Which of the following would represent a single displacement reaction between Potassium Bromide (K^+Br^-) and Iodine (I), which
1 answer:
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That will make a gold-202 nucleus.
<h3>Explanation</h3>Refer to a periodic table. The atomic number of mercury Hg is 80.
Step One: Bombard the with a neutron
. The neutron will add 1 to the mass number 202 of
. However, the atomic number will stay the same.
- New mass number: 202 + 1 = 203.
- Atomic number is still 80.
.
Double check the equation:
- Sum of mass number on the left-hand side = 202 + 1 = 203 = Sum of mass number on the right-hand side.
- Sum of atomic number on the left-hand side = 80 = Sum of atomic number on the right-hand side.
Step Two: The nucleus loses a proton
. Both the mass number 203 and the atomic number will decrease by 1.
- New mass number: 203 - 1 = 202.
- New atomic number: 80 - 1 = 79.
Refer to a periodic table. What's the element with atomic number 79? Gold Au.
.
Double check the equation:
- Sum of mass number on the left-hand side = 203 = 202 + 1 = Sum of mass number on the right-hand side.
- Sum of atomic number on the left-hand side = 80 = 79 + 1 = Sum of atomic number on the right-hand side.
A gold-202 nucleus is formed.