Answer: A quantitative observation
Explanation: Quantitative observations deal with numbers or amounts. As opposed to <em>qualitative</em> observations, which are observations made with the 5 senses.
Actually, we can answer the problem even without the first statement. All we have to do is write the reaction for the production of sulfur trioxide.
2 S + 3 O₂ → 2 SO₃
The stoichiometric calculations is as follows:
6 g S * 1 mol/32.06 g S = 0.187 mol S
Moles O₂ needed = 0.187 mol S * 3 mol O₂/2 mol S = 0.2805 mol O₂
Since the molar mas of O₂ is 32 g/mol,
Mass of O₂ needed = 0.2805 mol O₂ * 32 g/mol = 8.976 g O₂
Answer: The approximate equilibrium partial pressure of 
is 3.92 atm
Explanation:
Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.
The given balanced equilibrium reaction is,
![K_p=\frac{[H_2]^2\times [S_2]}{[H_2S]^2}](https://tex.z-dn.net/?f=K_p%3D%5Cfrac%7B%5BH_2%5D%5E2%5Ctimes%20%5BS_2%5D%7D%7B%5BH_2S%5D%5E2%7D)
![1.5\times 10^{-5}=\frac{[H_2]^2\times [S_2]}{[H_2S]^2}](https://tex.z-dn.net/?f=1.5%5Ctimes%2010%5E%7B-5%7D%3D%5Cfrac%7B%5BH_2%5D%5E2%5Ctimes%20%5BS_2%5D%7D%7B%5BH_2S%5D%5E2%7D)
On reversing the reaction:
initial pressure 4.00atm 2.00 atm 0
eqm (4.00-2x)atm (2.00-x) atm 2x atm
![K_p=\frac{[H_2S]^2}{[H_2]^2\times [S_2]}](https://tex.z-dn.net/?f=K_p%3D%5Cfrac%7B%5BH_2S%5D%5E2%7D%7B%5BH_2%5D%5E2%5Ctimes%20%5BS_2%5D%7D)
![0.67\times 10^5=\frac{2x]^2}{[4.00-2x]^2\times [2.00-x]}](https://tex.z-dn.net/?f=0.67%5Ctimes%2010%5E5%3D%5Cfrac%7B2x%5D%5E2%7D%7B%5B4.00-2x%5D%5E2%5Ctimes%20%5B2.00-x%5D%7D)
![[H_2S]=2x=2\times 1.96=3.92 atm](https://tex.z-dn.net/?f=%5BH_2S%5D%3D2x%3D2%5Ctimes%201.96%3D3.92%20atm)
Thus approximate equilibrium partial pressure of is 3.92 atm
Answer: 1.5
Explanation:
Just took the test on E2020!
