Since there is no weight, I would assume that this is a 100g of pure compound.
Okay so I would be changing the percentage to gram to solve for the mole.
So
40.0g C (1 mol C/12.01 g C) = 3.33 mol C
6.73g H (1 mol H/1.01 g H ) = 6.66 mol H
53.3g O (1 mol O/16.00 g O) = 3.33 mol O
With that, two of our moles is 3.33, so we consider that are our 1, as it is also the lowest. Therefore the empirical formula is CH2O
I believe it’s 2? Because it’s saying read the temperature. Which is basically saying look at what state the water is in. Ex: hot or cold
Answer:
Equilibrium concentrations of the gases are
Explanation:
We are given that for the equilibrium
Temperature, 
Initial concentration of
We have to find the equilibrium concentration of gases.
After certain time
2x number of moles of reactant reduced and form product
Concentration of
At equilibrium
Equilibrium constant
![K_c=\frac{product}{Reactant}=\frac{[H_2]^2[S_2]}{[H_2S]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7Bproduct%7D%7BReactant%7D%3D%5Cfrac%7B%5BH_2%5D%5E2%5BS_2%5D%7D%7B%5BH_2S%5D%5E2%7D)
Substitute the values
By solving we get
Now, equilibrium concentration of gases
Color they look the same hope this helps.
