Key for Unit II Practice Test
09-26-12
Multiple Choice
1. A (Law of Conservation of mass (types and kinds of atoms are constant).
2. A (also HOBr 1- ion and clearly an oxidation reduction, letter D the product is a hydrate)
3. D ( 20.0 g HF x 1.00 mol HF = 5.0 M (sig figs 2)
0.20 L
20.01 g HF
4. B (consider each M = mol and x L (volume) to determine the number of moles of each compound
L
Each compound results in a certain number of ions (NaCl = 2 ions), (CaCl2 = 3 ions), etc.
(Sucrose is bonded covalently so it contains no ions)
5. B [know your strong acids (HCl, HBr, HI, HNO3, H2SO4, HClO3, HClO4)
6. A (actually know as hydration see p. 128 in text)
7. D (ethanol C2H5OH has OH group so polar, while benzene ring C6H6 is non-polar) only the ethanol dissolves in
water, think about the lab (Vit A with methanol and then ether R—O—R)
8. D (Weak electrolyte (not strong acid or base and not soluble salt (ionic)) [know the solubility rules]
9. Write the double replacement (DR) reaction and decided if NaCl is a precipitate [soluble] and if CuS is a
precipitate [S2- sulfides are only slightly soluble, so weak electrolytes ( exception IA metals)]
10.
C (Ammonia (NH3)
11.
E (Solids, liquids, gases, weak acids, insoluble (slightly soluble) salts are written together, while strong
acids, bases, and soluble salts (strong electrolytes) are written apart)
12.
D
13.
D (Assume that you equal masses of each compound 100.0 g x 1.00 mol
and the same volume
1.0 L
Mm (molar mass )
14.
C
15.
D
16.
B (The oxidizing agent is the species reduced: PbO 2 (Pb4+) changes to Pb(SO4) (Pb2+)))
17.
E
18.
D
19.
E
20.
A M1V1 = M2V2 6.0 M (100 mL) = 0.54 M ( ??? mL) 1 111.111 mL = 1.11 L
21.
D Determine the moles that each contributes then divide by total volume (750.0 mL)
22.
D 0.500 mol NaOH x 0.0500 L x 1 mol HCl
x 1.00 L
= 0.250 L = 250 mL
1.00 L
1 mol NaOH
0.100 mol HCl
23.
24.
B
BaCl2(aq)
C
+
Molecular Equation:
2 Ag(NO3) (aq)
 Ba(NO3)2 (aq)
A
2 AgCl (s)
Complete Ionic Equation:
Ba2+(aq) + 2 Cl1-(aq) + 2 Ag1+(aq) + 2 (NO3)1-(aq)
25.
+
Net Ionic Equation:
2 Cl1-(aq) + 2 Ag1+(aq)
 2 AgCl (s)

Ba2+(aq) + 2 (NO3)1-(aq +
2 AgCl (s)
Reduces to :
Cl1-(aq) + Ag1+(aq)
 AgCl (s)
26.
A
27.
B (Remember that the Cl2 has a zero (0)charge as an element and in the compound CsCl the Cl has a 1charge, so the Cl2 is reduced or it is the oxidizing agent)
28.
B If it dissolves in H2O, which is polar it must also be polar (like dissolves like)
29.
30.
31.
A The solution contains H2SO4 (strong acid so it produces ions and the light bulb lights). AS the
compound is added the bulb grows dim so it has to be something that is causes fewer ions to be in
solution. Which one of these compounds will precipitate with either the H1+ ion or the (SO4)2- ion? The
only precipitate that will form is Ba(SO4) [p. 144 solubility rules] and this will cause fewer ions to be
present and it will be more dim. All others will cause the bulb to burn more brightly.
A
B 238U + 1n  1n + 144Cs + ______
32.
A Radiation is weakest (Alpha
33.
E Since there is 160.0 g to start each time a half-life passes (1.0 hour) there is half remaining 160/2 =
80, 80/2 = 40, 40/2 = 20 20/2 = 10, 10/2 = 5, 5/2 = 2.5, 2.5/2 = 1.25
This means 7 half-lives or 7 hours
mi x (1/2)n = mf
D
106
35.
A
198
36.
37.
38.
B
D
6 Zn +
12 H+
39.
H2O
+
3 SO32-
40.
H3X
+ 3 NaOH 
+
Po

Beta to Gamma) strongest
160 x (1/2)n = 1.25, solve for n = number of half-lives.
34.
Ag
to
0
1
4
2
e 
He +
+ As2O3 →
2 MnO4-
+
Na3X
106
194
Pd
Pb
6 Zn+ +
→
2 AsH3
3 SO42-
+
+
3 H2O
2 OH- +
2 MnO2
+ 3 H2O start with 0.307 g of the H3X
0.106 mol NaOH x 0.0352 L NaOH x 1 mol H3X = 0.00124 mol H3X
L of NaOH
3 mol NaOH
0.307 g H3X =
0.00124 mol
41.
246.8 g/mol
Get a 125 mL volumetric flask. Add the 25.5 mL of 12M CuSO4. Then add enough H2O to reach the etched
mark (total volume of 125 mL)
Calculations M1V1 = M2V2
42.
2.45 M (125 mL) = 12 M ( ??? mL)
25.5 mL
58.9 g ScxCly (aq) this is soluble so all the Cl1- ion is available to react with the Ag1+ from Ag(NO3)
So Cl1- + Ag1+  Precipitates into AgCl and 167.2 mg of AgCl is collected.
167.2 g AgCl x
35.453 g Cl = 41.41 g Cl
143.322 g AgCl
58.9 g ScCl
- 41.41 g Cl
17.49 g Sc
17.49 g Sc x 1 mol Sc = 0.3890 mol Sc
1
44.96 g Sc
Ratio
41.41 g Cl x 1 mol Cl = 1.168 mol Cl
3
so formula is ScCl3
35.453 g Cl
29.7% Sc; 70.3% Cl; ScCl3
43.
3.00 g alloy (contains Pb and Sn)
Metal + HNO3  Pb2+ ions and Sn2+ ions
all of the Pb is precipitated
Pb2+ + (SO4)2-  Pb(SO4) (s) and 2.93 g is collected
2.93 g Pb(SO4) x 207.2 g Pb = 2.00 g Pb so there is 1.00 g Sn since the total is 3.00g metal
303.26 g Pb(SO4)
1.00 g Sn
x 100 =
3.00 g total
44.
1.000 g MCl2 
1.286g AgCl x
33.3% Sn
M2+ + Cl1- + Ag1+  AgCl(s)
collect 1.286 g of AgCl
35.453 g Cl = 0.3181 g Cl
143.322 g AgCl
1.000 g MCl2
- 0.3181 g Cl
0.6819 g M
0.3181 g Cl x 1 mol Cl x 1 mol M = 0.00449 mol M
35.45 g Cl
2 mol Cl
45
a.
Sr(OH)2
(aq)
+ 2 HC2H3O2(aq) →
Sr(C2H3O2)2(aq) + 2 H2O(l)
2 OH- + 2 HC2H3O2 →
2 C2H3O2- +
b.
NaHCO3(s)
+ CO2(g)
c.
NaNO3(aq) + CsCl(aq) → NaCl(aq) + CsNO3(aq)
d.
LiCl(aq) +
e.
Li2CO3(s) + 2 HNO3(aq)
Li2CO3(s) + 2 HNO3(aq)
→
NaOH(s)
2 H2O2(l)
g.
NH4I(aq)
h.
2 KClO3(s)
i.
2 NaI(aq)
2 H2O (could be reduced)
AgNO3(aq) → LiNO3(aq) +
Cl- + Ag+ → AgCl
→
+
→
+
+
→
2 KCl(s)
Cl2(aq)
→
2 I- + Cl2
→
AgCl(s)
2 Li+ + H2O + CO2
2 H2O(l)
CsCl(aq)
No Reaction
2 LiNO3(aq) + H2CO3 [unstable in Acid H1+ ions])
2 LiNO3(aq) + H2O(l) + CO2(g)
→
→
Li2CO3 + 2 H+ →
f.
0.6819 g M = 151.99 g/mol
0.00449 mol M
Eu
O2(g)
NH4Cl(aq)
+
CsI(aq)
3 O2(g)
2 NaCl(aq)
2 Cl-
+
+
+
I2
I2(aq)
No Reaction