Answer:
c
Explanation:
I think it is right not sure but hope it helped
78.6 g (1mol/60.1 g)= 1.31 moles of isopropanol
Answer:
Amount left after 140 days is 38.125 g.
Explanation:
Given data:
Half life of Co-59 = 35 days
Total mass of sample = 610.0 g
Sample remain after 140 days = ?
Solution:
Number of half lives passes during 140 days:
Number of half lives = T elapsed / Half lives
Number of half lives = 140 days / 35 days
Number of half lives = 4
Amount left:
At time zero = 610 g
At first half life = 610 g/2 = 305 g
At second half life = 305 g/2 = 152.5 g
At 3rd half life = 152.5 g/2 = 76.25 g
At 4th half life = 76.25 g /2 =38.125 g
Amount left after 140 days is 38.125 g.
