After decaying for 48 hours, of the original mass of a radioisotope sample remains unchanged. what is the half-life of this radioisotope? 1. 3.0 h 2. 9.6 h 3. 12 h 4. 24 h

The question is incomplete.

The complete question is: After decaying for 48 hours, 1/16 of the original mass of a radioisotope sample remains unchanged. What is the half-life of this radioisotope?

First, calculate how many half-lives have passed to reach 1/16 of the original mass.

Each time that one half-life time passes the mass reduces to half: =>

1 half-life time => 1/2

2 half-life time => 1/4

3 half-life time => 1/8

4 half-life time = > 1/16

Then 4 half-life time correspond to 48 hours, and 1 half-life time corresponds to x:

 4           1
  =    
48h        x

=> x = 48 h * 1 / 4  = 12 h.

12 h.

Hno3(aq) → h+(aq) + no3−(aq). picture ii best represents the strong acid hno3. strong acids are strong electrolytes