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3 years ago

Calculate the number of moles contained in 550 ml of carbon dioxide at STP

1 answer:

5 0

Hello!

There is a useful relationship that we can use to determine the number of moles of an ideal gas at Standard Temperature and Pressure (0 °C; 1 bar).

It says that 1 mole of an ideal gas occupies 22,4 L. Knowing that we can apply the following conversion factor to go from mL of CO₂ to moles of CO₂

$550 mL CO_2* \frac{1 L}{1000 mL} * \frac{1 mol CO_2}{22,4 L CO_2}=0,0246 moles CO_2$

So, the number of moles contained in 550 mL of Carbon Dioxide at STP is 0,0246 moles

Have a nice day!

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2 years ago

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.

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2 years ago

What is the weight, on Earth, of a book with a mass of 16 kg?<br> 16N<br> 65N<br> 113N<br> 147N

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3 years ago

Element X has two isotopes. If 72.0% of the element has an isotope mass of 84.9 atomic mass units, and 28.0% of the element has

bija089 [108]

<h2>Answer:</h2>

Average atomic mass of an element is the sum of the masses of its isotopes each multiplied by its natural abundance

$\footnotesize \longrightarrow \: \rm Average \: atomic \: mass = \dfrac{ \sum\limits \: \% age \: of \: each \: isotope \times Atomic \: mass }{100} \\$

$\footnotesize \longrightarrow \: \rm Average \: atomic \: mass = \dfrac{ 72 \times84.9 + 28 \times 87 }{100} \\$

$\footnotesize \longrightarrow \: \rm Average \: atomic \: mass = \dfrac{ 6112.8 + 2436 }{100} \\$

$\footnotesize \longrightarrow \: \rm Average \: atomic \: mass = \dfrac{ 8548.8 }{100} \\$

$\footnotesize \longrightarrow \: \bf Average \: atomic \: mass = 85.488 \: amu \\$

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2 years ago