Description
 (a) If the concentration of toluene vapor in a polluted air is 92 ppm, how would that number change if we raise the temperature by 10 oC? If the concentration of toluene vapor is 92g/L how would the number change if we raise the temperature by 10 oC (at a constant pressure of 1 atm)? (1pt)
The concentration of Toluine would decrease according to Lechateliars principle in both cases.
 How much (in mg) lead (Pb) and cadmium (Cd) are in 1 kg of contaminated soil if the concentrations of Pb and Cd are 350 ppb and 5.6 g/g, respectively? (1pt)
1ppb = 1 g
350 ppb = 350 g of Pb in 1kg of soil
Therefore there are 0.35 mg in 1kg of soil
5.6 g of Cd are in 1 g of soil
There are 5600g of Cd in 1 kg of soil
Therefore, there are 5.6mg of Cd in 1kg of soil.
 The average surface temperature of the earth is about 15 o Convert that to oF, oK,and
oR. (1pt)
[°F] = [°C] × 9⁄5 + 32
[°F] = [15 °C] × 9⁄5 + 32
= 27 + 32
= 59^{0}F
[^{0}K] = [°C] + 273.15
[^{0}K] = [15 °C] + 273.15
= 288.15^{0}F
[°R] = ([°C] + 273.15) × 9⁄5
[°R] = ([15 °C] + 273.15) × 9⁄5
= 518.67 ^{0}R
 If the concentration of tetrachloroethylene (C2Cl4, a chlorinated solvent) in an industrial wastewater is 332 g/L how many mol of C2Cl4 is in a gallon of wastewater? (1pt)
mol of C_{2}Cl_{4} = 165.82 g
Concentration of C_{2}Cl_{4} in 1L = 1 × 332 /165.82
= 2.002 mol/L of C_{2}Cl_{4}
but 1 gallon = 3.78541L
therefore, there are 3.78541×2.002 mol in 1 gallon
= 7.5784 mol/Gal
 A water contains 90 ppb nitric oxide (NO). Express the concentration in (1pt)
1 ppb = 1g/L
90 ppb = 90 gL
But 1 mole of NO = 30g,
Therefore there are 90 × 10^{6}×1/30 moles of NO in the water
= 3 ×10^{6} M
But 10^{6} M = 1M
= 3M
 A gas contains 90 ppb NO. Express the concentration in g/L at STP. (1pt)
1 ppb = 1g/L
90 ppb = 90 gL
 A combustion exhaust contains 90 ppb NO. Express the concentration of NO in g/L if the exhaust is at 11.76 psi and 572 o (1pt)
11.76+2(572) =11155.76
 The National Ambient Air Quality Standard for ozone (O3) is 0.075 ppm. That standard
corresponds to how many O3 molecules in 1m
of air at STP? (1pt)
0.075ppm = 75ppb
For ozone at STP, 1 ppb = 2.00 μg/m3
Therefore, 75ppb =150 μg/m3
But 1 mole of Ozone = 48g
150 ×10^{6}g × 1/48 = 3.125 × 10^{6 }moles in 1M^{3} of air.
1mole = 6.023× 10^{23} molecules
= 1.8822 × 10^{18} molecules of ozone
 A chemist reports that the combined concentration of nitrate (NO3
—) plus nitrite (NO
—) in

a groundwater sample from an agricultural site is 0.85 mM. What is the sum of [NO3N] and [NO2N] in this groundwater (in ppm)? (1 pt)
Answer.
Using the Lechateliars principle we will be able to find out the above solution.

The table shows a typical chemical analysis of natural water. What is the hardness of this water, in mg/L CaCO3?(1pt)
SOLUTION
Ca+ Co3
35.8+131
=168.9*1000
=168900mg/L