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puteri [66]
3 years ago
12

Explain why components that are naturally found in air can be considered air pollutants.

Chemistry
1 answer:
poizon [28]3 years ago
6 0

Explanation:

Components naturally found in air can be considered air pollutants when there is an anomalous concentration of any of them.

When the level of any of the material or gases in air rises above normal, they become pollutants.

The general cause of this is most through anthropogenic sources on earth.

  • Carbon dioxide can become a pollutant if its concentration is too much in the atmosphere.
  • This is already leading to global climate change.
  • There should be some certain percentages of dusts in the air in order to form clouds.
  • When dusts becomes too much, air quality reduces drastically and this can lead to terrible health challenges.
  • This can be said of other air component.

learn more:

Pollution brainly.com/question/10743354

#learnwithBrainly

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How do the number of electrons in an energy level affect bond formation?
BigorU [14]

Concept:

When an atom has incomplete number of electron in its outermost orbit then it has great tendency to react with another atom which satisfies their octate either by sharing or by transferring their electrons. The involved electrons are called valence electrons. These electrons will effect the energy level because of the transition of these electrons from one energy level to another energy level.  

In case of electrovalent compound, the valance electron complete their octate by transferring their valence electrons while in the covalent compound, they complete their octate by the sharing of their valence electrons.

Hence, the valence electron of the atom effect the energy levels by the transition from one state to another state in the bond formation.

8 0
4 years ago
Suggest a method to liquefy atmospheric gases.
kicyunya [14]

Answer:

WASSUP BRO

Explanation:

The gases can be converted into liquids by bringing its particles closer so atmospheric either by decreasing temperature or by increasing pressure

8 0
3 years ago
Read 2 more answers
When solid water changes directly to water vapor without first becoming a liquid, the process is called?
Evgen [1.6K]

Sublimation. Sublimation is the change of a solid direct to gas or vapour without becoming liquid

4 0
3 years ago
Which acid could not be prepared by treating a Grignard reagent with CO2?
Anni [7]

Answer:

4-oxopentanoic acid.

Explanation:

In this case, we must remember that the Grignard reaction is a reaction in which <u>carbanions</u> are produced. Carboanions have the ability to react with CO2 to generate a new C-C bond and a carboxylate ion. Finally, the acid medium will protonate the carboxylate to produce the <u>carboxylic acid group. </u>

The molecules that can follow the mechanism described above are the molecules: p-methylbenzoic acid, cyclopentane carboxylic acid and 3-methylbutanoic acid. (See figure 1)

In the case of <u>4-oxopentanoic acid</u>, the possible carbanion <u>will attack the carbonyl group</u> to generate a cyclic structure and an alcohol group (1-methylcyclopropan-1-ol). Therefore, this molecule cannot be produced by this reaction. (See figure 2)

6 0
3 years ago
Does it take more, less, or the same amount of heat to melt 1.0 kg of ice at 0°C, or to bring 1.0 kg of liquid water at 0°C to t
Murljashka [212]

Answer : It takes less amount of heat to metal 1.0 Kg of ice.

Solution :

The process involved in this problem are :

(1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)

Now we have to calculate the amount of heat released or absorbed in both processes.

<u>For process 1 :</u>

Q_1=m\times \Delta H_{fusion}

where,

Q_1 = amount of heat absorbed = ?

m = mass of water or ice = 1.0 Kg

\Delta H_{fusion} = enthalpy change for fusion = 3.35\times 10^5J/Kg

Now put all the given values in Q_1, we get:

Q_1=1.0Kg\times 3.35\times 10^5J/Kg=3.35\times 10^5J

<u>For process 2 :</u>

Q_2=m\times c_{p,l}\times (T_{final}-T_{initial})

where,

Q_2 = amount of heat absorbed = ?

m = mass of water = 1.0 Kg

c_{p,l} = specific heat of liquid water = 4186J/Kg^oC

T_1 = initial temperature = 0^oC

T_2 = final temperature = 100^oC

Now put all the given values in Q_2, we get:

Q_2=1.0Kg\times 4186J/Kg^oC\times (100-0)^oC

Q_2=4.186\times 10^5J

From this we conclude that, Q_1 that means it takes less amount of heat to metal 1.0 Kg of ice.

Hence, the it takes less amount of heat to metal 1.0 Kg of ice.

5 0
3 years ago
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