Factor Label Method General Chemistry I and II Lab Manual Dakota State University page 222 of 232 GIVEN: gold bar 9"x5"x2", 19.3 g/mL FIND: pounds CONVERSION FACTORS: We will need 1 mL = 1 cm3, 2.54 cm/in,. 2.2046 lb/kg, 1,000 g/kg. In a problem, you would have a table of conversion factors that you can choose from. First, let's find the volume independently of the factor label method. We know we need volume, because mL, milliliter, is a volume. We have a height, width and length, and the volume of the gold bar is height times width times length, or 952903inchinchinchinch**=Notice that even here I used labels. Now that we have our volume, let's solve this problem using the factor label method. 902 5419 310002 2046628333inchcminchmLcmgmLkggpoundkgpound*.**.**..=A gold bar weighs over 60 pounds! Consider THAT the next time you see a movie where they are handling gold bars single handedly!! Notice above that we needed to convert from in3to cm3, but all we had was the conversion factor 2.54 cm/in.
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