CHM1045 Name_______________________________
Fall 1998
Dr. Michael Blaber SS#________________________________
Exam #2 100 points total
Tues June 29th 1999
D
T = 110-50 = 60°C (60K)mass: 233 x 103 moles * (55.8g/mole) = 13.0 x 106 g (13,000 kg)
heat transferred = specific heat * grams of substance *
DTheat transferred = 0.45 J g-1 K-1 * 13.0 x 106 g * 60K
heat transferred = 351 x 106 J
Note: 2 points for specific heat equation, 3 points for correct energy
D
T = quantity of heat transferred/(grams of substance * specific heat)D
T = 2.51 x 103 J/(150 mls * 1g/ml * 4.18 J g-1 K-1)D
T = 4KNote: 2 points for specific heat equation, 3 points for correct
DTC(s) + 2H2(g) + 1/2O2(g) -> CH3OH(l) DH = -239 kJ
C(s) + O2(g) -> CO2(g) DH = -394 kJ
H2(g) + 1/2O2(g) -> H2O(l) DH = -286 kJ
Calculate DH for the combustion of 15 grams of methanol (10 points)
2CH3OH(l) + 3O2(g) -> 2CO2(g) + 4H2O(l)
2C(s) + 2O2(g) -> 2CO2(g)
DH = -788 kJ4H2(g) + 2O2(g) -> 4H2O(l)
DH = -1144 kJ2CH3OH(l) -> 2C(s) + 4H2(g) + O2(g)
DH = 478 kJ2CH3OH(l) + 3O2(g) -> 2CO2(g) + 4H2O(l) DH = -1454 kJ
(-1454 kJ/2 moles CH3OH) * (15 g CH3OH) * (1 mole/32.0g) = -341 kJ
3 points balanced equation, 5 points correct combustion DH, 2 points for gram calculation

Freq=(RH /h)*(1/ni2 – 1/nf2) = (2.18 x 10-18 J/6.63 x 10-34 J s)*(1/16 – 1/1) = -3.08x1015s-1
Energy is EMITTED
Wavelength = c/freq = (3x108 m/s)/(-3.08x1015s-) = 9.73 x 10-8, or 97.3 nm
This is outside the visible spectrum, and is located in the ultraviolet spectrum.
1 point for Rydberg equation, 1 points for emission, 2 points for correct wavelength, 1 point for visibility
n=3 , l = 0,1,2, (s, p, d)
3s 1 orbital
3p 3 orbitals
3d 5 orbitals
9 orbitals total
1 point each orbital name, 2 points for correct # of orbitals
Calcium is [1s2 2s2 2p6 3s2 3p6] 4s2, with Z = 20
Zeff = 20-18 = 2+
1s2 2s2 2p4 O
[Ne] 3s2 Mg
1s2 2s2 2p6 3s2 3p6 Ar
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p4 Se
[Ar] 4s1 3d10 Cu
Fe

C

Zeff decreases as you move from right to left
Zeff unchanged as you move up
Kr
Na+ Cl- O2- Mg2+ Li
Li
13. Given the following information, determine the bond length for diatomic Oxygen (5 points)
C-O 1.43Å
C-C 1.54Å
C-N 1.43Å
N-N 1.47Å
C radius is ~ 1.54/2 = 0.77Å
O radius from C-O is therefore 1.43 - 0.77 = 0.66Å
N radius is ~ 1.47/2 = 0.735Å
O radius from C-N is therefore 1.43 - 0.735 = 0.70Å
O radius is therefore between 0.66-0.70Å, and O-O bond length is 1.32 - 1.40Å, with average value of 1.36Å
14. For the following pairs of elements indicate which one most likely has a more negative (i.e. greater) electron affinity (5 points)
Li
BeS Cl
F Ne
Mg Al
Ar K
15. Name five characteristics of metals (5 points)
a) Have a shiny luster
b) Malleable and ductile
c) Conductors of heat and electricity
d) Oxides are ionic solids that are basic
e) Tend to form cations in aqueous solution
16. Write the expected balanced equations for the following reactions involving metals (be sure to include the states of the reactants and products): (5 points)
a) Magnesium oxide plus water
MgO(s) + H2O(l) -> Mg(OH)2(aq)
b) Calcium oxide plus a dilute solution of hydrochloric acid
CaO(s) + 2HCl(aq) -> CaCl2(aq) + H2O(l)