Question

Determine the missing %abundance and mass of two isotope of Selenium given the atomic mass of Selenium from the periodic table and the other stable isotopes of Selenium that have the following %abundances and masses: (7 points)

      Isotope                                Isotope mass (amu)            %abundance                   

74Se 73.922477 ??????

76Se 75.919214 9.38

77Se 76.919915 7.63

78Se 77.917310 23.77

80Se 79.916522 49.61

82Se ???????? 8.73

b) Calculate the missing information for the isotopes of Selenium listed in Part A (Question 22) (5 points)

Isotope abbreviation	Isotopic symbol	protons	neutrons	electrons
Se-74	74          Se     34	34	40	34

c) Germanium has a melting of 428^oF

I. Calculate the melting point in Kelvin. (2 points)

II. If a sample of Germanium has a temperature 100 K higher than this melting point (K), what is the physical state (solid, liquid, or gas) of the germanium sample? (1 point)

III. Calculate the boiling point of Selenium (in K) if 2.269 x 10⁶ J of heat is needed to warm 20.0 kg of Selenium in Part II (Question#22), just to the boiling point. (sp Heat = 0.31 J/(g*^oC)) (5 points)

Answer 1

a)

From the given data

the % abundance of 74-Se isotope can be found by difference in %.

% of 74-Se = 100 - [ sum of % abundance of other isotopes]

                = 100 - [ 9.38 + 7.63 + 23.77 + 49.61 + 8.73]

               = 100 - 99.12

                = 0.88

Thus the % abundance of 74-Se is 0.88 %

The atomic mass of Se from periodic table is 78.96

Therefore the total of [7180.82607 + 8.73 M] / 100 = 78.96 where M is the mass of 82-Se isotope

                                   7180.82607 + 8.73 M = 7896

                                   8.73 M = 715.17393

                                     M = 715.17393 / 8.73

                                         = 81.92141

Thus the isotope mass of 82-Se is 81.92141 (amu)

-------------------------------------------------------------------------------------------------------

b)

The mass number = number of protons + number of neutrons.

The number of protons is always the same 34.

Hence the number of neutraons = mass number - 34.

The number of electrons for neutral atom = the number of protons

Isotope	isotope	Protons	neutrons	electrons
abbreviation	symbol
Se-74	74Se34	34	40	34
Se-76	76Se34	34	42	34
Se-77	77Se34	34	43	34
Se-78	78Se34	34	44	34
Se-80	80Se34	34	46	34
Se-82	82Se34	34	48	34

-----------------------------------------------------------------------------------------------------------------------

c)

(i)

Given the m.pt of Se = 428 deg F

K = 5/9 * [deg F - 32] + 273

      = 5/9 * ( 428 - 32) + 273

      = 220 + 273 = 493 K

Thus the m.pt of Se is 493 K

(ii)

At a temperature higher than 493 K is 493 + 100 = 593 K. At this point all the Se will be in molten state. That is in liquid state.

[ note that it is given as Germanium by mistake , it should be Selenium )

(iii)

Let the B.Pt of Se be T2 K

Therefore T = (T2 - 493 )

mass = 20 Kg = 20,000 g

Sp heat = 0.31 J / g-C

Heat Q = 2.269 x 10^6 J

Therefore Q = Heat to raise the tempr to b.pt

                  = mass x Heat of fusion + mass x sp.Heat x T

2.269 x 10^6    = 20,000 x 0.31 x ( T2-593)

                      =   6200 ( T2-593)

( T2-593) = 366

            T2 = 366 + 593 = 959 K

           Thus the B.Pt of Se = 959 K