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Elena-2011 [213]
3 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]3 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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Three kilograms of steam is contained in a horizontal, frictionless piston and the cylinder is heated at a constant pressure of
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Answer:

Final temperature: 659.8ºC

Expansion work: 3*75=225 kJ

Internal energy change: 275 kJ

Explanation:

First, considering both initial and final states, write the energy balance:

U_{2}-U_{1}=Q-W

Q is the only variable known. To determine the work, it is possible to consider the reversible process; the work done on a expansion reversible process may be calculated as:

dw=Pdv

The pressure is constant, so:  w=P(v_{2}-v_{1} )=0.5*100*1.5=75\frac{kJ}{kg} (There is a multiplication by 100 due to the conversion of bar to kPa)

So, the internal energy change may be calculated from the energy balance (don't forget to multiply by the mass):

U_{2}-U_{1}=500-(3*75)=275kJ

On the other hand, due to the low pressure the ideal gas law may be appropriate. The ideal gas law is written for both states:

P_{1}V_{1}=nRT_{1}

P_{2}V_{2}=nRT_{2}\\V_{2}=2.5V_{1}\\P_{2}=P_{1}\\2.5P_{1}V_{1}=nRT_{2}  

Subtracting the first from the second:

1.5P_{1}V_{1}=nR(T_{2}-T_{1})

Isolating T_{2}:

T_{2}=T_{1}+\frac{1.5P_{1}V_{1}}{nR}

Assuming that it is water steam, n=0.1666 kmol

V_{1}=\frac{nRT_{1}}{P_{1}}=\frac{8.314*0.1666*373.15}{500} =1.034m^{3}

T_{2}=100+\frac{1.5*500*1.034}{0.1666*8.314}=659.76 ºC

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