# objective questions

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1.* All above*

2.* Fine particulate matter, Carbon monoxide, Ozone, Nitrogen dioxide, Sulphur oxides*

3. *Carbon monoxide, Ozone, Nitrogen dioxide, Lead, Asbestos, Benzene*

4. Ds = (6x S particles/pi) 1/2 where pi=22/7
= (6x4x7/22)1/2)2
=3.7301×2
= *7.46micrometers*
5. We use the ideal gas law formula
PV=nRT
This will help us find the mass and we know the molecular mass of the NO2. Substituting in the formula above,
V = (mRT)/(MMP
Divide both sides by m to have
V/m = (RT)/ (MMP)
But density id given by m/V.
Therefore, m/V = (MMP)/ (RT)
MM of NO2 is 14+32= 46gram/mol
M/V= (46grams/mole * 2bar)/ (8.3144…kg.m^2.S-2.mol^-1 * 333K)
= 0.003323*1000
= *3.32kg/m3*

*6.* (1.5moles CO2/103) (44.01gCO2/1mol CO2) (103mg/1g)(1mol air/22.41L)
= (0.002745mg/1L air) (103L/1M3)
=*2.74mg/m3*

*7.* Yes, they can be detected by condensation particle counter

*8. *1atm = 101325Pa
0.0394 = 400Pa
(235x536x0.0394)/293×1=429.8977mg/Nm3
=*430mg/Nm3*

*9. All above*

*10* We know that 1L =1000mL
Therefore 4L= 4000mL
For an overnight, 30 CFU was left. Normally the CFU is calculated per milliliter.
This implies that 4000×15=60,000
600000X1000= 600000000
Finally, 600000000X3/4000 = 450,000
=*450,000CFU/M3*