Question 11- Two isotopes of the element Fa occur naturally, with 282145 Fa At 23.4% (281.7 amu) and 275145 Fa at 76.6% (274.6 amu). Calculate the atomic mass for Fa using the weighted average mass method. (Please look at attached images for the two isotopes in this question, also look at attached images for a special periodic table you may need to solve this question. Please show all your work so I can understand going forward) 1181A8A1Periodic Table of the ElementsH.1314151617Не1.0082AЗА4A5A6A7A4.003346.78.10LiBeFNe6.949.0110.8112.0114.0116.0019.0020.181112346.79.101112131415161718Na Mg3B4B5B6B7B8B1B2BAlSiCIAr22.9924.3126.9828.0930.9732.0735.4539.95192021222324252627282930313233343536KCaScTiVCrMnFeCoNiCuZnGaGeAsSeBrKr39.1040.0844.9647.8750.9452.0054.9455.8558.9358.6963.5565.3869.7272.6374.9278.9679.9083.80373839404142434445464748495051525354RbSrYZrNb MoTcRuRhPdAg CdInSnSbТеI.Xe85.4787.6288.9191.2292.9195.96(98)101.07 102.91106.42 107.87 112.41114.82 118.71121.76 127.60 126.90 131.29555657727374757677787980818283848586CsBaHfТаWReOsIrPtAu HgTlPbBiРоAtRn132.91137.33138.91174.97180.95 183.84 186.21190.23 192.22 195.08 196.97 200.59 204.38207.2208.98 (209)(210)(222)878889104105106107108109110111112113114115116117118FrRa Ac¢RfDbSgBhHsMtDsRgCnNhF1McLvTsOg(223)(226)(227)(267)(268)(271)(272)(277)(276)(281)(280)(285)(286)(289)(289)(293)(293)(294)120137145146147148149150Wn AaMyFaJeQu ApBg | Im295.8298.4341.3359.9362.5364.2366.1367.75859606162636465666768697071* CePrNd Pm SmEuGdTbDy HoErTm YbLu(145)150.36 151.96 157.25 158.93162.50 164.93 167.26 168.93 173.05 174.97140.12 140.91144.2492939495969798991001011021039091PuAm Cm BkCfEsFm Md NoLrNp(244)t ThPaU(252)(257)(258)(259)(262)232.04 231.04 238.03(237)(243)(247)(247)(247)2020Elements and Polyatomic Ions NOT Accepted by IUPACApochryphiteApochryphateApO,ApO,QuOQuOMythologiumQuackeryAaAlibabiumMyApochryphorBogusineAp2-ImImaginariumQuBgQuackeriteJesticonWnWonkiumFaFantasiumJeQuackerate8. 0282at 23.4/• (281.7 amu)and275Fa at 76.6- (274.6 amu)145

The atoms mass is the mass of neutrons and protons of an atom. The weighted average mass method can…

Question 11- Two isotopes of the element Fa occur naturally, with 282145 Fa At 23.4% (281.7 amu) and 275145 Fa at 76.6% (274.6 amu). Calculate the atomic mass for Fa using the weighted average mass method. (Please look at attached images for the two isotopes in this question, also look at attached images for a special periodic table you may need to solve this question. Please show all your work so I can understand going forward)

Question 11- Two isotopes of the element Fa occur naturally, with 282145 Fa At 23.4% (281.7 amu) and 275145 Fa at 76.6% (274.6 amu). Calculate the atomic mass for Fa using the weighted average mass method. (Please look at attached images for the two isotopes in this question, also look at attached images for a special periodic table you may need to solve this question. Please show all your work so I can understand going forward)

Chemistry & Chemical Reactivity

10th Edition

ISBN:9781337399074

Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Chapter25: Nuclear Chemistry

Section: Chapter Questions

Problem 77SCQ

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Shape of the D Orbital

Shapes of orbitals are an approximate representation of boundaries in space for finding electrons occupied in that respective orbital. D orbitals are known to have a clover leaf shape or dumbbell inside where electrons can be found.

Question

Question 11- Two isotopes of the element Fa occur naturally, with _{²⁸²145 Fa At} 23.4% (281.7 amu) and²⁷⁵_{145 Fa}at 76.6% (274.6 amu). Calculate the atomic mass for Fa using the weighted average mass method. (Please look at attached images for the two isotopes in this question, also look at attached images for a special periodic table you may need to solve this question. Please show all your work so I can understand going forward)

1
18
1A
8A
1
Periodic Table of the Elements
H.
13
14
15
16
17
Не
1.008
2A
ЗА
4A
5A
6A
7A
4.003
3
4
6.
7
8.
10
Li
Be
F
Ne
6.94
9.01
10.81
12.01
14.01
16.00
19.00
20.18
11
12
3
4
6.
7
9.
10
11
12
13
14
15
16
17
18
Na Mg
3B
4B
5B
6B
7B
8B
1B
2B
Al
Si
CI
Ar
22.99
24.31
26.98
28.09
30.97
32.07
35.45
39.95
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
39.10
40.08
44.96
47.87
50.94
52.00
54.94
55.85
58.93
58.69
63.55
65.38
69.72
72.63
74.92
78.96
79.90
83.80
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
Rb
Sr
Y
Zr
Nb Mo
Tc
Ru
Rh
Pd
Ag Cd
In
Sn
Sb
Те
I.
Xe
85.47
87.62
88.91
91.22
92.91
95.96
(98)
101.07 102.91
106.42 107.87 112.41
114.82 118.71
121.76 127.60 126.90 131.29
55
56
57
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
Cs
Ba
Hf
Та
W
Re
Os
Ir
Pt
Au Hg
Tl
Pb
Bi
Ро
At
Rn
132.91
137.33
138.91
174.97
180.95 183.84 186.21
190.23 192.22 195.08 196.97 200.59 204.38
207.2
208.98 (209)
(210)
(222)
87
88
89
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
Fr
Ra Ac¢
Rf
Db
Sg
Bh
Hs
Mt
Ds
Rg
Cn
Nh
F1
Mc
Lv
Ts
Og
(223)
(226)
(227)
(267)
(268)
(271)
(272)
(277)
(276)
(281)
(280)
(285)
(286)
(289)
(289)
(293)
(293)
(294)
120
137
145
146
147
148
149
150
Wn Aa
My
Fa
Je
Qu Ap
Bg | Im
295.8
298.4
341.3
359.9
362.5
364.2
366.1
367.7
58
59
60
61
62
63
64
65
66
67
68
69
70
71
* Ce
Pr
Nd Pm Sm
Eu
Gd
Tb
Dy Ho
Er
Tm Yb
Lu
(145)
150.36 151.96 157.25 158.93
162.50 164.93 167.26 168.93 173.05 174.97
140.12 140.91
144.24
92
93
94
95
96
97
98
99
100
101
102
103
90
91
Pu
Am Cm Bk
Cf
Es
Fm Md No
Lr
Np
(244)
t Th
Pa
U
(252)
(257)
(258)
(259)
(262)
232.04 231.04 238.03
(237)
(243)
(247)
(247)
(247)
2020
Elements and Polyatomic Ions NOT Accepted by IUPAC
Apochryphite
Apochryphate
ApO,
ApO,
QuO
QuO
Mythologium
Quackery
Aa
Alibabium
My
Apochryphor
Bogusine
Ap
2-
Im
Imaginarium
Qu
Bg
Quackerite
Jesticon
Wn
Wonkium
Fa
Fantasium
Je
Quackerate
8.

0282
at 23.4/• (281.7 amu)
and
275
Fa at 76.6- (274.6 amu)
145

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