Name:____answer key
Chemistry120 Exam 1 version 2 – October 15, 2010
Surname:__________________
Answer all questions on this exam paper. Pen or pencil can be used.
Total = 71 marks1
Part marks are awarded.
Marks
[2]
1.
Give molecular formula for the following names:
a)
Nickel (II) Sulfate
NiSO4
b)
Magnesium Nitride
Mg3N2
[4]
2.
For the ion pairs (cation and anion) give the molecular formula for the compound and
then provide a systematic name.
a)
Ca2+ and F
CaF2
b)
-
Calcium fluoride
Al3+ and O2
-
Al2O3
Marks
[4]
3.
For the following dimensional analysis, convert:
255.5
nm
(min)2
to
km per second squared (km)
s2
nm is nanometer (1.0 x 10-9 m = 1 nm, 1 km = 1000 m)
255.5
nm
x
2
(min)
1.0 x 10-9 m
1nm
x
1km x
1000 m
(1 min)2
(60 s)2
= 7.097 x10-14 km
s2
Chemistry120 Exam 1 version 2 – October 15, 2010
[5]
[3]
4.
5.
Fill the gaps in the following table: ( ½ mark for each box).
Symbol
Se2
Protons
U3+
Cl
34
92
17
Neutrons
46
143
20
Electrons
36
89
17
Net charge
2-
3+
0
Atomic mass
80
235
37
Cr2O72
-
b)
2x + 7(-2) = -2
Na2SO4
c)
2(+1) + x + 4(-2) =0
x = Chromium = +6
6.
-
Give the oxidation number for the (bolded, italicized, and underlined) atom (per atom if
there are more than one) in the following molecules or ions.
a)
[2]
2
x = Sulfur = +6
IO4
-
x + 4(-2) = -1
x = Iodine = + 7
Briefly describe what is meant by Na+(aq). Give a sketch for your explanation.
H
H
H
O
H
O
O
H
H
Na
H
H
O
O
H
O
H
H
H
Means that Na⊕ ion is surrounded
by water molecules through an
ion dipole -interaction
Chemistry120 Exam 1 version 2 – October 15, 2010
[5]
7.
3
A redox reaction is shown below.
Br2 (l) +
-
Sn (s)
2 Br (aq)
+
Sn2+(aq)
Answer the following questions:
a)
-
What is the half reaction involving the Br2 and 2Br couple( 2 marks)?
Br2 (l) + 2 e-
b)
-
2Br (aq)
Which reagent is oxidized (1.5 marks)
Sn (s)
c)
Which reagent is the oxidizing agent (1.5 marks)?
Br2 (l)
[6]
8
Give the required reactants or expected products for the following reactions. Balancing
the stoichiometric equation is not required, but balancing molecular formula for ionic
compound is (2 marks each)..
a)
HNO3(aq) + Ca(OH)2(aq)
b)
2H2O (l) + Ca(NO3)2 (aq)
Mg(NO3)2 (aq) + 2NaOH (aq)
Mg(OH)2(s) +
ignite
c)
C4H10 O2(l)
+ 11/2 O2
4CO2
+
5H2O
NaNO3 (aq)
Chemistry120 Exam 1 version 2 – October 15, 2010
[5]
9.
4
Caproic acid, the substance responsible for the aroma of dirty gym socks and running
shoes, contains carbon, hydrogen, oxygen. On combustion analysis, a 0.450 gram
sample of caproic acid gives 0.418 grams of H2O and 1.023 grams of CO2.
Caproic acid
0.450 grams
mass percent
of element
+ O2
CO2
1.023 grams
%C = 12/44 x 100 = 27.3%
C = (0.273)(1.023 g) =0.279 g
+
H2O
0.418 grams
%H = (2 x 1)/18 x100 =11.1% mass
H = (0.111)(0.418g) = 0.0464 g
mass of Oxygen = 0.450 grams (C, H, O in Carproic acid) - 0.279g (mass C) – 0.0464g (mass H)
= 0.1246 grams
a)
What is the mass of oxygen in 0.450 gram of Caproic acid (2 marks)?
mass of Oxygen = 0.450 grams (C, H, O in Caproic acid) - 0.279g (g C) – 0.0464g (g H)
= 0.1246 gram
b)
What is the empirical formula for Caproic acid if the molecular mass is 116.2 g/mol
( 3 marks)?
empirical formula
mole C = (0.279 gram C)(1 mole C) = 0.02325 mol C
12 g C
mole H = (0.0464 gram H) (1 mole H)
1gH
= 0.0464 mol H
mole O = 0.1246 gram O)(1 mole O) = 0.00779 mol O
16 g O
0.00779 is lowest number, divide each set by this number
C = 0.02325/ 0.00779 = 3 ; H = 0.0464/0.00779 = 6 ; O = 1
C3H6O is empirical formula. formula mass = 58 u
compound molecular mass is 116 u, then the common factor is 116/58 = 2.
C6H12O2 is the molecular formula of carproic acid.
Chemistry120 Exam 1 version 2 – October 15, 2010
[6]
10.
5
One step in the commercial process for converting ammonia to nitric acid is the
conversion of NH3 to NO:
4NH3 (g)
grams
3.75g
MM
17 g/mol
Mole avail 0.2206
Mol req’d
0.2062
+0.0144
+
5O2 (g)
8.25g
32 g/mol
0.2578
0.2578
0
4NO (g)
6.19 g
30 g/mol
0.2062
+
6H2O (l)
18 g/mol
In a certain experiment, 3.75 grams of NH3 is reacted with 8.25 grams Oxygen.
showing your calculations, determine the following:
a)
By
The limiting reagent. (1.5 marks)
O2 is limiting reagent. (see above calculation where NH3 is in excess)
b)
The excess reagent and amount in grams (1.5 marks)
NH3 grams = 0.0144 mole NH3 x 17 gram NH3 = 0.245 gram
1 mol NH3
c)
The amount of NO gas that can be theoretically obtained (1.5 marks).
mole NO produced theoretically = 0.2578 mol mol O2 x 4 mol NO = 0.2062 mol
5 mol O2
Grams NO = 0.2062 mol NO x 30 gram NO
1 mol NO
d)
= 6.19 gram
If the experiment was conducted on a pilot scale in the lab, and the amount of NO gas
that was obtained was only 3.35 grams. What is the percentage yield? (1.5 marks)
% Yield = actual
x 100
theoretical
3.35 g
6.19 g
x 100 = 54.1%
Chemistry120 Exam 1 version 2 – October 15, 2010
[6]
10
6
The diagram shown below is an energy level diagram for a hydrogen atom. Answer the
questions below it by filling the information in the space provided. Show all calculations
when necessary.
n= ∞
n=4
n=3
Y
n=2
ENERGY
X
n=1
a)
The process labeled X is called ___Relaxation or Emission_________(1 mark)
b)
What is the wavelength (in nanometers) of light for the process labeled X (show
calculations)? (2 marks)
1 = RH (1 - 1)
λ
ni2 nf2
1 = 1.0974 x 107
λ
(1 - 1)
32
12
= 1.0974 x 107 m-1 (-0.889)
λ =
1
9.752 x106
in nanometers
(1.025
= 9.752 x106 m-1
= 1.025 x 10-7 m
x 10-7 m) ( 1 x 109 nm) = 102.5 nm
1m
Chemistry120 Exam 1 version 2 – October 15, 2010
7
Question 10 continued,
c)
What is the energy required for process labeled Y (show calculations)? (2 marks)
E = - 2.18 x 10-18 (1 - 1 )
nf2 ni2
E = - 2.18 x 10-18 J(1 - 1 )
42 22
E = - 2.18 x 10-18J (-.1875)
Process X requires
= 4.088 x 10-19 J
E = (6.63 x10-34J s)(3.00 x108 ms-1) = 1.940 x10-18 J
1.025 x10-7m
E = hc
λ
Process X = more energetic
d)
Which of these two processes (X or Y) is the more energetic? (is there more energy
released or more energy required) (1 mark)
Process X requires
[6]
E = hc
λ
E = (6.63 x10-34J s)(3.00 x108 ms-1) = 1.940 x10-18 J
1.025 x10-7m
11.
Process X = more energetic
One type of sunburn occurs on exposure to UV light of wavelength at 325 nm.
a)
Calculate the energy of a photon at this wavelength (2 marks)
E = hc
λ
b)
E = (6.63 x10-34J s)(3.00 x108 ms-1) = 6.12 x10-19J
(325 nm) (1.0 x 10-9m)
photon
1 nm
Calculate the energy of a mole of these photons.(2 marks)
6.12 x 10-19 J
x
Photon
c)
6.022 x 1023 photons
1 mole
= 3.69 x 105 J/mol
How many photons are there in 1.00 mJ (1 mJ = 1 x10-3 J) of this radiation? (2 marks)
photons = 1.00 mJ ( 1J ) x
1000 mJ
1 photon
=
6.12 x10-19J
1.63 x 1015 photons
Chemistry120 Exam 1 version 2 – October 15, 2010
[6]
12.
In the 3-dimensionl Cartesian coordinate diagrams below, sketch the appropriate orbital.
Z
Z
Z
[2]
X
X
X
3 py orbital
3s orbital
Y
Y
Y
Marks
[3]
8
3 dx2-y2
13.
Give all possible combinations of the angular momentum quantum number, l and
magnetic quantum number ml) for each of the following principal quantum number, n:
a)
n=4
14.
What is the maximum possible number of orbitals and electrons for n = 4 in Q 13 (a)
l = 0, 1, 2,3
Ml
Ml
Ml
Ml
= 0 for l=0
= -1,0, 1 for l =1
= -2, -1, 0, 1, 2 for l =2
= -3, -2, -1, 0, 1, 2 ,3 for l =3
There are 4s, 4p, and 4 d, 4f orbitals that can hold a maximum 2,6, 10, and 14 electrons,
respectively. for a total of
16 orbitals (4s, three 4p, five 4d, and seven 4f)
32 electrons
Chemistry120 Exam 1 version 2 – October 15, 2010
Marks
[6]
15.
Give the full or abbreviated electron configuration for the following atoms or ions
(Assume the Aufbau principle is maintained).
a)
As atom
1s22s22p63s23p64s23d104p3 [Ar] 4s23d104p3
b)
Mn2+
1s22s22p63s23p63d5 or {Ar]3d5
c)
Sn2+
1s22s22p63s23p64s23d104p65s24d10 or [Kr] 5s24d10
9
Bonus (3 marks) - Briefly in a few sentences describe the photoelectric effect in the space below
Do not use extra space. .
:The photoelectric effect is explained by Einstein. Photons with a minimum amount of energy is
required to eject electron from a metal surface. Energy of photons lower than the
minimum will not be able to eject electrons from a metal surface.
Chemistry120 Exam 1 version 2 – October 15, 2010
-End of exam-
10