This should not matter because the pipet has gradations and usually more of the sample is taken up in the pipette than what is delivered into the flask the student should always rinse the container being used because they are contaminating the sample if they do not clean it out
Answer:
624510100
Explanation:
Doing a conversion factor:
![0,0006245101[km]*\frac{1000[m]}{1 km} *\frac{1x10^{9} nanometer}{1 m} =624510100 [nanometer]](https://tex.z-dn.net/?f=0%2C0006245101%5Bkm%5D%2A%5Cfrac%7B1000%5Bm%5D%7D%7B1%20km%7D%20%2A%5Cfrac%7B1x10%5E%7B9%7D%20nanometer%7D%7B1%20m%7D%20%3D624510100%20%5Bnanometer%5D)
For this problem we can use half-life formula and radioactive decay formula.
Half-life formula,
t1/2 = ln 2 / λ
where, t1/2 is half-life and λ is radioactive decay constant.
t1/2 = 8.04 days
Hence,
8.04 days = ln 2 / λ
λ = ln 2 / 8.04 days
Radioactive decay law,
Nt = No e∧(-λt)
where, Nt is amount of compound at t time, No is amount of compound at t = 0 time, t is time taken to decay and λ is radioactive decay constant.
Nt = ?
No = 1.53 mg
λ = ln 2 / 8.04 days = 0.693 / 8.04 days
t = 13.0 days
By substituting,
Nt = 1.53 mg e∧((-0.693/8.04 days) x 13.0 days))
Nt = 0.4989 mg = 0.0.499 mg
Hence, mass of remaining sample after 13.0 days = 0.499 mg
The answer is "e"
