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leva [86]
1 year ago
8

A 50.0 mg sample of iodine-131 was placed in a container 32.4 days ago. if its half-life is 8.1 days, how many milligrams of iod

ine-131 are now present?
Chemistry
1 answer:
sertanlavr [38]1 year ago
7 0

3.124mg of I-131 is present after 32.4 days.

The 131 I isotope emits radiation and particles and has an 8-day half-life. Orally administered, it concentrates in the thyroid, where the thyroid gland is destroyed by the particles.

What is Half life?

The time required for half of something to undergo a process: such as. a : the time required for half of the atoms of a radioactive substance to become disintegrated.

Half of the iodine-131 will still be present after 8.1 days.

The amount of iodine-131 will again be halved after 8.1 additional days, for a total of 8.1+8.1=16.2 days, reaching (1/2)(1/2)=1/4 of the initial amount.

The quantity of iodine-131 will again be halved after 8.1 more days, for a total of 16.2+8.1+8.1=32.4 days, to (1/4)(1/2)(1/2)=1/16 of the initial quantity.

If the original dose of iodine-131 was 50mg, the residual dose will be (50mg)*(1/16)=3.124mg after 32.4 days.

Learn more about the Half life of radioactie element with the help of the given link:

brainly.com/question/27891343

#SPJ4

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A 30.7 g sample of Strontium nitrate, Sr(NO3)2•nH2O, is heated to a constant mass of 22.9 g. Calculate the hydration number.
Elodia [21]

Answer:

  • <em>Hydration number:</em> 4

Explanation:

<u>1) Mass of water in the hydrated compound</u>

Mass of water = Mass of the hydrated sample - mass of the dehydrated compound

Mass of water = 30.7 g - 22.9 g = 7.8 g

<u>2) Number of moles of water</u>

  • Number of moles = mass in grams / molar mass

  • molar mass of H₂O = 2×1.008 g/mol + 15.999 g*mol = 18.015 g/mol

  • Number of moles of H₂O = 7.9 g / 18.015 g/mol = 0.439 mol

<u>3) Number of moles of Strontium nitrate dehydrated, Sr (NO₃)₂</u>

  • The mass of strontium nitrate dehydrated is the constant mass obtained after heating = 22.9 g

  • Molar mass of Sr (NO₃)₂ :  211.63 g/mol (you can obtain it from a internet or calculate using the atomic masses of each element from a periodic table).

  • Number of moles of Sr (NO₃)₂ = 22.9 g / 211.63 g/mol =  0.108 mol

<u>4) Ratio</u>

  • 0.439 mol H₂O / 0.108 mol Sr(NO₃)₂ ≈  4 mol H₂O : 1 mol Sr (NO₃)₂

Which means that the hydration number is 4.

4 0
4 years ago
Read 2 more answers
How many cm 3 are in 0.14 m 3?
nlexa [21]

Answer:

There are 14 cm in 0.14 m

5 0
3 years ago
What is the molarity of a solution that contains 1000.0 mg of AgNO3 that has been dissolved in 500 mL of water
Vaselesa [24]

0.012moldm⁻³

Explanation:

Given parameters:

Mass of AgNO₃  = 1000mg

Volume of water = 500mL

Unknown:

Molarity of solution  = ?

Solution:

The molarity of a solution is the number of moles of a solute dissolved in volume of solvent.

 Molarity = \frac{xnumber of moles}{Volume}

 

Number of moles of AgNO₃  = ?

   Number of moles = \frac{mass}{molar mass}

Molar mass of AgNO₃ = 108 + 14 + 3(16) = 170g/mol

   convert mass to g;

      1000mg = 1g

 Number of moles  = \frac{1}{170}  = 0.00588moles

   convert the given volume to dm³;

       1000mL  = 1dm³;

        500mL = 0.5dm³

Now solve;

  Molarity = \frac{0.00588}{0.5}  = 0.012moldm⁻³

learn more:

Molarity brainly.com/question/9324116

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4 0
4 years ago
What is a Control trial
AleksandrR [38]
It is a trial aimed to reduce bias during an experiment. An example would be a sugar pill, something that has no real effect so that the results of the true trial can accurately be compared. Its like a control group.
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3 years ago
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nordsb [41]

#1 B

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