Solutions to the Assigned Problems

Students generally committed three errors in this exercise.

1. Did not post the units of your answer.
2. Did not use engineering notation.
3. Did not maintain significant digits.

In addition there usually is a little faulty math and faulty logic here and there. Below I have written out the solutions to the problems exactly as I worked them out (and I did not use a calculator or computer). Get in the habit of working like this as it will save time in the long run!!!

1) By the odometer on my bicycle it is 4.1 miles from the Coast Guard Station at the south end of Wrightsville Beach to the northern tip of Shell Island, how far is that in feet?

        X = 4.1E0 miles

                1.0000E0 miles = 5.280E3 feet/mile

        X = (4.1E0 miles)(5.280E3 feet/mile)

        X = 2.2E4 feet

2) The porosity of a sandstone sample from an oil field reservoir was measured to be 11%. The field covers 640 acres and the reservoir has an uniform thickness of 28 feet. If all the pore space is filled with oil how many US stock tank barrels (standard unit of volume measure in the petroleum business of the U.S.A.) of oil are in the ground under the field?

        F = 11% = 1.1E-1

        A = 640 acres = 6.40E2 acres

        h = 28 feet = 2.8E1 feet

        Reservoir Volume V_r = Ah = (6.40E2 acres)(2.8E1 feet) = 1.8E4 acre-feet

        Fluid Volume V_f = FV_r = (1.1E-1)(1.8E4 acre-feet)

                1.0000E0 acre-feet = 1.2335E3 meters³

                1.0000E0 bbl = 1.59E-1 meters³

        V_f = (2.0E3 acre-feet)(1.2335E3 meters³/acre-foot) = 2.5E6 meters³

        V_f = (2.5E6 meters³)/(1.59E-1 meters³/bbl) = 4.0E7 bbl

3) An ore sample from a North Carolina gold mine was assayed at 5.1 ppm Au. The ore rock has a density of 2.68 grams/cm³. The vein has an average width of 1.1 feet. It has a surveyed length of 6783 feet. It has an explored depth of 125 feet. What is the amount of gold within the identified portion of the vein in Troy ounces?

        C_Au = 5.1ppm = 5.1E-6

        r = 2.68 gm/cm³ = 2.68E0 gm/cm³

        W = 1.1 feet = 1.1E0 feet

        L = 6783 feet = 6.783E3 feet

        D = 125 feet = 1.25E2 feet

        Q_Au = ?

        Vein Volume = V = WLD = (1.1E0 feet)(6.783E3 feet)(1.25E2 feet) = 9.3E5 feet³

                1.0000E3 feet³ = 2.83E1 meters³

        V = (9.3E5 feet³)(2.83E1 meters³/1.0000E3 feet³) = 2.6E4 meters³

                1.0000E0 meters³ = 1.0000E6 cm³

        V = 2.6E10 cm³

        Vein mass = M_v = rV = (2.68E0 gm/cm³)(2.6E10 cm³) = 7.0E10 grams

        Gold mass = M_Au = C_AuM_v = (5.1E-6)(7.0E10 grams) = 3.6E5 grams = 3.6E2 kg

                1.0000E0 ounce(Troy) = 3.110E-2 kg

        M_Au = (3.6E2 kg)/(3.110E-2 kg/ounce) = 1.2E4 ounce(Troy)

4) The sand down at Wrightsville Beach is mostly the mineral quartz (density 2.65000 grams/cm³). Loose sand has a porosity of 36%. How many tons (Avoirdupois) of sand are contained in 1.000000 cubic yard?

        r_q = 2.65000 grams/cm³ = 2.65000E0 grams/cm³

        F = 36% = 3.6E-1

        V_s = 1.000000 yard³ = 1.000000E0 yard³

        M_s = ? tons(Avoirdupois)

                1.00000E0 gm/cm³ = 1.00000E3 kg/m³

                1.00000E0 yard³ = 2.70000E1 feet³

                1.00000E0 feet³ = 2.8E-2 m³

        V_s = (2.70000E1 feet³)(2.8E-2 m³/feet³) = 7.6E-1 m³

        V_q = V_s(1.0000E0 - F) = (7.6E-1 m³)(1.0000E0 - 3.6E-1) = (7.6E-1 m³)(6.4E-1) = 4.9E-1 m³

        M_q = r_qV_q 

         r_q = 2.65000E3 kg/m³

        M_q = (2.65000E3 kg/m³)(4.9E-1 m³) = 1.3E3 kg

                1.0000E0 kg = 2.205E0 pounds (on Earth)

        M_q = (1.3E3 kg)(2.205E0 pounds/kg) = 2.9E3 pounds

                1.0000E0 ton(Avoirdupois) = 2.0000E3 pounds

        M_s = M_q = (2.9E3 pounds)/(2.0000E3 pounds/ton) = 1.4E0 tons(Avoirdupois)

5) You are to mix up 1.00 gallons of a 5.0% (by volume) solution of HCl. How much distilled water do you use and how much concentrated HCl do you use, and since all you have to work with in lab is a metric graduated cylinder you should express your answer in liters.

        V_T = 1.00 gallons = 1.00E0 gallons

        C_HCl = 5.0% = 5.0E-2

        V_HCl = ? liters

        V_H2O = ? liters

            1.0000E0 gallon = 3.785E0 liters

        V_HCl = C_HClV_T = (5.0E-2)(3.785E0 liters) = 1.9E-1 liters

        V_H2O = V_T - V_HCl = 3.78E0 liters - 0.19E0 liters = 3.59E0 liters = 3.6E0 liters

Notes: I have taken a few more steps than would have been necessary to derive these answers. I did this so that you could follow the route to the solution more easily. However, many times doing the extra steps really makes the whole problem solving thing easier.

Note that I have worked in engineering notation all the way through the problem. This makes it easier for one to visually check your work, especially since I did not use a calculator or computer to crunch the numbers but instead did them all in my head. Hope I have not screwed up anywhere. If I have, please let me know.

dockal