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3 years ago

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Write balanced equations for each of the processes described below. (Use the lowest possible coefficients. Omit states-of-matter

.)
a. Chromium-51, which targets the spleen and is used as a tracer in studies of red blood cells, decays by electron capture.
b. Iodine-131, used to treat hyperactive thyroid glands, decays by producing a β particle.
c. Phosphorus-32, which accumulates in the liver, decays by β-particle production.

1 answer:

Aleks [24]3 years ago

7 0

Answer:

d

Explanation:

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For the reaction shown, calculate how many grams of oxygen form when each quantity of reactant completely reacts.

Nikolay [14]

Explanation:

Molar mass of $KClO_{3}$ = 39.1 + 35.5 + 3(16.0) = 122.6 g

Molar mass of KCl = 39.1 + 35.5 = 74.6 g

Molar mass of $O_{2}$ = 32.0 g

According to the equation, 2 moles of $KClO_{3}$ reacts to give 3 moles of oxygen.

Therefore, 2 (122.6) = 245.2 g of $KClO_{3}$ will give 3 (32.0) = 96.0 g of oxygen. Thus, 245.2 g of $KClO_{3}$ gives 96.0 g of oxygen.

(a) Calculate the amount of oxygen given by 2.72 g of $KClO_{3}$ as follows.

$\frac {96g}{245.2g} \times 2.72 g = 1.06 g$ of $O_{2}$

(b) Calculate the amount of oxygen given by 0.361 g of $KClO_{3}$ as follows.

$\frac {96g}{245.2g} \times 0.361 g = 0.141 g$ of $O_{2}$

c) Calculate the amount of oxygen given by 83.6 kg $KClO_{3}$ as follows.

$\frac {96g}{245.2g} \times 83.6 g = 32.731 kg$ of $O_{2}$

Convert kg into grams as follows.

$\frac{32.731 kg}{1 kg} \times 1000 g$ = 32731 g of $O_{2}$

(d) Calculate the amount of oxygen given by 22.5 mg of $KClO_{3}$ as follows.

$\frac {96g}{245.2g} \times 22.5 mg = 8.79 mg$

Convert mg into grams as follows.

$\frac{8.79 mg}{1 mg}\times 10^{-3} g = 8.79 \times 10^{-3} g$ of $O_{2}$

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3 years ago

A molecule of glucose is comprised of 6 atoms of carbon, 12 atoms of hydrogen, and 6

Tanya [424]

Answer:

48

Explanation:

because you add 6 and 6 and 12 to get it

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2 years ago

What contains elements with similar properties in the periodic table?

Nataly [62]

A) a column

example: earth alkaline metals

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3 years ago

Read 2 more answers

Which properties determines the element that an atom Is?

balandron [24]

Answer:

The answer would be C. Number of protons in the atom.

Explanation:

On the periodic table, you see the element, with a big number at top, and a small number below the element name/abbreviation.

The big number is the amount of protons of the atom, which define each atom. The smaller number represents the atomic mass of the atom.

#teamtrees #WAP (Water And Plant)

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3 years ago

A compound is composed of C, H and O. A 1.621 g sample of this compound was combusted, producing 1.902 g of water and 3.095 g of

vlada-n [284]

Answer: The molecular of the compound is, $C_2H_3O$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=3.095g$

Mass of $H_2O=1.902g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in $3.095g$ of carbon dioxide, $\frac{12}{44}\times 3.095=0.844g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in $1.902g$ of water, $\frac{2}{18}\times 1.092=0.121g$ of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = $(1.621)-[(0.844)+(0.121)]=0.656g$

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.844g}{12g/mole}=0.0703moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.121g}{1g/mole}=0.121moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.656g}{16g/mole}=0.041moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is $0.041$ moles.

For Carbon = $\frac{0.0703}{0.041}=1.71\approx 2$

For Hydrogen = $\frac{0.121}{0.041}=2.95\approx 3$

For Oxygen = $\frac{0.041}{0.041}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 3 : 1

Hence, the empirical formula for the given compound is $C_2H_3O_1=C_2H_3O$

The empirical formula weight = 2(12) + 3(1) + 1(16) = 43 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{46.06}{43}=1$

Molecular formula = $(C_2H_3O_1)_n=(C_2H_3O_1)_1=C_2H_3O$

Therefore, the molecular of the compound is, $C_2H_3O$

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3 years ago