I would say D. Let me know if i am wrong.
<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:
The volume will be "2.95 L".
Explanation:
Given:
n = 0.104
p = 0.91 atm
T = 314 K
Now,
The Volume (V) will be:
= 
By putting the values, we get
= 
= 
= 
Radioactivity another name for radioactive decay. Radioactivity refers to particles emitted from nuclei as a result of nuclear instability.
Answer:
317 g
Explanation:
Cu + 2HCl --> CuCl2 +H2
1 : 2 1 : 1
1 mole of Cu = 63.5 g
1 mole of H2 = 2g
1 mole Cu produces = 1 mole of H2
63.5 g of Cu produces = 2 g of H2
So
10 g of H2 will be produced from = (63.5/2)*10 = 317 g of Copper