<u>Answer:</u> The concentration of 
required will be 0.285 M.
<u>Explanation:</u>
To calculate the molarity of 
, we use the equation:
Moles of = 0.016 moles
Volume of solution = 1 L
Putting values in above equation, we get:
For the given chemical equations:
![Ni^{2+}(aq.)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K_f=1.2\times 10^9](https://tex.z-dn.net/?f=Ni%5E%7B2%2B%7D%28aq.%29%2B6NH_3%28aq.%29%5Crightleftharpoons%20%5BNi%28NH_3%29_6%5D%5E%7B2%2B%7D%2BC_2O_4%5E%7B2-%7D%28aq.%29%3BK_f%3D1.2%5Ctimes%2010%5E9)
Net equation: ![NiC_2O_4(s)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K=?](https://tex.z-dn.net/?f=NiC_2O_4%28s%29%2B6NH_3%28aq.%29%5Crightleftharpoons%20%5BNi%28NH_3%29_6%5D%5E%7B2%2B%7D%2BC_2O_4%5E%7B2-%7D%28aq.%29%3BK%3D%3F)
To calculate the equilibrium constant, K for above equation, we get:
The expression for equilibrium constant of above equation is:
![K=\frac{[C_2O_4^{2-}][[Ni(NH_3)_6]^{2+}]}{[NiC_2O_4][NH_3]^6}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BC_2O_4%5E%7B2-%7D%5D%5B%5BNi%28NH_3%29_6%5D%5E%7B2%2B%7D%5D%7D%7B%5BNiC_2O_4%5D%5BNH_3%5D%5E6%7D)
As, is a solid, so its activity is taken as 1 and so for 
We are given:
![[[Ni(NH_3)_6]^{2+}]=0.016M](https://tex.z-dn.net/?f=%5B%5BNi%28NH_3%29_6%5D%5E%7B2%2B%7D%5D%3D0.016M)
Putting values in above equations, we get:
![0.48=\frac{0.016}{[NH_3]^6}}](https://tex.z-dn.net/?f=0.48%3D%5Cfrac%7B0.016%7D%7B%5BNH_3%5D%5E6%7D%7D)
![[NH_3]=0.285M](https://tex.z-dn.net/?f=%5BNH_3%5D%3D0.285M)
Hence, the concentration of required will be 0.285 M.
Answer:
what is the baby flowers color?
Explanation:
if the baby flower is purple the parents were purple
Answer:
is there a table or an equation for that ??
Answer:
9 : 8
Explanation:
Aluminum oxide has the following formula Al₂O₃.
Next, we shall determine the mass of Al and O₂ in Al₂O₃. This can be obtained as follow:
Mass of Al in Al₂O₃ = 2 × 27 = 54 g
Mass of O₂ in Al₂O₃ = 3 × 16 = 48 g
Finally, we shall determine the mass ratio of Al and O₂. This can be obtained as follow:
Mass of Al = 54 g
Mass of O₂ = 48 g
Mass of Al : Mass of O₂ = 54 : 48
Mass of Al : Mass of O₂ = 54 / 48
Mass of Al : Mass of O₂ = 9 / 8
Mass of Al : Mass of O₂ = 9 : 8
Therefore, the mass ratio of Al and O₂ in Al₂O₃ is 9 : 8
