Answers to selected problems, Chapter 1.
2. Which of the
following are examples of matter?
It's obvious that (f), an idea, is not matter. If you don't think that (b), air,
is matter, then remember that 1 cubic meter of air weighs 1 kg (2.2 lb)! Also,
red light (d) is not technically matter because its photons have no rest mass.
12. Which are
physical changes and which chemical?
(a) Shearing sheep and spinning the wool into yarn are both physical
changes because the chemical properties of the wool aren't changed by either
step; it's just cut and spun.
(b) Baking a cake from flour, baking powder, etc. creates a chemical
change. You mix everything so thoroughly to bring all the ingredients into
close contact with the others and facilitate the various reactions.
(c) Milk turning sour is a chemical change as the
organic molecules coagulate and turn acidic.
(d) Silkworms eating mulberry leaves to produce silk.
Obviously a chemical change because silk is a totally different substance
from mulberry leaves!
(e) Moving a lawn. At least the first stage of this change is
physical, as the blades just cut the grass. As the cut grass dries and
ripens, however, it changes chemically as well.
14. Which of the
following are substances, and which are mixtures? Explain.
(a), helium gas, is obviously a pure substance (element in this case) because
all its atoms are helium.
(d), salt, is a substance (a compound) because even down to the atomic scale,
one finds only Na and Cl ions in 1:1 proportions.
(b), lemon juice, will be a mixture because it is composed of the flavoring
agent(s), water, and pulp, all in variable proportions to one another depending
on how you squeezed the lemon. By the same token, the salt of (d) would be a
mixture when dissolved in water because the proportions of the NaCl and the
water could vary, with the strength of the solution.
(c) red wine is a mixture for reasons similar to those for lemon juice, except
that now the substances will include the coloring agent(s), the flavoring
agent(s), the alcohol(s), and the water.
19. Can a set
of measurements be precise without being accurate?
Yes, if they are all "off" to the same extent (a systematic error such
as improper calibration of an analytical device).
Can the average of a set of measurements be accurate even if the individual
measures are imprecise?
Yes, provided that the errors scatter randomly about the correct mean. (This
situation is very common in science.)
25. (a) 4.54 x 10-9 g = 4.54 ng (b) 3.76 x 103 m = 3.76 km (c) 6.34 x 10-6 g = 6.34 µg
29. (a) TC
= (136 °F
– 32)/1.8 = 57.8 °C
(b) TF = 1.8TC + 32 = 1.8(-120 °C)
+ 32 = -184 °F
33. The yardstick is three feet long. The 75-inch rope is about 6 feet long (6 ft = 72 in). The 1.21-m chain is about 20% more than 40 inches (really 39.37 inches), or about 48 inches, which is 4 feet. Thus the sequence of lengths is yardstick (3 0, rattlesnake (3.5 ft), chain (4 ft), and rope (6 ft).
35. Significant
figures.
(a) 8008 m has four significant figures because zeros between nonzero numbers
are significant.
(b) 0.00075. Remember that zeroes to the left of the other numbers, regardless
of the position of the decimal place, are not significant. Thus 0.00075 has only
two significant figures.
(c) 0.049300. From (b), the first two zeroes are not significant. But the two to
the right of the 3 are, so 0.049300 has five significant figures.
(d) 6.02 x 105 m. Same principle as (a): the zero is significant.
This gives three significant figures. (The power of ten doesn't count; only its
coefficient does.)
In the following numbers, the significant figures are underlined: (e) 4.200
x 105 s; (f) 0.1050 °C.
41. Adding and subtracting numbers with various levels of significant figures.
(a) 36.5 m
-2.16 m
3.452 m
37.792 Þ 37.8 m (because the sum may not be significant in more places than the least significant of its components)
45. The basic equation for density is D = M/V, where D is density, M is mass, and V is volume (all in appropriate units, of course). The density is then D = (78.0 g)/(25.0 mL) = 3.12 g mL-1. (The answer has three significant figures because both the mass and the volume do, too.)
49. D = M/V = (59.01 g)/(1.20 cm x 2.42 cm x 1.80 cm) = (59.01 g)/(5.206 cm3) = 11.3 g cm-3.
53.
Here is a simple example of solving an equation algebraically first and
numerically second. The first part of the problem uses the density equation to
find a volume. You just take D = M/V, transform it to V = M/D, and insert the
numbers with the proper units. The 898 kg must first be changed to 898,000 g, or
8.98 x 105 g. Then V = (8.98 x 105 g)/(7.76 g cm-3)
= 1.16 x 105 cm3.
For the second part, use the formula V = AH for the volume of
the cylindrical bar, where A is the area of the base and H is the height of the
bar. As in the first part, solve algebraically for H first, then substitute the
numbers. H = V/A = (1.16 x 105 cm3)/(1.50 cm2)
= 7.73 x 104 cm.
63. Note the misprint: the meter stick cannot be 1.005 mm long, for then you couldn't see it! It can be 1.005 m long or 1005 mm long, though, which amounts to being a little bit "stretched." Let us use the 1.005 m figure. The average length and width as (falsely) measured and reported in Table 1.4 were 1.827 and 0.762 m, respectively, which multiplied to produce an (false) area of 1.39 m2. (Note the three significant figures in the answer, from the three in 0.762 m.) To see how far off this area is, we find the real area by correcting the length and width and multiplying. Since the meter stick is too long, it is giving measurements that are too short, for it reads 1.000 m for a length that is really 1.005 m. Thus any measurement with this meter stick has to be scaled up by a factor of 1.005/1.000 = 1.005. Thus the true area of the poster board is A = LW = [1.827 m x 1.005] x [0.762 m x 1.005] = 1.836 m x 0.766 m = 1.41 m2, or 0.02 m2 greater than the original area of 1.39 m2.
71. I go through this simple problem to illustrate the ideal series of steps to take, which I think is as important to grasp as just getting the right answer. As we discussed in class a couple times, the basic procedure is to first write the equation and solve for the desired variable algebraically (including the correct units), and then put in the numbers. In this simple case, you begin with the equation for density, D = M/V, and convert it to one that contains the foil's thickness. That is done by recognizing that the volume of the "square" of aluminum is V = L2H, where L is the length of the sides of the square and H is the thickness of the square. The formula for density then becomes D = M/[L2H]. Now solve algebraically for H, the desired quantity in the problem: H = M/L2D. Since the problem uses the cgs system of units, H will be in cm. You can check this by inserting the units into the equation: [H] = [g]/[cm2 g cm-3] = [1/cm-1] = [cm]. (You now see the advantage of using negative exponents, as I mentioned in the first class.) So after you calculate H in cm, just multiply by 10 to get mm. (I suggest that for practice you derive the way to convert cm to mm.) Numerically, H = (1.762 g)/[(5.10 cm)2(2.70 g cm-3)] = 0.02509 cm = 0.2509 mm. The result should be expressed to three significant figures because that is the lowest of the three numbers in the calculation. Thus the thickness is 0.251 mm.
79. (The problem about the brass and the cork in the water.) The trick here is to recognize that the brass sinks because it is so much denser than the water. Thus, the brass cube can displace only its volume, 8.0 cm3. But how much water will the cork displace? A solid that floats is displacing a mass of water equal to its own mass. (So when you look at a ship in the water, you will know that the amount of water it is displacing, equivalent to the volume of its hull that is under water, just equals the mass of the total ship.) You can determine the mass of the cork by multiplying its volume (5.0 cm x 4.0 cm x 2.0 cm = 40 cm3) by its density (0.22 g cm-3) to get 8.8 g. Because water’s density is 1 g cm-3, the volume of water displaced by the cork will be 8.8 cm3, which is larger than the 8.0 cm3 displaced by the brass.
81. (a) The perimeter of any rectangular poster board is P
= 2L + 2W, where L and W are length and width, respectively. The values from
Table 1.4 we should use are the best estimates, which are the averages of the
sets of measurements. Thus our best estimate of the perimeter is P = 2(1.827 m)
+ 2(0.762 m) = 5.178 m. [To how many significant figures should we express this
answer? The 2L term should be cited to four significant figures, or to the third
decimal place, because 1.827 has four significant figures and 2 is perfectly
defined (i.e., not a matter of measurement). The 2W term should also be cited to
three decimal places because multiplying by exactly 2 changes nothing. The sum
of two numbers, each to the third decimal place, will also be significant to the
third decimal place. Thus, the stated value for the perimeter, 5.178 m, is given
to the correct number of significant figures.
The absolute error in the value for the perimeter will be, according to Problem
81, the sum of the absolute errors in the sum of the numbers defining it. This
can be viewed in two ways. The first way is to multiply the absolute error of L
(for example) by its coefficient of 2 (because the absolute error in 2L will be
twice the absolute error in L. This would give 2(0.003 m) = 0.006 m for 2L.
Similarly, the absolute error for 2W is 2(0.001 m) = 0.002 m. The total absolute
error for P would then be 0.006 m + 0.002 m = 0.008 m. The second way to get the
absolute error is to remember that the perimeter P = 2L + 2W is really P =
L + L + W + W. The absolute error for this version of P would then be the sum of
the four errors, or 0.003 m + 0.003 M + 0.001 m + 0.001 m = 0.008 m, which is
the same result as obtained the other way (as it has to be).
(b) To get the total area of the poster board we just
calculate A = LW = (1.827 m)(0.762 m) = 1.39217 m2. To how many
significant figures should this result be expressed? To the smallest number of
figures in any of the numbers in the multiplication, which is three significant
figures, for the width of 0.762 m. (Here the conventional rule starts to break
down, as explained below.) That means that 1.39217 m2 becomes 1.39 m2,
a harsh rounding indeed.
The (accumulated) percent error in the area will be,
according to the method stated in this problem, the sum of the percent errors,
or 0.2% + 0.1% = 0.3%. The absolute error in the area will then be the true area
multiplied the accumulated fractional area, or (0.3%/100) x 1.39 m2 =
0.004 m2.
(c) The absolute error in the area, 0.004 m2, is
in the third decimal place, which is the fourth significant figure. Thus to be
true to this error, four significant figures should be reported. Not all
calculations with error would require one extra significant figure to be
carried. It is a bit of a fluke here, involving mostly the 1 in the area. Had
the area been just under 1, say 0.953 m2, there would have been no
problem with the number of significant figures.