Step 1: Understanding the given data.
Coal burn rate = 1 kg/s
Sulfur content in coal = 3%
The amount of sulfur in the coal is \( 1 \, {kg/s} \times 0.03 = 0.03 \, {kg/s} \).
Step 2: Convert sulfur to SO2.
For each kg of sulfur in the coal, the equivalent mass of SO2 is given by the molecular weight ratio of SO2 to sulfur. The molecular weight of sulfur (S) is 32 g/mol, and for SO2, it is 64 g/mol. Hence, the ratio is: \[ \frac{64}{32} = 2. \] Therefore, for every kg of sulfur, 2 kg of SO2 is emitted.
Step 3: Calculate the SO2 emission rate.
The rate of SO2 emission is: \[ {SO2 emission rate} = 0.03 \, {kg/s} \times 2 = 0.06 \, {kg/s}. \] Step 4: Convert SO2 emission to kg/day.
To find the total emission in a day: \[ {SO2 emission per day} = 0.06 \, {kg/s} \times 86400 \, {seconds/day} = 5184 \, {kg/day}. \] Step 5: Final answer.
The SO2 emitted is \( \boxed{5184} \, {kg/day}. \)
A particle dispersoid has 1510 spherical particles of uniform density. An air purifier is proposed to be used to remove these particles. The diameter-specific number of particles in the dispersoid, along with the number removal efficiency of the proposed purifier is shown in the following table:
The overall mass removal efficiency of the proposed purifier is ________% (rounded off to one decimal place).