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DIA [1.3K]
3 years ago
6

Iodine-131 is administered orally in the form of NaI(aq) as a treatment for thyroid cancer. The half-life of iodine-131 is 8.04

days. How much time is required for the activity of a sample of iodine-131 to fall to 17.5 percent of its original value?
Chemistry
1 answer:
gregori [183]3 years ago
5 0

Answer:

24.12 days

Explanation:

First off, let's find out how many decays the compound ; iodine131 undergoes to get to 17.5 percent of its original value.

Half life is simply the time required for a quantity to reduce to half of its initial value.

The half-life of iodine-131 is 8.04 days.

100% - 50%  (First Half life)

50%  - 25%  (Second Half life)

25%  - 17.5%  (Third Half life)

This mwans i would take three halff lives;

Time requred = 3 * Half life = 3 * 8.04 days = 24.12 days

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Starting with the stock solution of 6.0 M, how many milliliters of 6.0 M sulfuric acid are needed to make 450 mL of 1.2 M soluti
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<u>Answer:</u> The volume of stock solution needed is 90 mL

<u>Explanation:</u>

To calculate the molarity of the diluted solution, we use the equation:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of the stock sulfuric acid solution

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We are given:

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Putting values in above equation, we get:

6.0\times V_1=1.2\times 450\\\\V_1=\frac{1.2\times 450}{6.0}=90mL

Hence, the volume of stock solution needed is 90 mL

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