SOLUTIONS - CHAPTER 1 Problems: 6, 10, 12, 13, 18, 24, 26, 30, 36, 43, 44, 48, 50, 54, 58, 64, 74, 78, 86, 110
6) Identify the element present in each of the following molecules (see Table 1.1)
hydrogen (white) carbon (black) fluorine nitrogen (blue)
oxygen (red) hydrogen (white) (light green)
hydrogen (white)
H2O CH4 HF N2
10) Give an example of an element and a compound. How do elements and compounds differ?
Element [examples: O2(g) molecular oxygen; Ar(g) (argon); Pb(s) lead]. A pure chemical substance that cannot be broken down into more simple substances by chemical methods.
Compound [examples: C2H6(g) ethane; C6H6() benzene; Co(NO3)2(s) cobalt II nitrate]. A pure chemical substance that can be broken down into two or more different elements by chemical means.
12) Give the names of the elements represented by the chemical symbols
LilithiumPtplatinum
FfluorineMgmagnesium
PphosphorusUuranium
CucopperAlaluminum
AsarsenicSisilicon
ZnzincNeneon
Clchlorine
13) Give the chemical symbols for the following elements:
a) potassium(K)f) plutonium(Pu)
b) tin(Sn)g) sulfur(S)
c) chromium(Cr)h) argon(Ar)
d) boron(B)i) mercury(Hg)
e) barium(Ba)
18) Name the SI units that are important in chemistry, and give the SI units for expressing the following:
a) length - meter
b) volume - m3 (though liters is the more commonly used unit)
c) mass - kilogram
d) time - second
e) temperature - Kelvin
24) The density of ethanol, a colorless liquid that is commonly known as grain alcohol, is 0.798 g/mL. Calculate the mass of 17.4 mL of the liquid.
Mass = 17.4 mL 0.798 g = 13.9 g
1 mL
26) a) Normally the human body can endure a temperature of 105. ºF for only a short period of time without permanent damage to the brain and other vital organs. What is this temperature in degrees Celsius.
ºC = 5/9 [ ºF - 32. ] = 5/9 [ 105. - 32. ] = 41. ºC
b) Ethylene glycol is a liquid organic compound that is used as an antifreeze in car radiators. It freezes at - 11.5 ºC. Calculate its freezing temperature in degrees Fahrenheit.
ºF = 9/5 ºC + 32 = 9/5 (- 11.5) + 32 = 11.3 ºF
c) The temperature at the surface of the sun is about 6300. ºC. What is this temperature in degrees Fahrenheit?
ºF = 9/5 ºC + 32 = 9/5 (6300) + 32 = 11400 ºF (Note that there is some ambiguity as to how many significant figures should be given in the answer).
30) Convert the following temperatures into degrees Celsius:
a) 77. K, the boiling point of liquid nitrogen
ºC = ºK - 273.15 = 77 - 273.15 = - 196. ºC
b) 4.22 K, the boiling point of liquid helium
ºC = ºK - 273.15 = 4.22 - 273.15 = - 268.93 ºC
c) 600.61 K, the melting point of lead
ºC = ºK - 273.15 = 600.61 - 273.15 = 327.46 ºC
36) Determine whether the following statements describe chemical or physical properties
a) Oxygen gas supports combustion
Chemical, as it deals with the reaction of oxygen with other substances
b) Fertilizers help support agricultural production
Chemical, as it deals with the effect of fertilizer on plant metabolism, a chemical process
c) Water boils below 100. ºC on top of a mountain
Physical, as it deals with a phase transition and not a chemical reaction
d) Lead is denser than aluminum
Physical, as density does not involve a chemical reaction
e) Uranium is a radioactive element
Physical, though this is a close call. The emission of radiation by most radioactive substances causes them to change into a different element, but this is not a chemical reaction.
43) Distinguish between the terms accuracy and precision. In general, explain why a precise measurement does not always guarantee an accurate result.
Precision refers to how close a series of measurements of the same thing are to one another (and so is a measure of the reproducibility or degree of scatter in the measurements). Accuracy refers to how close a measurement or average of several measurements is to the true value of the thing being measured.
If there is bias (systematic error) in a measurement the measurement might be precise (all the results are close together) but not accurate (the average result is far from the true value). An example would be a measurement of mass using an analytical balance that has not been correctly zeroed, and so has a systematic error of 1.5 g in the masses it displays. The results from successive measurements of the mass of the same object might be close together (precise) but they would be off from the true value for mass by 1.5 g because of the failure to correctly zero the balance.
44) Express the following numbers in scientific notation
a) 0.0000000272.7 x 10-8
b) 3563.56 x 102
c) 47,7644.7764 x 104
d) 0.0969.6 x 10-2
48) Determine the number of significant figures in each of the following measurements:
a) 4867 mi4 - all digits are nonzero and so significant
b) 56 mL2 - all digits are nonzero and so significant
c) 60,104 tons5 - the two zeros that appear are between nonzero digits,
and so all digits are significant
d) 2900 g2, 3, or 4 - there is no decimal point, and so no way to tell
whether the trailing zeros are significant or place-holders
e) 40.2 g/cm33 - the zero appears between two nonzero digits, and so all
digits are significant
f) 0.0000003 cm1 - the leading zeros are not significant
g) 0.7 min 1 - the leading zero is not significant
h) 4.6 x 1019 atoms2 - we only look at the mantissa (the number before the
power of 10) to count significant figures
50) Carry out the following operations as if they were calculations of experimental results, and express each answer in the correct units with the correct number of significant figures:
a) 5.6792 m + 0.6 m + 4.33 m = 10.6092 m = 10.6 m
b) 3.70 g - 2.9133 g = 0.7867 g = 0.79 g
c) (4.51 cm) x (3.6666 cm) = 16.536366 cm2 = 16.5 cm2
54) Carry out the following conversions
a) 22.6 m to decimeters
#decimeters = 22.6 m 10 dm = 226. dm
1 m
b) 25.4 mg to kilograms
# kilograms = 25.4 mg 1g 1 kg = 0.0000254 kg = 2.54 x 10-5 kg
1000 mg 1000 g
c) 556 mL to liters
# liters = 556 mL 1 L = 0.556 L
1000 mL
d) 10.6 kg/m3 to g/cm3
# g = 10.6 kg 1000 g 1 m 1 m 1 m = 0.0106 g = 1.06 x 10-2 g
cm3 m3 1 kg 100 cm 100 cm 100 cm cm3 cm3
Note that I could have done the volume conversion using 1 m3 = 106 cm3, but I wanted to show explicitly how you need to use the 1 m = 100 cm conversion three times when going from m3 to cm3.
58) How many minutes does it take light from the sun to reach Earth? (The distance from the sun to Earth is 93 million miles; the speed of light is 3.00 x 108 m/s)
While there is a direct conversion from miles to meters, I am going to do this calculation the long way to better show dimensional analysis in action. I have broken the calculation into two parts. I first find the distance between the Earth and the sun in meters, and then use the speed of light to convert from distance to time.
# meters = 93 x 106 miles 5280 ft 12 in 2.54 cm 1 m = 1.497 x 1011 m
1 mile 1 ft 1 in 100 cm
# minutes = 1.497 x 1011 m 1 s 1 min = 8.3 min
3.00 x 108 m 60 s
The number of significant figures in the final result is two, because 93 million miles has two significant figures. Note that I round to the correct number of significant figures at the end of the calculation, and not in the intermediate steps, to avoid roundoff error.
64) Carry out the following conversions:
a) 32.4 yards to centimeters
# cm = 32.4 yd 36 in 2.54 cm = 2.96 x 103 cm
1 yd 1 in
I have written the result in scientific notation to make clear that the final answer has three significant figures.
b) 3.0 x 1010 cm/s to ft/s
# ft = 3.0 x 1010 cm 1 in 1 ft = 9.8 x 108 ft/s
s s 2.54 cm 12 in
c) 1.42 light years to miles (a light year is an astronomical measure of distance - the distance light travels in a year, or 365 days; the speed of light is 3.00 x 108 m/s).
It is easier to do this calculation in two parts. We first find how many meters in 1.42 light years, and then convert from meters to miles.
# meters = 1.42 yr 365. day 24 hr 60 min 60 s 3.00 x 108 m
1 yr 1 day 1 hr 1 min 1 s
= 1.3434 x 1016 m
# miles = 1.3434 x 1016 m 100 cm 1 in 1 ft 1 mile
1 m 2.54 cm 12 in 5280 ft
= 8.35 x 1012 miles
All of the conversion factors are exact. Therefore, the number of significant figures in the final result is three, since both the distance travelled (in light years) and the speed of light are given to three significant figures.
74) In determining the density of a rectangular metal bar, a student made the following measurements: length = 8.53 cm; width = 2.4 cm; height = 1.0 cm; mass = 52.7064 g. Calculate the density of the metal to the correct number of significant figures.
volume = (length) x (width) x (height)
= (8.53 cm) (2.4 cm) (1.0 cm) = 20.47 cm3
density = mass = 52.7064 g = 2.575 g/cm3 = 2.6 g/cm3
volume 20.47 cm3
There are two significant figures in the final result because both width and height are given to two significant figures.
78) The speed of sound in air at room temperature is about 343 m/s. Calculate this speed in miles per hour (1 mi = 1609. m).
We will first convert to miles/s, and then to miles/hr.
# mile = 343 m 100 cm 1 in 1 ft 1 mile = 0.2131 mile/s
s 1 s 1 m 2.54 cm 12 in 5280 ft
# mile = 0.2131 mile 60 s 60 min = 767 mile/hr
hr s 1 min 1 hr
Since the speed m/s was given to three significant figures, the final result is also given to three significant figures.
86) A sheet of aluminum (Al) foil has a total area of 1.000 ft2 and a mass of 3.636 g. What is the thickness of the foil in mm? (density of Al = 2.699 g/cm3)?
We first convert the area from ft2 to cm2.
Area = 1.000 ft2 12 in 12 in 2.54 cm 2.54 cm = 929.03 cm2
1 ft 1 ft 1 in 1 in
The volume of the aluminum foil can be found from the mass of the foil and the density of aluminum.
Volume = 3.636 g 1 cm3 = 1.3472 cm3
2.699 g
But Volume = (Area) (thickness)
and so
thickness = Volume = 1.3472 cm3 = 0.001450 cm = 0.01450 mm
Area 929.03 cm2
= 1.450 x 10-2 mm
110) Bronze is an alloy made of copper (Cu) and tim (Sn). Calculate the mass of a bronze cylinder of radius 6.44 cm and length 44.37 cm. The composition of the bronze cylinder is 79.42 percent Cu and 20.58 percent Sn, and the densities of Cu and Sn are 8.94 g/cm3 and 7.31 g/cm3, respectively. What assumptions must you make in this calculation?
We first find the volume of the brass cylinder. For a cylinder
Volume = r2h = (6.44 cm)2 (44.37 cm) = 5781.1 cm3
If we assume that the volume of the alloy is additive (that is, that the volume of the alloy is equal to the volume of the copper + the volume of the tin used to make the alloy) we can then find the volume of a 1.0000 g sample of the alloy. A 1.000 g sample will contain 0.7942 g Cu and 0.2058 g Sn, and so
Volume of 1.000 g = (0.7942 g Cu) 1 cm3 + (0.2058 g Sn) 1 cm3 = 0.11699 cm3
8.94 g 7.31 g 1 gm
We can use this as a conversion factor to find the total mass of the cylinder
Mass cylinder = 5781.1 cm3 1 gm = 4.94 x 104 g = 49.4 kg
0.11699 cm3
As in some of the other problems, we have waited until the end of the calculation to round our results to the correct number of significant figures.
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