The half -life of the radioactive element unobtanium-29 is 5 seconds. If 80 grams of unobtanium-29 are initially present, how many grams are present after 5 seconds? 10 seconds? 15 seconds? 20 seconds? 25 seconds?

Step-by-step explanation:

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the remaining mass of unobtanium -41 after a 5 seconds, 10 seconds, 15 seconds and 20 seconds is 24 gr, 12 gr , 5.33 gr and 3 gr respectively

Step-by-step explanation:

the equation that governs the remaining mass m of unobtanium -41 after a time t is

m=m₀*2^(-t/T) , where t is in seconds ,m₀ represents initial mass and T=half-life

Therefore

a) for t=5 s

m=48 gr*2^(-t/5 s) = 48 gr*2^(-5 s/5 s) = 48 gr/2 = 24 gr

b)  for t=10 s

m=48 gr*2^(-t/5 s) = 48 gr*2^(-10 s/5 s) = 48 gr/4 = 12 gr

b)  for t=15 s

m=48 gr*2^(-t/5 s) = 48 gr*2^(-15 s/5 s) = 48 gr/9 = 5.33 gr

b)  for t=20 s

m=48 gr*2^(-t/5 s) = 48 gr*2^(-20 s/5 s) = 48 gr/16 = 3 gr

thus the remaining mass of unobtanium -41 after a 5 seconds, 10 seconds, 15 seconds and 20 seconds is 24 gr, 12 gr , 5.33 gr and 3 gr respectively

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it is a. 4.2 because 7.6-3.4 gets the remaining length of 4.2. i might be wrong but i am pretty sure this is the answer.