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Evgen [1.6K]
3 years ago
11

What is the pressure, in atmospheres, of a 0.108-mol sample of helium gas at a temperature of 20.0°C if its volume is 0.505 L?

Chemistry
1 answer:
SpyIntel [72]3 years ago
5 0

Answer:

P = 5.14ATM

Explanation:

Number of moles = 0.108moles

Temperature (T) = 20°C = 20 + 273.15 = 293.15K

Volume V = 0.505L.

Pressure (P) = ?

R = 0.082J/mol.K

From ideal gas equation,

PV = nRT

P = nRT / V

P = (0.108 * 0.082 * 293.15) / 0.505

P = 2.596 / 0.505

P = 5.14ATM

The pressure of the gas is 5.14ATM

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2.63 x 10^22 molecules of SO2.

To find this, start with what you know.

2.38g of SO2.

You need to first convert this into Moles since you cannot directly convert grams into molecules. In order to convert grams to moles, you need to find the molecular mass of SO2 - 64.066.

This is because Sulfur has the mass of 32.066 and Oxygen has a mass of 16, but since there are two Oxygen atoms, it's going to be 32.

Your equation should currently appear as so:

2.38g of SO2 = 1 Mole of SO2 / 64.066

Now, you need to convert this to molecules.

Whenever you are searching for molecules without a given amount, you always use Avogadro's number: 6.02 x 10^23

Now, your equation should appear as so:

2.38g SO2 = 1 Mole of SO2 / 64.066 next to a new fraction which is 6.02 x 10^23 / 1 Mole of SO2

Now, multiply across (2.38 x 1 x 6.02 x 10^23). When using Avogadro's number, don't forget to place parenthesis around it.

Then, divide that number by the bottom: 64.066.

Thus, your final answer is that there is 2.63 x 10^22 molecules of SO2.

Don't forget your units!

Hope this helps!

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