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C h e m i s t r y 3 0 S : C h e m i c a l R e a c t i o n s
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[A
1 H
hydrogen 1.008
11 N a
sodium 22.99
19
K [Ar]4s1
potassium 39.10 37
R b IKrUs1
rubidium 85.47 55
C s IXe]6s' cesium 132.9
87 F r [Rnj7s'
francium (223)
2A 4
B e {He]2s2
beryllium 9.012
12
Mg (Ne]3s2
24.31 20
C a [Ar]4s2
calcium 40.08
38 Sr [Kr]5s2
strontium 87.62
56 B a [Xe]6s2
barium 137.3
88 R a
[Rn]7s2
radium (226)
Lanthanide Series*
Actinide Series**
Los Alamos National Laboratory Chemistry Division
3B 4B 21 Sc [Ar)4s23d1
scandium 44.96
_ < r ~ .
Y [KrJ5s24d' yttrium 88.91
22 T i [Ar]4s23d2
titanium 47.88
21 Sc [Ar)4s23d1
scandium 44.96
_ < r ~ .
Y [KrJ5s24d' yttrium 88.91
40 Z r [Kr]5s24d2
zirconium 91.22
* 72 H f
[Xe]6s24f,45d2
hafnium 1 178.5
a|aje 104 R f
IRr,]7s25f146d2
rutherfordium (265)
/ 57
L a Xe]6s25d' lanthanum
138.9
"^58™"
C e fXeies f'sd'
cerium 140.1
89 A c [Rn]7s26d' actinium (227)
T h [Rn]7s26d2
thorium 232
S A I a m o s NATIONAL LABORATORY
[Rn]7s25f26d protactinium
231
tRr>17s-iSf',6dl
lawrencium (262)
( H E M S S T R Y
1
1
H +
Ihydrogenl
L i +
lithium
11
N a +
sodium
B e 2 +
beryllium
112 M g 2 +
acetate
arsenate
CH 3COO~ I TABLE OF POLYATOMIC IONS I oxalate C2O4 2-
C I O 4 -
acetate
arsenate A s 0 4 ^ dihydrogen phosphate H 2 P0 4 " perchlorate C2O4 2-
C I O 4 -arsenite AsOt hydrogen carbonate HC0 3 periodate I O 4 -benzoate QHsCOCT hydrogen oxalate HC2O4- permanganate M n 0 4 -borate B 0 3 ^ hydrogen sulfate H S O 4 - peroxide o 2
2 ~
bromate B r 0 3 - hydrogen sulfide HS~ phosphate
carbonate c o 3
2 ~ hydrogen sulfite HSO3- pyrophosphate PiO? 4"
chlorate c i o 3 - hydroxide O H - sulfate S0 4
2 ~ chloride c r hypochlorite c i c r sulfite so 3
2 ~ chlorite c i o r iodate i o 3 - thiocyanate SCN" chromate C r O ^ monohydrogen phosphate HPO4 2 " thiosulfate s 2 o 3
2 -cyanate CNO" nitrate N 0 3 " POSITIVE POLYATOMIC IONS
cyanide CN" nitrite N 0 2 " ammonium N H 4
+
H 3 0 + dichromate C r 2 0 7
2 - orthosilicate S i O A hydronium
N H 4
+
H 3 0 +
PERIODIC TABLE OF IONS KEY
atomic_^i number
symbol-
• "charge
I irorifll)
17 18
(iUPAC)
13 14 15 16
H" hydride
8 10 11 12
B boron
13
A l 3 +
aluminum
c carbon
14
Si silicon
N 3 " nitride
15
P 3 -
phosphide
8
O 2" oxide
F" fluoride
18
S 2" sulfiide
10
17
C I " chloride
H e helium
N e neon
118 A r
argon
19
K +
20
C a 2 +
calcium
21
S c 3 +
scandium
22 T i 4 +
M T i 3 +
«
23
vanadium(lll)
w (vanadium
C r 3 + 2 5 M f f 26 p e 3 +
iron (III)
""F iron l | ) _
2 7 C o 2 +
cobalt (II)
cobattjlll)
2 8 N « 2 +
nickel (II)
N i 3 +
nickel (111)
29 C u 2 +
copper (II)
C u +
C 0PP e rC)
30 31
Z n 2 +
zinc
G a 3 +
gallium
32
G e 4 +
germanium
33
As 3 " arsenide
34
S e 2 "
seleoide
35
Br "
bromide
36
Kr
krypton
37
R b H
I38
S r 2 +
39
Y 3 +
yttrium
40
Z r 4 +
zirconium
41 N b 5 +
Iniobium (V) N b 3 ;
niobium(lll)
43
M o 6 + T c 7 +
R u 3 +
Imolybdenuml technetium R u 4 + ruthenium(IV)|
45
R h 3 +
rhodium
4 6 P D 2 +
paladium(H)J
P d 4 +
paladium(IV)
47
A g +
silver
48
C d 2 +
cadmium
49
l n 3 +
indium
5 0 S n 4 +
tin (IV)
S n 2 +
tin (II)
5 1 S b 3 +
|antimony(lll)|
S b 5 +
52
T e 2 "
Itelluride
53
r iodide
54
X e
xenon
|55
C s +
cesium
|56
B a 2 +
barium
57
L a 3 +
lanthanum
72
Hf 4* hafnium
73
T a 5 +
tantalum
74
W 6 +
tungsten
75
R e 7 +
rhenium
76
O s 4 +
osmium
77
l r 4 +
iridium
78 p t 4 +
platinum(IV)| p t 2
|platinum(ll)|
7 9 A u 3 +
A u +
gold (i)
80 H g 2 +
[mercury
"Hg"+
mercury
81
(ll)|thallium T l +
(<) T l 3 +
(l)|thallium(lll)|
8 2 R b 2 +
lead (II)
P b 4 +
lead (IV)
83 g j 3 +
B i 5 +
bismuth(V)
8W1 Po<
poM«fft/)|
85
A t " astatide
86
R n radon
187 F r +
franciuml
88
R a 2 +
radium
89
A c 3 +
actinium 58
C e 3 +
cerium
59
P r 3 +
praseodymium
60
N d 3 +
neodymium
61
P m 3 +
pramethium
6 2 S m * samarium(lll)
S m 2 +
samarium(ll)
6 3 E u 3 +
europiumjlll)
E u 2 +
europium (II)
64
G d 3 +
gadolinium
65
T b 3 +
terbium
66
D y 3 +
dysprosium
67
H o 3 +
holmium
68
E r 3 +
erbium
69
T m 3 +
thulium
70 y b 3 +
ytterbium(lll)
Y b 3 +
ytterbium(ll)
71
L u 3 +
lutetium
90
T h 4 +
thorium
91 p a 5 +
protactiniumfV) •p a 4 ' + -
protaetiniumpv
92 y 6 +
uranium (VI)
uranium (IV)
93
N p 5 +
neptunium
94 p u 4 +
plutoniumJV)
P u 6 +
plutonium(VI)
9 5 A m 3 +
americium(lll)
" A m * americium(IV)
96
C m 3 +
9 7 B k 3 +
berkejium(lll)
98
C f 3 +
californium
99
E s 3 +
einsteinium F m 3 +
fermium
^ M . d 2 + 1 0 2 N O 2 ,
nobelium(ll)
103
L r 3 +
lawrencium
90
T h 4 +
thorium
91 p a 5 +
protactiniumfV) •p a 4 ' + -
protaetiniumpv
92 y 6 +
uranium (VI)
uranium (IV)
93
N p 5 +
neptunium
94 p u 4 +
plutoniumJV)
P u 6 +
plutonium(VI)
9 5 A m 3 +
americium(lll)
" A m * americium(IV)
96
C m 3 +
B k 4 +
berkelium(IV)
98
C f 3 +
californium
99
E s 3 +
einsteinium F m 3 +
fermium N o 3 +
nobelium(lll)
103
L r 3 +
lawrencium
A t o m i c S t r u c t u r e
A T O M COMPOSITION
Atoms are composed of positively charged protons, uncharged neutrons, and negatively charged electrons. The number protons and neutrons in the nucleus of each element is represented using an element's atomic mass (A.K.A. mass number). The number of
protons in an atom's nucleus is communicated using an atomic number. In a neutral atom the atomic number is also equal to the number of electrons orbiting the nucleus in shells or orbits. The outermost electron shell of an atom is called its valence shell.
IONS
Atoms may gain electrons or lose electrons and so become ions. Atoms which lose electrons have an excess of protons and are positively charged CA+IONS, whereas negatively-charged ions are called ANIONS.
B O H R DIAGRAMS
There are two main ways to represent atoms: the Bohr model, and the Lewis electron dot diagram. In the Bohr model, the number of protons and neutrons in the nucleus are recorded in the centre, and then circular shells are drawn around the nucleus. Electrons are placed into each shell according to the periods of the periodic table. Since the first period contains only two atoms, H and He, the first shell, called the K shell, can hold only two electrons. Therefore, H and He follow what is known as the duet rule. The B B —
second and third periods, representing the L and M shells, respectively, can accommodate 8 electrons. Atoms following this rule are said to require an octet of valence electrons in order to complete their outer shell.
When sketching Bohr diagrams of ions, typically the valence shell is complete with either two or eight electrons, depending on the structure of the atom. Furthermore, the atom must be enclosed in square brackets, and the sign and charge of the ion is included in the top right corner as a superscript.
L E W I S E L E C T R O N P O T DIAGRAMS
Lewis electron dot diagrams show only the valence electrons in an atom. The electrons are represented as dots around the element symbol in four quadrants—north, south, east, and west. As a rule, there can only be a maximum of two electrons in any quadrant. Generally, the number of valence electrons in an atom is the same as the group number in which it is placed on the periodic table, with the exception of He, the only noble gas to obey the duet rule. A A
A t o m i c M a s s - > 6 5
3 0 \ A t o m i c n u m b e r
P e r i o d i c T a b l e
Historical Development The periodic table was originally published in 1869 by Dimitri Mendeleev who classified the elements according to their regularly repeating physical and chemical properties as a function of their atomic mass—the number of protons and neutrons in an atom's nucleus. The modern periodic table was introduced in 1911 by A. van den Brock whereby elements are arranged according to their increasing atomic numbers—the number of protons in the nucleus, which, in a neutral atom, is equal to the number of orbiting electrons.
Organization The periodic table is represented as a grid separated into seven horizontal periods (A.K.. A. rows) and eighteen vertical groups (A.K.A. columns or families). Each period represents the filling of an outermost electron shell, called a valence shell. The groups are chunked into families with the following names.
Metals, Non-metals, and Metalloids All elements on the periodic table can be classified as metals, metalloids, or non-metals. The eight metalloids are found around a "staircase" beginning at boron. Metal atoms are located to the left of the staircase and non-metals to the right of the staircase. Hydrogen is a unique non-metal and is often placed with the alkali metals because of its valence of one.
Group Name Alkali metals
Alkaline earth metals Transition metals
Chalcogens or oxygen family Halogens
Noble gases
Group Number
B block
1A 2A
6A 7A 8A
Properties of Metals and Non-metals
M e t a l s N o n - M e t a l s
C o m p o u n d s a n d N a m i n g
Compounds are substances containing atoms of more than one element combined in fixed proportions. Compounds can be classified as either ionic or covalent. The following table highlights the main characteristics of each type of compound.
I o n i c C o m p o u n d s B i n a r y C o v a l e n t
C o m p o u n d s
• Composed of a metal cation (positively charged metal ion) and a non-metal anion or a complex ion (negatively charged).
• Electrons are completely transferred from the cation to the anion
Composed only of non-metals
Electrons are shared between atoms
• Can further be classified as Binary Ionic Compounds and Ternary Ionic Compounds
Both types of compounds are named simply by combining the names of both the cation and anion parts
If an atom has more than one possible combining capacity, then the Stock System of Naming or the Traditional System must be used
Prefix system of naming is used
Stock System uses roman numerals in brackets between the cation and anion names to indicate the oxidation state of the metal cation.
traditional System
» uses _ous and _ic suffixes to indicate the oxidation state of the metal cation.
\ Number of Atoms Prefix One mono Two </,•
Three ni Four tctra Five penta Six hexa
Seven hcpiu Eight acta Nine nona Ten deca
Diatoms, Tetra-atom, and Qcta-atom A number of atoms on the periodic table exhibit a unique characteristic in that they are only found paired with themselves in their free atomic state. That is, these elements are never found as single atoms. Al l of these atoms are covalently bonded in compounds, and, therefore, share electrons. Seven compounds are known as "the group of seven" or H-O-F-Br-I-N-Cl diatoms. In order to satisfy the octet or the duet rule, H, O, F, Br, I , N, and CI pair with themselves to become FF, C)2, F 2, Br 2, K N 2 . Cl 2 . Two other atoms on the periodic table; P, and S, combine to form the P4 tetra-atom, and S*, an octa-atom.
L a b : I o n i c B o n d i n g N a m e G a m e N a m e :
Objec t i ve : Your objective is to correctly write the chemical formulas and name ionic compounds.
Mate r ia ls : 6 ionic dice
P r o c e d u r e : 1) You will be placed in groups of four. Each group will be divided into
teams of two. 2) There are 5 different combinat ions of dice that you will be asked to roll. 3) For example, the 1 s t two dice you will roll are dice #1 and #4 4) You will roll the dice and then write down the positive and negative ion
on the appropriate sect ion of table 1. 5) Your goal is to determine the correct chemical formula and write the
correct name for the result ing ionic compound. 6) 1 person will roll the dice whi le the other will be responsible for
completing step 5. 7) One pair can start at table #1 and the other at table #5 .
T a b l e #1 Dice "+" H • u Chemical Formula Chemical Name ^ 1 i$4 ion ion
Rol l 1
Roll 2
Rol l 3
Rol l 4
Rol l 5
Rol l 6
T a b l e #2 Dice Chemical Formula Chemical Name
# 2 + #5 ion ion
Roll 1
Roll 2
Roll 3
Roll 4
Roll 5
Roll 6
T a b l e #3
D i c e # 3 + #5
" + "
ion
u u
ion
Chemical Formula Chemical Name
R o l M
R o l l 2
Ro l l 3
Ro l l 4
Ro l l 5
Ro l l 6
T a b l e #4 D i c e
U . If
+ U M
Chemica l Formula Chemical Name # 1 , + # £ ion ion
Ro l l 1
Ro l l 2
Ro l l 3
Ro l l 4
R o l l 5
Rol l 6
Tab le #5 Dice # 3 + #6
U . H +
ion
U M
ion Chemica l Formula I Chemical Name
R o l M
Rol l 2
Rol l 3
Rol l 4
R o l l S
Rol l 6
N a m i n g I o n i c C o m p o u n d s P r a c t i c e W o r k s h e e t
Name the following ionic compounds:
1) N H 4 C I
2) Fe (N0 3 ) 3
3) TiBr 3
4) Cu 3 P
5) SnSe 2 . .
6) GaAs ^
7) Pb (S0 4 ) 2
8) Be(HC0 3 ) 2 __
9) M n 2 ( S 0 3 ) 3 ^
10) AI(CN) 3
Write the formulas for the following compounds:
11) chromium (VI) phosphate
12) vanadium (IV) carbonate
13) tin (II) nitrite
14) cobalt (III) oxide
15) titanium (II) acetate
16) vanadium (V) sulfide
17) chromium (III) hydroxide
18) lithium iodide
19) lead (II) nitride
20 silver bromide
http ://www. chemf iesta. com
Binary Molecular Nomenclature
Rules for Binary Molecular Compounds Prefixes 1. The naming system is for compounds composed of two
nonmetallic elements. 1 - mono 2. The first element keeps its name 2 - d i
a. The first element gets a prefix if it has a subscript in the 3 - t r i formula 4 - tetra
3. The second element gets the -ide suffix (ending) 5 - penta a. The second element ALWAYS gets a prefix 6 - hexa
Compound Name Compound Formula Carbon dioxide
Carbon monoxide
Diphosphorus pentoxide
Dinitrogen monoxide
Silicon dioxide
Carbon tetrabromide
Sulfur dioxide
Phosphorus pentabromide
Iodine trichloride
Nitrogen triiodide
Dinitrogen trioxide
Compound Formula Compound Name N 2 0 4
S 0 3
NO
N 0 2
A s 2 0 5
P C I 3
C C I 4
H 2 0
SeF 6
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m e d i c a l s c a n s Where have all the medical isotopes gone? By Kelly Oowe, CBCNevws Posted: May 26, 2014 5:00 AM ET | Last Updated: May 26, 2014 10:43 PM ET
Isotope shortage
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About The Author
Albert Mann winced as he sat upright on the scanner bed to talk to me. A large camera had just processed an image of his body, reading the gamma rays radiating from his bones.
I asked him what he had to do to prepare for this test. "What did I have to do? Get cancer," he said.
Down the hall, seven other camera rooms were buza'ng with patients having similar scans. At this hospital in Brampton, Ont., the machines run six days a week.
Across Canada, about 20,000 patients undergo nuclear imaging procedures every week and the field of nuclear medicine is growing around the world.
But almost all of it rests on an increasingly fragile supply of radioactive isotopes, a short-lived medical product made mostly by small research reactors, and a looming shortage has specialists worried.
"I don't want to sound alarmist," said Dr. Norman Laurin, president of the Canadian Association of Nuclear Medicine. "But it's qoinq to have
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Kelly Crowe fvtodical science
Kelly Crowe is a medical sciences correspondent for CBC News, specializing in health and biomedical research. She joined CBC in 1991, and has spent 25 years reporting on a wide range of national news and current affairs, with a particular interest in science and medicine.
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medical consequences. There are people who are going to be denied care, or have a different kind of care that might not be the best for them.'
These patients probably won't know there's a problem until their test is cancelled, or they have to wait months for a diagnosis. But if they ask enough questions they'll find out the shortage of medical isotopes is largely a made-in-Canada problem.
Going out of business
Most Canadians don't realize it, but this country has been an international leader, the world's largest single supplier of medical isotopes used in nuclear imaging, for more than 50 years.
But all of that is about to end. Buried deep in the federal budget bill, now winding its way toward approval, is something called the Nordion and Theratronics Divestiture Authorization Act.
In a few short sentences that amendment removes all foreign ownership restrictions on Canada's medical isotope processor, Nordion, paving the way for the former Crown corporation to be sold to a U.S. firm.
The buyer, Sterigenics, is ready and waiting with an offer on the table. Shareholders will vote on May 27.
"The sale of Nordion represents the end of the road for the Canadian nuclear medicine file," says Dr. Norman Laurin said. But it is not so much the sale itself as the pending shutdown of the medical isotope production at the aging Chalk River research reactor in 2016 that will affect the diagnostic imaging business across this country.
What's an isotope
Amedical isotope is s adio<K;tive material that is injected into the body binding to specific tissue, and emitting gamma rays that are detected by special cameras.
These diagnostic tests have become an essential tool for helping diagnose cancer and heart disease, among other ailments, but the system will collapse without a reliable supply of the delicate isotopes, which decay quickly, losing radioactivity in -six hours.
That means they cannot be stockpiled. So every day fresh isotopes are produced in nuclear reactors and quickly shipped around the world before they e>pi re.
The federal government is determined to close Chalk River's isotope production as planned in less than two years, and when it does up to 40 per cent of the world's isotope supply will vanish, with no new supplier ready and waiting to fill the void.
Exit the business
Last week, Nordion reminded its investors about that problem. "Currently, the company does not have an alternative supply of reactor-based medical isotopes," it said in a news release. "It cannot be certain that it will be able to secure an alternate source of commercial supply that is viable."
Nordion says it is looking for a new source, but where? There are only a handful of other research reactors in the world producing isotopes, and most are also at the end of their life, and prone to the same repeated and unexpected shutdowns that have plagued Chalk River.
The new technologies for making the key isotope ingredient — technetium-99 — without a nuclear reactor are still in the research stage, with too many infrastructure and regulatory hurdles to be ready when Chalk River stops production.
Nordion bluntly warns that "if the company is unable to secure a long-term supply of medical isotopes it may exit the reactor-based medical isotope segment of its business." And that's exactly what some analysts expect will happen.
A recent report by the Nuclear Energy Agency of the Paris-based Organization for Economic Cooperation and Development is assuming Nordion's isotope processing capacity will
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"The loss of Canada's a r U i } S
processing capacity in
the second half of 2016 reduces current global processing capacity by approximately 25 per cent in that period." the report says.
It. also says there is "an increased risk" of supply shortages beginning as soon as next year.
Been down this road
That OECD warning triggered flashbacks for those Canadian nuclear medicine specialists who lived through the white knuckle crisis in 2009 when a leak shut down Chalk River for 15 months at the same time as an isotope-supplying Dutch reactor went down for repair.
• 'Patients wil l suffer* f rom 2009 shutdown, isotope shortages
In the political chaos that followed. Prime Minister Stephen Harper announced Canada would be getting out of the isotope business by 2016.
Ottawa would not follow the advice of its expert panel and build a new research reactor. Instead it offered $25 million in research money to encourage an alternative supply chain.
It would be up to the market to solve the problem of long-term isotope supply. And that likely means higher prices, because alternative processes like PET scans (positron emission tomography) are more expensive, and because private isotope suppliers will have to charge more to recover costs and make a profit.
"These isotopes are going to become more expensive, and I think it's right they should become more expensive," says OECD economist Ron Cameron. "In some cases they were sold at below the cost of producing them, that is economically unsustainable."
In other words, Canada helped create the world of nuclear medicine by making the supply of isotopes affordable, but without that low-cost option the situation changes.
It's the latest chapter in Canada's storied history of nuclear medicine, from the pioneering days in the 1950s, as this country moved from innovator and international leader to faltering player.
» From the archives: Chalk River's 'atom smasher"
There is a sadness among scientists and doctors as they watch this era of once ground-breaking science quietly come to an end.
"Getting out of it I think is a mistake," says Dr. Laurin. "It will mean a loss of Canadian know-how and expertise. Very good science will be lost, business and jobs are going to go elsewhere or be lost altogether, so I personally think it's a mistake.
"I may be biased because I'm a physician and I need what's being made at Chalk River for patient care. But if you look at it from a Canadian point of view, I think it's a loss for the entire community, not just Ontario, but all of Canada."
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Note: The Fact-Based and Issue-Based Article Analysis sheets must be copied back to back.
When you read the article, did it present a certain point of view about an issue under dispute? If so, use the other side of this sheet. If the article informed you but did not raise any concerns, use this side.
Key concept (written in a sentence). Write an article summary or definition in ^ " " ^ " 1 1 your own words. Do not list facts. Give
an overview.
Draw a figurative representation.
List your questions (at least two).
What are the facts? List at least five. |
List at least five key words.
Relevance to today: This is important or not important because . . .
6.114
Fact -Based Article Analysis: Used by permission of Lynda Matchullis and Bette Mueller, Nellie McClung Collegiate, Pembina Valley S.D. No. 27.
Chapter 6: Teaching and Learning Strategies
F a c t - B a s e d A r t i c l e A n a l y s i s
Key concept (written in a sentence):
The introduct ion of nylon 50 years ago
has to ta l l y revolutionized the way we live.
Draw a figurative representation.
T o o t h b r u s h
^ ^ T l M o n
-rent
CA £ ** tt/rvlS /
What are the scientific facts? List at least five.
1. Polymers are molecular chains of sub-unite.
2. Nylon can be heat set to make its yams coil and stretch much like telephone cords.
3. Chemists can string some of the polymer sub-units together in specific order.
4. Polymers can sometimes be made to fold up into molecular objects.
5. The simple rearrangement of molecules can transform air, water, and coal into nylon.
Chemistry 30S
Write an article summary or definition in your own words. Do not list facts. Give an overview.
Nylon was the first entirely synthetic fibre and
it totally changed industry and everyday life.
Nylon was used for many things (luggage,
toothbrushes, carpet, fishing line, surgical
sutures), but perhaps its best known use was
in fabrics. The popularity of nylon stockings led
to riots in the 19-406. The strength and
sheerness of nylon made it the ideal choice and
was also used in parachute cloth ropes and
tent fabric.
Nylon also led to other "unnatural" fibres
(Teflon, polyester, Spandcx) as well as work
with recombinant DNA and the biotechnology of
today. Work with protein polymers is likely to
create new methods of curing disease and to
serve as very small machines.
List your questions (at least two).
1. What are sub-units?
2. What is recombinant DNA?
Explain the technology presented.
Polymer technology was what created nylon. It involved stringing the sub-units of polymers together in a specific order to make a new substance. Polymer technology led to nanotechnology, which is folding up polymers into molecular objects which can serve as pieces of extremely small machines and electronic components. With molecular devices, one could make a microscopic computer.
List at least five key words.
• polymer • rearrangement
• revolutionized • molecular
• future • spin-offs
Relevance to today: This is important or not important because
Nylon is very Important today because it has created so many things and made so many changes in our lives, it has provided a far better material for many industries and Viae allowed the public to reap the benefits of these changes (e.g., no waiting at airports; stronger, sheerer stockings', affordable carpet; etc). Nylon has also opened up the way for new materials and new technologies such as Teflon, polyester, Spandex, recombinant DNA, biotechnology, and nanotechnology. Nanotechnology will surely be a great device in the future.
6.71
Name: Date:
Modelling Isotopes Lab
Purpose: Recent Canadian pennies (after 1978) consist of three different "isotopes." In this lab, you will mass quantities of each of these types of pennies so that you can determine both the mass of each "isotope" as well as the average "atomic mass" of a penny.
Materials: A container of Canadian pennies Balance
Procedure:
1. Take your pennies back to your desk and sort them into three groups—1996 and earlier, 1997-1999, and 2000 to the present.
2. Record the number of pennies in each group, as well as the total number of pennies.
3. Determine and record the mass of 10 pennies from each group.
Note: Canadian Pennies
Date Composition 1908-1920 95.5% copper, 3.0% tin, 1.5% zinc 1920-1941 95.5% copper. 3.0% tin, 1.5% zinc 1942-1977 98.0% copper, 0.5% tin, 1.5% zinc 1978-1979 98.0% copper, 1.75% tin, 0.25% zinc 1980-1981 98.0% copper, 1.75% tin, 0.25% zinc
(changed from round to 12-sided)
1982-1996 98.0% copper, 1.75% tin, 0.25% zinc 1997-1999 1.6% copper plating, 98.4% zinc 2000-present 4.5% copper plating, 1.5% nickel, 94.0% steel
Adapted from: Phillips, John S., Victor S. Strozak, and Cheryl Wistrom. Chemistry: Concepts and Applications. Ohio: Glencoe / McGraw-Hill, 2002. p. 102.
Chem 30S~Rubric For Modelling Isotopes Lab
Purpose (1 mark) Observations
Quantitative Isotope Number of pennies Mass of 10 pennies (g)
1996 and earlier 1997-1999 2000-present
Total
Isotope Average mass of the isotope (g)
Percent abundance Percent abundance x Average mass of the
isotope (a.m.u.) 1996 and earlier 1997-1999 2000-present
Total (7 marks)
Calculations (Show a sample calculation for each step) 1. Determine the average mass of the penny isotopes by dividing the mass of 10
pennies by 10. 2. Calculate the percent abundance of each isotope by dividing the number of
pennies from each isotope group by the total number of pennies in the container. 3. Calculate the atomic mass of a penny by
a) Multiplying the percent abundance by the average mass of the isotope. b) Adding the values from a).
(4 marks)
Conclusion —Statement of the "atomic mass" of a Canadian penny.
(1 marks)
Questions 1. Would the atomic mass be different i f you obtained a different container of
pennies containing a different number of pre and post 1997 pennies? Explain your answer.
2. Why did we use such a large sample size when determining the average atomic mass of each penny "isotope"?
(3 marks) Sources of error (not human errors!)
(1 mark)
T O T A L : 17 marks
Average Atomic Mass Worksheet
1) Rubidium has two common isotopes, 8 5 R b and 8 7 R b . If the abundance of 8 5 R b is 72.2% and the abundance of 8 7 R b is 27.8%, what is the average atomic mass of rubidium?
2) Uranium has three common isotopes. If the abundance of 2 J 4 U is 0.01 %, the abundance o f 2 3 5 U is 0 .71%, and the abundance of 2 3 8 U is 99.28%, what is the average atomic mass of uranium?
3) Titanium has five common isotopes: 4 6 T i (8.0%), 4 7 T i (7.8%), 4 8 T i (73.4%), 4 9 T i (5.5%), 5 0 T i (5.3%). What is the average atomic mass of titanium?
4) Explain why atoms have different isotopes. In other words, how is it that helium can exist in three different forms?
For chemistry help, visit www.chemfiesta.com © 2002 Cavalcade Publishing - All Rights Reserved
PERCENTAGE COMPOSITION N a m e
D e t e r m i n e t h e p e r c e n t a g e c o m p o s i t i o n o f e a c h o f t h e c o m p o u n d s b e l o w .
1. KMn0 4
K = _ _ _ _ _
_. .Mn = . . .
O = _ _ _ _ _ _
2. HCI
H = _ _ _ _ _ _ _
CI = _ _ _ _ _ _ _
3. M g ( N Q 3 ) 2
M g =
- N = _ _ _ _ _
O = _ _ _ _ _
4. ( N H 4 ) 3 P 0 4
N = _ _ _ _ _
H =
P =
O =
5. Ai 2(S0 4) 3
Al = _________
S =
O =
Solve t h e f o l l o w i n g p r o b l e m s .
6. H o w m a n y g r a m s o f o x y g e n c a n b e p r o d u c e d f r o m t h e d e c o m p o s i t i o n o f 100. g
o f K C I 0 3 ?
7. H o w m u c h i ron c a n b e r e c o v e r e d f r o m 25.0 g of F e 2 0 3 ?
8. H o w m u c h silver can b e produced f r o m 125 a of A C L S ?
i N a m e COMPOSITION OF HYDRATES
A hydrate Is an Ionic compound with wafer molecules loosely bonded to Its crystal * structure. The water Is In a specific ratio to each formula unit of the salt. For example, the formula Cu$O 4»5H 20 Indicates that there are five water molecules for every one formula unit of C u S 0 4 . Answer the questions below.
1. What percentage of water is found In CuSO 4 »5H 2 0?
2. What percentage of water Is found In Na 2S»9H 20?
3. A 5.0 g sample of a hydrate of BaCI2 was heated, and only 4.3 g of the anhydrous salt remained. What percentage of water was in the hydrate?
4. A 2.5 g sample of a hydrate of Ca(N0 3 ) 2 was heated, and only 1.7 g of the anhydrous salt remained. What percentage of water was in the hydrate?
5. A 3.0 g sample of Na 2CO 3»H 20 Is heated to constant mass. How much anhydrous salt remains?
6. A 5.0 g sample of Cu(NO3)2t>H20 Is heated, and 3.9 g of the anhydrous salt remains. What is the value of n?
M a r v i n D a M o l e S t r i kes A g a i n !
1. How many particles are there in one mole?
2. I t is estimated that a sample of matter contains 1.38 X 102 5 atoms. How many moles are present in the sample?
3. How many moles of barium are present in a sample having a mass of 22.3 grams?
4. A chemical reaction requires 3.7 moles of boron. What mass, in grams, of boron must be used in the reaction?
5. A sample of naturally occurring carbon has a mass of 1.732 grams. Calculate the number of moles of carbon in this sample.
6. A chemical reaction results in 57.2 grams of the gas carbon dioxide, C O 2 . How many molecules of gas were produced?
7. Calculate the mass of one trillion molecules of oxygen, O 2 .
8. Calculate the number of moles in:
a) 25 grams of oxygen, O 2
b) 0.27 g of ammonia, NH3
c) 10.5 g of sodium, Na
d) 347 g of ammonium nitrate, NH4NO3
9. Calculate the mass, in grams, of:
a) 1.24 moles of water, H 2 O
b) 0.269 moles of amonium chloride, NH4C1
c) 5.62 moles of sodium hydroxide, NaOH
d) 2.35 moles of sodium sulphate, N a 2 S C > 4
10. Calculate the number of molecules in:
a) 3.00 moles of chlorine, C I 2
b) 3.00 moles of uranium hexafluoride, UF^
e) 3.00 moles of hydrogen chloride gas, HCI
d) 3.00 moles of any kind of molecule
11. Calculate the number of atoms in:
a) 3.00 moles of chlorine, Ch
b) 3.00 moles of uranium hexafluoride, UF6
c) 3.00 moles of hydrogen chloride gas, HCI
d) 3.00 moles of ammonium sulphate, (NH^SC^
Answ er Menu
M a r v i n D a M o l e ! O u r H e r o
1. Calculate the mass of a) 2.00 moles of water, H 2 O b) 4.38 moles of chlorine, Cb c) 0.025 moles of ammonia, N H 3
d) 1.8 moles of oxygen, O2
2. Calculate the number of moles in a) 25 g of helium, He b) 12.5 g of methane, C H 4
c ) 0.364 g of iodine, 12 d) 40.0 g of sodium, Na
3. Calculate the number of particles in a) 2.50 moles of Neon, Ne b) 0.050 moles of iron, Fe
4. Calculate the number of moles in a) 9.03 X 1023 atoms of Cu
b) 3.76 X 102 5 molecules of S() 2
c ) 8.6 X 101 8 electrons
5. Calculate the number of molecules in a) 12.5 g of nitrogen, N 2 b) 0.76 g of ammonia, NH3 c) 0.60 g of hydrogen, H 2
6. Calculate the mass of a) 4.25 X 102 4 atoms of C
b) 6.02 X 102 1 molecules of H 2 0 c) one trillion atoms of Zn d) one atom of U
Answer Menu
Chemistry SOS Mole Calculations Lab
Objective
Solve problems requiring interconversions between moles, mass, and number of particles
Chemicals & Equipment
• Scupula • Electronic decigram • 100 mL beaker balance
• Zinc, mossy • Sodium chloride powder • magnesium sulfate powder
Procedure 1. In groups, students wil l answer one of the following questions showing all their
work, including units.
2. Once each group member has calculated and recorded their answers to the question they must have their calculations inspected by the teacher who wil l then permit them to weigh out the appropriate amount of substance.
3. After weighing, each group must have their substance inspected by the instructor.
4. The group must then return their massed substance to the original container and replace any equipment in their lab cupboard.
Conversions & Calculations 1. Mass out 2.01 x 102 2 atoms of zinc.
2. Mass out 0.1 mol of sodium chloride.
3. Obtain 0.02 mol of copper (II) sulphate.
5:6 Empfrlcal Formulas
EMPIRICAL FORMULA OF A COMPOUND
40.0% Carbon, 6.71% Hydrogen, 53.3% Oxygen
O
Percent Grams ins.
100 g sample
H 40 .0% ^ ••ml
• H 53 5% Cj|J:
• H i
¥
6.71% , ' f l i t
5 3 | s
Moles
12.0 J ^
1.01 b t H
3.33 16.0
Mole ratio
1.33 \
1.33 *
3L33 = 1 i 3.33 1
Empirical formula
CH^O
CHOH
E X A M P L E : E m p f r l c a l Formula What is the empirical formula of a compound which is 66.0% Ca and 34.0% P?
Solving process: Assume a 100-g sample so that we have 66.0 g Ca and 34.0 g P. Convert these quantities to moles of atoms.
FIGURE 5-4. This picture representation shows the steps In calculating an empirical formula from percentage composition.
66.0 g£a 1 mol Ca
34.0 g P
40.1
1 mol P
= 1.65 mol Ca
= 1.10 mol P
Dividing both results by 1.10, we obtain 1.50 to 1. This result is not close to a whole number. Substituting the fractional form of 1.5, we get 3/_. That ratio is 3 to 2. Thus, the ratio of Ca atoms to P atoms is 3 to 2 and the empirical formula is Ca 3 P 2 . Suppose we have an empirical formula problem which produces a ratio of 1 to 2.33. What is the correct whole number ratio? We can say 2.33 « 2Vs. Since 2V_ is %, the ratio is 7 to 3.
P R O B L E M S
Find the empirical formulas of the following compounds. 34. 1.67 g Ce, 4.54 g I 34. eel,
35. 31.9 g Mg, 27.1 g P 36. 4.04 gCs, 1.08 g CI £ JJ»JPa
37. 9.11 g Ni, 5.89 g F 37. NlF2
38. 6.27 g Ca, 1.46 g N 3 8 - C A 3 N '
DETERMINING EMPIRICAL FORMULAS
Name
I hat Is the empirical formula (lowest whole number ratio) of the compounds below?
1, 75% carbon, 25% hydrogen
2. 52.7% potassium, 47.3% chlorine
3. 22.1 % aluminum, 25.4% phosphorus, 52.5% oxygen
4. 13% magnesium, 87% bromine
5. 32.4% sodium, 22.5% sulfur, 45.1 % oxygen
6. 25.3% copper, 12.9% sulfur, 25,7% oxygen, 36,1% water
5:8 Hydrates
a expen-we need te of the dculated impound olecular lolecule. ;he ratio la. Then it in one unit has
e molec-
formula
d, if the i ( 29? B, if the s 184.5? mpirical
rentage 2.1.
a water s of the mtain a
to dry to
shows unit,
of the iply the NiS0 3-
1.A,*t.AV.^>:„i.; !i J M . ^
I S
FIGURE 5-5. The difference in color between the anhydrous and hydrated forms of a compound are shown. Heating drives off water molecules which causes the color change.
Tim Courlas
E X A M P L E : H y d r a t e C a l c u l a t i o n
We have a 10.407 gram sample of hydrated barium iodide. The sample is heated to drive off the water. The dry sample has a mass of 9.520 grams. What is the ratio between barium iodide, Ba l 2 , and water, H 2 0? What is the formula of the hydrate?
Solving process: The difference between the initial mass and that of the dry sample is the mass of water that was driven off.
10.407 - 9.520 = 0.887 g
The mass of water and mass of dry Ba l 2 are converted to moles.
9.520 ^Bafz 1 mol B a l 2
391.2j*£ai_-= 0.024 34 mol B a l 2
0.887 EJ*z& 1 mol H 2 0 18.0£jl_e-
= 0.0492 mol H 7 0
The ratio between B a l 2 and H 2 0 is seen to be 1 to 2. The formula for the hydrate is written as BaI 2 -2H 2 0.
P R O B L E M S
Find the formulas for the following hydrates. 44. 0.391 g L i 2 S i F 6 , 0.0903 g H 2 0 45. 0.737 g MgS0 3, 0.763 g H 2 0 46. 95.3 g LiN0 3 , 74.7 g H 2 0 47. 76.9% CaS0 3 , 23.1% H 2 0 48. 89.2% BaBr 2 , 10.8% H 2 0
44. L I 2 S i F 6 - 2 H 2 0
45. M g S 0 3 - 6 H 2 0 46. UN03 3H20 47. C a S 0 3 2 H 2 0 48. B a B r 2 - 2 H 2 0
The Mole
Molecular mass Is a whole number multiple of the empirical formula mass.
5:7 M O L E C U L A R F O R M U L A S
We have, thus far, calculated empirical formulas from experimental data. In order to calculate a molecular formula, we need one additional piece of data, the molecular mass. In one of the examples in the previous section, the empirical formula calculated was CH 2 0. If we know that the molecular mass of the compound is 180, how can we find the molecular formula? The molecular formula shows the number of atoms of each element in a molecule. Knowing that the elements will always be present in the ratio 1:2:1, we can calculate the mass of the empirical formula. Then we can find the number of these empirical units present in one molecular formula. In the substance CH 2 0, the empirical unit has a mass of
12 + 2(1) + 16, or 30.
It will, therefore, take six of these units to equal 180 or one molecular formula. Thus, the molecular formula is C 6 H 1 2 0 6 .
39. 40. 41. 42. 43.
C 6 H 6
C 2 H 2 O z C I 2
C 3 C I 3 N 3
T I 2 C 4 H 4 0 6
C 3 H A
P R O B L E M S
39. The molecular mass of benzene is 78 and its empirical formula is CH. What is the molecular formula for benzene?
40. What is the molecular formula of dichloroacetic acid, if the empirical formula is CHOCl and the molecular mass is 129?
41. What is the molecular formula of cyanuric chloride, if the empirical formula is CC1N and the molecular mass is 184.5?
42. What is the molecular formula of a substance with empirical formula T1C 2 H 2 0 3 and molecular mass 557?
43. Find the molecular formula for a compound with percentage composition 85.6% C, 14.4% H, and molecular mass 42.1.
Hydrates are crystals which contain water molecules.
5:8 H Y D R A T E S
There are many compounds which crystallize from a water solution with water molecules adhering to the particles of the crystal. These hydrates, as they are called, usually contain a specific ratio of water to compound. Chemists use heat to dry these compounds and then calculate the ratio of compound to water. An example of a hydrate is NiS0 3 -6H 2 0. The dot shows that 6 molecules of water adhere to 1 molecule of formula unit. To calculate the formula mass, we add the formula mass of the compound and water. For NiS0 3 we obtain 139. We multiply the 18 for water by 6 and add to the 139. The formula mass of NiS0 3 • 6H 2 0 is then 139 + 6(18), or 247.
DETERMINING MOLECULAR FORMULAS (TRUE FORMULAS)
Solve the problems below.
Name
1. The empirical formula of a compound Is N0 2 . Its molecular mass Is 92 g/mol. What is Its molecular formula?
2. The empirical formula of a compound Is CH 2 . Its molecular mass Is 70 g/mol. What is Its molecular formula?
3. A compound is found to be 40.0% carbon, 6.7% hydrogen and 53.5% oxygen. Its molecular mass is 60. g/mol. What is its molecular formula?
4. A compound is 64.9% carbon, 13.5% hydrogen and 21.6% oxygen. Its molecular mass is 74 g/mol. What is its molecular formula?
5. A compound is 54.5% carbon, 9.1% hydrogen and 36.4% oxygen. Its molecular mass is 88 g/mol. What is its molecular formula?
F O R M U L A O F A H Y D R A T E L A B
O B J E C T I V E S 1. Experimentally determine the percent composition by mass of water in a hydrate. 2. Calculate the formula of a hydrate. 3. Verify the Law of Definite Composition.
C H E M I C A L
Copper (II) sulfate hydrate (A.K.A. bluestone)
S A F E T Y • Wear eye protection over eyes • Keep a safe distance from hydrate when heating • Copper sulfate is toxic, rinse affected areas with copious quantities of water • Observe appropriate fire precautions • Clean up all spills immediately • Wash hands after performing the lab E X P E R I M E N T A L SETUP
DATA
1. mass of clean dry crucible . g 2. mass of blue hydrate . g 3. mass of dried salt and cool crucible . g 4. mass of dried salt . g 5. mass of water lost . g
OBSERVATIONS Sight, sound, and other sensory data (minimum of 3)
F O R M U L A O F A H Y D R A T E L A B
CALCULATIONS Your calculations must follow a logical and correct mathematical sequence, showing all steps and units. When you are finished, click here to check your calculations. I f you missed the lab, click here for sample data. 1. Percent composition of water by mass.
2. Moles of water.
3. Moles of dried salt.
4. Mole ratio of water to dried salt.
5. Experimental error of percent composition of water by mass.
GRAPHING 1. Construct a graph of moles of water versus moles of C U S O 4 in the hydrate from the data
sheet provided by the instructor. 2. Calculate the slope of the line of best fit. Include complete calculations including units on
the graph.
CONCLUSION Summarize the results of the experiment by answering the 3 objectives. Use the hydrate lab rubric to evaluate the soundness of your conclusion.
HAND IN title page, prelab, calculations, graph, and conclusion.
F O R M U L A O F A H Y D R A T E L A B
R E L A T I O N S H I P B E T W E E N T H E M O L E S O F W A T E R AND D R I E D S A L T IN A C O P P E E ( I I ) S U L F A T E H Y D E A T E I L L U S T R A T I N G T H E L A W O F D E F I N I T E COMPOSITION O F MASS
F O R M U L A O F A H Y D R A T E L A B
N A M E & PARTNERS: LAB GROUP NO.:
Pre Lab, Calculations, and Graphing Rubric (6 marks)
Criteria Novice Intermediate Expert
Prelab
Prelab has no additions from the lab outline (0) or contains less than 3
of the expert level criteria (1)
Prelab contains 3 of 4 expert level criteria (1.5)
Prelab contains all parts of the lab outline, including labelled sketch of
experimental setup, all recorded masses, and a minimum of 3 relevant
observations (2)
Calculations
No calculations submitted (0) or
significant errors or missing units (1)
Some units are nmsing and/or the mathematical sequence for
some calculations is incomplete or missing (1.5)
Correct answers to all 5 calculations, following a correct logical mathematical
sequence, showing all units (2)
Graphing Graph is missing (0) or less than 4 expert level criteria are present (1)
Graph contains at least 4 of six expert level criteria (1.5)
Graph contains labelled axes, including units and correct scales, correctly
plotted coordinates, a ruled trend line, and a correctly calculated slope (2)
Lab Report Conclusion Rubric (4 marks)
Criteria Novice Intermediate Expert
Conclusion is original
(no other criteria are evaluated i f this
criteria is assessed at the Novice level)
Aspects of the conclusion are the same
as those of another student or the entire
conclusion is the same as that of another
student
Intermediate level does not apply to this assessment criteria
Conclusion is written in the student's own words
Objectives summarized
Objectives are not summarized or clarified
(0)
Not all objectives are summarized, or objectives are
simply recopied using the same wording as the stated
objective^) (0.5)
Objective(s) is/are summarized or clarified in the student's own words (1)
Objectives answered by referencing
results
No conclusion present (0) or conclusion is
missing more than two of the expert level
criteria (0.5)
Conclusion contains all but two of the expert level criteria (1.5)
Each objective is correctly and concisely answered in paragraph format using results that are correct, including units and associated percent experimental error, and reference is made to the results from the graph and slope. (2)
Discrepancies between the
experimental results and the expected
results are explained
No explanation of the difference's) between
expected and experimental results (0)
Conclusion contains an explanation of the difference's)
between expected and experimental results or a
statement of explanation as to improvements to the lab design
that would not yield better results (0.5)
Conclusion contains an explanation of the difference(s) between expected and experimental results or a statement of explanation as to improvements to the lab design which demonstrates a self-reflection of the lab process and/or a deeper understanding of the concept
examined (1)
/ID
Part A:
Balance t h e f o l l o w i n g e q u a t i o n s us ing t h e l o w e s t w h o l e n u m b e r coe f f i c i en t s a n d
s ta te w h e t h e r t h e reac t i on is a syn thes is , d e c o m p o s i t i o n , s ingle d i s p l a c e m e n t ,
d o u b l e d i s p l a c e m e n t , o r c o m b u s t i o n r e a c t i o n .
Chemical Equation Type of Reaction
I . S 8 + 0 2 - > S 0 3 _ _ _ _ _ _ _
2- HgO -> Hg + _ 0 2
3. Na + H 2 0 NaOH + H 2
4 . C 1 0 H 1 6 + C l 2 - > C + HCI
5. . _FeS 2 + 0 2 - > F e 2 0 3 + S 0 2
6. C 5 H 1 0 + 0 2 - > , C 0 2 + H 2 0
7. K + Br 2 - » KBr
8. S i 0 2 + HF - > SiF 4 + H 2 0
9. KCIO3 - > KCI + 0 2
10. P 4 O 1 0 + H 2 0 - » H3PO4
I I . Sb + 0 2 - » S b 4 0 6
12. HCIO4 + P 4 O 1 0 - > H 3 P 0 4 + C l 2 0 7
13. C 2 H 5 O H + 0 2 C 0 2 + H 2 0
14. Ca 3 (P0 4 ) 2 + S i 0 2 + c - > CaSi0 3 + p4 + co not applicable
writing ana Baiancinl cnemlcal Equations
Part B:
Predict the products where appropriate and write the balanced chemical
equation for each word equation.
1. Sodium hydroxide - » sodium oxide + water
2. Iron + oxygen —» iron (lll)oxide
3. Carbon dioxide + water —> glucose (formula C 6 Hi 2 0 6 ) + oxygen
4. Iron (II) sulfide + hydrochloric acid (formula HCI) —»iron (II) chloride + hydrogen sulfide
5. Oxygen + hydrogen —»(predict the products of this synthesis reaction)
6. Chlorine + sodium iodide -^sodium chloride + iodine
7. Aluminum nitrate + sulfuric acid (formula H 2S0 4) ->aluminum sulphate + nitric acid
(formula HN0 3)
8. Silver oxide decomposes into its elements
9. Ammonium phosphate + barium hydroxide react in a double replacement reaction
10. Calcium hydroxide + nitric acid (formula HN0 3)
Stoichiometry Practice Problems (Level 1)
1. In the reaction shown here, what weight of iron is needed to react completely with 32.0 g of sulfur? F e r S - FeS
2. When zinc reacts with sulfuric acid, as shown here; what weight of hydrog e n is produced from 31.8 g of zinc?' Zn + H 2 S 0 4 — Z n S 0 4 + H 2 t
3. How much sulfurous acid can be produced when l2$-<jrof sulfur dioxide combines with water? S O f + ' H 2 0 — H 2 S 0 3 ' "
4. Silver bromide can be precipitated by the reaction of silver nitrate with sodium bromide. What weight of precipitate can be produced starting with 34.3 of sodium bromide? NaBr +• AgNOj - r NaN0 3 +• AgBr I
5. Hydrochloric acid is added to 50.0 g of iron (II) sulfide. What weight of hydrogen sulfide is produced? FeS t 2 HCI — FeGI 2 f H 2 S f
6. How much nitric acid is heeded to react completely with 25.0g of magnesium in the following reaction? Mg t 2 H N 0 3 r- Mg(N® s ) 2 +
F H^'r '
7. How much copper (I) chloride can be produced beginning with 75.0 g of copper (I) oxide? C u 2 0 + 2 HCI — 2 CuCf + H 2Q
8. What volume of oxygen gas is produced by the decomposition of 100.0 g / W t . = Z1A L of sodium nitrate? 2 NaN0 3 — 2 NaN0 2 + 0 2 t
9. What volume of oxygen is produced when 75.0 g of water is decomposed by electrolysis? 2 H 2 0 — 2 H 2 + O z f . • .
10. What volume of carbon dioxide is required to produce 50.0 L of carbon monoxide according to the following reaction? C 0 2 4- C — 2 CO f
Copyright 1964. Instructional Horizons. Inc. Published by J Weston Watch. Publisher. Portland. Maine 041040658
Stoichiometry Practice Problems (Level 2)
1. When aluminum is heated in oxygen, aluminum oxide is formed. What weight of the oxide can be obtained from 25.0 g of the metal?
2. When steam (hot water) is passed over iron, hydrogen gas and iron (III) oxide are formed. What weight of steam would be needed to react completely with 100.0 g of iron?
3. How much ammonium hydroxide is needed to react completely with 75.0 g of copper (II) nitrate in a double replacement reaction?
4. When ammonia is burned in oxygen, free nitrogen gas and water are produced. What volume of ammonia will react completely with 25.0 L of oxygen? What volume of nitrogen gas is formed?
5. When sodium carbonate reacts with hydrochloric acid, the carbonic acid that is formed immediately breaks down into carbon dioxide and water. What weight of sodium carbonate would have been present originally if 5.0 L of carbon dioxide were obtained in this way?
6. How much copper metal can be obtained by the single replacement reaction between copper (I) nitrate and 30.0 g of iron metal? (Iron [II) nitrate is formed.)
7. What weight of %uffuric acid will be needed to react completely with 35.5 g of ammonia in the production of ammonium sulfate?
8. Wh at weight of chlorine gas will be needed to react completely with 85.8 g of potassium iodide in a single replacement reaction? v
9. In the neutralization reaction between sulfuric acid and potassium hydroxide, how much potassium sulfate can be produced if you have 150.0 g of sulfuric acid to begin with?
10. What volume of nitrogen gas is needed to react completely with 150.0 L of hydrogen in the production of ammonia?
Copyright 1984. Instructional Horizons. Inc. Published by J. Weston Walch. Publisher. Ponland. Maine 04104 0658
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