Form A
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NAME: _ General Chemistry 115 Exam 1 Avagodro's number 6.022 x 1023 1 kg = 2.20 lb 1 lb = 454 g
SECTION:_
1. When you perform the following operation, how many significant figures should your answer have? Assume these are measured quanti
t1
Avogadros number
Electron charge (2)
Electron mass
Faraday cosmnt (F)
Gas constant (R)
Neutron mass
Plancks constant (k)
Proton mass
Rydberg constant (RH)
Speed of light in vacuum
1 lb = 453.6 g
1 gal : 3.785 L : 4 quarts
l in = 2.54 cm (exactly)
1 mi
Physical Properties (Units (SI)
Tuesday 6 September 2016
In Science there are 3 Fundamental Properties
- Length- Meter- Wavelength of Kr
- Mass- Kilogram
- Time- (Physical Property-?)- Second- 1/86400- Of a mean solar day- Jan 1st 1900
Speed of Light- C-
Some Practice Problems - Force
1. A box is being pulled with constant speed across a horizontal floor by a force of 60 N that is
making an angle of 37 with the horizontal. It is known that the coefficient of static friction
between the box and the floor i
Energy level
Type of orbital
Names and amount
number of electrons in each
1
s
one s, called 1s
Contains 2 electrons: 1s2
2
s
one s, called 2s
contains 2 electrons: 2s2
p
three p's, called 2px, 2py, 2pz
2 electrons each (6 e- altogether): 2p6
s
one s, call
principal
quantum
number, n
angular momentum
quantum number, l
can be 0 . n-1
magentic quantum
number ml can be
1
0
0
2
0
0
3
-l . 0 .+l
1
-1, 0, 1
0
0
1
2
-1, 0, +1
-2, -1, 0, 1, 2
types of orbital and
number of electrons
in each
one s, called 1s,
contai
Chapter 10
10.1 You need to know all the molecular geometries for molecules that have no lone pairs on central
atom. You also need to know the electronic geometry (the arrangement of electron pairs) and they are
one and the same if the central atom has no
Form B
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NAME: _ General Chemistry 115 Exam 2 h = 6.63 x 10-34 J . s RH = 2.18 x 10-18 R = 0.0821 L . atm/ K . mol 1 L . atm = 101.3 J April 5, 2004
SECTION:_
1 nm = 1 x 10 -9 m s of H2O = 4.184 J/g C = 8.314 J/ K . mol 1 atm = 760 torr
c = 3.00
1
NAME: _ General Chemistry 115 FINAL h = 6.63 x 10-34 J . s RH = 2.18 x 10-18 J R = 0.0821 L . atm/ K . mol 1 L . atm = 101.3 J E=h c= PV = nRT M = dRT/P M1V1 = M2V2 1. How many grams in 3.90 mole iron (III) carbonate? 1.14 x 103 g 215 g 591 g 78.4
master
1
NAME: _ General Chemistry 115 Exam 2 h = 6.63 x 10-34 J . s RH = 2.18 x 10-18 R = 0.0821 L . atm/ K . mol 1 L . atm = 101.3 J April 5, 2004
SECTION:_
1 nm = 1 x 10 -9 m s of H2O = 4.184 J/g C = 8.314 J/ K . mol 1 atm = 760 torr
c = 3.00
NavigationBar
MAIN
General
Quantum Numbers,
Atomic Orbitals, and
Electron Congurations
Contents:
Quantum Numbers and Atomic Orbitals
1. Principal Quantum Number (n)
2. Angular Momentum (Secondary, Azimunthal) Quantum Number (l)
3. Magnetic Quantum Numb