Chem 1151
Exam 1
100 points
Summer 2005
Name________KEY_______________
1 in = 2.54 cm
N = 6.022E+23
I.
Multiple Choice (16 questions @ 3 points = 48 points)
1.
The answer to
A.
B.
C.
**D.
E.
2.
C.
D.
E.
4.
1.9054
1.905
1.91
1.9
2.
Which of the following statements describes a chemical property of phosphorus?
A.
**B.
3.
(3.478 x 1.164) + 0.286 = 4.04839/2.5 + 0.286
2.5
= 1.6193 + 0.286 = 1.9053 = 1.9
Red phosphorus and white phosphorus are solids at room temperature.
When exposed to air, white phosphorus burns spontaneously but the red
form does not.
The white form is soluble in liquid CS2 but is insoluble in water.
Red phosphorus melts at 873 K and the white form melts at 317 K.
Red phosphorus is insoluble in water and CS2.
The volume of a milk carton is about 200. in3. What is the volume in m3?
A.
5.08E + 5 m3
B.
C.
**D.
E.
5.08 m3
0.787 m3
3.28E - 3 m3
2.00E - 4 m3
200. in3 x (2.54 cm/in)3 x (1 m/100 cm)3 = 3.28E - 3 m3
What is the total concentration of ions in a 0.420 M solution of K2CO3?
**A.
B.
C.
D.
E.
1.26 M
0.840 M
0.420 M
0.210 M
0.140 M
K2CO3  2K+ + CO32- ; one mol cmp  3 mol ions
3 x 0.420 M = 1.26 M
1
5.
How many protons, neutrons and electrons are present in 119Sn2+?
Sn has 50 protons. The 2+ ion has 48 electrons. 119-50 = 69 neutrons.
A
B
**C
D
E
6.
NaOH
NH3
Mg(OH)2
KOH
CH3CHO
see defn of base
metal and behave like Pb
Element 114 lies below Pb
metal and behave like Pt
nonmetal and behave like Hg
non metal and behave like Pb
be a metal and behave like Hg
The formulas for the hydroxide ion, nitrite ion and phosphate ion are
A.
B.
**C.
D.
E.
9.
#electrons
119
52
48
69
48
Element X (Z = 114) is expected to be a
**A.
B.
C.
D.
E.
8.
# neutrons
50
69
69
50
119
All of the following are bases except
A.
B.
C.
D.
**E.
7.
# protons
119
50
50
69
50
OH-, NO3-, PO43H-, NO3-, PO43OH-, NO2-, PO43O-, NO2-, PO33OH-, NO2-, P3-
See Table 2.3
0.5 mole of NH4Cl has
A.
B.
C.
**D.
E.
3.011E+23 H atoms
6.022E+23 H atoms
9.033E+23 H atoms
1.204E+24 H atoms = 0.5 mol cmp (4 mol H/1 mol cmp) N
1.084E+25 H atoms
2
10.
All of these compounds have ionic bonds except
A.
B.
C.
D.
**E.
11.
MnO2
KIO4
KI
Li3N
PCl3
P and Cl are nonmetals
When 5.0 mol copper reacts with 13. mol nitric acid, the number of moles of
water produced is
3Cu(s) + 8HNO3(aq)  3Cu(NO3)2(aq) + 2NO(g) + 4H2O(l)
5.0 mol Cu x (8 mol acid/3 mol Cu) = 13.33 mol acid needed. But only 13.0 mol acid
available, so nitric acid = LR.
13.0 mol acid x (4 mol water/8 mol acid) = 6.5 mol water
A.
B.
**C.
D.
E.
12.
2.25 mol of NaBr is dissolved in water to make a 350. mL solution. The molarity
of the solution is
A.
B.
**C.
D.
E.
13.
13. mol
6.7 mol
6.5 mol
5.0 mol
4.0 mol
6.25E-2 M
2.25 M
6.43 M = # mol solute/L soln = (2.25 g/0.350 L)
6.43E-3 M
0.789 M
The empirical formula of a compound is BH3. If its molar mass is 55.2 amu, then
the molecular formula is
BH3 molar mass = 13.8 g/mol. 55.2/13.8 = 4 so there are 4 units of BH3 in the
molecular formula
A.
BH3
B.
B2H6
C.
B3H3
**D. B4H12
E.
B5H15
3
14.
Molarity is defined as
A.
**B.
C.
D.
E.
15.
16.
#mol solute/liter solvent
# mol solute/liter solution
# mol solvent/liter solution
Total # mol/liter solvent
# mol solute/# mol solvent
Which statements are true when a solution is diluted?
1.
2.
3.
4.
5.
solvent is added
TRUE
the number of moles of solute stays the same
TRUE
the number of moles of solvent decreases
the volume of the solution increases
TRUE
the molarity increases
A.
B.
C.
**D.
E.
1 and 3
2 and 4
1, 2 and 5
1, 2 and 4
3, 4 and 5
The molar mass of the molecule C2H8N2 and its empirical formula are
A.
**B.
C.
D.
E.
30.0 amu and CH4N
60.0 amu and CH4N
30.0 amu and C2H8N2
60.0 amu and C2H8N2
none of these
II.
Problems
1.
(3)
C2H8N2 molar mass is 60.0 g/mol
Reduce subscripts by dividing by 2
Balance this chemical equation
Al4C3 +
2.
See defn of molarity
12H2O 
4Al(OH)3 +
3CH4
(5)
There are only two naturally occurring isotopes of chlorine.
They are Cl-35 with a mass of 34.968 amu and Cl-37 with a mass of 36.956 amu.
Calculate the abundances of these isotopes.
Let x = fractional abundance of Cl-35 and y = (1-x) = frac. abun. of Cl-37
Atomic mass of Cl (from Periodic Table) = 35.453 g/mol.
34.968x + 36.956 (1-x) = 35.453
This rearranges to give 1.998x = 1.503.
Then solve for x = 0.756 for Cl-35 and y = 0.244 for Cl-37
4
3.
(5)
Samples A and B contain carbon and hydrogen
Sampl
e
A
B
Mass carbon (g)
5.70 g
4.47 g
Mass hydrogen
(g)
1.90 g
0.993 g
Mass C/mass H
3
4.5
Are Samples A and B the same compound? What Law justifies your answer?
Now use these data to illustrate the Law of Multiple Proportions.
Since the mass ratios of Mass C/Mass H are not equal, the Law of Def. Proportions says
that Samples A and B are different.
The Law of Multiple Proportions says that 3/ 4.5 should be a ratio of small whole
numbers. 3/ 4.5 = 6/9 or 2/3.
4.
(5)
When a bright orange crystal was analyzed by mass, it was found to
contain 17.5% Na, 39.7% Cr and 42.8% O. What is the empirical formula for this
compound?
Na
17.5g/23g/mol = 0.76086; divide by 0.76086 to get 1 mol Na
Cr
39.7g/52 g/mol = 0.76346; divide by 0.76086 to get 1 mol Cr
O
42.8g/16 g/mol = 2.675; divide by 0.76086 to get 3.5 mol O
The mol ratios are 2: 2: 7 or Na2Cr2O7
2Ca3(PO4)2(s) + SiO2(s) + 10C(s)  P4(g) + 6CaSiO3(l) +10CO(g)
5.
(14)
a.
(7)
Name these reactants and products and tell what kind of bonding each one
contains.
Ca3(PO4)2
Name
Calcium phosphate
Ionic of Covalent bonding?
I
C
carbon
NA
P4
Phosphorus
C
CO
Carbon monoxide
C
5
b.
(5) How many grams of SiO2 (molar mass = 60.1 g/mol) are needed to produce
350. g of CaSiO3 (molar mass = 116.2 g/mol)?
(350. g/116.2g/mol) x (1 mol SiO2 /6 mol of CaSiO3) (60.1 g/mol) = 30.2 g SiO2
c.
(2)
If only 25.4 g of CaSiO3 are produced, what is the reaction yield?
% yield = 100 x (actual mass/theoretical mass) = 100 x 25.4g/350 g = 7.26%
Name____________________________
I.
_____________(48)
II.
_____________(32)
Total_______________(80) or ________________(100)
6