Answer:
17.5 g of manganese-56
Explanation:
Now we have to use the formula;
N/No= (1/2)^t/t1/2
Where
N= mass of manganese-56 left after time (t) = ????
No= initial mass of manganese-56 present = 20 g
t= time taken for N mass of manganese-56 to remain = 7.8 hours
t1/2= half-life of manganese-56 = 2.6 hours
Then we substitute values;
N/20 = (1/2)^7.8/2.6
N/20 = (1/2)^3
N= (1/2)^3 ×20
N= 2.5 g of manganese-56
Mass of manganese-56 that has disappeared after 7.8 hours= 20 -2.5 = 17.5 g of manganese-56
