Objects to resist motion hope this helps.
If the precipitate of CaSO4 was weighed before all of the
water has evaporated, therefore this will lead to a miscalculation on the
number of moles of CaSO4. During the calculation, the mass of the water will be
included hence there will be higher number of moles than what is correct.
Answer:
The fraction of Sr-90 left after 73 years is 0.1726
Explanation:
Initial mass of the isotope = ![N_o](https://tex.z-dn.net/?f=N_o)
Time taken by the sample, t = ![t_{\frac{1}{2}}](https://tex.z-dn.net/?f=t_%7B%5Cfrac%7B1%7D%7B2%7D%7D)
Formula used :
![N=N_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}](https://tex.z-dn.net/?f=N%3DN_o%5Ctimes%20e%5E%7B-%5Clambda%20t%7D%5C%5C%5C%5C%5Clambda%20%3D%5Cfrac%7B0.693%7D%7Bt_%7B%5Cfrac%7B1%7D%7B2%7D%7D%7D)
where,
= initial mass of isotope
N = mass of the parent isotope left after the time, (t)
= half life of the isotope
= rate constant
We have:
![t_{\frac{1}{2}}=28.8 years](https://tex.z-dn.net/?f=t_%7B%5Cfrac%7B1%7D%7B2%7D%7D%3D28.8%20years)
t = (2018 - 1945 )years = 73 years
So, on substituting the values:
![N=x\times e^{-(\frac{0.693}{28.8 years})\times 73 years}](https://tex.z-dn.net/?f=N%3Dx%5Ctimes%20e%5E%7B-%28%5Cfrac%7B0.693%7D%7B28.8%20years%7D%29%5Ctimes%2073%20years%7D)
Now put all the given values in this formula, we get
![N=N_o\times e^{-1.7566}](https://tex.z-dn.net/?f=N%3DN_o%5Ctimes%20e%5E%7B-1.7566%7D)
![N=N_o\times 0.1726](https://tex.z-dn.net/?f=N%3DN_o%5Ctimes%200.1726)
![\frac{N}{N_o}=0.1726](https://tex.z-dn.net/?f=%5Cfrac%7BN%7D%7BN_o%7D%3D0.1726)
The fraction of Sr-90 left after 73 years = 0.1726
Depending on what you are using the chemical could react in an explosion if the substance is not known