A scientist fills a large, tightly sealed balloon with 75,000 mL of helium at STP.

Chemistry

1 answer:

pav-90 [236]1 year ago

3 0

Considering the definition of STP conditions, 3.35 moles of helium are contained within the balloon.

<h3>Definition of STP condition</h3>

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.

<h3>Amount of moles of helium within the balloon</h3>

In this case, you know that scientist fills a large, tightly sealed balloon with 75,000 mL of helium at STP.

So, you can apply the following rule of three: if by definition of STP conditions 22.4 liters are occupied by 1 mole of helium, 75 L (75 L= 75000 mL, being 1 L= 1000 L) are occupied by how many moles of helium?

$amount of moles of helium=\frac{75 Lx1 mole}{22.4 L}$

<u><em>amount of moles of helium= 3.35 moles</em></u>

Finally, 3.35 moles of helium are contained within the balloon.

Learn more about STP conditions:

brainly.com/question/26364483

brainly.com/question/8846039

brainly.com/question/1186356

You might be interested in

A containing vessel holds a gaseous mixture of nitrogen and butane. Thepressure in the vessel at 126.9 Cis 3.0 atm. At 0 C, the

Viefleur [7K]

A vessel that contains a mixture of nitrogen and butane has a pressure of 3.0 atm at 126.9 °C and a pressure of 1.0 atm at 0 °C. The mole fraction of nitrogen in the mixture is 0.33.

A vessel contains a gaseous mixture of nitrogen and butane. At 126.9 °C (400.1 K) the pressure is due to the mixture is 3.0 atm.

We can calculate the total number of moles using the ideal gas equation.

$P \times V = n \times R \times T\\\\n = \frac{P \times V}{R \times T} = \frac{3.0 atm \times V}{0.082 atm.L/mol.K \times 400.1 K} = 0.091 mol/L \times V$

At 0 °C (273.15 K), the pressure due to the gaseous nitrogen is 1.0 atm.

We can calculate the moles of nitrogen using the ideal gas equation.

$P \times V = n \times R \times T\\\\n = \frac{P \times V}{R \times T} = \frac{1.0 atm \times V}{0.082 atm.L/mol.K \times 400.1 K} = 0.030 mol/L \times V$