We are asked to fill up the third row of the table below:
Symbol | 79Br | ||||
Protons | 25 | 82 | |||
Neutrons | 30 | 64 | |||
Electrons | 48 | 86 | |||
Mass no. | 222 | 207 |
1st column
Given: 79Br mass no = 79
For 79Br :
Atomic number of Br = no of protons= no of electrons (neutral atom) = 35
Symbol | 79Br | ||||
Protons | 25 | 82 | |||
Neutrons | 30 | 64 | |||
Electrons | 48 | 86 | |||
Mass no. | 222 | 207 |
Complete the third row of the table.
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