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3 years ago

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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A 17.6-g sample of ammonium carbonate contains ________ mol of ammonium ions.

postnew [5]

These problems are a bit interesting. :)

First let's write the molecular formula for ammonium carbonate.

NH4CO3 (Note! The 4 and 3 are subscripts, and not coefficients)

17.6 gNH4CO3

Now to convert to mol of one of our substances we take the percent composition of that particular part of the molecule and multiply it by our starting mass. This is what it looks like using dimensional analyse.

17.6 gNH4CO3 * (Molar Mass of NH4 / Molar Mass of NH4CO3)

Grab a periodic table (or look one up) and find the molar masses for these molecules! Well. In this case I'll do it for you. (Note: I round the molar masses off to two decimal places)

NH4 = 14.01 + 4*1.01 = 18.05 g/mol
NH4CO3 = 14.01 + 4*1.01 + 12.01 + 3*16.00 = 78.06 g/mol

17.6 gNH4CO3 * (18.05 molNH4 / 78.06 molNH4CO3)
= 4.07 gNH4

Now just take the molar mass we found to convert that amount into moles!

4.07 gNH4 * (1 molNH4 / 18.05 gNH4) = 0.225 molNH4

4 0

2 years ago

What is the [H+] concentration of blood, given the ph is 2.5?

belka [17]

Answer:

0.00316

Explanation:

You have to use the following equation:

$pH=-log[H^+]$

You are given the pH and need to find the concentration of H+. Plug in the given components and solve.

$2.5=-log[H^+]\\H^+ = 10^{-2.5}\\H^+=0.00316$

The concentration of H is 0.00316.

8 0

2 years ago

Does it make a difference if the jar is square or round? What about the size of the jar or glass?

morpeh [17]

no it doesn't make differance at all ...only the material used make difference

4 0

3 years ago

Based on the balanced chemical equation for the preparation of malachite, what is the composition of the bubbles formed when sod

spin [16.1K]

Answer:

Carbon Dioxide = CO2

Explanation:

The synthesis of Malachite is seen in the chemical formula:

CuSO 4 . 5H2O(aq) + 2NaCO3(aq) --> CuCO 3 Cu(OH) 2 (s) + 2Na 2 SO 4 (aq) + CO 2 (g) + 9H 2 O(l)

The bubbles mentioned in the question hints that our interest is the compounds in their gseous phase (g).

Upon examining the chemical equation, only CO2 is in the gaseous state and hence the only one that can be formed as bubbles,

3 0

3 years ago

Arrange the molecules H2O, NH3, Ar, NaCl in order of expected increasing boiling points. 1. None of these 2. NaCl, H2O, NH3, Ar

Aleks04 [339]

Answer:

Option (4) is correct

Explanation:

NaCl is an ionic solid. $Na^{+}$ and $Cl^{-}$ ions are held together by strong coloumbic attractive force. Hence NaCl has highest boiling point'
Ar is a monoatomic molecule. Hence only weak london dispersion force exists between Ar atom/molecules. Hence it requires lowest amount of energy to boil and thereby possess lowest boiling point.
$NH_{3}$ and $H_{2}O$ are polar protic molecules. Hence they possess london dispersion force, dipole-dipole force and hydrogen bonding as intermolecular forces.
Dipole-dipole force is stronger in $H_{2}O$ than $NH_{3}$ due to more polar O-H bond than N-H bond. Also molecular weight of $H_{2}O$ is higher than $NH_{3}$ . So, more energy is required to boil $H_{2}O$ . So, $H_{2}O$ has higher boiling point than $NH_{3}$

Hence order of boiling point : $NaCl>H_{2}O>NH_{3}>Ar$

4 0

3 years ago