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kakasveta [241]
3 years ago
12

the _ of an element is the average mass of an element naturally occurring atom, or isotopes,taking into account the _ of each is

otopes​
Chemistry
1 answer:
NARA [144]3 years ago
7 0

Answer:

the<u> atomic mass</u> of an element is the average mass of an element naturally occurring atom, or isotopes, taking into account the <u>percentage</u> of each isotopes​

Explanation:

The atomic mass of an element is obtained by obtaining the relative abundances (in percentages) of naturally occurring atoms and the masses of the isotopes. The atomic mass can also be defined as the sum of the protons and the neutrons in the nucleus of an element.

In the periodic table, the atomic mass is indicated below the symbol of each of the elements and is usually in the decimal form.

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Answer:

\Delta t = 13\,h

Explanation:

Half time is period required to desintegrating the half of the initial number of atoms. Then, the total time is:

\Delta t = 5\cdot (2.6\,h)

\Delta t = 13\,h

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Which statement best reflects a change in weather ?
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This element is a liquid at room temperature.<br> (A) Hg<br> (B) Th<br> (C) Na<br> (D) Cl<br> (E) Co
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3 years ago
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A mixture of 15.0 g of the anesthetic halothane (C2HBrClF3 197.4 g/mol) and 22.6 g of oxygen gas has a total pressure of 862 tor
AlexFokin [52]

Answer : The partial pressure of C_2HBrClF_3 and O_2 are, 84 torr and 778 torr respectively.

Explanation : Given,

Mass of C_2HBrClF_3 = 15.0 g

Mass of O_2 = 22.6 g

Molar mass of C_2HBrClF_3 = 197.4 g/mole

Molar mass of O_2 = 32 g/mole

First we have to calculate the moles of C_2HBrClF_3 and O_2.

\text{Moles of }C_2HBrClF_3=\frac{\text{Mass of }C_2HBrClF_3}{\text{Molar mass of }C_2HBrClF_3}=\frac{15.0g}{197.4g/mole}=0.0759mole

and,

\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{22.6g}{32g/mole}=0.706mole

Now we have to calculate the mole fraction of C_2HBrClF_3 and O_2.

\text{Mole fraction of }C_2HBrClF_3=\frac{\text{Moles of }C_2HBrClF_3}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.0759}{0.0759+0.706}=0.0971

and,

\text{Mole fraction of }O_2=\frac{\text{Moles of }O_2}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.706}{0.0759+0.706}=0.903

Now we have to partial pressure of C_2HBrClF_3 and O_2.

According to the Raoult's law,

p^o=X\times p_T

where,

p^o = partial pressure of gas

p_T = total pressure of gas

X = mole fraction of gas

p_{C_2HBrClF_3}=X_{C_2HBrClF_3}\times p_T

p_{C_2HBrClF_3}=0.0971\times 862torr=84torr

and,

p_{O_2}=X_{O_2}\times p_T

p_{O_2}=0.903\times 862torr=778torr

Therefore, the partial pressure of C_2HBrClF_3 and O_2 are, 84 torr and 778 torr respectively.

6 0
3 years ago
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3 0
3 years ago
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