REVIEW ANSWERS …EXAM 2
GENERAL CHEMISTRY I
These questions represent only a random sample of possible types of exam
questions. Students should be knowledgeable about all material presented during
lecture and on the topics list.
1)
Predict the products resulting from these aqueous reactions
(include phase labels):
a) KNO3 + NaI  KI (aq) and NaNO3 (aq)
b) Na2SO3 + HCL  NaCL (aq) and H2O (l) and SO2 (g)
c) BaCL2 + Na2SO4  NaCL (aq) and BaSO4 (s)
d) HNO3 + KOH  H2O (l) and KNO3 (aq)
2)
List the three gases that can typically form during a “metathesis” reaction.
3)
H2S, CO2 and SO2
Assign oxidation numbers and determine what is oxidized, reduced,
oxidizing agent, and reducing agent for each of the following reactions:
a) BaBr2 + 2 NaNO3  Ba(NO3)2 + 2 NaBr
Atom
Ba
Br
Na
N
O
Reactant side Product side
+2
-1
+1
+5
-2
+2
-1
+1
+5
-2
NO REDOX (no change in oxidation numbers)
b) H2SO4 + Na  Na2SO4 + H2
Atom
H
S
O
Na
Reactant side
+1
+6
-2
0
Product side
0
+6
-2
+1
H atom changed from +1  0 (reduced); H2SO4 oxidizing agent
Na atom changed from 0  +1 (oxidized); Na reducing agent
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4)
Look at the Activity Series to answer this question. An industrial chemist
wants to purchase a large pot to use to brew very acidic beer. This pot will
routinely hold H+ ions. Which pot would be less likely to dissolve with
continued use, a nickel pot or a copper pot? Explain.
Use the copper pot. The Activity Series shows that H+ ions do not
appear below the Cu metal on the table, so copper will not be
oxidized/dissolved by acid H+ ions.
5)
How many moles of CaS are in 255 mL of a 1.25 M CaS solution?
M = mol solute / L solution; rearrange to mol solute = ML
mol solute = (1.25 mol/L) (0.255 L) =0.319 mol / L or M
(Did you remember to convert mL to L?)
6)
How many liters of 3.00 M HI would be needed to neutralize 50.0 mL of
8.00 M Mg(OH)2 ? The balanced equation is Mg(OH)2 + 2 HI  2 H2O + MgI2
8.00 mol Mg(OH)2
0.0500 L OH ________________
-
1 L Mg(OH)2
7)
2 mol HI
------------------
1 L HI
___________
1 mol Mg(OH)2
3.00 mol HI
=
0.267 L
(3 sf data)
Consider the following thermochemical equation:
2 Mg + O2  2 MgO
ΔH = -1204 kJ
What would be the enthalpy change if 24.31 g Mg reacts ?
24.31 g Mg ( 1 mol Mg / 24.31 g ) ( -1204 kJ / 2 mol Mg) = -602.0 kJ (4 sf)
Did you relate the ΔH to the 2 moles of Mg as expressed in the equation?
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8)
Calculate the Specific Heat of a 245 gram piece of metal that required
77 Joule to heat up by 12.0 °C.
q = C m ΔT ; rearrange to C = q / m ΔT
77 J
C = --------------------------(245 g) ( 12.0 °C)
9)
= 0.026 J / g °C (2 sf)
Calculate the ΔHrxn for CH4 + 2 O2  CO2 + 2 H2O by using the
following:
2 H2 + C  CH4
ΔH = -74.81 kJ/mol REVERSE (to get CH4 on left)
[+74.81]
2 H2 + O2  2 H2O ΔH = -571.66 kJ/mol SAME
[-571.66]
C + O2  CO2
ΔH = -393.52 kJ/mol SAME
[-393.52]
CH4 + 2 H2 + O2 + C + O2  2 H2 + C + 2 H2O + CO2
subtract common items
CH4 + 2 H2 + C + O2 + C + O2  2 H2 + C + CO2 + 2 H2O sum
ΔHrxn -890.37 (2 dp)
10)
Calculate the heat of formation for
C2H2 (g) + H2 (g)  C2H4 (g)
where the following heats of formation apply:
C2H2 (g) = 227 kJ/mol and C2H4 (g) = 52 kJ/mol
∆H°rxn = ∑ ∆H° f products - ∑ ∆H° f reactants
[ ∆H° f C 2 H 4 (g) ] - [ ∆H° f C 2 H 2 (g) + ∆H° f H 2 (g)]
Recall the ∆H° f for an element is zero.
[ 52 ] – [ 227 + 0 ] = -175 kJ/mol
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11)
Ultraviolet radiation has a higher frequency than microwave radiation.
Which of these two types of radiation would have the longer wavelength?
Microwaves. Wavelength is inversely related to frequency, so the lower
frequency radiation would have the longer wavelength.
12)
Calculate the frequency of 746 nm light, THEN calculate the energy of
1 photon of that light.
Know your Greek! You are given wavelength (λ ), but need frequency (ν) to
use Planck's relationship.
3.00 X 10 8 m s-1
1 x 10 9 nm
ν = c / λ = ------------------------------------------ = 4.02 X 10 14 s-1
746 nm
1 m
THEN…
E = n h ν = (1) (6.626 X 10 -34 J • s) (4.02 X 10 14 s-1 ) = 2.66 X 10 -19 J
13)
The symbol ψ (psi) represents what concept?
ψ (psi) represents the wavefunction; this describes the path that an
electron can travel around a nucleus.
14)
Is the following set of four quantum numbers possible? EXPLAIN.
n
l
ml
ms
3
1
0
- 1/2
This set CAN exist. When n = 3, l can be 0, 1, or 2. THEN, when l =1, ml
can be -1, 0 or -1. The value -1/2 is allowed for ms.
15)
Which one of the following is the ground state electron configuration of the
element magnesium?
a) 1s22s22p63s1
b) 1s22s22p63s13p1
c) 1s22s22p63s2
(c) 1s22s22p63s2 This configuration shows the lowest to highest
orbitals filled in order for a twelve electron system.
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16)
(a)
(b)
(c)
(d)
(e)
For the ground state of the element antimony (Sb)…
a) write the electron configuration
b) draw the orbital diagram
c) count the number of valence electrons
d) indicate the number of unpaired electrons
e) indicate whether it is diamagnetic or paramagnetic
[Kr] 5s24d105p3
↑
↑
↑
___ ___ ___
5p
5p
5p
5 valence electrons (the highest "n" electrons)
3 unpaired electrons (they do not pair unless forced to)
This is a paramagnetic substance since it has at least one
unpaired electron.
coffee cup calorimeter
quantum "s" orbital
Bumper sticker seen on a car: (can't be sure if this is true)
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