Zellmer
Time: Dr. 30 mins
Chemistry 125
Winter Quarter 2012
Pre-quiz
Name
KEY
Thursday
January 5, 2012
Rec. TA/time.
1.
(7 pts) The amount of NO2 in the air on a particular day is 1.55 x 10!6 kg/m3. How many
pounds of NO2 are present in 2.00 ft3? (1 in = 2.54 cm, 1.000 lb = 453.6 g) (You MUST use
dimensional analysis to receive full credit!)
2.
(8 pts) A quantity of heat equal to 600.0 cal is ADDED to 40.0 g of water at an initial
temperature of 35.0oC. Calculate the FINAL temperature (in oC). (The specific heat of water
is 1.00 cal/gCC) Show your work.
q=m
heat
C
c
C
)T
(mass) C (sp. heat) C (temp. change)
q = + 600.0 cal
(heat is added or required so q is positive; endothermic)
Want final temp. so must find )T first
q
600.0 cal
)T = -------- = ----------------------------- = + 15.0C
(40.0 g)(1.00 cal/gCC)
(3 s.f.)
mCc
)T = Tf & Ti
Tf = Ti + )T
= 35.0C + (15.0C)
= 50.0C (3 s.f.)
Copyright R. J. Zellmer, January 5, 2012
35.0 (only known to tenths)
+ 15.0
-----50.0 (to tenths place)
3.
(3 pts) Correctly name or write the formulas for the following compounds. Also, are they ionic
or molecular?
Chromium (VI) oxide
Cr6+
O2&
==>
The roman numeral gives the charge on the chromium ion (+6).
CrO3
IONIC
For future reference, you should know how many particles (ions) you get when you put an ionic
substance in water. First of all it dissociates (comes apart). Thats the easy part. You also need
to remember that these polyatomic ions stay together in solutions (i.e. act as a single particle).
H2 O
CrO3 (s) ----------> Cr6+ (aq) +
K3PO4
3 O2& (aq)
potassium phosphate
IONIC
H2 O
K3PO4 (s) ----------> 3 K+ (aq) +
P2O5
4.
(you get 4 particles in solution per
formula unit)
PO43& (aq)
diphosphorus pentoxide
(you get 4 particles in solution per
formula unit)
MOLECULAR
(2 pts) Determine the oxidation number of the underlined element in the following compound.
(Show all work.)
Chg. on HSO4& is -1
HSO4&
5.
1(+1)H + XS + 4(-2)O = -1;
XS ! 7 = -1;
XS = +6
(2 pts) In the following reaction, which element is being oxidized and which is being reduced?
N: reduced
HNO3: oxidizing agent
Copyright R. J. Zellmer, January 5, 2012
Cu: oxidized
Cu: reducing agent
6.
(2 pts) Internal energy, E, is a state function. What does this mean?
Internal energy is a state function. A state function depends only on the present conditions and not how
the system got to its present state. It doesnt depend on the path taken. Other examples of state
functions are potential energy, temperature, enthalpy, pressure, volume and others. A difference in state
functions is also a state function (i.e. )E = Ef - Ei and since internal energy is a state function the change
in internal energy for a process is state function).
7.
(3 pts) The standard enthalpies of formation, )Hf, at 25C for SO2(g) and SO3(g) are -296.9
kJ/mol and -395.2 kJ/mol, respectively. What is the enthalpy of reaction, )H is AND the
reaction endothermic or exothermic?
2 SO2(g) + O2(g)
)Hf (kJ/mol)
H
- 296.9
0
=
nH
v
0.0
0
products
2 SO3(g)
()Hf for elements in their standard
states is zero by definition)
- 395.2
products
mH
0
reactants
reactants
)H = [(2 mol)(- 395.2 kJ/mol)] - [(2 mol)(- 296.9 kJ/mol) + (1 mol)(0.0 kJ/mol)]
= [-790.4 kJ] - [-593.8 kJ]
= - 196.6 kJ
Exothermic (a negative )H; energy is released, as a product)
8. See next page.
9.
(1 pt) What volume (in mL) of a 5.0 M NaCl solution is needed to make 500.0 mL of a 2.0 M
NaCl?
Dilution problem so use
M1 = 5.0 M
M2 = 2.0 M
M2*V2 = M1*V1
V1 = ? mL
V2 = 500.0 mL
M2*V2
(2.0 M) * (500.0 mL)
V1 = ----------- = --------------------------- = 200.0 mL = 2.0 x 102 mL
(5.0 M)
M1
Copyright R. J. Zellmer, January 5, 2012
8.
(12 pts) Consider the following molecules and list their molecular shapes (NOT the electron
domain geometries) and whether they are polar or nonpolar. Provide a short explanation for
your choices.
1) H2S
bent (. 109.5 angles)
4 things around S and not all identical
(2 H and 2 lpe&)
..
..
..
Polar (like H2O:, H2Se:, :SCl2, etc.)
2) BH3
trigonal planar (= 120 angles, exactly)
3 things around B and all identical
(3 H and NO lpe!)
Nonpolar (as is BCl3, BF3 AlCl3 etc.)
3) NF 3
trigonal pyramidal (. 109.5 angles)
4 things around N and not all identical
(3 F and 1 lpe!)
Polar (like :NH3,:PH3, :PCl3, etc.)
4) CH3Cl
tetrahedral (. 109.5 angles)
All 4 atoms on C are not identical
Polar (as is CHCl3, CH3F, CH2F2, ect.)
NOTE:
The 5 basic symmetric molecular geometries (linear, trigonal planar, tetrahedral,
trigonal bipyramidal and octahedral) can give nonpolar but only if all surrounding atoms
(or groups of atoms) are identical to each other.
See Tables 9.2 and 9.3 on pages 347 and 350 in the textbook.
* NOTE: If there are 1 or more lpe- on the central atom the molecule will generally be polar.
There are two shapes that can be exceptions to this rule. (See Table 9.3 on page 350 of the textbook.)
The linear shape resulting from the trigonal bipyramidal e--pair geometry will be nonpolar if the 2 atoms
around the central atom are identical to each other (e.g. XeF2 has 3 lpe- in a trigonal planar geometry
around the Xe and the 2 F atoms linear to each other).
The square planar shape resulting from the octahedral e--pair geometry will be nonpolar if the 4 atoms
around the central atom are identical to each other (e.g. XeF4 has 4 F atoms in a square planar
arrangement with 2 lpe- linear to each other, 1 above and 1 below the square planar arrangement).
For BH3 (#2) all atoms surrounding B are identical so the molecule is nonpolar.
H2S and NF3 (1 & 3) with lpe! on the central atom are polar. In general, molecules that have 1 or more
lpe! on the central atom are polar. (See discussion above about exceptions.)
For CH3Cl (#4) all surrounding atoms are NOT identical so the molecule is polar.
Copyright R. J. Zellmer, January 5, 2012