lecture_slides_chem_141

lecture_slides_chem_141 - Chemistry basics Physical...

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12/7/10 Chemistry basics Physical properties Force energy
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12/7/10 Physical properties Chemistry- the study of matter and the changes it undergoes Matter- anything that has mass and occupies space States of matter- solid, liquid and gas
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12/7/10 Physical properties cont. Physical properties- A characteristic that can be observed without changing the identity of the substance. Chemical property- the ability of a substance to change into another substance
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12/7/10 Physical properties cont. Base Units Unit Symbol measurement meter m length Kilogram Kg mass second s time
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12/7/10 Physical properties cont. SI prefixes Prefix Symbol factor pico p 1/1012 nano n 1/109 micro 1 1/106 milli m 1/1000 centi c 1/100 kilo k 1000
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12/7/10 Physical properties cont. Unit conversions- converting between units Conversion factor = given units required units
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12/7/10 Physical properties cont. Dimensional analysis Example- how many microseconds are in one year sec 10 15 . 3 sec 1 sec 10 min 1 sec 60 1 min 60 1 24 1 365 1 13 6 μ μ x hr d hr yr d yr =
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12/7/10 Physical properties cont. Practice example- Calculate the volume in cm3 for a box having the dimensions 1.0 ft x 2.0 ft x 3.0 ft
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12/7/10 Physical properties cont. Volume of the box is 6.0 ft3 ( 29 ( 29 ( 29 ( 29 3 3 3 3 3 3 901 , 169 0 . 1 54 . 2 0 . 1 0 . 12 0 . 6 cm in cm ft in ft =
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12/7/10 Force and energy Force = mass x acceleration (F=ma) Units: F = kg x m/s2 = N (newton)
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12/7/10 Force and energy cont. Energy - capacity to do work Work - motion against an opposing force Joule (J) – unit of energy
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12/7/10 Force and energy cont. Calorie – is the unit of energy required to raise one gram of water by 1.0 degrees Celsius from 14.5 oC to 15.5 oC. 1 Cal = 4.184 J
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12/7/10 Force and energy cont. Kinetic energy – energy in motion units = kg x m2/s2 = N x m 2 2 1 mv E k =
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12/7/10 Force and energy cont. Potential energy – energy based on current position or state. m = mass , g = gravity , h = height units again = kg x m/s2 x m = N x m mgh E p =
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12/7/10 Force and energy cont. Coulomb potential energy- attraction or repulsion due to electric charges. q1 and q2 = charge of particle in coulombs 10= vacuum permittivity = 8.854 x 10- r q q E p 0 2 1 4 πε =
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12/7/10 Coulomb potential energy Calculate the Coulomb potential energy found (in eV) in a hydrogen atom where the proton and electron are separated by 53 pm? given 1 eV = 1.602e-19 J
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12/7/10 Force and energy cont. Therefore and k = 0 4 1 : let πε r q kq E p 2 1 = 10 99 . 8 2 9 C m J x k × =
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12/7/10 Total energy 2 2 1 mv E k = 2 2 1 Force of Law s Coloumb' r q kq f = = r mv ma f 2 also = =
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12/7/10 Total energy cont. And Therefore r q kq mv r q kq r mv 2 1 2 2 2 1 2 so, = = r q kq mv E k 2 1 2 2 1 2 1 = =
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12/7/10 Total energy cont.
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