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Unformatted text preview: (M) We will use the density of vvater, and its mass in the pycnometer, to ﬁnd the volume
of liquid held by the pycnometer. ,
1 mL t 1 =35.552 —25'.601 x=——=9.969mL
pycnome ervo ume ( g g) 0.99821g The mass of the methanol and the pyncnometer's volume determine liquid density. 33.490 g—~25.601g__ 07914 mi ' th 1 i: W,
density of me ano 9969 mL (M) First, calculate the total mass of ice in the Antarctic, Which yields the total mass of
water which is obtained if all the ice melts: 5 3 ' .
3.Ol><107 km3 ice x MX 992mg? = 2.769><1022 g ice
 1km 1 cm Ice all of which converts to water. The volume of this extra water is then calculated. ' 3 3.
2.769x1022 g H20 >< wx w = 2.769x107 kma H20
_ lg H20 (1x10 cm) H20
Assuming that Vol (H20 on Earth) = X h # 3.62X108 krnz, the total increase in the
height of sea levels with the addition of the melted continental ice will be:  h = 2.759x107 an / 3.62><108 m2 =' 0.0765_km =' 76.4 m. w v (M)." (a) We can determine that carbon dioxide ha’s‘a ﬁxed composition by ﬁnding the % C
in each sample. (In the calculations below, the abbreviation “cmpd” is short for compound.)
%c=—933§C—x100%=27.3% c %C=Mx100%=27.3% c
13.26 g cmpd . 21.66 g cmpd
%c'= 7'07 g C x100% = 27.3% c 25.9] g cmpd I
Since all three samples have the same percent of carbon, these data do establish that carbon dioxide has a fixed composition. (b) Carbon dioxide contains only carbon and oxygen. As determined above, carbon dioxide is
27.3 % C by mass. The percent of oxygen in carbon dioxide is obtained by difference. %0 = 100.0 % {27.3 %C) = 72.7 %0 18, (M) Assuming the intermediate is “halfway” between CO (oxygencarbon mass ratio = .
‘ 16:12 or 1.333) and C02 (oxygencarbon mass ratio = 32:12 or 2.6667), then the
oxygen‘ carbon ratio would be 2:1, or O:C = 24: 12. This mass ratio gives a mole ratio of
02C 5 1.5:1. Empirical formulas are ,simple whole number ratios of elements; hence, a
formula of C302 must be the correct empirical formula for this carbon oxide. (Note: C302
is called tricarbon dioxide or carbon suboxide). ‘ Fundamental Charges and Massto—Charge Ratios _ g, (M) We can calculate the charge on each drop, express each in terms of 10’]9 C,Iand r' 11 1 I 1n—lq A ‘ 58. 78. 72.64 u =(14.5792 u) + (19.8073 u) + (26.8260 11) + (72.92346 ux) + [11.6312 u  75.92140 ux)] (M) The number of protons is the same as the atomic number for iodine, Which is 53.
There is one more electron than the number of protons because there IS a negatlve charge
on the ion. Therefore the number of electrons is 54. The number of neutrons IS equal to 70, mass number minus atomic number. {17\ na‘:nJ:—A ’7— E‘) "2...! 6 541.1. ('1 .___n;,,2_ n 71 (M) Since the three percent abundances total 100%, the percent abundance of 4“K is found by
difference. % 40K :100.0000% —93.2581% 4 6.7302% : 0.0117% .  Then the expression for the weightedaverage atomic mass is used, with the percent
abundances converted to fractional abundances by dividing by 100. Note that the average
atomic mass of potassium is 39.0983 11. 39.0983u = (0.932581 x 38.963707u)+(0.000117 x 39.963999 u)+(0.067302 x 4‘K) : 36.3368u + 0.00468u +(0.067302 x 41K) 39.098311436336811 +0.00468u) _
0.067302 I  mass of 41K = 40.962 u (E) ix”
1000 mg Kr 83.80 g Kr
= 3.77 ><10l9 atoms KI ‘ 1 mol Kr X 6.022x1023 atoms Kr a b Krt =5.25m er
(_) nurn er aoms g 11m)le (b) Molar mass is deﬁned as the mass per mole of substance. mass = 2.09 g. This calculation requires that the number of moles be determined. moles = 2.80x1022 atoms x+ =: 0.0465 mol 6.022 x10 atoms
molar mass = mass = = 44.9 g/mol The element is So, scandium.
moles 0.0465 mol _
l P . (c) mass 9:44.75 gngn—linilMix “no xwzﬂﬂ?) gP 24.3050 g Mg 1 mol Mg 1 mol P
Note: Thesame answer is obtained if you assume phosphorus is P4 instead of P. . (D) The sum of the percent abundances of the two minor isotopes equals 100.00% — 84.68%
15.32%. Thus, we denote the fractional abundance of 73 Ge as x, and the other as (0.1532 4x). These fractions are then used in the expression for average atomic mass. 
atomic mass = 72.64 u = (69.92425 u X 0.2085) + (71.92208 u >< 0.2754) + (73.92118 u >< 0.3629) +(72.92346 u x x) + [75.92140 11 x (0.1532 x)] 0.2037 : 2.99794x Hence: 7: i 0.068. 73 Ge has 6.8% natural abundance and 76Ge has 8.5% natural abundance. From the calculations, we can see that the number of signiﬁcant ﬁgures drops from four in the percen_ natural abundances supplied to only two signiﬁcant ﬁgures owing to the imprecision in the supplie
values of the percent natural abundance. It "90. (D) One way to determine the common factor of which all 13 numbers are multiples is to
1 ﬁrst divide all of them by the smallest number in the set. The ratios thus obtained may be
either integers or rational numbers whose decimal equivalents are easy to recognize. Obs.123456'78910111213
Quan. 19.6 24.60 29.62 34.47 39.38 44.42 49.41 53.91 59.12 63.68 68.65 78.34 83.22 Ratio 1.00 1.251 1.507 1.753 2.003 2.259 2.513 2.742 3.007 3.239 3.492 3.984 4.233
Mult. 4.00 5.005 6.026 7.013 8.012 903810.05 10.97 12.03 12.96 13.97 15.94 16.93
‘_"'_.Int. '4 5 6 7 8 '9 10‘ 11 12 13 14 16 17
The row labeled “Mult.” is obtained by multiplying the row “Ratio” by 4.000. In the row
labeled “Int.” we give the integer closest to each of these multipliers. It is obvious that each of the 13 measurements is exceedingly close to a common quantity multiplied by an
integer. ' (26) 3. (4) "‘ f__ﬁ7___s b. n the making of the initial atomic bomb, two compounds were considered for the gas (12) 4 a. Describe the experiment by Rutherford, conducted in the early 1900's. that dispelled the “plum pud ' 9 model" of the atom. _ {L1
l '1} alparibdé’ chi'eéttCJ 5L1. nucleus  {Hm fin ‘II >m i _\ J u
l o l '2) (21' fDAr‘tICl'f'S uridr'r'ﬁtr trialDY Crerﬂecﬂbonb) “ldwa‘hng , II ‘ 7
“Itﬂacﬁon Wi—f“ r'fmigﬁ’Itra‘ﬁOH 011' diary? , Le.) t1 FMCIELLS. phase separation of 23592U from 238QZU: UFa and UCle. Given the following nuclide masses and % abundances:
“nor 34.968852 75.77% ‘%F 18.9984 100%
“nor 36.965902 24.23% Which compound was chosen for the separation? Why? "‘l'D be GE. OHMS om: need ‘5? F nucleus means cowE 22 maSSeS
1&5 :  ~ ' 1" " ‘ c
‘12 UHF)» (Mid U. (‘q‘ Fla . [
UClig ‘. 1‘1l dilﬁfiﬂﬁ'lc‘ﬂ't be’ff‘lt's is mam ounmippinc magstﬂ; . Dr‘H‘ICLLH SFPaf(de‘
c. in a mass spectrometer, the ratio of an unknown isotope of an element, 2, to 3115P (mass = 30.973761) was 1.2579585. What are the atomic mass and the elemental symbol of Z? M1 C\:15”}q555 )( 30. 63:? 576') ': 38‘ "“ "— l 25'1‘1585
M33? Sepcwcdrd; Frcbabij d. The compound PC3 is a common starting material for the synthesis of various
phosphoruscontaining compounds, including many usad in insecticides. PhOSphorus is 100%
3115P (30.973761). Using a mass number of 31 for P. and mass numbers (%’s) 35 (76%) and 37
(24%) for 35170l and 37170I, respectively, predict what the mass spectrum of PCI3 would look like by giving the % abundances and masses of all moiecularions. 0/5,
r8915 —~—————
‘ O
3"? (310‘s w~~u——~8 r42 {0.24)S v. 100% —. l838/2
‘i 5'} 3‘5 ‘ __.,___.r.._.::‘.> ‘ ‘) y ‘ 5 H C01 M) Ma 84.86.24? 3x (0.287103%) 8 100% ‘ t3“ 0
35,37,31
3“Pec~ieo>z was . a w W
:2 21.3? ‘51 (0.2141(051‘62 i! 100% ’ 4" °
‘3" V) ’39 l 1 ——~——— ' . 293.35.“.
L C )‘J a at“ 48.80%
(c ‘tccig vi 100% "’
W rod00% ...
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 Spring '07
 ZAX,D
 Chemistry

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