Answer:
c. 40.4 grams of Ne
Explanation:
Let's consider the ideal gas law.
P × V = n × R × T
V = n × R × T/P
As we can see, when the samples have the same temperature and pressure, the volume is directly proportional to the number of moles. The higher the number of moles, the larger the volume.
a.
1 mole of N₂ contains 6.02 × 10²³ molecules of N₂. The moles corresponding to 5 × 10²³ molecules of N₂ are:
5 × 10²³ molecules × (1 mol/6.02 × 10²³ molecules) = 0.8 mol
b. 0.306 moles COâ‚‚
c. The molar mass of Ne is 20.18 g/mol. The moles corresponding to 40.4 g of Ne are:
40.4 g × (1 mol/20.18 g) = 2.00 mol
The sample of Ne is the one with the higher number of moles, therefore it has the largest volume.
The gas sample which would have the largest volume is 40.4 grams of Neon (Ne).
Given the following data:
To find which gas samples would have the largest volume, if they are all at the same temperature and pressure:
Using the ideal gas equation:
[tex]PV = nRT[/tex]
Where;
Making V the subject of formula, we have;
[tex]V = n\frac{RT}{P}[/tex]
Hence, we can deduce that the volume of all gas samples is directly proportional to the number of molecules of a sample at the same temperature and pressure.
Also, Avogadro's law states that, equal volume of all gas samples have the same number of molecules at the same temperature and pressure.
For Nitrogen gas ([tex]N_2[/tex]):
1 mole of [tex]N_2[/tex] = [tex]6.02 \times 10^{23}[/tex] molecules
X mole of [tex]N_2[/tex] = [tex]5 \times 10^{23}[/tex] molecules
Cross-multiplying, we have:
[tex]X = \frac{5 \times 10^{23}}{6.02 \times 10^{23}}[/tex]
X = 0.8306 moles
For 40.4 grams of Ne:
[tex]Number\;of\;moles = \frac{mass}{molar\;mass}\\\\Number\;of\;moles = \frac{40.4}{20.18}[/tex]
Number of moles = 2.00 moles
Therefore, the gas sample which would have the largest volume is 40.4 grams of Neon (Ne).
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