How many new generations will be required before

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Unformatted text preview: even perform a calculation without requiring that you have covered that subject in class! For instance, suppose you have a gold bar that measures about 9 inches by 5 inches by 2 inches. Given the density of gold is 19.3 g/mL, how many pounds will this bar weigh? Dakota State University page 221 of 232 Factor Label Method General Chemistry I and II Lab Manual 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 9inch * 5inch * 2inch = 90inch 3 Notice that even here I used labels. Now that we have our volume, let's solve this problem using the factor label method. 3 mL 19.3g kg 2.2046 pound 2.54cm 90inch * * * = 62.8 pound * 3* inch cm mL 1000g kg 3 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 in3 to cm3 , but all we had was the conversion factor 2.54 cm/in. If we had multiplied 90 in3 * (2.54 cm/in), we would have seen that only ONE of the "inch" units would have canceled, leaving us with 228.6 in2 cm, a meaningless number. To cancel all three inches in the in3 term, we had to cube our conversion factor 2.54 cm/in, as shown above. Another approach would have been to cube this number before hand and use the conversion factor 16.387 cm3 /in3 . Carry out this calculation and be sure that your answer matches that shown in order to be sure that you are "cubing" correctly! Example 3: The nearest star is about 4.5 light years away (one light year is the distance light will travel in a vacuum in one year). The Apollo spacecraft travelled about as fast as a bullet, which we can approximate as 1,200 ft/second. Suppose that we stock a craft which travels at 1,200 ft/sec with a crew of men and women who are to create a new generation to take control of the ship every 20 years. How many new generations will be required before reaching Alpha-Centauri? GIVEN: 4.5 light years, 1,200 ft/sec, 1 generation/20 years FIND: generations CONVERSION FACTORS: we will need 3.0x1010 cm/s (speed of light), 365.25 days/year, 24 hr/day, 60 min/hr, 60 sec/min, 2.54 cm/in, 12 in/ft OK, first, let's see what distance we are talking, thereby converting this into a standard distance/speed problem. Dakota State University page 222 of 232 Factor Label Method 4.5year * General Chemistry I and II Lab Manual 365.25day 24hour 60 min 60 sec 3.0x1010 cm * * * * = 4.3x1018 cm year day hour min sec Now, let's finish up the problem. 4.3x1018 cm * inch ft sec min hr day year generation * * * * * * * = 2.54cm 12inch 1,200 ft 60 sec 60 min 24hour 365.25day 20year 186,000generations This problem was inspired by an old episode of Star Trek....
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This note was uploaded on 09/18/2012 for the course CHEMISTRY 1010 taught by Professor Kumar during the Fall '11 term at WPI.

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