Chem101,
Term 082,
First Major Exam solutions,
Tuesday, 7.4.2009, 7 pm
0 version, so all correct choices are A
1. The temperature at 444 K is equivalent to,
A)340. oF
D)270. oC
B)171 oF
E)-247 oC
C)95.0 oF
Since we have choices in Fahrenheit and Celcius, we must calculate both of them:
T ( o C) = (444 - 273.15 )o C = 170. 85o C = 171o C
T ( o F) =  1.8  170.85 + 32 o F = 340 .o F
Neither D nor E are 171 oC, but 340. oF corresponds to A.
So, right choice: A
2. The correct formula for calcium nitrate is,
A)Ca(NO3)2
E)Ca2NO2
B)Ca(NO2)
C) Ca(NO2)3
D)CaNO3
There are two oxoanions of nitrogen, nitrate, NO3-, and nitrite, NO2-.
That the latter one cannot be a nitrate, rules out choices B, C, and E.
That A is correct and D is wrong follows from the charge neutrality condition:
Ca is in main group II and thus has only 2+ cations, Ca2+, and therefore a Ca2+ ions needs to NO3- ions
to form a charge neutral formula. Further the naming is as in type I ionic compounds, because Ca
has only 1 cation: (Ca2+)(NO3-)2, choice A is correct.
3. Which one of the following is hydrosulfuric acid?
A)H2S
as molecular gas it would be dihydrogen sulfide, but in aqueous solution - in analogy to hydrochloric
acid (HCl(aq)) - it would be hydrosulfuric acid, correct choice.
B)H2SO4, which is sulfuric acid
C)H2SO3, which is sulfurous acid
D)HS-, which is the hydrogen sulfide ion
E)HSO4-, which is the hydrogen sulfate ion
4. The ion 8939Y3+ contains,
A)39 protons, 50 neutrons, 36 electrons
B)39 protons, 50 neutrons, 42 electrons
C)50 protons, 50 neutrons, 50 electrons
D)39 protons, 50 neutrons, 50 electrons
E)50 protons, 39 neutrons, 36 electrons
The number of protons, p, is the atomic number, the subscript left of the element symbol: p = 39
protons (charge +1)
The superscript left of the element symbol is the mass number, A=n+p, n being the number of
neutrons: n = A - p = 89 - 39 = 50 neutrons (charge 0)
3 positive charges indicate that there are three negative electrons less than there are positive protons:
e = p - 3 = 39 - 3 = 36 electrons (charge -1)
So, the correct choice is A.
5. The number of protons contained in 5.00 grams of an Yttrium (Y) sample is,
A)1.32 x 1024
D)1.95 x 102
B)3.39 x 1022
E)2.68 x 1026
C)2.35 x 1025
The number of moles of Y in 5.00 g is
n=
5.00 g
= 0.05624 mol
g
88.91
mol
N p = 39  n  N A = 39  0.05624 mol  6.022  10 mol
23
-1
= 1.32  10 24
There are Avogadro's number, NA, of Y atoms in each mol of Y, and there are 39 protons (atomic
number) in each Y atom.
6. How many moles of phosphorus atoms are present in 0.0320 g of Ca3(PO4)2?
A) 2.06 x 10-4
E) 1.24 x 1021
B) 1.24 x 1020
C) 1.03 x 10-4
D) 2.07 x 10-3
Molar mass of calcium phosphate:
MM Ca3( PO4 )2 = (3  40.08 + 2  30.97 + 8  16.00)
= 310.18
g
mol
g
mol
Number of moles mass of calcium phosphate:
nCa3( PO4 )2 =
0.0320 g
= 1.0317  10-4 mol
g
310.18
mol
There are 2 moles of P atoms in each mole of calcium phosphate:
-4
-4
nP = 2  1.0317  10 mol = 2.06  10 mol
So, the correct choice is A.
7. For which compound does 0.265 mol weigh 12.92 g?
Here we have to calculate the molar mass for each choice and multiply with n = 0.265 mol. Then the
correct choice is the one that gives 12.92 g:
A) CH3Cl, m = (12.01 + 3 x 1.008 + 35.45)(g/mol) x 0.256 mol = 12.92 g
So, A is the correct choice
B) C2H4O, m = (2 x 12.01 + 4 x 1.008 + 16.00)(g/mol) x 0.256 mol = 11.28 g
C) CO2, m = (12.01 + 2 x 16.00)(g/mol) x 0.256 mol = 11.27 g
D) BH3, m = (10.81 + 3 x 1.008)(g/mol) x 0.256 mol = 3.54 g
E) H2O, m = (16.00 + 2 x 1.008)(g/mol) x 0.256 mol = 4.61 g
8. Complete combustion of 1.000 g of an organic compound gave 3.35 g carbon dioxide (CO2). What
is the mass percent of carbon in the compound?
A)91.4%
E)25.5%
B)12.0%
C)26.7%
D)86.7%
nC O 2 =
3.35 g
(12.01 + 2  16.00)
mC = 12.01
g
mol
= 0.076119 mol
g
 0.076119 mol = 0.91419 g
mol
_(mass)%C=
0.91419 g
 100% = 91.4%
1.000 g
So, choice A is correct.
9. A 2.25 g sample of scandium (Sc) metal is reacted with excess hydrochloric acid (HCl) to produce
0.1502 g hydrogen gas (H2). If the balanced equation is,
2 Sc(s) + 2x HCl(aq)  2 ScClx(aq) + x H2(g),
what is the formula of the scandium chloride produced in the reaction?
A)ScCl3
D)ScCl
B)ScCl2
C)ScCl4
E)ScCl5
Molenumbers of H2 and Sc:
nH 2 =
0.1502 g
= 0.07450 mol H 2
g
2  1.008
mol
n Sc =
2.25 g
= 0.05004 mol Sc
g
44.96
mol
With a unit factor from the equation, we can determine x:
0.05004 mol Sc 
x mol H 2
= 0.0745 mol H 2
2 mol Sc
x 0.07450
=
; x = 2.978 = 3
2 0.05004
So, choice A is correct.
10. What is the mass of NH3 that will be formed when 1.5 x 102 g H2 are allowed to react with 1.7 x
102 g N2 according to
3 H2(g) + N2(g)  2 NH3(g).
A)2.1 x 102 g
B)1.8 x 102 g
D)8.5 x 102 g
E)6.4 x 102 g
C)3.0 x 102 g
Limiting reactant calculation:
If all hydrogen could react, we would obtain
nH 2 =
150 g
2  1.008
74.405 mol H 2 
g
mol
= 74.405 mol
2 mol NH 3
= 49.6032 mol NH 3
3 mol H 2
If all nitrogen could react, we would obtain
nN 2 =
170 g
g
2  14.01
mol
6.06709 mol N 2 
= 6.06709 mol
2 mol NH 3
= 12.1342 mol NH 3
1 mol N 2
Since N2 can yield the smallest amount of product, N2 is the limiting reactant and when all N2 has
reacted and formed 12.1342 mol NH3 the reaction must stop and some H2 is left over. Thus the
amount of NH3 we have to calculate from the 12.1342 mol NH3:
m NH 3 = 12.1342 mol  (14.01 + 3  1.008)
g
=
mol
= 206.7 g = 2.1  10 2 g
So, choice A is correct.
11. Which of the following will not dissociate into ions when it dissolves in water?
A) C2H5OH (ethanol): correct, because we have learned that ethanol dissolves as molecules in water
(it is polar)
B) Ca(OH)2  Ca2+(aq) + 2 OH-(aq)
C) HC2H3O2 (acetic acid)  H+(aq) + Ac-(aq) (acetate)
D) HNO3  H+(aq) + NO3-(aq)
E) HOCl  H+(aq) + ClO-(aq)
12. Which of the following two reagents will give a precipitate when they react?
A) K2S(aq) + Ni(NO3)2(aq)  NiS(s) + 2 KNO3(aq)
correct: NiS precipitate (most sulfides are insoluble and NiS is no exception)
B) Al(s) + H2SO4 Al will be oxidized tp Al2(SO4)3(aq), 2 H+ will be reduced to H2(g)
C) KBr(aq) + MgCl2(aq), nothing happens: most chlorides and bromides are soluble and those of K+
and Mg2+ are no exceptions.
D) Na2S(aq) + K3PO4(aq), nothing happens: all alkali salts are soluble
E) Mg(OH)2(aq) + HNO3(aq): neutralization, yielding Mg(NO3)2(aq) and water
13. When 38.0 mL of 0.1250 M H2SO4 are added to 100. mL of a solution of PbX2, a precipitate of
PbSO4 forms. The PbSO4 is filtered from the solution, dried, and weighed. If the recovered PbSO4 is
found to have a mass of 0.0471 g, what was the concentration of the X ions in the original solution?
A)3.11 x 10-3 M
D)1.55 x 10-3 M
B)1.55 x 10-4 M
E)2.05 x 10-2 M
C)6.20 x 10-3 M
The number of moles of SO42- ions in the original H2SO4 solution together with the number of moles
of PbSO4 ensures us, that sulfate was actually in excess and that all lead ions are precipitated:
nSO42- = 0.1250
n PbSO4 =
mol
 0.0380 L = 4.75  10-3 mol SO42L
0.0471 g
g
(207.20 + 32.07 + 4  16.00)
mol
=
_ = 1.5531  10-4 mol PbSO 4
Thus sulfate was in excess and thus all lead precipitated.
The number of moles of X ions in the 100. mL PbX2 solution is double the number of moles of
PbSO4 and gives the molarity of X ions in the original 100. mL solution:
-4
-4
n X = 2  1.5531  10 mol = 3.1062  10 mol
MX=
3.1062  10-4 mol
= 3.11  10-3 M
0.100 L
14. Balance the half reaction, NO3-(aq)  NH4+(aq), taking place in acidic medium. How many
electrons are needed to balance this half-reaction?
A)8 electrons, left side
C)4 electrons, left side
E)8 electrons, right side
B)3 electrons, right side
D)3 electrons, left side
Oxidation number of O is -2, of H is +1.
NO3-, nitrate: x = oxidation number of nitrogen: x-6 = -1, x=+5
NH4+, ammonium ion: x = oxidation number of nitrogen: x+4 = +1, x=-3
So N is reduced from +5 to -3, thus 8 electrons are involved to change the oxidation number by 8
steps, since it is a reduction half-reaction, the 8 electrons must be gained and thus on the left side
The 3 O atoms in nitrate we must place in 3 H2O on the right and for this provide 6 H+ + four further
H+ on the right to form the NH4+(aq):
NO3-(aq) + 10 H+(aq)  NH4+(aq) + 3 H2O(l)
Now we have a charge of 9+ on the left and one of 1+ on the right. To reduce 9+ on the left to the
same 1+ as on the right we must add 8 electrons on the left:
8 e- + NO3-(aq) + 10 H+(aq)  NH4+(aq) + 3 H2O(l)
So, choice A is correct.
15. What is the reducing agent in the following oxidation-reduction reaction?
VO43-(aq) + SO2(g) + 2 H+(aq)  VO2+(aq) + SO42-(aq) + H2O(l)
A)SO2(g)
D)H+(aq)
B)VO43-(aq)
E)SO42-(aq)
C)VO2+(aq)
An agent must be a reactant, so we can exclude from the beginning C) and E) which are products.
D), H+, is a reactant, but hydrogen has the oxidation number +1 in H+ as well as in H2O. So H is
neither reduced nor oxidized and thus neither oxidizing nor reducing agent.
oxidation number x of (all O atoms here have -2)
V in VO43-: x-8=-3, x=+5
V in VO2+: x-2=+2, x=+4 (V is reduced from +5 to +4 and thus VO43- is the oxidizing agent)
S in SO2: x-4=0, x=+4
S in SO42-: x-8=-2, x=+6 (S is oxidized from +4 to +6 and thus SO2 is the reducing agent)
So, choice A is correct.
16. A mixture contains 1.50 g of NaCl and NaNO3. When this mixture is dissolved in 100. mL of
water and precipitated with 0.500 M AgNO3. A 0.641 g of white solid precipitated out. What is the %
by mass of NaCl in the original mixture?
A)17.4%
D)18.4%
B)0.261%
E)0.447%
C)0.277%
The white precipitation can only be AgCl(s). The number of moles of AgCl is thus
n AgCl =
0.641 g
g
(107.87 + 35.45)
mol
= 4.474  10 -3 mol = n NaCl
-3
m NaCl = 4.474  10 mol  (22.99 + 35.45)
g
=
mol
= 0.26148 g
_%NaCl =
0.26148 g
 100% = 17.4%
1.50 g
So, choice A is correct.
17. What is the molar volume of N2 gas at -50.0oC and 0.0100 atm? (Assumeg N2 to be an ideal gas)
Solution:
The ideal gas law applies and can be solved for the molar volume (volume divided by number of
moles:
PV = nRT V m =
V m = 0.0821
V RT
=
n
P
L atm (-50.0 + 273.15) K
L

= 1832
K mol
0.0100 atm
mol
_
L
= 1.83  103
mol
A) 1.83 x 103 L/mol
B) 1.832 x 103 L/mol
C) 4.11 x 102 L/mol
D)-4.11 x 102 L/mol
E) 1.85 x 105 L/mol
A) is correct with the correct three significant figures; B) has one figure too much
C) is calculated with 50.0 (oC); D) is calculated with -50.0 (oC) instead of K
E) is calculated with R = 8.31 [J/(K mol)]
18. A 8.99 g sample of a gas occupies a volume of 3.67 L and exerts a pressure of 1.98 atm at 3.00oC.
Which of the following is its molecular formula?
Solution:
First we use the ideal gas equation to calculate the number of moles, n, of the gas and its molar mass,
Mm:
PV = nRT n =
n=
PV
RT
1.98 atm  3.67 L
= 0.3205 mol
L atm
0.0821
 276.15 K
K mol
M m=
8.99 g
g
= 28.05
0.3205 mol
mol
Now only the molar masses of the choices have to be compared with that:
A) C2H4
M m = 2  12.01
g
g
g
+ 4  1.008
= 28.052
mol
mol
mol
so this is the correct choice
B) N2H4
M m = 2  14.01
g
g
g
+ 4  1.008
= 32.052
mol
mol
mol
M m = 2  12.01
g
g
g
+ 3  1.008
= 27.044
mol
mol
mol
C) C2H3
D) NH3
M m = 14.01
g
g
g
+ 3  1.008
= 17.034
mol
mol
mol
E) N2H2
M m = 2  14.01
g
g
g
+ 2  1.008
= 30.036
mol
mol
mol
19. A sample of xenon (Xe) gas was collected in a 5.0 L flask by displacing water at 30oC and 790
mmHg. Calculate the number of xenon atoms in the flask. The vapor pressure of water at 30oC is
29.6 mmHg.
A) 1.2 x 1023
B) 1.21 x 1023
C) 1.2 x 1024
D) 1.22 x 1024
E) 1.2 x 1021
The partial pressure of Xe is obtained from the total pressure and the vapor pressure
PT = P Xe + P H 2O
P Xe = PT - P H 2O
P Xe = (790 - 29.6) torr = 760.4 torr
=
760.4 torr
= 1.0005 atm
torr
760.0
atm
The ideal gas law yields the number of moles of Xe, which multiplied by Avogadro's number gives
the number of Xe atoms in the flask, NXe:
P Xe V = n Xe RT
n Xe =
n Xe =
P Xe V
RT
1.0005 atm  5.0 L
= 0.2010 mol
L atm
0.0821
 303.15 K
K mol
N Xe = n Xe  N A = 0.2010 mol  6.022  10 mol
23
-1
= 1.21  10 23 = 1.2  10 23
A) is correct with 2 significant figures as in 5.0 L, B) is also correct, but with 1 figure too much
C) is calculated with 30oC, D) also with 1 figure too much
for E) R was used in the wrong units
20. What mass of potassium periodate must be decomposed to yield 211 L of oxygen gas at 100oC
and
0.900 atm pressure:
2 KIO4(s)  2 KI(s) + 4O2(g) ?
A) 713 g
B) 712.9 g
C) 2.66 kg
D) 2.660 kg
E) 7.04 g
The number of moles of oxygen follows from the ideal gas law:
nO 2 =
_ nO 2 =
n KIO4 =
P O2 V
RT
0.900 atm  211 L
= 6.1987 mol
L atm
0.0821
 373.15 K
K mol
2 mol KIO 4
 6.1987 mol O 2 = 3.09935 mol KIO 4
4 mol O 2
M m =  39.10 + 126.90 + 4  16.00

g
g
= 230.00
mol
mol
_
g
 3.09935 mol = 712.85 g
m KIO4 = M m  n KIO4 = 230.00
mol
= 713 g
A) is correct with 3 significant figures, B) is also correct, but with 1 figure too much
C) is calculated with 100oC, D) also with 1 figure too much
for E) R was used in the wrong units