Notes
1) Average score = 2.3
a) two points
Mol. wt. = (1.3717 grams / (0.2211 mol/L) x (0.0523 L)
= 1.3717 grams / 7.789 x 10¯5 mole = 176.1 grams/mol
b) three points
pH = 4.23 therefore [H+] = 5.9 x 10¯5 M
[A¯] = [(0.01523 L) (0.2211 mol/L)] / (0.07000 L)
= 0.004422 mol / 0.07000 L = 0.06317 M
[HA] = [(0.01523 L) (0.2211 mol/L)] / (0.07000 L)
= 0.00367 mol / 0.07000 L = 0.04810 M
K = ([H+] [A¯]) / [HA] = [(5.9 x 10¯5) (0.06317)] / (0.04810) = 7.7 x 10¯5
c) one point
A¯ + H2O = HA + OH¯
K= ([HA] [OH¯]) / [A¯] = Kw / Ka
= (1.0 x 10¯14) / (7.7 x 10¯5) = 1.3 x 10¯10
d) three points
[A¯] at equiv. point = 7.789 x 10¯3 mol / 0.08523 L = 9.14 x 10¯2 M
[OH¯]2 = (1.3 x 10¯10) (9.14 x 10¯2) = 1.2 x 10¯11
[OH¯] = 3.4 x 10¯6 M
pOH = - log (3.4 x 10¯6) = 5.47
pH = 14 - 5.47 = 8.53
2) Average score = 3.3
a) two points
No. moles S2O32¯ = (0.224 M) (0.0373 L) = 8.39 x 10¯3 mol
1 mole I2 = 1/2 mole S2O32¯
moles I2 = 4.20 x 10¯3 mol
b) three points
Phydrogen = Ptotal - Pwater
= (752 - 24) mm Hg = 728 mm Hg
1 mole H2 = 1 mole I2
PV = nRT
V = [(4.2 x 10¯3) (0.0821 L atm) (298K)] / [(728/760 atm) ( mol-K)] = 0.107 L
c) one point
At anode: 2 I¯ ---> I2 + 2e¯
d) three points
4.20 x 10¯3 mole I2 = 8.40 x 10¯3 mole e¯
(8.40 x 10¯3 F) (96500 coul / F) = 811 coul
amperes = 811 coul / 180 sec
3) Average score = 3.2
a) two points
[delta]G° = - RT lnK
= - (8.31 J mol¯1 K¯1) (298 K) (ln 0.281)
= 3.14 x 103 J / mol
b) four points
[delta]H° = (193 J/g) (160. g/mol) = 3.08 x 104 J / mol
[delta]G° = [delta]H° - T [delta]S°
[delta]S° = ([delta]H° - [delta]G°) / T
= [(3.084 x 104) - (3.14 x 103)] / 298
= (2.770 x 104 J / mol) / 298 K = 92.9 J / mol K
c) two points
At boiling pt, [delta]G° = 0 and thus,
T = [delta]H° / [delta]S° = 3.08 x 104 / 92.9 = 332 K
d) one point
Vapor pressure = 0.281 atm
Note added for student's benefit: this comes directly from the fact that Kp = PBr2.
4) Average score = 5.2
3 points per reaction: 1 point for correct reactants, 2 points for correct products. Partial credit patterns were developed, taking into account the nature of each reaction. For some of the reactions, formulas other than those shown below received credit.
a) Zn2+ + PO43¯ ---> Zn3(PO4)2
b) Ag+ + Br¯ ---> AgBr
c) Cl2 + OH¯ ---> OCl¯ + Cl¯ + H2O
d) H+ + SO32¯ ---> H2O + SO2 (or H2SO3)
e) Sn2+ + H+ + MnO4¯ ---> Sn4+ + Mn2+ + H2O
f) Fe3+ + SCN¯ ---> Fe(SCN)2+ (or Fe(SCN)63¯)
g) BCl3 + NH3 ---> Cl3BNH3
h) CS2 + O2 ---> CO2 + SO2 (or SO3)
5) Average score = 2.7
a) 1 point for each correct diagram, 1 point for any two structure names, 1 more point for all three structure names.
b) three points. A maximum of 1 point was deducted if repulsion of atoms rather than of electron pairs was mentioned.
CF4 - 4 bonding pairs around the C at corners of regular tetrahedron to minimize repulsion (maximize bond angles).
XeF4 - 4 bonding pairs and 2 lone pairs give octahedral shape with lone pairs on opposite sides of Xe atoms.
ClF3 - 3 bonding pairs and 2 lone pairs give trigonal bipyramid with lone pairs in equatorial positions 120° apart.
6) Average score = 1.6
a) two points
Alakali metals have metallic bonds: cations in a sea of electrons.
As cations increase in size (Li to Cs), charge density decreases and attractive forces (and melting points) decreases.
b) two points
Halogen molecules are held in place by dispersion (van der Waals) forces: bonds due to temporary dipoles caused by polarization of electron clouds.
As molecules increase in size (F2 to I2), the larger electrons clouds are more readily polarized, and the attractive forces (and melting points) increase.
c) four points
Melting point order: LiF > NaCl > KBr > CsI
Compounds are ionic
Larger radii of ions as listed
Larger radii leads to smaller attraction and lower melting points.
7) Average score = 2.5
a) any two parts = 1 point; all three parts = 2 points
Sodium is softest of the three.
Na added to H2O leads to gas and base.
Na + H2O ---> H2 + NaOH
b) any two parts = 1 point; all three parts = 2 points
Magnesium reacts with HCl
Mg + 2H+ ---> Mg2+ + H2
Redn. potentials: Mg = - 2.37 V; Ag = 0.80 V
Mg, but not Ag, reacts with HCl.
c) unbalanced equation = 1 point; balancing adds another point
Ag + 4 H+ + NO3¯ ---> 3 Ag+ + NO + 2 H2O
OR
Ag + 2 H+ + NO3¯ ---> Ag+ + NO2 + H2O
d) two points
A white precipitaion forms.
Ag+ + Cl¯ ---> AgCl(s)
8) Average score = 2.7
a) two points
Effective concentration are increased
So collision frequency is increased.
b) two points
Slight increase in collision frequency occurs.
More molecules have enough energy that many more collisions have the necessary activation energy. Raises reaction rate a great deal.
c) two points
Catalytic nickel lowers the activation eneryg needed for a reaction.
More often molecules have the needed energy for a reaction.
d) two points
Greater surface area with powdered Ni.
More catalytic sites means a greater rate.
9) Average score = 3.4
a) two points
positron decay
OR
electron capture
b) two points
beta decay
c) two points
Gamma rays have no mass or charge (or they are energy) so they need not be shown in nuclear equations.
d) two points
Measure the amount of C-14 in the dead wood.
Compare with the amount of C-14 in a similar living object.
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