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1 year ago

Which of the following are examples of useful hydrocarbons?

Chemistry

1 answer:

kari74 [83]1 year ago

7 0

Answer:

the most useful hydrocarbon is fuel.

Explanation:

some examples like: natural gases,coal,diesel fuel, kerosene etc.

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What is the electron configuration of an element with atomic number 15?

Ghella [55]

1s to the second power, 2s to the second power, 2p to the 6th power, 3s to the second power and 3p to the third power.

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2 years ago

What happens to the chemical identity of a substance during a physical change?

klio [65]

Nothing happens because a physical change only changes the appearance of the substance, whereas a chemical change transforms the substance into something completely different.
i.e. (chemical change) wood + fire = ash
(physical change) wood + axe = smaller pieces of wood

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2 years ago

How do you solve for the Atomic Mass in chemistry?

loris [4]

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3 years ago

Read 2 more answers

A tank of 0.1m3 volume contains air at 25∘C and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at

Dmitriy789 [7]

Answer:

Amount of Energy = 23,467.9278J

Explanation:

Given

Cv = 5/2R

Cp = 7/2R wjere R = Boltzmann constant = 8.314

The energy balance in the tank is given as

∆U = Q + W

According to the first law of thermodynamics

In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

The Internal energy to keep the tank's constant temperature is given as

∆U = Cv((45°C) - (25°C))

∆U = Cv((45 + 273) - (25 + 273))

∆U = Cv(20)

∆U = 5/2 * 8.314 * 20

∆U = 415.7 J/mol

Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

The Initial mole is calculated as

(P * V)/(R * T)

Where P = P1 = 101.33kPa = 101330Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

n = (1500000 * 0.1)/(8.314*298)

n = 60.54314465936812

n = 60.543

So, tue moles that entered the tank is ∆n

∆n = 60.543 - 4.089

∆n = 56.454

Amount of Energy is then calculated as:(∆n)(U)

Q = 415.7 * 56.454

Q = 23,467.9278J

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2 years ago

Describe what golgi looks like.

Margaret [11]

Here is what Golgi looks like, so that you can look at the picture and describe it.
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Glad I could help, and good luck!
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7 0

2 years ago