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Otrada [13]
2 years ago
7

Iodine-131 is one of the most important isotopes used in the diagnosis of thyroid cancer. One atom has a mass of 130.906114 amu.

Calculate the binding energy(c) per mole in kJ.
Chemistry
1 answer:
emmainna [20.7K]2 years ago
5 0

Iodine-131 is one of the most important isotopes used in the diagnosis of thyroid cancer. One atom has a mass of <u>130.906114</u> amu.\

<h3>What is thyroid cancer?</h3>

Cancer that originates in the tissues of the thyroid gland is known as thyroid cancer. It is a condition where cells develop improperly and are susceptible to spreading to different bodily regions. A bump in the neck or swelling are examples of symptoms. Thyroid cancer is not always diagnosed because it can move from other parts of the body to the thyroid.

Young age radiation exposure, having an enlarged thyroid, and family history are risk factors. Papillary thyroid cancer, follicular thyroid cancer, medullary thyroid cancer, and anaplastic thyroid cancer are the four primary kinds. Ultrasound and tiny needle aspiration are frequently used in diagnosis. As of right now, it is not advised to screen those who are healthy and at normal risk for the disease.

To learn more about thyroid cancer from the given link:

brainly.com/question/11880360

#SPJ4

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Draw a structural formula for the alkene you would use to prepare the alcohol shown by hydroboration/oxidation.
Nastasia [14]

Answer:

See explanation

Explanation:

The question is incomplete because the image of the alcohol is missing. However, I will try give you a general picture of the reaction known as hydroboration of alkenes.

This reaction occurs in two steps. In the first step, -BH2 and H add to the same face of the double bond (syn addition).

In the second step, alkaline hydrogen peroxide is added and the alcohol is formed.

Note that the BH2 and H adds to the two atoms of the double bond. The final product of the reaction appears as if water was added to the original alkene following an anti-Markovnikov mechanism.

Steric hindrance is known to play a major role in this reaction as good yield of the anti-Markovnikov like product is obtained with alkenes having one of the carbon atoms of the double bond significantly hindered.

5 0
3 years ago
Which of these zones of the ocean is most hospitable to life?
Nat2105 [25]
The answer is epipelagic zone
3 0
3 years ago
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A gas cylinder initially contains 463 L of gas at a pressure of 159 atm. If the final volume of gas is 817 L, what is the final
Agata [3.3K]

Answer:

The final pressure is 90.1 atm.

Explanation:

Assuming constant temperature, we can solve this problem by using <em>Boyle's Law</em>, which states:

  • P₁V₁=P₂V₂

Where in this case:

  • P₁ = 159 atm
  • V₁ = 463 L
  • P₂ = ?
  • V₂ = 817 L

We <u>input the given data</u>:

  • 159 atm * 463 L = P₂ * 817 L

And <u>solve for P₂</u>:

  • P₂ = 90.1 atm

The final pressure is 90.1 atm.

6 0
3 years ago
Which two structures would provide a postive idenfitacation of a plant cell under a microscope
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8 0
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25 points please help
maxonik [38]

Answer : Option (A) Accelerator 2 model has the lowest percentage of energy lost as waste.

Solution : Given,

For Accelerator 1 model,

Input energy = 2078.3 J

Wasted energy = 663.1 J

Output energy = 1415.2 J

For Accelerator 2 model,

Input energy = 7690.0 J

Wasted energy = 2337.5 J

Output energy = 5353.5 J

For Accelerator 3 model,

Input energy = 4061.9 J

Wasted energy = 2259.6 J

Output energy = 1802.3 J

Formula used for lowest percentage of energy lost as waste is:

% energy lost as waste = (Total energy wasted / Total input energy )  ×  100

For Accelerator 1 model,

% energy lost as waste = \frac{663.1}{2078.3}\times100 = 31.90%

For Accelerator 2 model,

% energy lost as waste = \frac{2337.5}{7690.0}\times100 = 30.39%

For Accelerator 3 model,

% energy lost as waste = \frac{2259.6}{4061.9}\times100 = 55.62%

So, we conclude that the Accelerator 2 model has the lowest percentage of energy lost as waste.



6 0
3 years ago
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