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rosijanka [135]
3 years ago
9

Iodine-131 is used to

Chemistry
2 answers:
tekilochka [14]3 years ago
7 0
<span>
Radioactive iodine(l-131) an isotope of iodine that emits radiation</span>
Rufina [12.5K]3 years ago
6 0

Answer: iodine - 131 is used to treat hyperthyroidism.

Explanation:

Iodine - 131, a radioactive iodine salt, is an important diagnostic tool and medical therapeutic tool used to treat hyperthyroidism due to nodules in thyroid gland or graves' diseases.

Iodine-131 is usually administered orally as a single dose. Once taken, it emits both beta and gamma radiation. The beta radiations (the active agent) kills the overactive thyroid cells. The Beta rays travel small distances, but not usually exiting from the body of such patient. While, the gamma radiation can travel long distances, existing from the patient's body which can affect people in such surroundings.

Iodine-131 alters iodine absorption mechanism in the thyroid glands.

It can also be used in diagnostic imaging techniques for neuroblastoma.

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If 2.0 g of copper(II) chloride react with excess sodium nitrate, what mass of sodium chloride is formed in this double replacem
cluponka [151]

Taking into account the reaction stoichiometry, 1.729 grams of NaCl is formed.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

CuCl₂ + 2 NaNO₃ → Cu(NO₃)₂ + 2 NaCl

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • CuCl₂: 1 mole
  • NaNO₃: 2 moles
  • Cu(NO₃)₂ : 1 mole
  • NaCl: 2 moles

The molar mass of the compounds is:

  • CuCl₂: 134.44 g/mole
  • NaNO₃: 85 g/mole
  • Cu(NO₃)₂ : 187.54 g/mole
  • NaCl: 58.45 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • CuCl₂: 1 mole ×134.44 g/mole= 134.44 grams
  • NaNO₃: 2 moles ×85 g/mole= 170 grams
  • Cu(NO₃)₂ : 1 mole ×187.54 g/mole= 187.54 grams
  • NaCl: 2 moles ×58.45 g/mole= 116.9 grams

<h3>Mass of NaCl formed</h3>

The following rule of three can be applied: If by stoichiometry of the reaction 134.44 grams of CuCl₂ form 116.9 grams of NaCl, 2 grams of CuCl₂ form how much mass of NaCl?

mass of NaCl=\frac{2 grams of CuCl_{2}x116.9 grams of NaCl }{134.44grams of CuCl_{2}}

<u><em>mass of NaCl= 1.739 grams</em></u>

Finally, 1.729 grams of NaCl is formed.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

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If the distance between two objects is decreased to - of the original
Charra [1.4K]

Answer:

The new force will be \frac{1}{100} of the original force.

Explanation:

In the context of this problem, we're dealing with the law of gravitational attraction. The law states that the gravitational force between two object is directly proportional to the product of their masses and inversely proportional to the square of a distance between them.

That said, let's say that our equation for the initial force is:

F = G\frac{m_1m_2}{R^2}The problem states  that  the distance decrease to 1/10 of the original distance, this means:[tex]R_2 = \frac{1}{10}R

And the force at this distance would be written in terms of the same equation:

F_2 = G\frac{m_1m_2}{R_2^2}

Find the ratio between the final and the initial force:

\frac{F_2}{F} = \frac{G\frac{m_1m_2}{R_2^2}}{G\frac{m_1m_2}{R^2}}

Substitute the value for the final distance in terms of the initial distance:

\frac{F_2}{F} = \frac{G\frac{m_1m_2}{(\frac{R}{10})^2}}{G\frac{m_1m_2}{R^2}}

Simplify:

\frac{F_2}{F} = \frac{\frac{1}{100R^2}}{\frac{1}{R^2}}=\frac{1}{100}

This means the new force will be \frac{1}{100} of the original force.

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