Answer : The limiting reagent is 
Solution : Given,
Moles of methane = 2.8 moles
Moles of = 5 moles
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
From the balanced reaction we conclude that
As, 2 mole of react with 1 mole of 
So, 5 moles of react with 
moles of
From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.
Hence, the limiting reagent is
Answer:
The factor that will change the volume of the diver's lungs upon reaching the surface is 4
Explanation:
Given data:
Pressure increases 1 atm = 101.325 kPa
34 ft = 10.3632 m
Depth of 102 ft = 31.0896 m
Question: What factor will the volume of the diver's lungs change upon arrival at the surface, V₂/V₁ = ?
The pressure at 31.0896 m:
The factor will the volume of the diver's lungs change upon arrival at the surface:
Answer:
The new equilibrium concentration of HI: <u>[HI] = 3.589 M</u>
Explanation:
Given: Initial concentrations at original equilibrium- [H₂] = 0.106 M; [I₂] = 0.022 M; [HI] = 1.29 M
Final concentrations at new equilibrium- [H₂] = 0.95 M; [I₂] = 0.019 M; [HI] = ? M
<em>Given chemical reaction:</em> H₂(g) + I₂(g) → 2 HI(g)
The equilibrium constant (
) for the given chemical reaction, is given by the equation:
![K_{c} = \frac {[HI]^{2}}{[H_{2}]\: [I_{2}]}](https://tex.z-dn.net/?f=K_%7Bc%7D%20%3D%20%5Cfrac%20%7B%5BHI%5D%5E%7B2%7D%7D%7B%5BH_%7B2%7D%5D%5C%3A%20%5BI_%7B2%7D%5D%7D)
<u><em>At the original equilibrium state:</em></u>
<u><em>Therefore, at the new equilibrium state:</em></u>
![K_{c} = \frac {[HI]^{2}}{(0.95\: M) \times (0.019\: M)}](https://tex.z-dn.net/?f=K_%7Bc%7D%20%3D%20%5Cfrac%20%7B%5BHI%5D%5E%7B2%7D%7D%7B%280.95%5C%3A%20M%29%20%5Ctimes%20%280.019%5C%3A%20M%29%7D)
![\Rightarrow K_{c} = 713.59 = \frac {[HI]^{2}}{0.01805}](https://tex.z-dn.net/?f=%5CRightarrow%20K_%7Bc%7D%20%3D%20713.59%20%3D%20%5Cfrac%20%7B%5BHI%5D%5E%7B2%7D%7D%7B0.01805%7D)
![\Rightarrow [HI]^{2} = 713.59 \times 0.01805 = 12.88](https://tex.z-dn.net/?f=%5CRightarrow%20%5BHI%5D%5E%7B2%7D%20%3D%20713.59%20%5Ctimes%200.01805%20%3D%2012.88)
![\Rightarrow [HI] = \sqrt {12.88} = 3.589 M](https://tex.z-dn.net/?f=%5CRightarrow%20%5BHI%5D%20%3D%20%5Csqrt%20%7B12.88%7D%20%3D%203.589%20M)
<u>Therefore, the new equilibrium concentration of HI: [HI] = 3.589 M</u>
