parte da energia térmica fornecida ao sistema(Q1) é transformada em trabalho mecânico (τ) e o restante (Q2) é dissipado, perdido para o ambiente.

sendo:

τ: trabalho realizado (J) [Joule]
Q1: energia fornecida (J)
Q2: energia dissipada (J)

temos: τ = Q1 - Q2

O rendimento (η) é a razão do trabalho realizado pela energia fornecida:

η= τ/Q1

Exercícior resolvido:
Uma máquina térmica cíclica recebe 5000 J de calor de uma fonte quente e realiza trabalho de 3500 J. Calcule o rendimento dessa máquina térmica.

solução:

τ=3500 J
Q1=5000J

η= τ/Q1
η= 3500/5000
η= 0,7 ou seja 70%

Energia dissipada será:

τ = Q1 - Q2
Q2 = Q1- τ

Q2=5000-3500
Q2= 1500 J

Exercicio: Qual seria o rendimento se a máquina do exercicio anterior realizasse 4000J de trabalho com a mesma quantidade de calor fornecida ? Quanta energia seria dissipada agora?

obs: Entregar foto da resolução ou o cálculo passo a passo na mensagem

4 0

3 years ago