Chem 127 Class Notes - Week 1
Chem 127 Class Notes - Week 1 CHEM 160
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This 2 page Class Notes was uploaded by Aenea Mead on Sunday September 25, 2016. The Class Notes belongs to CHEM 160 at California Polytechnic State University San Luis Obispo taught by unknown in Fall 2016. Since its upload, it has received 7 views. For similar materials see General Chemistry for Agriculture and Life Science I in Chemistry at California Polytechnic State University San Luis Obispo.
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Date Created: 09/25/16
Dimensional Analysis Dimensional analysis is a convenient way to convert quantities from one unit to another. It would be used in an instance where two volumes need to be added, but they are in different units. Dimensional analysis would help to put them in the same unit, so that they can then be added. It is like how fractions cannot be added when they have different denominators, so in order to add them a common denominator needs to first be found. Dimensional analysis could also be used to solve more complex problems such as determining the density (in g/cm )of a solid when only the diameter (in nm) of each atom and its mass (in amu) is known ahead of time. Above is the what and why, n ere is ow: Question: How far (in miles) will a person run during a 5 kilometer race? 5km 0.62mi Solution #1: * = 3.1 mi /race race 1 km Solution #2: km 1mi = 3.1 mi /race race * 1.61 km **Note: You want your old units to cancel diagonally so that you end with the new units. **Note: It does not matter which way you write the conversio1mfor 0.01m as long as they are on the correct side of the fraction line. ***Tip: It can be helpful when organizing your dimensional analysis to put the beginning and end units at the top of the page. It may not be necessary for an problem like the above example, however when the questions get more complex it is a good way to stay organized and headed in the right direction. 7 Question: A city uses 1.2 x 10 gallons of water /day. How many liters per hour needs to be pumped from the lake in order to supply the city? (1 gal = 3.785L) gallons/day → liters/hour 7 Solution: 1.2*10 gal 1 day 3.785L = 1,892,500L/hr day * 24 hours * 1 gal **Note: Notice how days and gallons cancel diagonally in order to end with L/hr. **Note: Oftentimes a conversion factor will be needed (in this case gallons to liters), which will always be provided in a classroom setting. Question: A tank of gas holds 25.8 gallons. If the density of gasoline is 0.9315 g/mL, find the mass (in lbs) of the fuel in a full tank. (1 gal = 3.785L, 1g = 0.002205lb) Solution: 25.8gal 3.785L 1000mL 0.9315g 0.002205lb = 200.6lb/tank tank * 1gal * 1 L * 1 mL * 1 g
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