Two nitro (NO_2) groups are chemically bonded to a patch of surface. They can't move to another location on the surface, but the
y can rotate (see sketch at right). It turns out that the amount of rotational kinetic energy each NO_2 group can have is required to be a multiple of epsilon, where epsilon = 1.0 times^-24 J. In other words, each NO_2 group could have epsilon of rotational kinetic energy, or 2 epsilon, or 3 epsilon, and so forth - but it cannot have just any old amount of rotational kinetic energy. Suppose the total rotational kinetic energy in this system is initially known to be 39 epsilon. Then, some heat is added to the system, and the total rotational kinetic energy rises to 59 epsilon. Calculate the change in entropy. Round your answer to 3 significant digits, and be sure it has the correct unit symbol.
1 answer:
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Answer:
8.625 grams of a 150 g sample of Thorium-234 would be left after 120.5 days
Explanation:
The nuclear half life represents the time taken for the initial amount of sample to reduce into half of its mass.
We have given that the half life of thorium-234 is 24.1 days. Then it takes 24.1 days for a Thorium-234 sample to reduced to half of its initial amount.
Initial amount of Thorium-234 available as per the question is 150 grams
So now we start with 150 grams of Thorium-234
So after 120.5 days the amount of sample that remains is 8.625g
In simpler way , we can use the below formula to find the sample left
Where
is the initial sample amount
n = the number of half-lives that pass in a given period of time.