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Minchanka [31]
3 years ago
8

A West Virginia coal is burned at a rate of 8.02 kg/s. The coal has a sulfur content of 4.40 % and the bottom ash contains 2.80

% of the input sulfur. What is the annual rate (in kg/yr) of stack emissions of SO2
Chemistry
1 answer:
aleksley [76]3 years ago
5 0

Answer: The annual emission rate of SO2 is 1.08 × 10^{7} kg/yr

Explanation:

  • The rate <em>r</em> at which the coal is been burnt is 8.02 kg/s.
  • Amount of sulphur in the burning coal is given as 4.40 %

i.e., 4.4/100 × 8.02 = 0.353 kg/s. Which is equivalent to the rate at which the sulphur is been burnt.

  • Since the burning of sulphur oxidizes it to produce SO2, it follows that the non-oxidized portion of the sulphur will go with the bottom ash.

  • The bottom ash is said to contain 2.80 % of the input sulphur.

  • Hence the portion of the SO2 produced is 100 — 2.80 = 97.20 %.

  • The rate of the SO2 produced is percentage of SO2 × rate at high sulphur is been burnt.

  = 97.20/100 × 0.353 kg/s.

  = 0.343 kg/s.

  • To get the annual emission rate of SO2, we convert the kg/s into kg/yr.

1 kg/s = 1 kg/s × (60 × 60 × 24 × 365) s/yr

1 kg/s = 31536000 kg/yr

  • Therefore, 0.343 kg/s = 0.343 × 31536000 kg/yr

     = 10816848 kg/yr

     = 1.08 × 10^7 kg/yryr.

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   This is done in order to find the mols of NaOH to convert to mols of HCl.
2. 8.885x10^-3mol NaOH x (1 mol HCl/1mol NaOH) = 8.885 x 10^-3mol HCl
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From the mols of HCl all we have to do is divide by the amount of liters in the solution. Since we started with 10mL HCl and added 24.75mL NaOH, the total volume is 34.75mL = 0.03475L. So:
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M1 = the molarity of our HCl in the titrated mixture (2.557 x 10^-1M HCl)
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Simply solving for M2 gives us:
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Such principle is widely applied to the electrons; i.e., the electrons posses wave and particle propeties, which must be understodd as that some of their properties may be explained as if they were particles and others as if they were waves.

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