THE ANSWER TO THE QUESTION IS A
<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:
Explanation:
You would have to add up the atomic masses of all the compounds in the compound, making sure you include how many molecules of each are in the compound
For example, in CuSOA we have 1 molecule of Cu and S, as 4 molecules of O
The atomic masses are as follows:
Cu = 63.55 u
S = 32.065 u
O = 15.99 units
This is how we would add it up:
(Atomic mass of Cu) + (Atomic mass of S) + 4(Atomic Mass of O)
(63.55) + (32.065) + 4(15.99)
(63.55) + (32.065) + 63.96
= 159.575 u
B because A is for radios of course and C is thermal energy and D is radioactive
Explanation:
When experimenting, the best hypothesis to develop would be a null hypothesis (H₀). A null hypothesis is a statement indicating no change or effect. In this case, it would be;
“There is no effect of temperature on the solubility of a solid in a liquid”
An alternative hypothesis (Hₐ) would be;
“There is an effect of temperature on the solubility of a solid in a liquid”
In this experiment, the null hypothesis would be rejected and the alternative would be accepted. This is because the experiment would show that increased temperatures of the liquid increases solubility of the solid in the liquid.
