CHAPTER 1 MATTER—ITS PROPERTIES AND MEASUREMENT PRACTICE EXAMPLES
1A (E) Convert the Fahrenheit temperature to Celsius and compare.
1B (E) We convert the Fahrenheit temperature to Celsius.
5C5C 9F9F C=F32 F=15F32 F=26 C
. The antifreeze only protects to 22 C and thus it will not offer protection to temperatures as low as 15 F=26.1 C .
2A (E) The mass is the difference between the mass of the full and empty flask. 291.4 g108.6 g density==1.46 g/mL 125 mL
2B (E) First determine the volume required. V = (1.000 × 103 g) (8.96 g cm-3) = 111.6 cm 3 . Next determine the radius using the relationship between volume of a sphere and radius.
V = 4 3 r 3 = 111.6 cm 3 = 4 3 (3.1416)r3 r = 3 111.63 4(3.1416)
= 2.987 cm
3A (E) The volume of the stone is the difference between the level in the graduated cylinder with the stone present and with it absent.
mass28.4 g rock density===2.76 g/mL volume44.1 mL rock &water 33.8 mL water = 2.76 g/cm3
3B (E) The water level will remain unchanged. The mass of the ice cube displaces the same mass of liquid water. A 10.0 g ice cube will displace 10.0 g of water. When the ice cube melts, it simply replaces the displaced water, leaving the liquid level unchanged.
4A (E) The mass of ethanol can be found using dimensional analysis.
1000 mL0.71 g gasohol10 g ethanol1 kg ethanol ethanol mass=25 L gasohol
1 L1 mL gasohol100 g gasohol1000 g ethanol =1.8 kg ethanol
4B (E) We use the mass percent to determine the mass of the 25.0 mL sample.
100.0 g rubbing alcohol
rubbing alcohol mass=15.0 g (2-propanol)=21.43 g rubbing alcohol 70.0 g (2-propanol)
21.4 g
rubbing alcohol density==0.857 g/mL 25.0 mL
5A (M) For this calculation, the value 0.000456 has the least precision (three significant figures), thus the final answer must also be quoted to three significant figures.
5B (M) For this calculation, the value 1.310 3 has the least precision (two significant figures), thus the final answer must also be quoted to two significant figures. 8.21101.310 0.002364.07110
6A (M) The number in the calculation that has the least precision is 102.1 (+0.1), thus the final answer must be quoted to just one decimal place. 0.236+128.55102.1=26.7
6B (M) This is easier to visualize if the numbers are not in scientific notation.
INTEGRATIVE EXAMPLE
A (D) Stepwise Approach: First, determine the density of the alloy by the oil displacement. Mass of oil displaced = Mass of alloy in air – Mass of alloy in oil = 211.5 g – 135.3 g = 76.2 g
VOil = m / D = 76.2 g / 0.926 g/mL = 82.3 mL = VMg-Al
DMg-Al = 211.5 g / 82.3 mL = 2.57 g/cc
Now, since the density is a linear function of the composition,
DMg-Al = mx + b, where x is the mass fraction of Mg, and b is the y-intercept.
Substituting 0 for x (no Al in the alloy), everything is Mg and the equation becomes:
1.74 = m · 0 + b. Therefore, b = 1.74
Assuming 1 for x (100% by weight Al):
2.70 = (m × 1) + 1.74, therefore, m = 0.96
Therefore, for an alloy:
2.57 = 0.96x + 1.74
x = 0.86 = mass % of Al
Mass % of Mg = 1 – 0.86 = 0.14, 14%
B (M) Stepwise approach:
Mass of seawater = D • V = 1.027 g/mL × 1500 mL = 1540.5 g
2.67 g NaCl39.34 g Na
1540.5 g seawater = 16.18 g Na
100 g seawater100 g NaCl
Then, convert mass of Na to atoms of Na
1 kg Na1 Na atom
16.18 g Na = 4.23910 Na atoms
1000 g Na3.81710 kg Na
Conversion Pathway:
1540.5 g seawater
2.67 g NaCl39.34 g Na1 kg Na1 Na atom
100 g seawater100 g NaCl1000 g Na3.817510 kg Na
EXERCISES
The Scientific Method
1. (E) One theory is preferred over another if it can correctly predict a wider range of phenomena and if it has fewer assumptions.
2. (E) No. The greater the number of experiments that conform to the predictions of the law, the more confidence we have in the law. There is no point at which the law is ever verified with absolute certainty.
3. (E) For a given set of conditions, a cause, is expected to produce a certain result or effect. Although these cause-and-effect relationships may be difficult to unravel at times (“God is subtle”), they nevertheless do exist (“He is not malicious”).
4. (E) As opposed to scientific laws, legislative laws are voted on by people and thus are subject to the whims and desires of the electorate. Legislative laws can be revoked by a grass roots majority, whereas scientific laws can only be modified if they do not account for experimental observations. As well, legislative laws are imposed on people, who are expected to modify their behaviors, whereas, scientific laws cannot be imposed on nature, nor will nature change to suit a particular scientific law that is proposed.
5. (E) The experiment should be carefully set up so as to create a controlled situation in which one can make careful observations after altering the experimental parameters, preferably one at a time. The results must be reproducible (to within experimental error) and, as more and more experiments are conducted, a pattern should begin to emerge, from which a comparison to the current theory can be made.
6. (E) For a theory to be considered as plausible, it must, first and foremost, agree with and/or predict the results from controlled experiments. It should also involve the fewest number of assumptions (i.e., follow Occam’s Razor). The best theories predict new phenomena that are subsequently observed after the appropriate experiments have been performed.
Properties and Classification of Matter
7. (E) When an object displays a physical property it retains its basic chemical identity. By contrast, the display of a chemical property is accompanied by a change in composition.
(a) Physical: The iron nail is not changed in any significant way when it is attracted to a magnet. Its basic chemical identity is unchanged.
(b) Chemical: The paper is converted to ash, CO2(g), and H2O(g) along with the evolution of considerable energy.
(c) Chemical: The green patina is the result of the combination of water, oxygen, and carbon dioxide with the copper in the bronze to produce basic copper carbonate.
(d) Physical: Neither the block of wood nor the water has changed its identity.
8. (E) When an object displays a physical property it retains its basic chemical identity. By contrast, the display of a chemical property is accompanied by a change in composition.
(a) Chemical: The change in the color of the apple indicates that a new substance (oxidized apple) has formed by reaction with air.
(b) Physical: The marble slab is not changed into another substance by feeling it.
(c) Physical: The sapphire retains its identity as it displays its color.
(d) Chemical: After firing, the properties of the clay have changed from soft and pliable to rigid and brittle. New substances have formed. (Many of the changes involve driving off water and slightly melting the silicates that remain. These molten substances cool and harden when removed from the kiln.)
9. (E) (a) Homogeneous mixture: Air is a mixture of nitrogen, oxygen, argon, and traces of other gases. By “fresh,” we mean no particles of smoke, pollen, etc., are present. Such species would produce a heterogeneous mixture.
(b) Heterogeneous mixture: A silver plated spoon has a surface coating of the element silver and an underlying baser metal (typically iron). This would make the coated spoon a heterogeneous mixture.
(c) Heterogeneous mixture: Garlic salt is simply garlic powder mixed with table salt. Pieces of garlic can be distinguished from those of salt by careful examination.
(d) Substance: Ice is simply solid water (assuming no air bubbles).
10. (E) (a) Heterogeneous mixture: We can clearly see air pockets within the solid matrix. On close examination, we can distinguish different kinds of solids by their colors.
(b) Homogeneous mixture: Modern inks are solutions of dyes in water. Older inks often were heterogeneous mixtures: suspensions of particles of carbon black (soot) in water.
(c) Substance: This is assuming that no gases or organic chemicals are dissolved in the water.
(d) Heterogeneous mixture: The pieces of orange pulp can be seen through a microscope. Most “cloudy” liquids are heterogeneous mixtures; the small particles impede the transmission of light.
11. (E) (a) If a magnet is drawn through the mixture, the iron filings will be attracted to the magnet and the wood will be left behind.
(b) When the glass-sucrose mixture is mixed with water, the sucrose will dissolve, whereas the glass will not. The water can then be boiled off to produce pure sucrose.
(c) Olive oil will float to the top of a container and can be separated from water, which is more dense. It would be best to use something with a narrow opening that has the ability to drain off the water layer at the bottom (i.e., buret).
(d) The gold flakes will settle to the bottom if the mixture is left undisturbed. The water then can be decanted (i.e., carefully poured off).
12. (E) (a) Physical: This is simply a mixture of sand and sugar (i.e., not chemically bonded).
(b) Chemical: Oxygen needs to be removed from the iron oxide.
(c) Physical: Seawater is a solution of various substances dissolved in water.
(d) Physical: The water-sand slurry is simply a heterogeneous mixture.
Exponential Arithmetic
16. (E) (a) 173 thousand trillion watts W =173,000,000,000,000,000=1.731017
(b) one ten millionth of a meter = 7 110,000,000 m=110 m
(c) (trillionth = 1 10-12) hence, 142 10-12 m or 1.42
10-10 m
Significant Figures
17. (E) (a) An exact number—500 sheets in a ream of paper.
(b) Pouring the milk into the bottle is a process that is subject to error; there can be slightly more or slightly less than one liter of milk in the bottle. This is a measured quantity.
(c) Measured quantity: The distance between any pair of planetary bodies can only be determined through certain astronomical measurements, which are subject to error.
(d) Measured quantity: the internuclear separation quoted for O2 is an estimated value derived from experimental data, which contains some inherent error.
18. (E) (a) The number of pages in the text is determined by counting; the result is an exact number.
(b) An exact number. Although the number of days can vary from one month to another (say, from January to February), the month of January always has 31 days.
(c) Measured quantity: The area is determined by calculations based on measurements. These measurements are subject to error.
(d) Measured quantity: Average internuclear distance for adjacent atoms in a gold medal is an estimated value derived from X-ray diffraction data, which contain some inherent error.
19. (E) Each of the following is expressed to four significant figures.
20. (E) (a) 450 has two or three significant figures; trailing zeros left of the decimal are indeterminate, if no decimal point is present.
(b) 98.6 has three significant figures; non-zero digits are significant.
(c) 0.0033 has two significant digits; leading zeros are not significant.
(d) 902.10 has five significant digits; trailing zeros to the right of the decimal point are significant, as are zeros flanked by non-zero digits.
(e) 0.02173 has four significant digits; leading zeros are not significant.
(f) 7000 can have anywhere from one to four significant figures; trailing zeros left of the decimal are indeterminate, if no decimal point is shown.
(g) 7.02 has three significant figures; zeros flanked by non-zero digits are significant.
(h) 67,000,000 can have anywhere from two to eight significant figures; there is no way to determine which, if any, of the zeros are significant, without the presence of a decimal point.
The average speed is obtained by dividing the distance traveled
miles) by the elapsed time (in hours). First, we need to obtain the elapsed time, in hours.
(b) First compute the mass of fuel remaining mass gal
Next determine the mass of fuel used, and then finally, the fuel consumption. Notice that the initial quantity of fuel is not known precisely, perhaps at best to the nearest 10 lb, certainly (“nearly 9000 lb”) not to the nearest pound.
0.4536 kg mass of fuel used= (9000 lb82 lb)4045 kg 1 lb 25,012 mi1.609344 km fuel consumption==9.95 km/kg or ~10 km/kg 4045 kg1 mi
26. (M) If the proved reserve truly was an estimate, rather than an actual measurement, it would have been difficult to estimate it to the nearest trillion cubic feet. A statement such as 2,911,000 trillion cubic feet (or even 310183 ft ) would have more realistically reflected the precision with which the proved reserve was known.
31. (E) Express both masses in the same units for comparison.
32. (E) Express both masses in the same units for comparison.
smaller than 3257
33. (E) Conversion pathway approach:
34. (M) A mile is defined as being 5280 ft in length. We must use this conversion factor to find the length of a link in inches.
35. (M) (a) We use the speed as a conversion factor, but need to convert yards into meters.
The final answer can only be quoted to a maximum of two significant figures.
(b) We need to convert yards to meters. 100 yd36 in.2.54 cm1 m speed==9.83
(c) The speed is used as a conversion factor.
38. (D) Here we must convert pounds per cubic inch into grams per cubic centimeter:
40.
First we will calculate the radius for a typical red blood cell using the equation for the
sphere. V = 4/3
43. (M) Let us determine the Fahrenheit equivalent of absolute zero.
of 465 F
cannot be achieved because it is below absolute zero.
44. (M) Determine the Celsius temperature that corresponds to the highest Fahrenheit temperature, 240 F .
Because 116 C is above the range of the thermometer, this thermometer cannot be used in this candy making assignment.
45. (D) (a) From the data provided we can write down the following relationship:
-38.9 C = 0 M and 356.9 C = 100 M. To find the mathematical relationship between these two scales, we can treat each relationship as a point on a twodimensional Cartesian graph:
Therefore, the equation for the line is y = 3.96x - 38.9 The algebraic relationship between the two temperature scales is t(C) = 3.96(M) - 38.9 or rearranging, t(M) = t(C) + 38.9 3.96
Alternatively, note that the change in temperature in °C corresponding to a change of 100 °M is [356.9 – (-38.9)] = 395.8 °C, hence, (100 °M/395.8 °C) = 1 °M/3.96 °C. This factor must be multiplied by the number of degrees Celsius above zero on the M scale. This number of degrees is t(°C) + 38.9, which leads to the general equation t(°M) = [t(°C) + 38.9]/3.96.
The boiling point of water is 100 C, corresponding to t(M) = 100 38.9
(b) t(M) = -273.15 38.9
= -59.2
M would be the absolute zero on this scale.
46. (D) (a) From the data provided we can write down the following relationship:
= 0
A and -33.35 C = 100 A. To find the mathematical relationship between these two scales, we can treat each relationship as a point on a twodimensional Cartesian graph.
Therefore, the equation for the line is y = 0.444x - 77.75 The algebraic relationship between the two temperature scales is
49. (M) The mass of acetone is the difference in masses between empty and filled masses.
55. (M) The calculated volume of the iron block is converted to its mass by using the provided density.
56. (D) The calculated volume of the steel cylinder is converted to its mass by using the provided density.
57. (M) We start by determining the mass of each item.
In order of increasing mass, the items are: iron bar aluminum foil water. Please bear in mind, however, that, strictly speaking, the rules for significant figures do not allow us to distinguish between the masses of aluminum and water.
59. (D) First determine the volume of the aluminum foil, then its area, and finally its thickness.
61. (D) Here we are asked to calculate the number of liters of whole blood that must be collected in order to end up with 0.5 kg of red blood cells. Each red blood cell has a mass of 90.0 10-12 cm
For 0.5 kg or 5 102 g of red blood cells, we require = 5 102 g red blood cells 1 mL of blood
62. (D) The mass of the liquid mixture can be found by subtracting the mass of the full bottle from the mass of the empty bottle = 15.4448 g 12.4631 g = 2.9817 g liquid. Similarly, the total mass of the water that can be accommodated in the bottle is 13.5441 g 12.4631 g = 1.0810 g H2O. The volume of the water and hence the internal volume for the bottle is equal to
1.0810 g H2O 2 2
1 mL HO 0.9970 g HO = 1.084 mL H2O (25 C)
Thus, the density of the liquid mixture = 2.9817 g liquid
1.084 mL = 2.751 g mL-1
Since the calcite just floats in this mixture of liquids, it must have the same density as the mixture. Consequently, the solid calcite sample must have a density of 2.751 g mL-1 as well.
Percent Composition
63. (E) The percent of students with each grade is obtained by dividing the number of students with that grade by the total number of students.
22 B's %B=100%=28.9%B
76 students
8 D's %D=100%=11%D 76 students
7 A's %A=100%=9.2%A 76 students
37 C's %C=100%=48.7%C 76students
2 F's %F=100%=3%F 76 students
Note that the percentages add to 101% due to rounding effects.
64. (E) The number of students with a certain grade is determined by multiplying the total number of students by the fraction of students who earned that grade.
no. of 18 A's A's=84 students=15 A's 100 students
25B's no. of B's=84 students=21 B's 100 students
13D's no. of D's=84 students=11 D's 100 students
32 C's no. of C's=84 students=27 C's 100 students
12 F's no. of F's=84 students=10 F's 100 students
65. (M) Use the percent composition as a conversion factor.
Conversion pathway approach: 3 1000 mL1.118 g soln28.0 g sucrose mass of sucrose=3.50 L=1.1010 g sucrose 1 L1 mL100 g soln
66. (D) Again, percent composition is used as a conversion factor. We are careful to label both the numerator and denominator for each factor.
