Answer:
M=0.15
Explanation:
138 g AgNO -> 1 mol AgNO
10 g AgNO -> x
x= (10 g AgNO * 1 mol AgNO)/138 g x=0.07 mol AgNO
450 mL=0.45 L
M= mol solute/L solution
M= 0.07 mol AgNO/0.45L
M=0.15
Answer:
4 g OF IODINE-131 WILL REMAIN AFTER 32 DAYS.
Explanation:
Half life (t1/2) = 8 days
Original mass (No) = 64 g
Elapsed time (t) = 32 days
Mass remaining (Nt) = ?
Using the half life equation we can obtain the mass remaining (Nt)
Nt = No (1/2) ^t/t1/2
Substituting the values, we have;
Nt = 64 * ( 1/2 ) ^32/8
Nt = 64 * (1/2) ^4
Nt = 64 * 0.0625
Nt = 4 g
So therefore, 4 g of the iodine-131 sample will remain after 32 days with its half life of 8 days.
