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A sample of carbon dioxide gas has a pressure of 1.89 atm and a volume of...

A sample of carbon dioxide gas has a pressure of 1.89 atm and a volume of 21.5 mL at a temperature of 76.2 ℃. How many molecules of carbon dioxide gas are in the sample?

a) 0.0142

b) 8.54 × 1020

c) 8.54 × 1023

d) 70.7

e) 4.26 × 1025

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Answer #2

To calculate the number of molecules of carbon dioxide gas in the sample, we can use the ideal gas law equation:

=

where: P = Pressure of the gas (in atm) V = Volume of the gas (in liters) n = Number of moles of the gas R = Ideal gas constant (0.0821 L.atm/mol.K) T = Temperature of the gas (in Kelvin)

First, we need to convert the given values to the appropriate units:

Pressure (P) = 1.89 atm Volume (V) = 21.5 mL = 0.0215 L Temperature (T) = 76.2 ℃ = 76.2 + 273.15 = 349.35 K (converted to Kelvin)

Now, let's calculate the number of moles (n):

=

=1.89×0.02150.0821×349.35

0.0142

The number of molecules can be calculated using Avogadro's constant, which is approximately 6.022×1023 molecules/mol.

Number of molecules = 0.0142mol×6.022×1023molecules/mol

Number of molecules ≈ 8.54×1021 molecules

So, the correct answer is option (b) 8.54×1021 molecules.

answered by: Hydra Master
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Answer #3

To find the number of molecules of carbon dioxide gas in the sample, we can use the ideal gas equation:

PV = nRT

where: P = pressure (in atm) V = volume (in L) n = number of moles R = ideal gas constant (0.08206 L atm / (mol K)) T = temperature (in Kelvin)

First, we need to convert the given temperature from Celsius to Kelvin: T(K) = 76.2 ℃ + 273.15 = 349.35 K

Next, we can rearrange the ideal gas equation to solve for the number of moles (n):

n = PV / RT

Now, let's plug in the given values:

P = 1.89 atm V = 21.5 mL = 21.5 / 1000 L (convert from mL to L) R = 0.08206 L atm / (mol K) T = 349.35 K

n = (1.89 atm * 21.5 / 1000 L) / (0.08206 L atm / (mol K) * 349.35 K)

n ≈ 0.0142 mol

Finally, to find the number of molecules (N), we can use Avogadro's number:

N = n * NA

where NA is Avogadro's number (6.022 x 10^23 mol^-1).

N ≈ 0.0142 mol * (6.022 x 10^23 mol^-1)

N ≈ 8.54 x 10^21

So, the correct answer is b) 8.54 x 10^20 molecules of carbon dioxide gas in the sample.


answered by: Mayre Yıldırım
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