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acid mine drainage poses sever environmental pollution problems due to its high acidity, toxic metals and sulphate contents. the effluent is reported to contain 200 ppb of manganese (mass basis). the effluent flow rate is 500L/hr with a density of 7440 kg/m*3. calculate the mass of Mn that is released from the effluent.

Question

acid mine drainage poses sever environmental pollution problems due to its high acidity, toxic metals and sulphate contents. the effluent is reported to contain 200 ppb of manganese (mass basis). the effluent flow rate is 500L/hr with a density of 7440 kg/m*3. calculate the mass of Mn that is released from the effluent.

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Solution

To calculate the mass of Mn released from the effluent, we need to follow these steps:

  1. Convert the concentration of Mn from ppb (parts per billion) to kg/kg. 1 ppb is equivalent to 1e-9 kg/kg. So, 200 ppb is 200e-9 kg/kg.

  2. Convert the flow rate from L/hr to m^3/hr. 1 L is equivalent to 1e-3 m^3. So, 500 L/hr is 0.5 m^3/hr.

  3. Calculate the mass flow rate of the effluent in kg/hr. This is done by multiplying the flow rate by the density. So, 0.5 m^3/hr * 7440 kg/m^3 = 3720 kg/hr.

  4. Calculate the mass of Mn in the effluent in kg/hr. This is done by multiplying the mass flow rate of the effluent by the concentration of Mn. So, 3720 kg/hr * 200e-9 kg/kg = 0.000744 kg/hr.

So, the mass of Mn that is released from the effluent is 0.000744 kg/hr.

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