Explanation:
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Http://www.geologyin.com/2014/11/how-to-identify-minerals-in-10-steps.html
1070 hours.
1 mole of iron-59 would mass 59 grams, so 0.133 picograms would be 0.133x10^-12 / 59 = 2.25x10^-15 moles of iron-59. Multiplying by Avogadro's number, we can determine the number of atoms of iron-59 we have, so: 2.25x10^-15 * 6.02214x10^23 = 1.35x10^9
Since we have 242 decays over a period of 1 second, we can divide the
number of atoms left by the original number of atoms
(1350000000 - 242)/1350000000
= 1349999758/1350000000
= 0.999999820740741
And calculate the logarithm to base 2 of that quotient.
ln(0.999999820740741)/ln(2)
= -1.79259275281191x10^-7/0.693147180559945
= -2.58616467481524x10^-7
The reciprocal of this number will be the half life in seconds. So
-1/2.58616467481524x10^-7
= -3866729.79388461
And dividing by 3600 (number of seconds in an hour) will give the half-life in
hours.
-3866729.79388461 / 3600 = -1074.091609
So the half life in hours to 3 significant figures is 1070 hours.
Dividing that figure by 24 gives a half life of 44.58 days which is in pretty close agreement to the official half-life of 44.495 days for iron-59.
Answer:
![[SO_2Cl_2]=0.0175M](https://tex.z-dn.net/?f=%5BSO_2Cl_2%5D%3D0.0175M)
Explanation:
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In this case, considering that the decomposition reaction of SO2Cl2 is first-order, we can write the rate law shown below:
![r=-k[SO_2Cl_2]](https://tex.z-dn.net/?f=r%3D-k%5BSO_2Cl_2%5D)
We also consider that the integrated rate law has been already reported as:
![[SO_2Cl_2]=[SO_2Cl_2]_0exp(-kt)](https://tex.z-dn.net/?f=%5BSO_2Cl_2%5D%3D%5BSO_2Cl_2%5D_0exp%28-kt%29)
Thus, by plugging in the initial concentration, rate constant and elapsed time we obtain:
![[SO_2Cl_2]=0.0225Mexp(-2.90x10^{-4}s^{-1}*865s)](https://tex.z-dn.net/?f=%5BSO_2Cl_2%5D%3D0.0225Mexp%28-2.90x10%5E%7B-4%7Ds%5E%7B-1%7D%2A865s%29)
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