Considering the ideal gas law, a sample weighing 9.49 g occupies 68.67 L at 353 K and 2.00 atm.
Ideal gases are a simplification of real gases that is done to study them more easily. It is considered to be formed by point particles, do not interact with each other and move randomly. It is also considered that the molecules of an ideal gas, in themselves, do not occupy any volume.
An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:
P× V = n× R× T
In this case, you know:
- P= 2 atm
- V= ?
- n= being 2g/mole the molar mass of H2, that is, the amount of mass that a substance contains in one mole.
- R= 0.082
- T= 353 K
Replacing:
2 atm× V = 4.745 moles× 0.082× 353 K
Solving:
V = (4.745 moles× 0.082× 353 K)÷ 2 atm
<u><em>V= 68.67 L</em></u>
Finally, a sample weighing 9.49 g occupies 68.67 L at 353 K and 2.00 atm.
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I believe it is density but i'm not for certain try looking it up somewhere else
Answer:
To do this question. we first have to find the mass% of nitrogen in N2O and then, using that percentage, we can simply find the number of moles of N from the number of moles of N2O
<u></u>
<u>Mass % of Nitrogen:</u>
Mass% of nitrogen = (Molar mass of N2 / Molar mass of N2O)*100
Mass% nitrogen = (28 / 44)*100
Mass% of Nitrogen = 0.63 * 100
Mass% nitrogen = 63%
<u>Mass of Nitrogen:</u>
So, now we can say that in any given mass of N2O. 63% of the total mass is the mass of Nitrogen
Hence, total mass * 63/100 = Mass of Nitrogen
Replacing the variables
40 * 0.63 = Mass of Nitrogen
Mass of Nitrogen = 25.2 grams