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tamaranim1 [39]
3 years ago
13

5. Write the two resonance hybrids for the carbocation that would be formed by protonation at C-1 of 2-methyl-1,3-pentadiene. Wi

thout doing a calculation, would you expect C-2 or C-4 (the two end carbons of the allylic cation) to have the most positive charge on it
Chemistry
1 answer:
kupik [55]3 years ago
7 0

Answer:

Follows are the solution to this question:

Explanation:

Please find the complete solution in the attached file.

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What products would you expect from the reaction of ammonia and sulfuric acid in aqueous solution
Lemur [1.5K]

Answer:

(NH4)2SO4

Explanation:

2NH3(aq) + H2SO4 (aq) ---> (NH4)2SO4(aq)

Ammonia reacts with sulfuric acid to give the fertilizer ammonium sulfate.

6 0
2 years ago
In an experiment, the molar mass of the compound was determined to be 118.084 g/mol. What is the molecular formula of the compou
Mrrafil [7]

Answer:

need more info

Explanation:

i think so we need more info

4 0
2 years ago
Silver chloride, AgCl (Ksp = 1.8 x 10‒10), can be dissolved in solutions containing ammonia due to the formation of the soluble
Sergeeva-Olga [200]

Explanation:

The given reaction will be as follows.

           AgCl(s) \rightarrow Ag^{+}(aq) + Cl^{-}(aq) ............. (1)

     K_{sp} = [Ag^{+}][Cl^{-}] = 1.8 \times 10^{-10}

Reaction for the complex formation is as follows.

          Ag^{+}(aq) + 2NH_{3}(aq) \rightleftharpoons [Ag(NH_{3})_{2}]^{+}(aq) ........... (2)

          K_{f} = \frac{[Ag(NH_{3})_{2}]}{[Ag^{+}][NH_{3}]^{2}} = 1.0 \times 10^{8}

When we add both equations (1) and (2) then the resultant equation is as follows.

             AgCl(s) + 2NH_{3}(aq) \rightarrow [Ag(NH_{3})_{2}]^{+}(aq) + Cl^{-}(aq) ............. (3)

Therefore, equilibrium constant will be as follows.

                       K = K_{f} \times K_{sp}

                          = 1.0 \times 10^{8} \times 1.8 \times 10^{-10}

                          = 1.8 \times 10^{-2}

Since, we need 0.010 mol of AgCl to be soluble in 1 liter of solution after after addition of NH_{3} for complexation. This means we have to set

               [Ag^{+}] = [Cl^{-}]

                          = \frac{0.010 mol}{1 L}

                          = 0.010 M

For the net reaction, AgCl(s) + 2NH_{3}(aq) \rightarrow [Ag(NH_{3})_{2}]^{+}(aq) + Cl^{-}(aq)

Initial :                             0.010         x                     0                           0

Change :                    -0.010         -0.020             +0.010                +0.010

Equilibrium :                   0            x - 0.020           0.010                 0.010

Hence, the equilibrium constant expression for this is as follows.

              K = \frac{[Ag(NH_{3})^{+}_{2}][Cl^{-}]}{[NH_{3}]^{2}}

     1.8 \times 10^{-2} = \frac{0.010 \times 0.010}{(x - 0.020)^{2}}

             x = 0.0945 mol      

or,          x = 0.095 mol (approx)

Thus, we can conclude that the number of moles of NH_{3} needed to be added is 0.095 mol.

3 0
4 years ago
If a precipitation reaction occurs, what will be the products of the unbalanced
levacccp [35]

Answer:

{ \sf{B.  \: AgCl _{(s)}  + NaNO _{3(aq)}}}

Explanation:

Silver chloride precipitates out because it has a low value of Ksp.

7 0
3 years ago
How does the gas propellant move when an aerosol can is used?
Rudik [331]
From high pressure to lower pressure.
4 0
3 years ago
Read 2 more answers
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