This question is missing the part that actually asks the question. The questions that are asked are as follows:
(a) How much of a 1.00 mg sample of americium remains after 4 day? Express your answer using 2 significant figures.
(b) How much of a 1.00 mg sample of iodine remains after 4 days? Express your answer using 3 significant figures.
We can use the equation for a first order rate law to find the amount of material remaining after 4 days:
[A] = [A]₀e^(-kt)
[A]₀ = initial amount
k = rate constant
t = time
[A] = amount of material at time, t.
(a) For americium we begin with 1.00 mg of sample and must convert time to units of years, as our rate constant, k, is in units of yr⁻¹.
4 days x 1 year/365 days = 0.0110
A = (1.00)e^((-1.6x10^-3)(0.0110))
A = 1.0 mg
The decay of americium is so slow that no noticeable change occurs over 4 days.
(b) We can simply plug in the information of iodine-125 and solve for A:
A = (1.00)e^(-0.011 x 4)
A = 0.957 mg
Iodine-125 decays at a much faster rate than americium and after 4 days there will be a significant loss of mass.
Answer:
if i got it right i think it woud be a
Yes they are stable because they follow octet rule but am not sure if they are exist separetly
Explanation:
The given data for case (1) is as follows.
h = 20 cm = 0.2 m
Assuming that a rectangular slab is placed above the pipe and we will calculate the heat transfer as follows.
Q = 
where, A = area
L = length
k = thermal conductivity = 0.8 W/m
= change in temperature.
Therefore, putting the given values into the above formula as follows.
Q =
= 
= 168 W
For case (2), h = 180 cm = 1.8 m
Therefore, heat lost will be calculated as follows.
Q =
= 
= 18.67 W
Thus, we can conclude that 18.67 W heat lost if the pipe was buried at a depth of 180 cm.
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Here is the link to a great video that explains your question nicely, hope this helps.
