Answer:
The correct order of increasing reactivity toward nucleophilic acyl substitution is E < D < C < A < F < B.
Explanation:
The stability of the leaving group best determines the manner of reactivity of carboxylates to nucleophilic substitution after the substitution of the nucleophile to the leaving group. The leaving group should, therefore, be protonated with hydrogen ion in the solution to form a stable molecule. From the given list: The leaving group for A, Ethyl thioacetate will be ethanethiol. For B, Acetyl chloride will be Hydrochloric acid. For C, Sodium acetate will be Sodium Hydroxide. For D, Ethyl acetate will be Ethanol. For E, Acetamide will be Ammonia, and for F, Acetic anhydride will be Ethanoic acid. The reactivity of the substitution reaction is dependent on the stability of these leaving groups. The stability of these leaving groups depends on their pKa, and the more the pKa, the lesser the acidity of the leaving group, and the lower the reactivity. Therefore, considering their pKa: A is 8.5, B is -7, C is 13.8, D is 15.9, E is 36, and F is 4.8. When we rearrange this pKa in descending order, we have E, D. C, A, F, B. Which is also the increased reactivity of the nucleophilic acyl substitution.
Is this a multiple choice? Anyways I will just give you a written answer and hopefully it helps.
A: Helium is made up of two electrons held by electromagnetic force to two protons that are inside of a nucleus along with one or two neutrons.
The question is incomplete, complete question is;
A solution of 
is added dropwise to a solution that contains
of 
and 
and 
.
What concentration of 
is need to initiate precipitation? Neglect any volume changes during the addition.
value 
value 
What concentration of is need to initiate precipitation of the first ion.
Answer:
Cadmium carbonate will precipitate out first.
Concentration of is need to initiate precipitation of the cadmium (II) ion is 
.
Explanation:
1) 
The expression of an solubility product of iron(II) carbonate :
![K_{sp}=[Fe^{2+}][CO_3^{2-}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BFe%5E%7B2%2B%7D%5D%5BCO_3%5E%7B2-%7D%5D)
![2.10\times 10^{-11}=0.58\times 10^{-2} M\times [CO_3^{2-}]](https://tex.z-dn.net/?f=2.10%5Ctimes%2010%5E%7B-11%7D%3D0.58%5Ctimes%2010%5E%7B-2%7D%20M%5Ctimes%20%5BCO_3%5E%7B2-%7D%5D)
![[CO_3^{2-}]=\frac{2.10\times 10^{-11}}{1.15\times 10^{-2} M}](https://tex.z-dn.net/?f=%5BCO_3%5E%7B2-%7D%5D%3D%5Cfrac%7B2.10%5Ctimes%2010%5E%7B-11%7D%7D%7B1.15%5Ctimes%2010%5E%7B-2%7D%20M%7D)
![[CO_3^{2-}]=1.826\times 10^{-9}M](https://tex.z-dn.net/?f=%5BCO_3%5E%7B2-%7D%5D%3D1.826%5Ctimes%2010%5E%7B-9%7DM)
2) 
The expression of an solubility product of cadmium(II) carbonate :
![K_{sp}=[Cd^{2+}][CO_3^{2-}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BCd%5E%7B2%2B%7D%5D%5BCO_3%5E%7B2-%7D%5D)
![1.80\times 10^{-14}=0.58\times 10^{-2} M\times [CO_3^{2-}]](https://tex.z-dn.net/?f=1.80%5Ctimes%2010%5E%7B-14%7D%3D0.58%5Ctimes%2010%5E%7B-2%7D%20M%5Ctimes%20%5BCO_3%5E%7B2-%7D%5D)
![[CO_3^{2-}]=\frac{1.80\times 10^{-14}}{0.58\times 10^{-2} M}](https://tex.z-dn.net/?f=%5BCO_3%5E%7B2-%7D%5D%3D%5Cfrac%7B1.80%5Ctimes%2010%5E%7B-14%7D%7D%7B0.58%5Ctimes%2010%5E%7B-2%7D%20M%7D)
![[CO_3^{2-}]=3.103\times 10^{-12} M](https://tex.z-dn.net/?f=%5BCO_3%5E%7B2-%7D%5D%3D3.103%5Ctimes%2010%5E%7B-12%7D%20M)
On comparing the concentrations of carbonate ions for both metallic ions, we can see that concentration to precipitate out the cadmium (II) carbonate from the solution is less than concentration to precipitate out the iron (II) carbonate from the solution.
So, cadmium carbonate will precipitate out first.
And the concentration of carbonate ions to start the precipitation of cadmium carbonate we will need concentration of carbonate ions greater than the concentration.
Answer:
Explanation:
Hello!
In this case, since the average rate of reaction is computed as a change given by:
![r=\frac{\Delta [NH_4NO_2 ]}{\Delta t}](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B%5CDelta%20%5BNH_4NO_2%20%5D%7D%7B%5CDelta%20t%7D)
In such a way, given the concentrations at the specified times, we plug them in to obtain:
Whose negative sign means the concentration decreased due to the decomposition.
Best regards!
