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
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- 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.
Learn more:
Answer:
The net chemical equation is: 6 H2O(g) + C3H8(g) → 10 H2(g) + 3 CO2(g)
Explanation:
First equation:
The reactants propane and water react to form the products CO and H2
C3H8(g) + 3H2O(g) → 3CO(g) + 7H2(g)
Second equation:
The products CO and H2 will react
CO(g) + H2O(g) → H2(g) + CO2(g)
We should multiply the equation by 3 (to equal the products of the first equation)
3CO(g) + 3H2O(g) → 3H2(g) + 3CO2(g)
Add the second to the first equation:
C3H8(g) + 3 H2O(g) + 3 CO(g) + 3 H2O(g) → 3 H2(g) + 3CO2(g) + 3 CO(g) + 7 H2(g)
The net chemical equation is: 6 H2O(g) + C3H8(g) → 10 H2(g) + 3 CO2(g)