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Chemistry LibreTexts

0.E: Exercises

  • Page ID
  • 0.1-0.3

    9. 1.74 g/cm3


    1. Write a single equation to show how to convert
      1. \(cm/min\) to \(km/h\);
      2. \(cal/oz\) to \(J/g\)
      3. \(lb/in^2\) to \(kg/m^2\) and
      4. \(°C/s\) to \(K/h\).
    2. How many Calories are contained in an 8.0 oz serving of green beans if their fuel value is 1.5 kJ/g?
    3. Gasoline has a fuel value of 48 kJ/g. How much energy in joules can be obtained by filling an automobile’s 16.3 gal tank with gasoline, assuming gasoline has a density of 0.70 g/mL?


    1. Converting from one compound unit to another
      1. \[\left(\dfrac{\cancel{cm}}{\cancel{min}}\right)\left(\dfrac{1\;\cancel{m}}{100\; \cancel{cm}}\right)\left(\dfrac{1\; km}{1000\;\cancel{m}}\right)\left(\dfrac{60\;\cancel{min}}{1\; h}\right)= km/h\]
      2. \[\left(\dfrac{\cancel{cal}}{\cancel{oz}}\right) \left(\dfrac{4.184 \;J}{1\; \cancel{cal}}\right) \left( \dfrac{16\; \cancel{oz}}{1\; \cancel{lb}}\right) \left(\dfrac{1\; \cancel{lb}}{453.59 \;g}\right)= J/g \]
      3. \[\left(\dfrac{\cancel{lb}}{\cancel{in^2}}\right)\left(\dfrac{16 \;\cancel{oz}}{\cancel{lb}}\right) \left(\dfrac{28.35\; \cancel{g}}{\cancel{oz}}\right)\left(\dfrac{1\; kg}{1000\; \cancel{g}}\right) \left[\dfrac{(36\; \cancel{in.})^2}{(1\;\cancel{yd})^2}\right] \left[\dfrac{1.09 \;\cancel{yd^2}}{1 \;m^2}\right] =kg/m^2\]
      4. \[\left(\dfrac{°C}{\cancel{s}}\right)\left(\dfrac{60\;\cancel{s}}{1\;\cancel{min}}\right)\left(\dfrac{60\;\cancel{min}}{h}\right)+273.15 K = K/h\]
    2. Our goal is to convert 1.5 kJ/g to Calories in 8 oz:\[\left(\dfrac{1.5 \; \cancel{kJ}}{1\; \cancel{g}}\right)\left(\dfrac{1000\; \cancel{J}}{1\; \cancel{kJ}}\right)\left(\dfrac{1\; \cancel{cal}}{4.184\; \cancel{J}}\right)\left(\dfrac{1\; Cal}{1000\; \cancel{cal}}\right)\left(\dfrac{28.35 \;\cancel{g}}{1\; \cancel{oz}}\right)\left(8.0\; \cancel{oz}\right)= 81\; Cal\]
    3. Our goal is to use the energy content, 48 kJ/g, and the density, 0.70 g/mL, to obtain the number of joules in 16.3 gal of gasoline: \[\left(\dfrac{48\; \cancel{kJ}}{g}\right)\left(\dfrac{1000\; J}{\cancel{kJ}}\right)\left(\dfrac{0.70\; \cancel{g}}{\cancel{mL}}\right)\left(\dfrac{1000\; \cancel{mL}}{\cancel{L}}\right)\left(\dfrac{3.79 \;\cancel{L}}{\cancel{gal}}\right)\left(16.3 \;\cancel{gal}\right)= 2.1 \times 10^9 J\]
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