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Mason - CHEM 211 - CHEM 211, Week 6 Notes - Class Notes

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Mason - CHEM 211 - CHEM 211, Week 6 Notes - Class Notes

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background image CHEM 211  MW 10:30-11:45 AM  Prof. Gregory Foster  Page 1 of 4     Notes for Week of 10/10  I.  Redox reactions  a.  Combination reactions  i.  2Ca + O2  2CaO  1.  2  1 ratio  b.  Decomposition  i.  2NaCl  2Na + Cl2  1.  1  2 ratio  c.  Displacement  i.  Ni +H2SO4  NiSO4 +H2  1.  2  2 ratio  II.  Basic properties of gases  a.  Gases are compressible 
b.  Gas volumes vary greatly with temperature 
c.  Most gases have low densities, roughly 1 g/L 
d.  Gases have low viscosity 
e.  Gases form solutions in any proportions 
f.  Gas Pressure (P) 
i.  P = Force/area  1.  Force = ma  a.  Kgm/s^2 
b.  (Kgm/s^2)/m^2 
c.  Kg/ms^2 = Pa = Pascal 
2.  In the US pressure is defined in units of atmospheres (atm)  a.  This is defined as the air pressure at sea level on an average  day  3.  We can convert between the two  a.  1 atm = 101,325 Pa 
b.  1 atm = 760 mmHg  
i.  760 torr  ii.  Mercury in a manometer  ii.  Sample question  1.  In an open-end monometer, the height of the column of mercury  was 152 mm lower on the right arm. Patm = 752 mmHg. What is 
the P of the gas in the flask of the monometer 
a.  Pg <Patm 
b.  Pg = 752 mmHg – 152 mmHg = 600mmHg 
2.  At a pressure of 0.85 atm, what is the height of the column of  liquid water in a barometer (d water is 1000 kg/m^3 and 
gravitational constant is 9.81 m/s^2) 
a.  Patm = gdh 
b.  G=9.81m/s^2 
c.  D=1000 kg/m^3 
background image CHEM 211  MW 10:30-11:45 AM  Prof. Gregory Foster  Page 2 of 4     d.  H= height, meters 
e.  Use the atm to convert to pascals 
i.  1atm = 101,325 pascal  ii.  .85*101325=86126 Pa  iii.  Divide that by G and density of water to find H  iv.  Gives 8.78 meters of water as the hight of the water  in the tube  g.  Boyles Law  i.  Used a u-tube to find difference in height of the sides and find pressure  ii.  As pressure increases for a fixed amount of gas at a fixed temperature then  volume decreases  1.  Pressure times volume gives us a constant  iii.  Sample problem  1.  A sample of gas occupies 3.65 L at 745 torr. What is the volume at  210 torr (T and n are constant)?  a.  PV(T,n) = k(constant) 
b.  P1V1=P2V2 
c.  745torr*3.65L/210 torr = 12.94 L 
h.  Charles Law  i.  Effect of volume and temperature  ii.  There is a direct relationship between volume of a gas and temperature  iii.  All plots of volume vs temperature will pass through -273 degrees Celsius  which led to the creation of the Kelvin Scale and absolute zero  iv.  Sample Problem  1.  A 93.0 L sample of dry air cools from 145 degrees to -22.0 degrees  Celsius. What is the final volume (P and n are constant)?  a.  V/T = k(constant) 
b.  V1/T1 = V2/T2 
c.  T2V1/T1=V2 
d.  -22.0C*93.0L/145C =  
i.  We have to convert to kelvin bc this is negative  e.  251K*93.0L/418K = 55.8 L  i.  Avogadros Law  i.  Mass and volume are directly proportional  ii.  Sample question  1.  A sample of gas is 0.285 L for 0.0116 mol in a variable volume  container. An additional 0.500 mol of gas is added to the container. 
What is the new volume (P&T are constant)? 
a.  V/n= k 
b.  V1/N1 = V2/n2 
c.  V2 = n2V1/n1 
d.  0.5116mol*0.285 L/ .0116 L = 12.6 L 
j.  STP 

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School: George Mason University
Department: Chemistry
Course: General Chemistry 1
Professor: Paul Cooper
Term: Summer 2015
Tags: General Chemistry
Name: CHEM 211, Week 6 Notes
Description: These notes cover redox reactions
Uploaded: 10/22/2017
4 Pages 12 Views 9 Unlocks
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