<span><span>S is for soil,</span><span>cl (sometimes c) represents climate,</span><span>o organisms including humans,</span><span>r relief,</span><span>p parent material, or lithology, and</span><span>t time.</span></span>
Hello!
For this problem, we will be applying <em>Charles' Law</em>:
V1/T1 = V2/T2
Now that we have the formula, let's convert the temperature to Kelvin.
27 + 273 = 300K
Let's plug everything in now!
10/300 = 12.0/x
Simplified:
1/30 = 12.0/x
Cross-multiply:
1x = 30*12.0
<u>x = 360</u>
<em>Check!</em>
10/300 = 12/360
300*12 = 360*10
3600 = 3600
Therefore, you would have to heat the gas at a temperature of 360K in order to raise the volume to 12.0L.
5L O2 x 1 mol O2/ 22.4L O2 x 2 mol H20/ 1 mol O2 x 22.4L H20/ 1 mol H20 = 10 L H2O
For future reference though, since its at STP that means that the coefficient are in proportion. Since oxygen has a coefficient of 1 and water has a coefficient of 2 for every 1 liter of oxygen there is 2 liters of water. Hence you started with 5 liters and ended with 10
The given question is incomplete. The complete question is
If 1.0 M HI is placed into a closed container and the reaction is allowed to reach equilibrium at 25∘C∘C, what is the equilibrium concentration of H2 (g). Given the equilibrium constant is 62.
Answer: The equilibrium concentration of 
is 0.498 M
Explanation:
Initial concentration of 
= 1.0 M
The given balanced equilibrium reaction is,
initial (1.0) M 0 0
At eqm (1.0-2x) M (x) M (x) M
The expression for equilibrium constant for this reaction will be,
![K_c=\frac{[H_2]\times [I_2]}{[HI]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BH_2%5D%5Ctimes%20%5BI_2%5D%7D%7B%5BHI%5D%5E2%7D)
Now put all the given values in this expression, we get :
By solving we get :
Thus the equilibrium concentration of is 0.498 M
Stoichiometry measures these quantitative relationships, and is used to determine the amount of products/reactants that are produced/needed in a given reaction.
