Download Chemistry Workbook 1-1A

Chem 1A
Dr. White
Fall 2013
1 Chem istry W orkbook 1: Problem s For Exam 1
Section 1-1: Significant Figures
How many significant figures do the following numbers have?
1)
1234 _____
2)
0.023 _____
3)
890 _____
4)
91010 _____
5)
9010.0 _____
6)
1090.0010 _____
4
-3
7)
0.00120 _____
8)
3.4 x 10 _____
9)
9.0 x 10 _____
-2
10)
9.010 x 10 _____
11)
0.00030 _____
12)
1020010 _____
13)
780. _____
14)
1000 _____
15)
918.010 _____
Solve the following mathematical problems such that the answers have the correct number of significant figures:
16) 334.54 grams + 198 grams = ________
17) 34.1 grams / 1.1 mL = _______
3
18) 2.11 x 10 joules / 34 seconds = ______
19) 0.110 meters – 0.10 m = ______
20) 349 cm + 1.10 cm + 100 cm = ________
21) 450 meters / 114 seconds = _____
22) 298.01 kilograms + 34.112 kilograms = _____
23)
84 m/s x 31.221 s = ________
24) (5.159 + 82.3) x (0.024 + 3.00) = _____
25) 2.34 x 4.4391 / 3.465 = _____
26) (0.00015 x 54.6) + 1.0020 = _____
27) Read the following measurements
Section 1-2: Dimensional Analysis and Significant Figures
Round numerical answer to contain the correct number of significant digits. If the answer is smaller than 0.01 or
larger than 1000, use scientific notation
1. Convert 0.0456 nm to cm.
2. Convert 0.03030 kg to ng.
3. Convert 72.6°C to °F.
4. Convert 14.5°F to Kelvin.
5. A piece of metal alloy weighs 12.0000g and contains iron, cobalt, and nickel. What is the weight of the
nickel is the iron and cobalt are determined to be 11.53g and 0.1233g respectively?
6. If 18.57 g of water is combined with 15.3 mL of water, what is the total volume? The density of water is
0.998 g/mL.
3
3
3
7. The density of iron is 7.20 g/cm . Calculate the mass, in kilograms, of 7.6 x 10 cm or iron?
3
3
8. What is the volume, in liters, of 544 g of iron? The density of iron is 7.20 g/cm . (Recall that one cm is
exactly 1 mL).
Chem 1A
Dr. White
Fall 2013
2 9. Light travels at 186,000 miles per second (3 sig figs). How many kilometers does light travel in one
day?
3
10. The speed of sound in water is 4.7 x 10 feet per second. How long, in minutes, does it take for sound
to travel 19,000 km?
11. A solution contains 4.52 g of sugar per liter. What volume of the solution, in liters is needed to supply
2.0 kg of sugar?
12. A solution contains 144.1 mg of NaCl per liter. What volume of the solution, in milliliters, is needed to
supply 27.8µg of NaCl?
3
13. A cylindrical redwood tree is 255 ft tall and 16 ft in diameter. The density of the redwood is 0.9 g/cm .
2
Calculate the mass of the tree in pounds. (Vcylinder = πr h)
3
2
14. The density of Saran Wrap is 0.80 g /cm . Calculate the mass, in grams, of a roll that is 2.00 x 10 ft
long, 12.00 inches in diameter. (assume it is a cylinder).
15. What is the volume, in cubic centimeters, of a rectangular block whose dimensions are 3.88 in by 5.0 in
by 36 in?
1
3
16. What is the diameter, in millimeters, of a red blood cell whose volume is 9.0 x 10 µm ? Assume that
3
the cell is spherical (Recall that the volume of a sphere is V=4/3πr ).
17. A piece of metal with mass 34.65 is placed into a 50-mL graduated cylinder partially filled with water,
raising the water level from 22.5 mL to 30.7 mL. What is the density of the metal?
18. The mass of a vial is 18.41 g. After a 10.00 mL sample of methanol is pipetted into it, the vial and
sample have a mass of 26.34 g. Calculate the density of methanol. Remember that the volume of a
pipet is accurate to 0.01 mL.
19. The mass of a vial is 28.20 g. After five milliliters of benzene is pipetted into the vial, the vial and its
contents have a mass of 32.55 g. What is the density of benzene?
20. If 50 mL of water is pipetted into a flask weighing 90.45 g, what will be the total weight of the flask with
the water? The density of water is 0.998g/mL.
Section 1-3: Atomic Structure Intro
1. In the early 1900’s a common model for the structure of the
atom was the plum pudding model. In this model, negatively
charged electrons reside in the atom surrounded by a diffuse,
continuous medium of positive charge (like plums in a pudding,
or for a more modern analogy, like chocolate chips in a cookie).
(a) In 1911 Earnest Rutherford carried out an experiment
that changed the way we view the atom. Briefly describe
this gold foil experiment.
(b) How is the plum pudding model inconsistent with
Rutherford’s experimental findings? Approach this by
considering the plum pudding hypothesis, and then predict what you would have expected to happen in
Rutherford’s experiment if the plum pudding model was true.
(c) Draw a picture of Rutherford’s model of the atom. Use your picture to describe why most of the alpha
particles pass through the gold foil in Rutherford’s experiment.
Chem 1A
Dr. White
Fall 2013
3 2. Atoms X, Y, Z, and R have the following nuclear compositions:
Which two are isotopes? Explain.
3. Fill in the following table for neutral atoms of the given isotopes.
4.
Silicon, which makes up about 25% of Earth's crust by mass, is used widely in the modern electronics
28
29
30
industry. It has three naturally occurring isotopes, Si, Si, and Si. Calculate the atomic mass of
silicon. (28.09 amu)
Isotope
Isotopic Mass (amu)
Abundance %
28
Si
27.976927
92.23
29
Si
28.976495
4.67
30
Si
29.973770
3.10
Section 1-4: Wave and Particle Behavior
1. Consider the following types of electromagnetic radiation:
i) microwave
ii) ultraviolet
iii) radio waves iv) infrared
a) arrange them in order of increasing wavelength
b) arrange them in order of increasing frequency
c) arrange them in order of decreasing energy
d)
2. In the diagram below, draw in each of the following. Make sure you label each new wave!
a) A wave with shorter wavelength
b) A wave with smaller frequency
c) A wave with the same wavelength and frequency but less amplitude.
v) x-ray
3. Some of the new cordless phones are said to operate at 900.MHz. Calculate the wavelength and energy of
these waves.
Chem 1A
Dr. White
Fall 2013
4 4. Cobalt-60 is a radioactive isotope used to treat cancers of the brain and other tissues. A gamma ray emitted
-13
by an atom of this isotope has a wavelength of 9.32 x 10 m.
a. What is this wavelength in nm?
b. What is the frequency of this light in Hz
c. Does radiation with λ = 242 nm have greater or lesser energy than the radiation emitted by the
gamma ray above? What about radiation with λ = 2200 Å
-10
(1 Å = 1 x 10 m)? (You shouldn’t need to do calculations to answer these questions)
5. Molybdenum metal will eject an electron if it absorbs a photon with an energy of at least 7.22 x 10
is irradiated with UV light with a wavelength of 120. nm, could an electron be ejected?
–19
J. If Mo
6
6. Calculate the de Broglie wavelengths for (a) an electron with a speed of 3.0 x 10 m/s (mass of an electron is
–31
9.109 x 10 kg) and (b) a 12 g bullet whose speed is 200 m/s.
7. Two objects are moving at the same speed. Which (if any) of the following statements about them are true?
a. The wavelength of the heavier object is longer.
b. If one object has twice as much mass as the other, its wavelength is one-half the wavelength of
the other.
c. Doubling the mass of one of the objects will have the same effect on its wavelength as does
doubling its speed.
6
8. The speed of the electron in the ground state of the hydrogen atom is 2.2 x 10 m/s. What is the wavelength
of the electron?
9. At what speed must a human weighing 150. lb. be traveling in order to have a wavelength in the visible
region, say at 650 nm? Does it seem likely that a human would attain such a speed? (1 kg = 2.20 lb)
10. Identify the following transitions as either absorption or emission:
a) n = 3 → n = 1
b) n = 6 → n = 4
c) n = 1 → n = 5
d) n = 1 → n = 2
11. Which of these transitions correspond to absorption and which to emission of radiation (or energy in
general)? Illustrate each transition on the provided energy level diagram.
a) n = 2 to n = 4
n=5
n=4
b) n = 3 to n = 1
n=3
c) n = 5 to n = 2
d) n = 3 to n = 4
E
n=2
n=1
(note that as n increases, the levels get closer and closer together)
e) Of the transitions listed above, which corresponds to emission of radiation with the longest
wavelength? Explain. You should not need to use your calculator to answer this!
f) Of the transitions listed above, which corresponds to absorption of radiation with the highest
frequency? Explain. You should not need to use your calculator to answer this!
Chem 1A
Dr. White
Fall 2013
5 12. Calculate the wavelength (in nm) of the photon emitted when a hydrogen atom undergoes a transition from
n = 5 to n = 2. Is this visible light?
13. An electron in a hydrogen atom undergoes a transition from the n = 3 level to the n = 6 level. To
accomplish this, energy, in the form of light, must be absorbed by the hydrogen atom.
a) Calculate the energy of the light (in kJ/photon) associated with this transition. Watch the sign of
your answer!
b) In order to move to the n = 5 level, would this same electron need to absorb more or less
energy? Explain.
14. An electron in a hydrogen atom in the n = 5 level emits a photon with wavelength 1281 nm.
a) To what energy level does this electron move?
b) In order to move to the n = 1 level, will this electron emit a photon with longer or shorter wavelength
than 1281 nm? Explain. Again, you should not need to use your calculator here.
15. An electron in Hydrogen in an unknown energy level emits a photon with a wavelength of 1005 nm. The
nd
final state of the electron is the 2 excited state. What was the initial state of the electron?
16. What is the minimum energy needed to ionize a hydrogen atom with the electron in the first excited state? Report your answer is J and kJ/mol.
Section 1-5: Atomic Structure and Electron Configurations
1. Fill in the following table:
Quantum Number
Orbital or Electron Property
Allowed Values
ml
energy level and size
0 to n-1
ms
2. a) If the principal quantum number n of an atomic orbital is 4, what are the possible values of l ?
b) If the angular momentum quantum number l is 3, what are the possible values of ml ?
3. There is something wrong with each set of quantum numbers given below. Make a correction so that each
set is “valid.”
a) n = 1, l = 1, ml = 0, ms = 1/2
b) n = 2, l = 1, ml = -1, ms = 1
c) n = 4, l = 2, ml = -3, ms = -1/2
4. Identify the pictured orbital as s, p or d (write your answer next to the picture) and then answer the
following questions:
a. What is the l value for this type of orbital?
b. Can this type of orbital be found in the n = 2 energy level? If not, explain why.
c. List one complete and valid set of allowed quantum numbers for an electron occupying this type of orbital
in the n = 4 level.
Chem 1A
Dr. White
Fall 2013
5. Write the electron configuration of Osmium.
2
10
2
6. Name the element with the electron configuration [Kr]5s 4d 5p .
7. Write a set of quantum numbers that could describe a 5d electron.
8. Which one of the following equations correctly represents the process relating to the ionization energy
of X?
+
A.
X(s) → X (g) + e
+
B.
X2(g) → X (g) + X (g)
C.
X(g) + e → X (g)
D.
X (g) → X(g) + e
+
E.
X(g) → X (g) + e
9. Give the full electronic configurations for the following. Which are paramagnetic?
(a)
Be
(b)
N
(c)
Ar
(d)
Sc
(e)
Cr
10. Write a set of quantum numbers (n, l, ml, ms) for one of the highest energy electrons in europium (Eu) (i.e.,
one of the last electron to fill in)
11. Arrange strontium, cesium, iodine and tin in order of increasing atomic size (radii)
12. Use the figure below to answer the following questions.
a. Given the representation of a chlorine atom, which circle might represent an atom of sulfur?
b. Given the representation of a chlorine atom, which circle might represent an atom of argon?
c. Given the representation of a chlorine atom, which circle might represent an atom of fluorine?
13. The energy required to remove an electron from an atom is known as _____________
14. Which would you predict to be more reactive, potassium or rubidium? Explain your answer.
15. State the relationship between the group number in the chart and the number of valence or outermost
electrons in an atom.
2
4
2
7
16. Name the element with the outermost electron configuration 4s 4p .
17.
a. Name the element with the electron configuration [Kr]5s 4d .
b. How many valence electrons does it have?
c. Draw an orbital box diagram for the valence electrons.
18. Choose the element or ion in each pair that has the greater electron affinity.
a. Iodine or Chlorine
b. Al or P
6 Chem 1A
Dr. White
Fall 2013
7 19. Choose the element or ion in each pair that has the larger radius.
a. Oxygen or Selenium
b. Sc or Zn
20. Choose the element or ion in each pair that has the smaller ionization energy.
a. Boron or Neon
b. K or Ge
21. Which of the following atoms has the smallest atomic radius?
a) Se
b) S
c) Te
d) O
22. Put the following elements (N, O, F, Be) in order of decreasing ionization energy. Circle the correct answer.
a. N > O > Be > F
b. O > N > F > Be
c. Be > N > O > F
d. F > O > N > Be
e. F > Be > N > O
f.
23. Which of the following atoms has the smallest electron affinity?
Br, Se, Ca, K
24. Fill in the following orbital box diagram for the 2p subshell with 4 electrons total: (use arrows to indicate
electrons – arrows can point up or down)
a)
b) How many unpaired electrons are there?
c) Which atom is represented here? (only one correct answer)
+
25. Would you expect the ionization energy for Li be less than or greater than the ionization energy for Li?
Briefly explain your answer.
26. Fill in the following table for each of the following atoms and ions.
Full Electron Configuration
,
2+
+
Abbreviated Electron
Configuration
# core
electron
s
# valence
electrons
Paramagnetic or
Diamagnetic?
3-
N, Cr, Li, Ga, Zn, Ag Rb, Fe , K , P , The two common ions of Pb*
27. Use your electron configurations above to answer the following:
a. A set of quantum numbers (not all possible values) for the first electron removed in the formation of
2+
Zn
b. A set of quantum numbers (not all possible values) for the first electron removed in the formation of
one of the lead ions.
Chem 1A
Dr. White
Fall 2013
8 Answers: Section 1-1: Significant Figures
1. 4
2. 2
3. 2
4. 4
5. 5
8. 2
9. 2
10. 4
11. 2
12. 6
15. 6
16. 533 g
17. 31 g/mL
18. 62 J/s
19. 0.01 m
3
22. 332.12 kg 23. 2.6 x 10 m 24. 264
25. 3.00
26. 1.0102
27. Left figure – 16.8°C
Center Figure – 22.21 mL Right Figure
1.
2.
3.
Symbol
6. 8
13. 3
20. 500 cm
7. 3
14. 1
21. 3.95 m/s
– 11.4 mL
Section 1-2: Dimensional Analysis and Significant Figures
-9
10
21. 4.56 x 10 cm
22. 3.030 x 10 ng
24. 263.43K
23. 163°F
25. 0.35 g
26. 33.9 mL
-2
27. 55 kg
28. 7.56 x 10 L
10
2
29. 2.59 x 10 km/day
30. 2.2 x 10 min or 220 min
2
31. 4.4 x 10 L
32. 0.193 mL
6
6
33. 3 x 10 lb
34. 3.6 x 10 g
4
3
-3
35. 1.1 x 10 cm
36. 5.6 x 10 mm
37. 4.2 g/mL
38. 0.793 g/mL
39. 0.870 g/mL
40. 140.4 g
Section 1-3: Atomic Structure Intro
(a) Rutherford used positive alpha particles. These alpha particles were directed at the gold foli. The
paths of these particles were examined.
(b) In the plum pudding model, negatively charged electrons reside in the atom surrounded by a
continuous medium of positive charge. At the time, it was believed that this continuous medium of
positive charge was very diffuse both in terms of charge and mass. The “pudding” was viewed as
having the same amount of charge as the electrons, but that charge was spread out over the whole
atom, as was the mass of the atom. Thus, Rutherford’s hypothesis was that the alpha particles would
pass right through the foil. As they did not all do this, but many were deflected or bounced back, he had
to revise the initial hypothesis to account for this. He did this by proposing that the atom has a positively
charged nucleus that contained most of the mass of the atom.
(c)
Since most of the atom is empty space, the particles will pass through the atom. Those that hit the
dense nucleus will bounce off the nucleus and are deflected.
Isotopes have different mass numbers due to a difference in the number of neutrons. They have the
same number of protons. Thus, X and Z are isotopes. They have the same number of protons (186)
but different mass numbers (412 and 410).
Atomic
Number
Mass Number
Number of
protons
Number of
neutrons
Number of
electrons
18
38
18
20
18
Chem 1A
Dr. White
15
7N
€
4.
Fall 2013
7
15
7
8
7
1
2
1
1
1
16
34
16
18
16
3
7
3
4
3
38
84
38
46
38
9 28.09 amu
Section 1-4: Wave and Particle Behavior
1.
a. x-ray < ultraviolet < infrared < microwave < radio waves
b. radio waves < microwaves < infrared < ultraviolet < x-ray
c. x-ray > ultraviolet > infrared > microwave > radio waves
2.
.
−25
3. 0.333m, 5.96 x 10 J
-4
20
4. a. 9.32 x 10 nm
b. 3.22 x 10 Hz
c. Energy is inversely proportional to wavelength, so a larger
-4
wavelength means a smaller Energy and vice versa. A wavelength of 242 nm is larger than 9.32 x 10 nm, so it
would have a smaller energy. A wavelength of 2200 Å is equal to 220 nm, so it also has a smaller energy.
-34
8
-9
-18
5. yes, an electron can be ejected E = hc/λ = (6.626 x 10 Js)(2.998 x 10 m/s)/120 x 10 m = 1.66 x 10 J>7.22
–19
x 10 J, so an electron could be ejected.
-10
-34
6. 2.4 x 10 m, 2.8 x 10 m
7. a. FALSE, wavelength of heavier object is shorter
b. TRUE
c. TRUE
-12
8. 3.31 x 10 m
-29
9. 1.49 x 10 m/s
- It doesn’t seem likely a human could be at this speed, we would have to be moving
VERY slowly, but not stopped.
10. a. emission
b. emission
c. absorption d. absorption
11. a. absorption
b. emission
c. emission
d. absorption
e. Longest wavelength corresponds to lowest change in energy. So, the EMISSION with the longest
wavelength is n=5 to n=2.
f. Highest frequency corresponds to highest change in energy, so the ABSORPTION with the highest frequency
is n=2 to n=4.
12. 433.9 nm
-22
13. a. 1.815 x 10 kJ b. The electron would need to absorb less energy since it is moving to a lower energy
level.
14. a. n = 3
b. To move to the n=1 level, the electron would emit a photon with more energy. Energy is
inversely proportional to wavelength, so it would emit a photon with a shorter wavelenth
15. nf = 7
-19
16. 5.445 x 10 J and 327.9 kJ/mol
Chem 1A
Dr. White
Fall 2013
10 Section 1-5: Atomic Structure and Electron Configurations
1. Quantum Number
Orbital or Electron Property
Orientation of orbital in 3-D
space
ml
Allowed Values
energy level and size
Can range from –l to + l,
including 0
Positive Integers from
n = 1 to ∞
orbital shape
0 to n-1
Electron spin
+1/2 or –1/2
n
l
ms
2. a) If n = 4, l = 3, 2, 1, 0
b) If l = 3, m = 3, 2, 1, 0, -1, -2, -3
3.
a. n = 1, l = 1, m = 0, ms = 1/2
if n = 1, l = 0
b. n = 2, l = 1, m = -1, ms = 1
ms = 1/2 or –1/2 only
c. n = 4, l = 2, m = -3, ms = -1/2
if l = 2, m = -2, -1, 0, 1 and 2, NOT -3!!
4.
a. this is a d orbital, L = 2
b. a d orbital cannot be found in the n = 2 level, when n = 2, l can only be either 1 or 0, for l to equal 2,
must be in n = 3 or higher
c. n = 4, l = 2 (we’re talking about d orbitals), m = 1, ms = 1/2 (these last two, just pick an allowed
value.)
2 14
6
5. [Xe]6s 4f 5d
6. Tin (Sn)
7. n=5, l=2, ml=-2, ms=1/2
(ml=-1, 0, 1, 2 also valid and ms=-1/2 also valid)
8. E
9. Give the full electronic configurations for the following. Which are paramagnetic?
2
2
a. Be 1s 2s
2
2
3
b. 1s 2s 2p
paramagnetic
2
2
6
2
6
c. Ar
1s 2s 2p 3s 3p
2
2
6
2
6
2
1
d. Sc
1s 2s 2p 3s 3p 4s 3d paramagnetic
2
2
6
2
6
1
5
e. Cr
1s 2s 2p 3s 3p 4s 3d paramagnetic
10. n=4, l=3, ml=+3 (could also be -3, -2, -1, 0, 1, or 2), ms=1/2
11. I<Sn<Sr<Cs
12. a. D
b. B
c. B
13. ionization energy
14. Which would you predict to be more reactive, potassium or rubidium? Explain your answer.
Rb would be more reactive/ It is larger, so it has a smaller ionization energy and hence is more reactive.
15. The group number tells the number of valence electrons for main group atoms.
16. There is no such atoms – it is missing the 3d electrons.
17.
a. Rh
b. 7 valence electrons
c.
l
l
l
l
l
l
5s
18. a. Chlorine b. P
19. a. Selenium
20. a. Boron
b. K
21. d
22. F > O > N > Be
b. Sc
4d
Chem 1A
Dr. White
Fall 2013
23. K
24.
a)
b. How many unpaired electrons are there?
2
c. Which atom is represented here? (only one correct answer) Oxygen
25. I expect the ionization energy for Li+ to be greater. Li+ has a full octet in its outer shell (it has the same
nd
electron configuration as a noble gas), therefore it is very stable. It would be much harder to remove the 2
electron from the very stable octet.
11 Chem 1A
Dr. White
Fall 2013
12 26. Full Electron Configuration
Abbreviated
Electron
Configuration
# core
electrons
# valence
electrons
P or D
N
1s2 2s2 2p3
[He]2s2 2p3
2
5
P
Cr
1s2 2s2 2p6 3s2 3p6 4s1 3d5
[Ar]4s1 3d5
18
6
P
Li
1s2 2s1
[He]2s1
2
1
P
Ga
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p1
[Ar]4s2 3d10 4p1
28
3
P
Zn
1s2 2s2 2p6 3s2 3p6 4s2 3d10
[Ar]4s2 3d10
28
2
D
Ag
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s1 4d10
(like Cu, full d subshell is very stable!)
[Kr]5s1 4d10
46
1
P
Rb
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s1
[Kr]5s1
36
1
P
6
18
Fe
2+
2
2
6
2
6
1s 2s 2p 3s 3p 3d
6
[Ar]3d
18
+
1s2 2s2 2p6 3s2 3p6
[Ar]
P3-
1s2 2s2 2p6 3s2 3p6
[Ar]
10
Pb2+: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2
4d105p66s24f145d10
[Xe]6s24f145d10
78
[Xe]4f145d10
78
K
Pb
2+
Pb4+
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2
4d105p64f145d10
6 (we had 8
valence and
removed 2)
0 (we
removed the
2 valence
electrons!)
8 (we had 5
valence and
added 3)
2 (we had 4
valence and
removed 2)
0 (we had 4
valence and
removed 4)
27. a. Since the 4s electron is removed: n= 4; l = 0; ml = 0, and ms = ½ [ or another possible answer is n= 4; l =
0; ml = 0, and ms =- ½]
b. Since the 6p electron is removed first: n = 6 ; l = 1; ml = -1; ms = ½ [other answers possible for ml and ms; ml
can be -1, 0, or 1 and ms can be + or – ½)
P
D
D
D
D