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ICE Princess25 [194]
3 years ago
8

When fracking liquid waste is left in pools on the surface, _ can evaporate into the air and contribute to pollution

Chemistry
2 answers:
Rom4ik [11]3 years ago
8 0
<span>When fracking liquid waste is left in pools on the surface, Toxic Waste or Toxic Chemicals can evaporate into the air and contribute to pollution. </span><span>A new study shows that these spills have left surface waters in the area carrying radium, selenium, thallium, lead, and other toxic chemicals that can continue for years at unsafe levels. </span>
maks197457 [2]3 years ago
5 0

Answer:

The answer is "toxic chemicals".

Explanation:

Hydraulic fracture uses compounds such as sand to prevent fractures from closing, chemical additives, and several hundred chemical products, some very toxic and carcinogenic, that perform the function of preventing gas and oil from becoming contaminated. They are also used to prevent corrosion. By depositing fracking liquid waste in the pools, the most volatile products tend to evaporate contributing to air pollution. Some of the elements contained in this wastewater are selenium, radio, and lead.

Have a nice day!

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A 2. 75-l container filled with co2 gas at 25°c and 225 kpa pressure springs a leak. When the container is re-sealed, the pressu
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The number of moles of gas lost is  0.0213 mol. It can be solved with the help of Ideal gas law.

<h3>What is Ideal law ?</h3>

According to this law, "the volume of a given amount of gas is directly proportional to the number on moles of gas, directly proportional to the temperature and inversely proportional to the pressure. i.e.

PV = nRT.

Where,

  • p = pressure
  • V = volume (1.75 L = 1.75 x 10⁻³ m³)
  • T =  absolute temperature
  • n = number of moles
  • R =  gas constant, 8.314 J*(mol-K)

Therefore, the number of moles is

n = PV / RT

State 1 :

  • T₁ = (25⁰ C = 25+273 = 298 K)
  • p₁ = 225 kPa = 225 x 10³ N/m²

State 2 :

  • T₂ = 10 C = 283 K
  • p₂ = 185 kPa = 185 x 10³ N/m²

The loss in moles of gas from state 1 to state 2 is

Δn = V/R (P₁/T₁ - P₂/T₂ )

V/R = (1.75 x 10⁻³ m³)/(8.314 (N-m)/(mol-K) = 2.1049 x 10⁻⁴ (mol-m²-K)/N

p₁/T₁ = (225 x 10³)/298 = 755.0336 N/(m²-K)

p₂/T₂ = (185 x 10³)/283 = 653.7102 N/(m²-K)

Therefore,

Δn = (2.1049 x 10⁻⁴ (mol-m²-K)/N)*(755.0336 - 653.7102 N/(m²-K))

    = 0.0213 mol

Hence, The number of moles of gas lost is 0.0213 mol.

Learn more about ideal gas here ;

https://brainly.in/question/641453

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