Skills to develop
The CHEM120/121 final examination usually consists of two parts.
Part A. Written Questions: Usually you can choose 5 of 6
Part B. Multiple choice questions,
The examination covers the entire course, Chapters 3 to 11 including Survey 1. more or less evenly, but again, it depends on the prof's mood!
The DIALOGUE questions are some questions from old examinations. Note that these question are old, and they may not reflect the content of this term.
If too many of your answers are incorrect, you should stop practice, study hard, and then come back to PRACTICE again.
Press key to practice for yourself, NOT for anyone else.
This is neither a quiz, nor exam. Written questions for practice
are in another module.
If your answer is...I'm lost! If your answer is...e If your answer is...I'm lost! If your answer is...e If your answer is...I'm lost! If your answer is...e If your answer is...I'm lost! If your answer is...a If your answer is...I'm lost! If your answer is...b If your answer is...I'm lost! If your answer is...b If your answer is...I'm lost! If your answer is...c If your answer is...I'm lost! If your answer is...c If your answer is...I'm lost! If your answer is...b
Zn(s) + 1/2 O2(g) --> ZnO(s) Hf = -348.1 kJ/mol
Zn(s) + H2(g) + O2(g) --> Zn(OH)2(s) Hf = -642.2 kJ/mol
H2(g) + 1/2 O2(g) --> H2O(l) Hf = -285.8 kJ/mol
a. +8.3 kJ/mol b. -704.6 kJ/mol
c. +1276 kJ/mol d. -8.3 kJ/mol
e. -1276 kJ/mol
If your answer is...I'm lost! If your answer is...d If your answer is...I'm lost! If your answer is...c If your answer is...I'm lost! If your answer is...e If your answer is...I'm lost! If your answer is...b If your answer is...I'm lost! If your answer is...c If your answer is...I'm lost! If your answer is...b If your answer is...I'm lost! If your answer is...b If your answer is...I'm lost! If your answer is...c If your answer is...I'm lost! If your answer is...c If your answer is...I'm lost! If your answer is...b If your answer is...I'm lost! If your answer is...d If your answer is...I'm lost! If your answer is...d If your answer is...I'm lost! If your answer is...c If your answer is...I'm lost! If your answer is...c If your answer is...I'm lost! If your answer is...c If your answer is...I'm lost! If your answer is...e If your answer is...I'm lost! If your answer is...a If your answer is...I'm lost! If your answer is...e If your answer is...I'm lost! If your answer is...d If your answer is...I'm lost! If your answer is...27.5%Example Multiple Choice Questions
a. 3.63g
b. 2.22g
c. 2.00g
d. 4.00g
e. 3.27g
Consider...
Concepts: limiting reactant, and stoichiometry.
Excellent...
Good!
35.00mL of 0.284 M KCl
45.00mL of 0.196 M AlCl3
Assuming that the volumes are additive, what is the concentration
of Cl- in the resulting solution?
a. 0.124M b. 0.235M
c. 0.382M d. 0.436M
e. 0.455M
Consider...
35*0.284 + 45*3*0.196 / (35+45) = ?
Concepts: concentration of solutions.
Excellent...
Good!
a. 0.384g b. 1.78g
c. 2.67g d. 8.00g
e. 0.889g
Consider...
Concepts: stoichiometry and concentration.
Excellent...
Good!
a. 2.79M b. 2.45M
c. 1.14M d. 6.48M
e. 11.6M
Consider...
Practice unit conversion in concentration.
Excellent...
Good!
a. Fe3Cl b. FeCl3
c. Fe2Cl d. FeCl2
e. FeCl
Consider...
Chemically, FeCl2 and FeCl3 exist. But here, you better calculate.
Excellent...
Good!
a. 4.55 * 1023 molecules
b. 3.82 * 1023 molecules
c. 4.97 * 1023 molecules
d. 9.48 * 1023 molecules
e. 3.76 * 1023 molecules
Consider...
Calculate n in PV = nRT, and then n * Avogadro's number N.
Excellent...
Good!
a. 3 b. 5
c. 7 d. 9
e. 12
Consider...
Practice balance redox reactions.
Excellent...
Good!
a. work is done by the system and heat flows out of the system
b. work is done on the system and heat flows out of the system
c. work is done on the system and heat flows into the system
d. work is done by the system and heat flows into the system
e. work is done by the system and there is no heat flows
Consider...
Recall definition of internal energy.
Excellent...
Good!
a. 22.4 g/L
b. 0.839 g/L
c. 12.3 g/L
d. 20.1 g/L
e. 1.68 g/L
Consider...
Derive the formula: PM = dRT. Then d = ? Mol. wt. of Ne = 20.18
Excellent...
Good!
ZnO (s) + H2O (l) ---> Zn(OH)2 (s).
Consider...
dH = Sum of Hf of products - Sum of Hf of reactants
Excellent...
Good!
a. 5.00 L
b. 3.75 L
c. 7.50 L
d. 30.0 L
e. 15.0 L
Consider...
SO2 + O2 = SO3 balance it to find out.
Excellent...
Good!
a. -5.29 kJ/mol
b. -50.3 kJ/mol
c. -572 kJ/mol
d. -660 kJ/mol
e. -1.41 * 10 ^ 3 kJ/mol
Consider...
Energy is released. Note the units.
Excellent...
Good!
Note: TREAT " / as spin up and \ as spin down "
/\ /\ /\ /\ /\ /\ /\ /\ /\ /\ / /
-- -- -- -- -- -- -- -- -- -- -- -- --
1s 2s 2p 3s 3p 4s 4p
a. Pauli exclusion principle
b. Aufbau principle
c. Hund's rule
d. Heisenberg's Uncertainty Principle
e. None, the configuration shown is correct.
Consider...
3d orbitals should be filled before 4p according to the Aufbau principle.
Excellent...
Good!
B --> A dH2
B --> C dH3
For the transformation A --> C the enthalpy change, dH1, is
a. dH1 = dH2 + dH3
b. dH1 = dH2 - dH3
c. dH1 = dH3 - dH2
d. dH1 = 2dH2 - dH3
e. dH1 = dH3 + dH2
Consider...
Practice calculation using Hess law.
Excellent...
Good!
a. HBr
b. HF
c. HI
d. HCl
e. All are strong
Consider...
HF is a weak acid due to very strong bond H-F.
Excellent...
Good!
a. Na
b. Rb
c. Be
d. Sr
e. Li
Consider...
A problem from Survey 1.
Excellent...
Good!
1. When light strikes a metal in a vacuum, photoelectrons are emitted,
provided the lambda of the light striking the metal is longer than
the (lambda 0) required for photoemission.
2. According to the de Broglie theory the wavelength associated with a
proton travelling at 1.0*106 m/s would be longer than that of an
electron travelling at the same speed.
3. One of the valence electrons in a ground state phosphorous atom could
be described using the four quantum numbers, as 3, 1, -1, +1/2.
4. For any n value equal to or greater than 4, there are seven available
f orbitals.
5. Atoms of all of the elements in Group VI (the oxygen group) of the
periodic table, in their ground state, are paramagnetic.
a. 1, 3, 4, 5
b. 2, 3, 4, 5
c. 3, 4, 5
d. 1, 4, 5
e. 1, 2, 3
Consider...
3, 4, 5 are correct statements.
Excellent...
Good!
a. The lattice energy of MgF2(s) is larger than the lattice energy
of NaF(s)
b. Based on electronegativity differences you would expect lithium
fluoride to be less ionic than cesium fluoride.
c. The formal charge on the central oxygen in the ozone molecule, O3
is -1.
d. The bond length of the ClF molecule is shorter than the bond length
in Cl2.
e. The boron trifluoride molecule, BF3, does not have a dipole moment.
Consider...
Develop critical thinking ability.
Excellent...
Good!
a. 13%
b. 77%
c. 58%
d. 43%
e. 31%
Consider...
Calculate percent of charge on the nucleus.
Excellent...
Good!
Choose the one INCORRECT statement.
a. Oxygen is the second most electronegative element.
b. Most metal oxides are basic
c. NO2 is a brown paramagnetic gas which forms the colorless diamagnetic
dimmer, N2O4(g).
d. For a given halogen, the oxo acid strength decreases with increasing
oxidation number of the halogen atom.
e. Nitrogen can have oxidation states of -3, -2, -1, 0, +1, +2, +3, +4, +5.
Consider...
This problem comes from Survey 2.
Excellent...
Good!
What is the correct electronic configuration for the ion Zn2+
a. [Ar] 4s2 3d10 4p0
b. [Ar] 4s2 3d8 4p0
c. [Ar] 4s2 3d10 4p2
d. [Ar] 4s0 3d10 4p0
e. [Kr] 4s2 3d10 4p0
Consider...
The two s electron are removed in this ion.
You may also need a periodic table to answer this problem.
Excellent...
Good!
Most sodium is prepared on an industrial scale by:
a. the reduction of sodium oxide and hydroxide by hydrogen.
b. the reaction of sodium hydroxide with aluminum.
c. the electrolysis of fused molten sodium chloride.
d. neutralization of sodium hydroxide with concentrated nitric acid.
e. electrolysis of brine (aqueous sodium hydroxide).
Consider...
Sodium is too reactive for any chemical reduction.
Excellent...
Good!
a. sp3
b. sp2
c. sp3d
d. sp3d2
e. (sp2)(sp)
Consider...
Check the shape of ClF3, and count the number of atomic orbitals used.
Excellent...
Good!
a. an antibonding molecular orbital always has lower energy than
the bonding molecular orbital formed from the same atomic orbitals.
b. the sigma(3s) molecular orbital has higher energy than the sigma*(3s)
molecular orbital.
c. the pi(2py) molecular orbital has the same energy as the pi(2pz)
molecular orbital.
d. the pi(2py) molecular orbital has the same energy as the pi(3pz)
molecular orbital.
e. all bonding orbitals are filled before any antibonding orbitals
are filled.
Consider...
This requires some thinking. Study chemical bonding.
Excellent...
Good!
a. (sig1s)2 (sig*1s)2 (sig2s)2 (sig2px)2 (pi2py)2 (pi2pz)2
b. (sig1s)2 (sig2s)2 (sig2pz)2 (pi2py)1 (pi2px)1
c. (sig1s)2 (sig*1s)2 (sig2s)2 (sig*2s)2 (sig2pz)2 (pi2py)1 (pi2px)1
d. (sig1s)2 (sig*1s)2 (sig2s)2 (sig*2s)2 (sig2pz)2 (pi*2px)2
e. (sig1s)2 (sig*1s)2 (sig2s)2 (sig*2s)2 (pi2py)2 (pi2px)2
Consider...
Would have been simpler if we use symbols.
Review Molecular Orbital approach to chemical bonding.
Excellent...
Good!
Cr2O72-(aq) + CH3OH (aq) ---> Cr3+ + CH2O (aq)
a. 8 on the left side of the equation
b. 7 on the left side of the equation
c. 14 on the left side of the equation
d. 6 on the right side of the equation
e. 4 on the right side of the equation
Consider...
Balance Redox Equation.
Excellent...
Good!
a CH3-CH2-O-O-H
b H2O2
c Na2O2
d BaO2
e H2SO4
Consider...
Sulfuric acid is not a peroxide.
Excellent...
H2SO4 is the formula for sulfuric acid
Which orbital do the following figures represent?
|y
Psi^2 | X
| XXX
| XX V XX
| ** --
Consider...
The two figures are associated with a d orbital.
Excellent...
These are crude figures
Consider...
The answer is Hummmmm...
Try using the concept of "mole": n BaCl2 + n BaI2 = n BaSO4
Excellent...
Good!