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Tamiku [17]
3 years ago
7

What is the percent composition of Sb2F3.What is the percent composition of F in this molecule? Round the answer to five signifi

cant figures.
Chemistry
1 answer:
poizon [28]3 years ago
3 0

Answer:

\%F=18.967\%

Explanation:

Hello,

In this case, a percent composition is computed by considering the atomic mass of each element in the compound as well as the number of atoms and the molar mass of the compound. Thus, Sb2F3 has a molar mass of 300.52 g/mol, antimony has an atomic mass of 121.76 g/mol and fluorine 19.0 g/mol, therefore, the percent composition of fluorine which has three atoms with five significant figures is:

\% F=\frac{3*19.0g/mol}{300.52g/mol}*100\%\\ \\\%F=18.967\%

Best regards.

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Match each statement with a layer of the atmosphere. There will be two statements for each layer.
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E. It is heated by the ozone layer beneath it. 

D. This layer is where most meteor showers occur. 

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Depending on the Earth's  temperature the atmosphere can be separated into layers. The troposphere, the stratosphere, the mesosphere and the thermosphere are those layers. The lowest layer is named as Troposphere (0-10 km from the Earth outer surface), it comprises about 75% of the atmosphere's total air and nearly most the water vapor.

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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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