All cations are assumed to be in aqueous solution and

All cations are assumed to be in aqueous solution, and the common anion is the
nitrate ion. (8 points)
A- Separation of K + from Ag + ions when we added chloride ions Cl - then AgCl is
insoluble and will precipitate out while KCl being soluble will remain in the
solution.
2
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B- Add carbonate ions OH-, Pb(OH)2 being insoluble will from precipitate. The
Ba(OH)2 is soluble and remains in the solution.
C- (NH4)2 CO3 being soluble will remain in the solution whereas CaCO3 is an
insoluble and will precipitate out.
D- Add sulfate ions. The 2+ ions will react to from different compounds with the 2-
sulfate ion.
(Reference: Chang 4.24)
Excellent work here! Each of your choices shows the selective precipitation result.
4. Give the oxidation number of the underlined atoms in the following molecules
and ions: (7 points )
a. Cl F
= x + (-1)=0
= x = +1
= Cl = +1
b. I F
7
x + 7(-1) =0
x+7
I=+7
c. C H
4
x + 4(+1)=0
x = -4
C = -4
d. C
2
H
2
2(x) + 2 (+1)=0
x= -1
C= -1
e. C
2
H
4
2 (x) + 4 (+1) = 0
x = -2
C = -2
f. K
2
Cr O
4
2 (+1) + 2x + 7(-7) = 0
x = + 6
Cr = + 6
g. K
2
Cr
2
O
7
2(+1) + 2x + 7(-2) =0
3
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2x=12
X=+6
Cr = + 6
h. KMn O
4
X+ 1 + (-2*4)=0
X= +7
Mn = +7
i. NaHC O
3
(+1)+(+1)+X+(3x-2)=0
+2 -6 + X = 0
-4+X=0
X= +4
C= + 4
j. O
2
O= 0
k. NaI O
3
3(-2) + 1+X =0
X=+5
I= +5
l. K
2
O
2(+1) + x = 0
X=-2
O=-2
m. P F
6
X + 6(-1)= 0
P = +6
n. Au Cl
4
X+4(-1)=0
X= 4
Au= +4
(Reference: Chang 4.47)
Excellent! Each of these are correctly identified
5. For the following reactions, identify the type (combination, combustion,
decomposition, displacement, methathesis [double-displacement)]
neutralization), balance the reaction, and for the redox reactions identify the
oxidizing agent and reducing agent. (12 points)
4
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(a) CH
4
+ O
2
= CO
2
+ H
2
O
combustion / CH
4
+ 2O
2
→ CO
2
+ 2 H
2
O
Reaction and Redox Reaction
Oxidizing agent = O
2
Reducing agent = CH
4
(b) HCl + NaOH = H
2
O + NaCl
Neutralization / HCl + NaOH → H2O + NaCl correct!
(c) H
2
+ O
2
= H
2
O
Redox reaction / 2H2 + O2 → 2H2O
Oxidizing agent=O2 and reducing agent = H2 yes, and combination
(d) CaCO
3
+ HCl = CaCl
2
+ CO
2
+ H
2
O
Methathesis (double-displacement) / CaCO3 + 2HCl → CaCl2 + CO2 + H2O correct
(e) Zn + HCl = ZnCl
2
+ H
2
Displacement + redox / Zn + 2HCl → ZnCl
2
+ H
2
Oxidizing agent=Zn and reducing agent = H
Oxidizing agent = HCl
Reducing agent = Zn
(f) NaCl = Na + Cl
2
Decomposition / 2NaCl =2Na + Cl
2
and Redox Reaction
Oxidizing agent = Na
+
Reducing agent = Cl -
6. Calculate the molarity of each of the following solutions: (4 points)
a. 29.0 g of ethanol (C
2
H
5
OH) in 545 mL of solution
Moles of C2H5OH=29.0g x 1mol C2H5OH/ 46.024g C2H5OH
=.630 moles of C2H5OH
=.630 mol/.545L
= 1.15M correct!