Methane composes most of the natural gas.
Answer:
V₂ = 5.97 L
Explanation:
Given data:
Initial temperature = 9°C (9+273 = 282 K)
Initial volume of gas = 6.17 L
Final volume of gas = ?
Final temperature = standard = 273 K
Solution:
Formula:
The Charles Law will be apply to solve the given problem.
According to this law, 'the volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure'
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 6.17 L × 273K / 282 k
V₂ = 1684.41 L.K / 282 K
V₂ = 5.97 L
Answer:
The following relationship makes this possible: 1 mole of any gas at standard temperature and pressure (273 K and 1 atm) occupies a volume of 22.4 L.
Explanation:
Note that this is occurring at STP, where 22.4L of any gas is equal to 1mol of that gas.
First, convert the liters of O₂ to moles of O₂ using the conversion factor 22.4LO₂ = 1molO₂.
8.6LO₂ × 1molO₂/22.4LO₂
= 8.6/22.4
≈ 0.3839molO₂
Next, convert moles of O₂ to moles of H₂O. In the balanced equation, the coefficients show that there are 2 moles of H₂O for every mole of O₂. So, use the conversion factor 1molO₂ = 2molH₂O.
0.3839molO₂ × 2molH₂O/1molO₂
= 0.3839 × 2
= 0.7678molH₂O
Finally, convert the moles of H₂O to liters of H₂O using the same conversion factor from before, 22.4LH₂O = 1molH₂O.
0.7678molH₂O × 22.4LH₂O/1molH₂O
= 0.7678 × 22.4
≈ 17LH₂O
So, the answer is 17 liters of gaseous water is collected! Note that its rounded to 17 because the measurement given in the problem has 2 sig figs. Hope that helps! :)
