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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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What is the volume of a gas if 0.182 moles of the gas is at 1.99 atm and 83.4oC?
Sunny_sXe [5.5K]

Answer:

2,67 L

Explanation:

8 0
2 years ago
Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho
Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3

c) The leftover is computed as follows:

m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0
3 years ago
Scientific notation is
nataly862011 [7]

Scientific Notation Is Basically The Shorthand For Writing Numbers. Scientists Use It When They Want To Write Numbers That Are Very Big Or Very Small.  

4 0
3 years ago
Read 2 more answers
What is the molarity of a 10 L solution containing 5.0 moles of solute?
Setler79 [48]
Molarity = Moles of solute/ L(liters) of solution

So let's plug in the information. 

5.0 moles/10L = 0.5 M


3 0
3 years ago
Read 2 more answers
Please help ASAP! Will mark brainiest if correct
Scilla [17]

Answer: 1 is A

2 is C

3 is D

Hope this help

7 0
3 years ago
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