The answer is B or the second option
Explanation: bond length or bond distance is defined as the average distance between nuclei of two bonded atoms in a molecule. It is a transferable property of a bond between atoms of fixed types, relatively independent of the rest of the molecule.
Answer:
The equilibrium concentration of hydrogen gas is 0.0010 M.
Explanation:
The equilibrium constant of the reaction = 
}
Moles of hydrogen sulfide = 0.31 mol
Volume of the container = 4.1 L
![[concentration]=\frac{moles}{volume (L)}](https://tex.z-dn.net/?f=%5Bconcentration%5D%3D%5Cfrac%7Bmoles%7D%7Bvolume%20%28L%29%7D)
![[H_2S]=\frac{0.31 mol}{4.1 L}=0.076 M](https://tex.z-dn.net/?f=%5BH_2S%5D%3D%5Cfrac%7B0.31%20mol%7D%7B4.1%20L%7D%3D0.076%20M)
Initially
0.076 M
At equilibrium
(0.076-2x) 2x x
The expression of an equilibrium constant :
![K_c=\frac{[H_2]^2[S_2]}{[H_2S]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BH_2%5D%5E2%5BS_2%5D%7D%7B%5BH_2S%5D%5E2%7D)
Solving for x:
x = 0.00051
The equilibrium concentration of hydrogen gas:
![[H_2]=2x=2\times 0.00051 M=0.0010 M](https://tex.z-dn.net/?f=%5BH_2%5D%3D2x%3D2%5Ctimes%200.00051%20M%3D0.0010%20M)
i think its a right tringle
Answer:
The answer to your question is 203.8 g
Explanation:
Data
Average atomic mass = ?
Centium-200
Centium-203
Centium-209
Equal amounts
Process
1.- Determine the abundance of each isotope.
If they are in equal amounts in nature, just divide 100 by 3
100/3 = 33.33 % or 0.333
2.- Determine the Average atomic mass
Average atomic mass = Atomic mass 1 x abundance + Atomic mass 2 x
abundance + Atomic mass 3 x abundance
- Substitution
Average atomic mass = 200 x 0.333 + 203 x 0.333 + 209 x 0.333
- Simplification
Average atomic mass = 66.6 + 67.6 + 69.6
Result
Average atomic mass = 203.8 g
