2 '‫תרגיל בית מס‬
1. Complete the following table. What minimum amount of information is
required to characterize completely an atom or ion? (Note that not all rows can
be completed.)
Name
Symbol
Number
Number
Number
Mass
Protons
electrons
neutrons
number
Sodium
23
11Na
11
11
12
23
Silicon
28
14Si
14
14
14
28
Rubidium
85
37Rb
37
-----
48
85
Potassium
40
19K
19
-----
21
40
--------------------------------------
33
42 --------------
Neon
20
10
8
10
20
Bromine
80
35
-----
45
80
+2
10Ne
35Br
-------------------------------------------------------------------- 126 ---------------
2. Arrange the following species in order of increasing: a) number of electrons
b) Number of neutrons c) mass.
112
50Sn
40
18Ar
122
52Te
59
29Cu
a)
40
18Ar
39
19K
58
27Co
59
b)
39
19K
40
59
29Cu
58
c)
39
19K
40
18Ar
58
18Ar
27Co
120
48Cd
58
27Co
39
19K
120
48Cd
112
50Sn
122
52Te
27Co
112
50Sn
122
52Te
120
48Cd
59
29Cu
112
50Sn
120
48Cd
122
29Cu
3. Given the following species: 24Mg+2
47
Cr
60
Cr+3
35
Cl-
90
Sr.
a) Has equal numbers of neutrons and protons?
24
Mg+2
b) Has protons contributing more than 50% of the mass? 47Cr
c) Have 50% more neutrons than protons?
60
Cr+3
124
52Te
Sn+2
226
Th
4. What is the total number of atoms in each of the following samples?
a) 12.7mol Ca. b) 0.00361mol Ne. c) 1.8X10-12 mol Pu
a) 1 mol Ca
12.7 mol Ca
6.023*1023 atoms
X
atoms
12.7*6.023*1023 = 7.64*1024 mol
1
The total number of atoms in 12.7 mol Ca is: 7.64*1024
b) 1 mol Ne
0.00361 mol Ne
6.023*1023 atoms
X
atoms
0.00361*6.023*1023 = 2.17*1021 mol
1
The total number of atoms in 0.00361 mol Ne is: 2.17*1021
c) 1 mol Pu
1.8*10-12 mol Pu
6.023*1023 atoms
X
atoms
1.8*10-12*6.023*1023 = 1.08*1012 mol
1
The total number of atoms in 1.8*10-12 mol Pu is: 1.08*1012
5. Calculate the quantities indicated.
a) The number of moles represented by 2.18X1026 Fe atoms.
b) The mass in grams of 7.71 mol Kr.
c) The mass in mg of a sample containing 6.15X1019 Au atoms.
a) 1 mol Fe
X mol Fe
6.023*1023 atoms
2.18*1026 atoms
1*2.18*1026 =361.94 atoms
6.023*1023
The number of moles represented by 2.18X1026 Fe atoms is: 361.94 moles.
b) 1 mol Kr
7.71 mol Kr
83.80 gr
X gr
7.71*83.80 = 646.1 gr
1
The mass of 7.71 mol Kr is: 646.1 gr.
c) 6.023*1023 atoms Au
6.15*1019 atoms Au
196.97 gr
X
gr
6.15*1019*196.97 = 0.02 gr
6.023*1023
0.02*1000=20.11
The mass of a sample containing 6.15X1019 Au atoms is: 20.11 mg.
6. Hydrogen and chlorine atoms form simple diatomic molecules with H and Cl in a
1:1 ratio that is HCl. The natural abundances of the chlorine isotopes are 75.53%
35
Cl and 24.47% 37Cl. The natural abundances of 2H and 3H are 0.015% and less
than 0.001% respectively. How many different HCl molecules are possible and
what are their mass numbers (i.e. the sum of the mass numbers of the H and Cl
atoms)?
35
Cl + 1H =
36
HCl
35
37
HCl
35
38
HCl
37
38
HCl
37
39
HCl
37
40
HCl
Cl + 2H =
Cl + 3H =
Cl + 1H =
Cl + 2H =
Cl + 3H =