Chemistry 121
Mines, Fall 2016
Answer Key, Problem Set 1 (complete solutions)
1. NT1; 2. 1.34; 3. 1.73; 4. 1.82 ([unofficial added part]: Also be able to report the quantities in (a) and (d) in standard scientific
notation [after the rounding] AND be able to answer this Q: In (b), before rounding, the “4” means four _______s?); 5. 1.76 (Add:
e) 83400 mi; f) 350.0 kg); 6. 1.78; 7. 1.80; 8. NT2; 9. 1.84; 10. 1.86; 11. NT3; 12. 1.72; 13. 1.147; 14. NT4; 15. 1.56; 16.
1.58; 17. 1.60; 18. 1.100; 19. 1.128; 20. 1.104; 21. 1.41 & 1.42; 22. 1.48; 23. 1.49
---------------------------------------------1. NT1. (a) Discuss the fundamental difference between a (scientific) law and a (scientific) theory. These are NOT the same
thing! (b) Discuss whether or not each of the following is "proven" and thus "true": (i) an observation, (ii) a scientific law, (iii) a
scientific theory. (c) Does knowledge in science need to be "proven" to be useful? Explain.
(NOTE: To save space, I have not copied the links or that part of the question in the answer key.)
(a) In short, a law describes what happens in nature, whereas a theory attempts to explain nature.
That is, laws are general statements of observation (often mathematical relationships between
quantities that appear to hold true); they are in the realm of “what happens”. They do NOT explain
why things behave the way they do. Theories are possible explanations for why things in nature
behave as they do. They are typically “models”. They are usually in the realm of “interpretation” in
my opinion. Once a certain law appears to hold true, scientists will often try to come up with a
theory to explain that law. Science cannot definitively prove theories. The best that can be done is
to gather so many observations and other experimental “data” that are consistent with the
proposed theory, that more and more people will be convinced that it is correct. But it is always
possible that one day an observation that is inconsistent with some part of a theory could be made,
and then if that observation is repeatable, at least part of the theory cannot be “correct” and must
be modified. Theories NEVER become laws, and laws NEVER become theories. They are
different “beasts”. Neither one is ever “proven”, just “supported” by evidence (observations and
measurements).
(b) As noted above, neither laws nor theories can be proven "absolutely" and can be said to be "true"
without qualification. Laws, which generally express simple mathematical relationships between
physical quantities, can be shown to hold true as best as our limits of measurement can
indicate or within specific well defined conditions or limitations, but that doesn't strictly make
them "true" absolutely. Do you see why?
Theories are even harder to pin a "true" label on. Since they are, in most cases, "models" of how
things "work", they, too, necessarily have limitations. The best we can say about "prevailing"
theories is that they are consistent with (and presumably reasonably "explain") all currently
available observations/measurements. They are supported by those data. But "support" and
"consistency" are not the same as "truth" or "proof". In fact, most of the time in science, theories
continue to get modified as more raw data are gathered. So the "main ideas" of theories often (but
not always!) remain essentially intact (i.e., considered to be correct) but not "100%" so. Dalton's
atomic theory is a great example of how a scientist can get certain "big" ideas seemingly correct
(i.e., we still believe them to be valid models 200 years later), even though many details of the
statements as they were initially made have been shown to be incorrect.
For example, Dalton stated that all atoms of a given element are identical. We now have raw data that
indicate they are not [there are different isotopes; see PS2]! However, the way in which they are
different (# of neutrons in each atom's nucleus) has such a tiny effect on chemical properties that it is
essentially undetectable in most chemical reactions and experiments.
Observations can be considered to be "true" if they are repeatable by other scientists. (However,
even here one must be careful because as the web author noted, we are human beings and our
senses can be fooled (as he demonstrates with an optical illusion)!).
(c) No. As long as observations are repeatable, laws hold true within well defined limits, and theories
are well supported by available data, a tremendous amount can be done that is useful (even
though these bits of scientific knowledge are not "proven" to be true in an absolute sense. The
laws of motion hold true within certain limits, and have been used, for example, to send probes to
other planets. Scientists of the 19th century learned a tremendous amount about the nature of
PS1-1
Answer Key, Problem Set 1
matter and chemical changes by applying Dalton's atomic theory (postulates), even though the
notion that atoms are "indestructible", as well as several other of his assumptions, were later
shown to be incorrect. Science knowledge is, almost by definition, always expanding and
changing. The goal of science is not "absolute truth"—it is a never-ending pursuit of trying to
determine how the physical universe behaves, and rationalize that behavior with models that we
can visualize.
2. 1.34. Classify each statement as (being closest to) an observation, a law, or a theory.
NOTES: 1. Observations and Laws are both in the “What happens?” realm. Theories (and
hypotheses) are in the “Why?” realm.
(Technically, in this problem you could not distinguish between a hypothesis and a theory since all theories
begin as hypotheses and “develop into” theories as data are acquired to support them. You are not told in
this kind of problem whether the “why?” statements given have a lot of support or not! I think that’s why the
authors left out “hypothesis” as an option in this problem.)
2. Observations describe a specific instance of “what happens?” whereas laws are general
statements or relationships (i.e., equations) about how nature behaves.
As such, statements that are laws will generally include the words “all” or “any” (or
similar words) or imply that the behavior is general, not specific.
(a) Chlorine is a highly reactive gas. Law (describes “what happens” (reacts); may at first appear to be
specific [only to chlorine], but implies it reacts with [almost]
any other substance)
NOTE: If the statement had been “Chlorine reacts with sodium to yield a white crystaline solid.”, it
would be an observation. But “being highly reactive” is not specific enough to be an
observation—it implies that it will react with “most substances”, which means some
substances that it has never been “tried” with. It is thus more predictive of future behavior
with “new” substances than an observation (which only states what happens with specific
substances). I hope this makes sense! This one “fooled” me initially. In fact, nobody
questioned my (wrong) key from 2011 until 2016, so I was not aware of my error until I
started noticing low scores on this problem in Mastering and thought about it further!
(b) If elements are listed in order of increasing mass of their atoms, their chemical reactivity follows a repeating
pattern.
Law (describes a relationship or pattern of behavior; implies “all” elements [not specific to one])
(c) Neon is an inert (or nonreactive) gas. Law (like (a)). Since no specific substance is listed as being
attempted to react with neon, the statement is general, implying that it won’t react with “anything”
(generalization—it has not been “tried” with everything in the universe! Just a large number of
substances, and it didn’t react with any of them.)
(d) The reactivity of elements depends on the arrangement of their electrons.
Theory (suggests a reason why some elements are reactive and others are not. Also, electrons
are not directly “observable”; they are part of a model [theory] of what comprises an
atom)
3. 1.73. Read each measurement to the correct number of significant figures. Note: Laboratory glassware should
always be read from the bottom of the meniscus [the often thick, slightly curved “border” at the top of a sample
of a liquid]
(a) 73.3, 73.4, or 73.5 mL (my best guess would be 73.4  0.1 mL )
The meniscus appears as a black line here, whose bottom lines up pretty close to (but, to me,
slightly below) the midpoint between the 73 mL mark and the 74 mL mark (obviously there is
uncertainty in any “reading”!)
PS1-2
Answer Key, Problem Set 1
NOTE: The answer in the back of the book is clearly wrong! There is clearly space between
the bottom of the black line and the 73.0 mL mark, and the answer should be reported as
a range here because of the uncertainty in the estimated digit.
(b)88.2, 88.3, or 88.4C. Probably even 88.5C.
The distance between 88 and 89 is so small on this scale, that I think the uncertainly is probably
 0.2 rather than just  0.1 C. My “best guess” would be 88.3 rather than 88.2 (the answer in
the back of the book)
(c) 643, 644, 645, 646, or even 647 are all acceptable to me (i.e., about 645  2 mL)
It looks to me like the top of the meniscus (black line) is right at the 650 mL mark, so the bottom
must be less than that. Since the mark is not actually alligned with the bottom of the meniscus
(they really should have rotated the glassware a bit to be most accurate!), it is hard to tell
precisely how far down it “goes”. It looks to be right around 645 mL where the mark is, but I think
it is a bit lower in the middle of the cylinder.
4. 1.82 Round each number to three significant figures. (Added in key [you are responsible for this skill on Exam 1]: Also
report each quantity in standard scientific notation [after the rounding] AND answer this Q: In (b), before
rounding, the “4” means four _______s?).
(a) 79,845.82
7
Rounded (3 SF)
In Scientific Notation (3 SF)
79800
7.98 x 10
4
7
(b) 1.548937 x 10
1.55 x 10
(c) 2.3499999995
2.35
(d) 0.000045389
0.0000454
-5
4.54 x 10
In (b), before rounding, the 4 means four “100,000’s” (one hundred thousands)
Reasoning
(a) The digit after the 8 is a 4, so you “round down”, meaning “keep the 8 an 8”. (Note: 79,845.82
is closer to 79,800 than it is to 79,900). The decimal point was moved 4 places. Since 7.98 is
smaller than 79,800, you must multiply it (7.98) by a number larger than 1 to make it equal
79,800. So the exponent is +4 not -4.
(b) The digit after the 4 is an 8, so you “round up”, meaning “make the 4 a 5”. (Note: 1.548937 is
closer to 1.55 than it is to 1.54)
The 4 can be seen to be in the 100,000’s place if it is expressed not in scientific notation:
7
15,489,370. Also, one could reason it this way: The 1 is in the 10 ’s place (from the scientific
7
6
5
notation “x 10 ”, the 5 is thus in the 10 ’s place (next one over), and the 4 is in the 10 ’s place.
5
10 = 100,000.
(c) The digit after the 4 is a 9, so you “round up”, meaning “make the 4 a 5”.
NOTE: This problem would have been a bit more challenging had they asked you to round the
value to four SF’s instead of three. To make sure you know how to do this, I’ll answer that
question here as well: 2.3499999995 would round to 2.350 (not 2.35!). You would need to
put that zero there to show that there are 4 SF, not 3.
(d) The digit after the 3 is a 8, so you “round up”, meaning “make the 3 a 4”. The decimal point
was moved 5 places. Since 4.54 is greater than 0.0000454, you must multiply it (4.54) by a
number smaller than 1 to make it equal 0.0000454. So the exponent is -5 not +5.
PS1-3
Answer Key, Problem Set 1
(extra Q): The “4” is in the hundredths place of 1.54…, but the quantity is in scientific notation and
7
the “x 10 ” needs to be accounted for in order to determine the true place of the “4”. 0.01 x
7
5
10 = 10 = 100,000. So the 4 is in the 100,000’s place.
5. 1.76. For each number, underline the zeroes that are significant and draw an x through the zeroes that are not:
NOTE: As noted in class (and PowerPoint), I think it is most meaningful to look at it this way: Start
counting significant figures from the first nonzero digit on the left. That digit, and every digit after
that, all the way until the end, regardless of whether it is a zero or not, is significant unless there is
no decimal point and there are trailing zeroes. In other words, as long as you start counting from
the first non-zero digit on the left, the only digits that are not significant are “trailing zeroes when
there is no decimal point”.
(a) 180,701 mi
Answer: 180,701 (i.e., the qty overall has 6 sig. figs. (SF, for short)
All digits are significant here, since the leftmost digit is not a zero, and there are no trailing zeros.
x xx
(b) 0.001040 m Answer: 0.001040
(i.e., the qty overall has 4 SF)
Every digit (starting) from the “1” until the end is significant since there is a decimal point.
x xx
(c) 0.005710 km Answer: 0.005710
(i.e., the qty overall has 4 SF)
Every digit (starting) from the “5” until the end is significant since there is a decimal point.
(d) 90,201 m Answer: 90,201 (i.e., the qty overall has 5 SF)
As in (a), All digits are significant here, since the leftmost digit is not a zero, and there are no
trailing zeros.
x x (i.e., the qty overall has 3 SF)
(e) 83,400 mi Answer: 83400
In this case, there is no decimal point, so the “trailing zeros” are not significant.
(f) 350.0 kg Answer: 350.0 (i.e., the qty overall has 4 SF)
In this case, there is a decimal point, so the “trailing zeros” are significant.
6. 1.78. How many significant figures are in each number [I’d say “quantity”]?
(a) 0.1111 s Answer: 4 SF (Decimal pt. present  start with the first nonzero digit and count till end)
(b) 0.007 m Answer: 1 SF (same as (a); you never count “preceding zeros”)
(c) 108,700 km Answer: 4 SF (No decimal pt. present  trailing zeroes are not significant)
11
(d) 1.563300 x 10
m Answer: 7 SF (same as (a))
NOTE: For a quantity in scientific notation, the 10 and its power are not included in counting SFs
(at all) [they only allow you to identify the true “decimal place” (meaning) of each digit in the
quantity (i.e., the “1” here really is in the “hundred billions” place)]
(e) 30,800 (should have a unit here?) Answer: 3 SF (same as (c); no decimal pt.  trailing zeroes
not significant)
7. 1.80. Indicate the number of sifnificant figures in each number (I’d say “quantity”). If the number is an exact number
(quantity), indicate an unlimitted number of significant figures.
(a) 305435087 (2008 U.S. population) Answer: unlimited # of SF (“counted” numbers are “exact”;
implied unit is “people”)
(b) 2.54 cm = 1 in Answer: unlimited # of SF
(This is a bit tricky; this particular “equivalence” is a defined quantity [i.e., an inch is defined to be
exactly 2.54 cm], although you could not know this just by looking at it. You’d have to have looked
at a proper table of equivalences, in which it should be indicated as being “exact” (as in Table 1.3
in your textbook. All equivalences within a given unit system (i.e., in the SI system) are exact, but
PS1-4
Answer Key, Problem Set 1
those between unit systems are usually not exact, and are often shown in different tables rounded
to a different number of significant figures because of “convenience”.)
(c) 11.4 g/cm (density of lead) Answer: 3 SF (Measured [experimentally determined] quantity 
not exact)
3
(d) 12 = 1 dozen Answer: unlimited # of SF (see explanation in (b). This is a defined qty)
8. NT2. Two teams of students measured the length of a standard object for a lab experiment on making measurements. The
measurements from the two groups are shown below, along with the average of each team's five measurements. The
manufacturer of the object guaranteed its length to be 0.28671  0.00002 m.
AVG:
Team 1
0.301m
0.29 m
0.27 m
0.29 m
0.28 m
Team 2
0.242 m
0.246 m
0.248 m
0.244 m
0.245 m
0.29 m
0.245 m
(a) For each group’s average, state (i) the uncertain but significant digit, (ii) the decimal place of that digit, (iii) the
approximate value of the uncertainty in the quantity, and (iv) the number of significant digits.
Answers:
Average:
(i) Uncertain but significant digit (underlined in row 1)
(ii) Decimal place of the uncertain but significant digit
(iii) Approximate value of the uncertainty in average
(iv) Number of significant figures (digits) in average
Team 1
0.29 m
the 9
hundredths
 0.02 m
two
Team 2
0.245 m
the 5
thousandths
 0.003 m
three
Explanation: Team 1’s average has an uncertainty of about  0.02 m. That is, their
measurements indicate that the length of the object is somewhere between about 0.27 m and
0.31 m. A way to indicate this directly is to report the value as: 0.29 ± 0.02 m (see Experiment 1
handout). Team 2’s average has an uncertainty of about 0.003 m since their individual
measurements are all no more than about 0.003 m above or below the average value of 0.245 m.
It could be reported as 0.245 ± 0.003 m. The uncertain but significant digit is always the digit
in the decimal place that matches the approximate uncertainty. That is, if the uncertainty is 
0.02 m, then that means that the digit in the hundredths place is not certain (but is still significant).
NOTE: The number of significant figures is not generally equal to the number of decimal places, although they
happen to be numerically equal in this problem. So just to clarify, if another object were measured and
Team 1 measured it to be 23.12  0.02 m , and Team 2 got 23.073  0.003 m, the values of the
uncertainty would still be 0.02 m and 0.003 m respectively, but the number of significant figures would be
four and five respectively.
(b) Assuming all the students are equally competent at measuring, offer an explanation as to why each team's
measurements are reported to a different number of digits. Be specific.
Answer: Each team must have been using a different ruler. A proper, precise measurement
is one in which the last digit is estimated. Since Team 1’s values were all reported to the
hundredths of a m, we can infer that the ruler they were using had markings only every tenth of a
meter (and they estimated the hundredths place value). Since Team 2’s values were reported to
the thousandths place, their ruler must have had markings every hundredth of a meter, allowing
them to estimate the value of the digit in the thousandths place.
(c) Which group’s final result (average) is more accurate? Which is more precise? Give your reasoning.
PS1-5
Answer Key, Problem Set 1
Team 1’s average is more accurate, because its value (0.29 m) is closer to the presumed
“actual value” given by the manufacturer as 0.28671  0.00002 m (0.29 m is 0.00329 m
greater than the manufacturer’s reported value, and 0.245 m is 0.04171 m below; the latter
value is much farther away).
Team 2’s result is more precise, because their measurements were closer to one another (all
within 0.003 m of their average) than Team 1’s measurements were (only within 0.02 m of
their average). Another way to look at this is that the value of the uncertainty is smaller (
0.003 m <  0.02 m) for Team 2.
NOTE: You might wonder how a more precise ruler can give a less accurate result. Simple! Perhaps the ruler was
marked improperly at the factory. Perhaps the markings that were supposed to be 0.01 m apart were actually
0.009 m apart instead. The ruler would still “look” fine, and you could make very precise measurements with it (to
within what appears to be about plus or minus 0.001 m, with estimation), but the value of length that you’d obtain
would be “off”, right? That is, it would be inaccurate (even though precise).
(d) Which group's ruler would you use if you were to make another measurement of a different object? Why?
I’d use Team 1’s ruler. It is basically pointless to use an inaccurate measuring tool unless you
know how to correct for the inaccuracy. It’s better to use a less precise tool that is accurate.
(Note: You might try to use the above set of measurement to “calibrate” Team 2’s ruler (correct for the
inaccuracy) so that you’d then have a precise AND ACCURATE ruler, but that would involve more work
and several assumptions that may or may not be true.).
9. 1.84 (only b-d shown in key). Calculate to the correct number of significant figures.
5
2
2
(b) (5.01 x 10 ) ÷ (7.8 x 10 ) = 642.30…(on calculator*) = 640 (or better yet, 6.4 x 10 ) [2 SF]
*NOTE: Although I hope you all know your algebra, I strongly suggest that you use the EE or EXP button on your
calculator whenever you have calculations that involve values in scientific notation. That is precisely what that
button is for. If you are unsure of what this button does or how to use it, please either see the “Calculator_EEButton” handout at
http://www.oakton.edu/~gmines/CHM121/PS01-Wk1/NotHandedOut/
or ask me in person. (Also, the “…” above indicates that there were extra digits on my calculator that I chose
not to write down since I knew they would not affect the final result.)
Explanation (SF’s): This calculation involves only multiplication and division. Thus use the x / ÷
5
2
rule: 5.01 x 10 has 3 SF, 7.8 x 10 has 2 SF. Thus, the rule estimates
that the result should have only 2 SF.
(c) 4.005 x 74 x 0.007 = 2.07459 = 2 [1 SF]
Explanation: Same as (b). Use x / ÷ rule: 4.005 has 4 SF, 74 has 2 SF, and 0.007 has just 1 SF.
So the rule estimates that the result should have only 1 SF.
NOTE: If you redo the calculation using 0.006 instead of 0.007, the result is1.77822. Thus, 2.1 (i.e., 2 SF) would
underestimate the uncertainty. Although 2 (1 SF) slightly overestimates the uncertainty, that is better than
the other way around. Remember, using the SF rules is only a way of estimating the uncertainty—it is not
“perfect”.
(d) 453 ÷ 2.031 = 223.04… = 223 [3 SF]
Explanation: Use x / ÷ rule: 453 has 3 SF, 2.031 has 4 SF, so result has only 3 SF.
10. 1.86 (only a,c,d shown in key). Calculate to the correct number of significant figures.
(a) 0.004 + 0.09879 = 0.10279 = 0.103 (uncertainty in the thousandths place)
Explanation (SF’s): This calculation involves only addition or subtraction. Thus use the + / - rule:
rd
0.004 has uncertainty in the thousandths (3 decimal) place; 0.09879 has
th
uncertainty in the hundred thousandths (5 decimal) place. So the result is
rd
uncertain in the thousandths (3 decimal) place.
(c) 2.4 – 1.777 = 0.623 = 0.6 (uncertainty in the tenths place)
PS1-6
Answer Key, Problem Set 1
st
Explanation (SF’s): Same as (a). Use + / - rule: 2.4 has uncertainty in the tenths (1 decimal)
rd
place; 1.777 has uncertainty in the thousandths (3 decimal) place. So the
st
result is uncertain in the tenths (1 decimal) place.
(d) 532 + 7.3 – 48.523 = 490.777 = 491 (uncertainty in the units place)
Explanation (SF’s): Use + / - rule: 532 has uncertainty in the units place; 7.3 has uncertainty in the
tenths place; 48.523 has uncertainty in the thousandths place. So the result is
uncertain in the units place.
11. NT3. Perform the following mathematical operations, and express the results of the following calculations with the
correct number of significant figures.
REMEMBER:
1) You should not round (to indicate the proper number of significant figures) until the end of a calculation. (But
you do not need to write all of the digits throughout. See Note below part (a).)
2) Use underline marks to indicate which digit is the uncertain but significant one throughout the calculation.
Answers: (a) 0.125 g; (b) 3.1 g/mL
Work/Reasoning:
(a)
1)
3.41 g - 0.23 g
3.18 g
x 0.205 mL =
x 0.205 mL
5.233 mL
5.233 mL
1) apply + / - rule; decimal place dictates
uncertain digit [hundredths here b/c of 0.23]
2)
 0.607682 g/mL x 0.205 mL = 0.1245748 g = 0.125 g
2) apply x / ÷ rule [twice]; 3 SF only
NOTE: If you keep just ONE extra digit after the underlined one in the intermediate step, you get the same result:
0.6077 g/mL x 0.205 mL = 0.12458 g = 0.125 g. In general, if you do these on your calculator, you can carry
all of the digits forward, but when your WRITE the results of each step, you need only carry one or two digits
beyond the uncertain one (to avoid rounding error). You will see in part (b) below that I often use a series of
dots (.....) to show that my calculator showed more digits than I have written
3)
1)
(b)
5.556 mL x 2.3 g/mL
12.7788 g
=
 3.0844... = 3.1 g/mL
4.223 mL - 0.08 mL
4.143 mL
3) apply + / - rule last; 12.7788, with
two SF limits the result to 2 SF
2)
1) apply x / ÷ rule in numerator; 2.3 limits result to 2 SF;
2) apply + / - rule in denominator; 0.08 limits result to hundredths place])
Note: Again, if you keep only 1 digit after the underlines, you get the same final result after rounding:
12.8 g
 3.09  3.1 g/mL
4.143 mL
12. 1.72. Human fat has a density of 0.918 g/cm3. How much volume (in cm3) is gained by a person who gains 10.0 lbs of pure fat?
3
Answer: 4940 or 4.94 x 10 cm
3
Reasoning/Work:
10.0 lb x
453.59g
1 cm3
x
 4941.06...  4940 or 4 . 9 4 x 1 03 cm 3
lb
0.918 g
OR
10.0 lb x
453.59 g
4535.9 g
3
3
 4535.9 g ;
 4941.06... 4940 or 4 . 9 4x 1 0 cm
lb
0.918 g/cm3
PS1-7
Answer Key, Problem Set 1
13. 1.147. For each box, examine the blocks attached to the balances. Based on their positions and sizes, determine
which block is more dense (the dark block or the lighter-colored block), or if the relative densities cannot be
determined. (Think carefully about the information being shown.)
greater density
greater density
(b)
(a)
cannot tell
(d)
(c)
[added in key for practice only]
NOTE: This was in a Mastering problem, but you should be able to explain your reasoning in such a
problem. Be extra careful to distinguish “mass” from “density” as well.
(a) Answer: dark block is more dense (greater density)
Reasoning: The dark block has greater mass (pulled lower on balance) in a smaller volume
(looks smaller).
Conceptually, more mass "squished" into a smaller space means "greater density".
Mathematically, a greater numerator (mass) and smaller denominator (V) both make the
quotient (m/V) larger.
(b) Answer: lighter block is more dense
Reasoning: The volumes are equal (based on visual inspection), but the lighter block has the
greater mass (pulled lower on balance).
Conceptually, more mass "squished" into the same amount of space means "greater density".
Mathematically, for a given value in the denominator (V), a greater numerator (mass) makes
the quotient (m/V) larger.
(c) Answer: cannot tell
Reasoning: The dark block has greater mass (pulled lower on balance), but it is in a larger
volume (looks larger). Thus without quantitative information about how many times larger the
mass and volume are compared to the lighter one, there is no way to know which block has the
greater density.
For example, if the mass were 2 x larger, but the volume were 3 x larger (than the lighter block), the white block
would have a smaller density. But if its mass were 3 x larger and the volume 2 x larger (than the ligher block), the
the dark block would have the greater density.
In general, if the numerator (mass) AND the denominator (V) are both larger (or both smaller), you
would need to know which one is bigger (or smaller) by the larger factor in order to make a
conclusion about relative densities.
----------------------(d) [was not assigned; just added in key for completeness and practice] Answer: lighter block is more dense
Reasoning: The masses are equal (in balance), but the lighter block has the smaller volume
(looks smaller).
Conceptually, a given amount of mass "squished" into a smaller space means "greater density".
PS1-8
Answer Key, Problem Set 1
Mathematically, for a given value in the numerator (mass), a smaller denominator (V) makes
the quotient (m/V) larger.
14. NT4. A 25.00-g sample of a solid is placed in a graduated cylinder and then the cylinder is filled with benzene until
the reading is 50.0 mL. The mass of benzene and solid together is 58.80 g. Assuming that the solid is
insoluble in benzene and that the density of benzene is 0.880 g/cm3, calculate the density of the solid.
I think it is useful to make a picture here (certainly at least in your head, if not on actual paper):
Benzene (enough to make the reading
on the cylinder to be 50.00 mL)
Total mass of
benzene + solid = 58.80 g
Total volume of
benzene + solid = 50.00 mL
25.00 g sample of solid
Strategy:
1) It is often useful to sort out the information in a complex problem and write down the "givens"
and the "asked for". In a sense, the givens ended up in my picture above (this is one reason
that pictures are helpful!)
Givens:
Asked for:
 mass of solid 

 volume of solid 
density of solid  
mass of solid = 25.00 g
mass of benzene plus solid = 58.80 g
3
volume of benzene plus solid = 50.0 mL= 50.0 cm
3
density of benzene = 0.880 g/cm
2) You are asked for the density of the solid. Thus you will need to determine both the mass and
the volume of the sample of solid in this problem. The density will be the mass divided by the
volume since density is "mass per a given volume".
3) The mass of solid is given! It's 25.00 g! Thus the hard part of this problem is figuring out the
volume of the solid (which is NOT given in the problem).
4) Recognize that the total volume of the benzene + solid is 50.0 mL (given). To get the volume of
the solid, you need to figure out the volume of the benzene and subtract it from the total
(because there is nothing in the problem that can directly get you the volume of the solid!).
NOTE: You must realize that the volume of the benzene is not given in this problem (it is NOT 50.0
mL!). From the picture (or a careful analysis of the wording in the problem), hopefully you can see that
50.0 mL equals the total volume of benzene + solid. If not, ask me.
5) So how do you get Vbenzene? Recognize that you can get the mass of the benzene by
subtracting the mass of the solid (given) from the mass of the total (also given). And once you
have the mass of the benzene, you can get its volume from knowing its mass and its density
(which is given)!
6) Go back and calculate Vsolid = 50.0 – Vbenzene, and then dsolid
Execution of Strategy:
mbenzene = 58.80 g (total) – 25.00 g (of solid) = 33.80 g
Vbenzene =
33.80 g benzene
0.880 g/cm3 benzene
3
 38.409... cm 3
Or you could use dimensional analysis:
33.80 g benzene x
3
3
Vsolid = 50.0 cm (total) – 38.409.. cm (benzene) = 11.59... cm solid
PS1-9
1 cm3
 38.409... cm3 benzene
0.880 g benzene
Answer Key, Problem Set 1
dsolid =
25.00 g solid
 2.156...  2.16 g/cm 3 (solid)
11.59... cm3 solid
15. 1.56 (only a,b shown in key). Use prefix multipliers to express each measurement without any exponents.
5
(a) 38.8 x 10 g
-10
(b) 55.2 x 10
Answer: 3.88 Mg (megagrams)
s Answer: 5.52 ns (nanoseconds)
Explanation/Strategy: First find the “nearest” power of ten that corresponds to an SI prefix (Table 1-2).
Then either stepwise (if you are not as math “comfortable” as you’d like) or “in your head” (if you’re
really confident about such things) take into account how many powers of ten you need to change the
power that is given in order to keep your quantity’s value the same. For example:
6
5
(a) “10 ” (closest to 10 ) is associated with the prefix “mega”, with abbreviation “M”. So, multiply the 10
6
by 10 to get 10 , and divide 38.8 by 10 to keep the magnitude of the product the same:
5
38.8 x 10 =


38.8
5
x 105 x 10  3.88 x 106  3.88 "mega" . Thus, 38.8 x 10 g = 3.88 megagrams
10
-9
-10
(b) “10 ” (closest to 10 ) corresponds to “nano” (n). So multiply 10
-10
55.2 x 10
5
=

-10
by 10 and divide 55.2 by 10:

55.2
-10
x 10-10 x 10  5.52 x 10-9  5.52 "nano" . Thus, 55.2 x 10 s = 5.52 ns.
10
16. 1.58 (only a,d shown in key). Use scientific notation to express each quantity with only the base units (no prefix
multipliers).
(a) 35 L Answer: 35 x 10 L = 3.5 x 10 L
-6
-5
-6
( means “micro” and “10 of”)
-2
-2
(d) 1.5 cg Answer: 1.5 x 10 g (cmeans “centi” and “10 of”)
17. 1.60 (only a,b shown in key). Complete the table.
7
(a) 355 km/s
35,500,000 (or 3.5 x 10 )_cm/s
_355_ m/ms
(b) 1228 g/L
_1.228_ g/mL
_1,228,000 (or 1.228 x 10 )_ kg/ML
6
Reasoning/Work:
“In your head” reasoning (recommended only for those really “understand” units already):
5
(a) A km is 1000 x a meter, and a meter is 100 x a cm, so a km is 1000 x 100 (=10 x) a cm. 355
5
2
5
7
x 10 = (3.55 x 10 ) x 10 = 3.55 x 10 .
A km is 1000 x a meter, and a s is 1000 x a millisecond, so the quotients are the same here
(numerator and denominator are both 1000 times smaller in m/ms).
(b) A L is 1000 x a mL. So the number of grams in a mL must be 1000 x smaller than the
amount in a L.
A g is 1000 x smaller than a kg, so when you convert to kg, the number must be 1000
6
smaller. However, a L is 10 x smaller than a ML, so the number of “things” in a ML must be
6
6
3
3
10 x greater. 10 / 1000 = 10 , so the number is 10 times the original
Dimensional Analysis approach (for those that want a more “regimented” approach):
3
-2
-3
(a) 1 km = 10 m; 1 cm = 10 m; 1 ms = 10 s
#
Thus:
cm 355 km
103 m
1 cm

x
x
 355 x 105  3.55 x 107 cm/s
s
s
1 km
10-2 m
PS1-10
Answer Key, Problem Set 1
#
m 355 km
103 m 10-3 s

x
x
 355 m/ms
ms
1 ms
s
1 km
-3
6
3
(b) 1 mL = 10 L; 1 ML = 10 L; 1 kg = 10 g
Thus:
#
g
1228 g 10-3 L

x
 1.228 g/mL
mL
1 mL
L
#
kg 1228 g
1 kg
106 L

x
x
 1.228 x 106 kg/ML
3
ML
1 ML
L
10 g
18. 1.100 (only a,c shown in key). A bedroom has a volume of 115 m3. What is its volume in:
(a) km3?
#km3  115 m3 x
1 km
1 km
1 km
x
x
 1.15 x 10-7 km3
1000 m
1000 m
1000 m
(c) cm3?
# cm3  115 m3 x
1 cm
1 cm
1 cm
x
x
 1.15 x 108 cm3
2
2
2
10 m
10 m
10 m
You could do these multiple ways.
Tro has you create the conversion
factor with cubes in it first. Either
way is fine.
19. 1.128. The world’s record in the 100 m dash is 9.69 s and in the 100 yard dash is 9.21 s. Find the speed in mi/hr of the
runners who set these records.
(a) #
mi 100 m
60 s
60 min
1 km
0.62137 mi

x
x
x
x
 23.084...  23.1 mi/hr
3
hr 9.69 s
1 hr
1 min
10 m
1 km
(b) #
mi 100 yd
60 s
60 min
1m
1 km
0.62137 mi

x
x
x
x
x
 22.209...  22.2 mi/hr
3
hr
1
hr
9.21s
1 min
1.0936 yd
10 m
1 km
NOTE: The “100 m” & “100 yd” here have been interpretted to mean something like 100.0 units (i.e. 4 SF rather than 1 SF (!)).
Clearly the distance of a race would be known to a pretty high degree of precision even though the SF rules would say
that “100” should have only 1 SF. Sometimes you have to use common sense!!
20. 1.104. An ibuprofen suspension for infants contains 100 mg/5.0 mL suspension. The recommended dose is 10 mg/kg
body weight. How many mL of this suspension should be given to an infant weighing 18 lb? (Assume 2 SF.)
NOTE: Be aware that you do not need to convert mg to kg here (in fact, it will be wrong)!! Why?
Because the mg refers to the medicine and the kg refers to “body weight” (BW). You can’t
convert lbs of apples into kg of oranges, right? The “things” are different.
#mL  18 lb BW x
1kg BW
2.2046 lb BW
x
10 mg ibup.
1 kg BW
x
5.0 mL suspension
 4.082...  4.1 mL suspension
100 mg ibup
21. 1.41 & 1.42. Determine whether each molecular diagram represents a pure substance or a mixture. If it represents a
pure substance, classify the substance as an element or a compound. If it represents a mixture, classify
the mixture as homogeneous or heterogeneous.
NOTE: Pictures have been omitted due to time contraints. Consult text for pictures.
1.41 (a) pure substance; all six molecules shown are identical (“one kind of basic unit”)
compound; Each molecule contains more than one kind of atom (black and red)
(b) mixture (of compounds); there are two different kinds of molecule (basic units) in the sample
PS1-11
Answer Key, Problem Set 1
heterogeneous; the molecules are not randomly mixed up at the nanoscopic level (there are
“clumps” of each kind in different regions (specifically, 4 molecules “on top” in the picture and 24
molecules “below”)
(c) mixture (of compounds); there are two different kinds of molecule (basic units) in the sample
homogeneous; the molecules are randomly mixed up at the nanoscopic level (there are no
“clumps” of each kind in different regions)
(d) pure substance; all eight atoms (basic units) shown are identical (“one kind of basic unit”)
element; Each basic unit contains only one kind of atom (white)
1.42 (a) pure substance; all 24 atoms (basic units) shown are identical (“one kind of basic unit”)
element; Each basic unit contains only one kind of atom (red)
(b) mixture (of compounds); there are two different kinds of molecule (basic units) in the sample
homogeneous; the molecules are randomly mixed up at the nanoscopic level (there are no
“clumps” of each kind in different regions) NOTE: Those huge “possibly looking like clumps”
things are actually individual molecules with a large number of atoms in each.
(c) pure substance; all 7 molecules shown are identical (“one kind of basic unit”)
compound; Each molecule contains more than one kind of atom (black and white)
(d) pure substance; all 8 molecules shown are identical (“one kind of basic unit”)
compound; Each molecule contains more than one kind of atom (red, black and white)
NOTE: This problem unfortunately does not give an example of a molecular element. You are
responsible for distinguishing these from compounds. I will plan to put such a problem on PS2 so
that you can have practice with this distinction. Also, see note in problem 1.49 →
22. 1.48. Classify each change as physical or chemical.
(a) Sugar burns when heated on a skillet. Answer: Chemical.
New substances are made (indicated by the color change, odor change, etc. Also, “burning” by
definition is “[chemical] reaction with oxygen”)
(b) Sugar dissolves in water. Answer: Physical.
No new substances are made (you still have sugar and water after “dissolution”)
(c) A platinum ring becomes dull because of continued abrasion. Answer: Physical.
No new substances are made (abrasion just means “scratching” or “roughening” of the surface, not
changing its composition)
(d) A silver surface becomes tarnished after exposure to air for a long period of time. Answer: Chemical.
New substances are made (indicated by the reference to “air”. Some component of the air (oxygen)
chemically reacts with the surface to create a new substance “tarnish”)
23. 1.49. Based on the molecular diagram, classify each change as physical or chemical.
NOTE: Pictures have been omitted due to time contraints. Consult text for pictures.
(a) physical change; The same four molecules of a pure substance exist both before and after the
change. Hence no new substances have formed during this change, making it a
physical change. The molecules have merely moved apart from one another,
PS1-12
Answer Key, Problem Set 1
indicating a phase change only (probably a liquid to a gas, although they didn’t
make it 100% clear.
(b) chemical change; Although there happen to be six molecules before and after the change, they are
not the same molecules that you started with. Hence new substances have formed
during this change, making it a chemical change.
Note (additional practice like problem 1.41): The starting substances in this reaction are an element (represented by the four
molecules with two red atoms each [O2]) and a compound (represented by the two molecules each having one black
atom and [what is supposed to be] four white atoms [CH4]. The ending substances are both compounds (one
represented by the 4 molecules with red and white atoms [H2O] and one represented by the two molecules with red
and black atoms [CO2]
(c) physical change; Same as in (a) [Except five identical molecules are shown]
PS1-13