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Elena-2011 [213]
2 years ago
5

Which anthropogenic pollutants are implicated in the formation of most acidic precipitation? 1. carbon oxides 2. ozone and carbo

n monoxide 3. phosphoric acid and hydrochloric acid 4. nitrogen oxides and sulfur oxides
Chemistry
1 answer:
VashaNatasha [74]2 years ago
3 0

Answer:

nitrogen oxides and sulphur oxides

Explanation:

Factors responsible for antropogenic pollution are:

  • burning of fossil fuels
  • deforestation
  • mining
  • sewage
  • industrial effluent
  • pesticides, fertilizers, etc.  

The primary air pollutants released from burning of fossil fuels are oxides of nitrogen, sulfur oxides and carbon monoxide.

Out of which the main pollutants that are responsible for acidic precipitation are oxides of nitrogen and sulfur oxides.

Sulfur oxides and Oxides of nitrogen reacts with moisture present in the air to form sulfuric acid and nitric acid respectively.

These acids get mixed with rain and cause acidic precipitation.

Therefore, the correct option is oxides of nitrogen and sulfur oxides.

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It is transpiration.
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What’s the answer pls hrlp
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The answer is A

Explanation:

5 0
3 years ago
A gaseous mixture contains 443.0 Torr H 2 ( g ) , 369.9 Torr N 2 ( g ) , and 82.7 Torr Ar ( g ) . Calculate the mole fraction, χ
mr_godi [17]

Answer:

χH₂ = 0.4946

χN₂ = 0.4130

χAr = 0.0923

Explanation:

The total pressure of the mixture (P) is:

P = pH₂ + pN₂ + pAr

P = 443.0 Torr + 369.9 Torr + 82.7 Torr

P = 895.6 Torr

We can find the mole fraction of each gas (χ) using the following expression.

χi = pi / P

χH₂ = pH₂ / P = 443.0 Torr/895.6 Torr = 0.4946

χN₂ = pN₂ / P = 369.9 Torr/895.6 Torr = 0.4130

χAr = pAr / P = 82.7 Torr/895.6 Torr = 0.0923

3 0
3 years ago
A 36-gram sample of water has an initial temperature of 22°c. After the sample absorbs 1200 joules of heat energy, the final tem
professor190 [17]

Answer:

≈29.94 [°C].

Explanation:

all the details are in the attachment, the answer is underlined with orange colour.

7 0
2 years ago
A sample of iron is put into a calorimeter (see sketch at right) that contains of water. The iron sample starts off at and the t
Mariulka [41]

Answer:

Therefore, the specific heat capacity of the iron is 0.567J/g.°C.

<em>Note: The question is incomplete. The complete question is given as follows:</em>

<em>A 59.1 g sample of iron is put into a calorimeter (see sketch attached) that contains 100.0 g of water. The iron sample starts off at 85.0 °C and the temperature of the water starts off at 23.0 °C. When the temperature of the water stops changing it's 27.6 °C. The pressure remains constant at 1 atm. </em>

<em> Calculate the specific heat capacity of iron according to this experiment. Be sure your answer is rounded to the correct number of significant digits</em>

Explanation:

Using the formula of heat, Q = mc∆T  

where Q = heat energy (Joules, J), m = mass of a substance (g)

c = specific heat capacity (J/g∙°C), ∆T = change in temperature (°C)

When the hot iron is placed in the water, the temperature of the iron and water attains equilibrium when the temperature stops changing at 27.6 °C. Since it is assumed that heat exchange occurs only between the iron metal and water; Heat lost by Iron = Heat gained by water

mass of iron  = 59.1 g, c = ?, Tinitial = 85.0 °C, Tfinal = 27.6 °C

∆T = 85.0 °C - 27.6 °C = 57.4 °C

mass of water = 100.0 g, c = 4.184 J/g∙°C, Tinitial = 23.0 °C, Tfinal = 27.6 °C

∆T = 27.6°C - 23.0°C = 4.6 °C

Substituting the values above in the equation; Heat lost by Iron = Heat gained by water

59.1 g * c * 57.4 °C  = 100.0 g * 4.184 J/g.°C * 4.6 °C

c = 0.567 J/g.°C

Therefore, the specific heat capacity of the iron is 0.567 J/g.°C.

5 0
2 years ago
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