# physics questions

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1. (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 92g/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 5600g 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

= 590F

[0K] = [°C] + 273.15

[0K] = [15 °C] + 273.15

= 288.150F

[°R] = ([°C] + 273.15) × 9⁄5

[°R] = ([15 °C] + 273.15) × 9⁄5

= 518.67 0R

• 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 C2Cl4 = 165.82 g

Concentration of C2Cl4 in 1L = 1 × 332 /165.82

= 2.002 mol/L of C2Cl4

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 = 1g/L

90 ppb = 90 gL

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 = 1M

= 3M

• A gas contains 90 ppb NO. Express the concentration in g/L at STP. (1pt)

1 ppb = 1g/L

90 ppb = 90 gL

• 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 ×10-6g × 1/48 = 3.125 × 10-6 moles in 1M3 of air.

1mole = 6.023× 1023 molecules

= 1.8822 × 1018 molecules of ozone

• A chemist reports that the combined concentration of nitrate (NO3

—) plus nitrite (NO

—) in

 2

a groundwater sample from an agricultural site is 0.85 mM. What is the sum of [NO3-N] and [NO2-N] in this groundwater (in ppm)? (1 pt)

Using the Lechateliars principle we will be able to find out the above solution.

 (1

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