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iris [78.8K]
3 years ago
12

Iodine-131 has a half-life of 8.10 days. In how many days will 50 grams of Iodine-131 decay to one-eighth of its original amount

?
Chemistry
2 answers:
shtirl [24]3 years ago
5 0
What you need to do is find 1/8 of 50
you can just divide 50 by 8 to get 6.25
so now you have to find how many days it will take till there are 6.25 grams of iodine left
every 8.1 days its mass is split in half 
so start splitting it in half and every time you do, you add 8.1 days
50/2 =25                                               8.1
25/2 =12.5                                        +  8.1
12.5/2= 6.25                                      +8.1
now you have reached 1/8 of the original amount of Iodine-131
so to find how long it took just add 8.1+8.1+8.1
(this is the same as 8.1x3)
which equals 24.3
it will take 24.3 days for Iodine 131 to decay to 1/8  of its original mass.

(good luck on the regent if thats what your studying for :)

Murljashka [212]3 years ago
3 0
It doesn't matter what the original mass is. 1/8 is (1/2)-cubed, and 3 half-lives is 8.3 days.
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Procedures involved:

The cations or anions may transfer positions in this twofold replacement/displacement reaction, which results in AB + CD AD + CB. In such a reaction, water, an insoluble gas, or an insoluble solid must be one of the byproducts (precipitate). The reaction in question has the following molecular equation:

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N76 [4]

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No of moles = mass/molar mass

We can find mass from above formula.

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Also,

No of moles = no of molecules/Avogadro number

N=n\times N_A\\\\N=8.437\times 10^{-2}\times 6.023\times 10^{23}\\\\N=5.08\times 10^{22}\ \text{molecules}

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