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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 chemical reaction was used to produce 2.95 moles of copper(II) bicarbonate, Cu(HCO3)2.

BARSIC [14]

Answer:

About 547 grams.

Explanation:

We want to determine the mass of copper (II) bicarbonate produced when a reaction produces 2.95 moles of copper (II) bicarbonate.

To do so, we can use the initial value and convert it to grams using the molar mass.

Find the molar mass of copper (II) bicarbonate by summing the molar mass of each individual atom:

$\displaystyle \begin{aligned} \text{MM}_\text{Cu(HCO$_3$)$_2$} &= (63.55 + 2(1.01)+2(12.01)+6(16.00))\text{ g/mol} \\ \\ &=185.59\text{ g/mol} \end{aligned}$

Dimensional Analysis:

$\displaystyle 2.95\text{ mol Cu(HCO$_3$)$_2$}\cdot \frac{185.59 \text{ g Cu(HCO$_3$)$_2$}}{1 \text{ mol Cu(HCO$_3$)$_2$}} \Rightarrow 547 \text{ g Cu(HCO$_3$)$_2$ }$

In conclusion, about 547 grams of copper (II) bicarbonate is produced.

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3 years ago

Calculate the Ph of:

andrey2020 [161]

Answer:

pH= 3.82

Explanation:

Sodium ethanoate or sodium acetate (CH3COONa) ionises completely.

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3 years ago

A major component of gasoline is octane (C8H18). When octane is burned in air, it chemically reacts with oxygen gas (O2) to prod

Olin [163]

Answer:

The answer is 6.25g.

Explanation:

First create your balanced equation. This will give you the stoich ratios needed to answer the question:

2C8H18 + 25O2 → 16CO2 + 18H2O

Remember, we need to work in terms of NUMBERS, but the question gives us MASS. Therefore the next step is to convert the mass of O2 into moles of O2 by dividing by the molar mass:

7.72 g / 16 g/mol = 0.482 mol

Now we can use the stoich ratio from the equation to determine how many moles of H2O are produced:

x mol H2O / 0.482 mol O2 = 18 H2O / 25 O2

x = 0.347 mol H2O

The question wants the mass of water, so convert moles back into mass by multiplying by the molar mass of water:

0.347 mol x 18 g/mol = 6.25g

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Answer:

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