Question

In: Chemistry

1. The free energy available from a 1.78V Fe(III) – Ag cell under standard conditions is:...

1. The free energy available from a 1.78V Fe(III) – Ag cell under standard conditions is:

a. -515 KJ

b. +515 KJ

c. +172 KJ

d. -172 KJ

2. What is the nuclear binding energy for a mole of lithium-7 (Li-7), whose mass is 7.01600? (mass of a proton = 1.00728 AMU; neutron = 1.00866)

a. 6.34 x 1017J

b. 6.34 x 1014J

c. 3.64 x 1012J

d. 3.44 x 1015J

3.  Tc-99 is commonly used in nuclear medicine studies. It has a half-life of 6 hours. How long does it take for less than 1% of the original injected isotope to remain?

a. 24 hours

b. 30 hours

c. 36 hours

d. 42 hours

Solutions

Expert Solution

ANSWER

Question 1

  • The cell reactions are

  • This cell has a potential (Eº) of 1.78 V
  • The free energy of acell could be defined as

where n = number of electrons transferred and F = Faraday's constant = 96485 J / mol.V

  • Then

  • The free energy for the cell is -172 kJ (option d)

​​​​​​​----------------------------------------------------------

Question 2

  • The nuclear binding energy is defined as

where c = speed light = 3.00x108 m/s and m = mass defect of nucleus in kg

  • Then, we need to find the vale of for Li-7

  • The nucleus Li-7 is composed by

  • then

  • This is the mass for one atom of Li-7, then for one mole of Li-7

  • Finally, the binding energy for one mole of Li-7 is

  • The correct answer is the option c​​​​​​​

​​​​​​​----------------------------------------------------------

Question 3

  • The formula that relates the half-life of an isotpe an the amount is

where A = final amount of isotope

A0 = initial amount of isotope

t = time

t1/2 = half-life

  • The number of hours required to decrease the amount of isotope to 1% is

  • Answer: after 42 hours (option d) it will remains less than 1% of Tc99

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