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Marta_Voda [28]
3 years ago
14

Suggest a reason why the equilibrium between succinic acid and acetic anhydride lies on the side favoring formation of acetic ac

id and succinic anhydridse
Chemistry
1 answer:
Sergio [31]3 years ago
7 0
Due to <span>succinic anhydride is cyclic 5 membered anhydride which is stable compared to acetic anhydride which is open chain anhydride. hence equilibrium favors one side towards succinic anhydride.
Also: </span><span>Since, acetic acid is stronger acid than succinic acid, it will form its salt rather than succinic acid</span>
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When a nonmetal bonds with a nonmetal it is a _ bond
Karolina [17]

It is a covalent bond

3 0
3 years ago
Which set of coefficients will balance this chemical equation?
Montano1993 [528]

Answer:

Option B is correct = 1,3

Explanation:

Chemical equation:

C₂H₄ + O₂     →    CO₂ + H₂O

Balanced chemical equation:

C₂H₄ + 3O₂     →    2CO₂ + 2H₂O

Step 1:

Left side                      Right side

C = 2                           C = 1

H = 4                           O = 3

O = 2                           H = 2

Step 2:

C₂H₄ + O₂     →    2CO₂ + H₂O

Left side                      Right side

C = 2                           C = 2

H = 4                           O = 5

O = 2                           H = 2

Step 3:

C₂H₄ + O₂     →    2CO₂ + 2H₂O

Left side                      Right side

C = 2                           C = 2

H = 4                           O = 6

O = 2                           H = 4

Step 4:

C₂H₄ + 3O₂     →    2CO₂ + 2H₂O

Left side                      Right side

C = 2                           C = 2

H = 4                           O = 6

O = 6                           H = 4

5 0
3 years ago
Write the net ionic equation for the reaction between hydrochloric acid and potassium cyanide. Do not include states such as (aq
zhannawk [14.2K]

Answer: H^++CN^-\rightarrow HCN, product favoured

Explanation:

Complete ionic equation : In complete ionic equation, all the substance that are strong electrolyte and present in an aqueous are represented in the form of ions.

Net ionic equation : In the net ionic equations, we do not include the spectator ions in the equations.

When hydrochloric acid react with potassium cyanide, then it gives potassium chloride and hydrocyanic acid as products.

The complete ionic equation will be:

H^++Cl^-+K^++CN^-\rightarrow K^++Cl^-+HCN

The net ionic equation will not contain spectator ions which are K^+ and Cl^-:

H^++CN^-\rightarrow HCN

The reaction is product favoured.

 

5 0
3 years ago
Which is the solubility product expression for pbcl2(s)
Sever21 [200]
The answer is [Pb^2+][Cl-]^2
4 0
3 years ago
Consider the following system at equilibrium. CaCO3(s) Double headed arrow. Ca2 (aq) CO32â€""(aq) The addition of which compound
Dominik [7]

The compound that has been capable to shift the equilibrium with common ion effect has been  \rm \bold{Na_2CO_3}.  Thus, option D is correct.

The complete equation for the decomposition of calcium carbonate has been:

\rm CaCO_3\rightleftharpoons Ca^2^+\;+\;CO_3^2^-

The equilibrium has been the state in the reaction, when the number of products and has been equivalent to the number of reactants. The increase or decrease in the concentration of any of the reactant or product results in the shift in the equilibrium.

For the given reaction, the addition of Calcium or carbonate ions to the solution results in the increase in the product concentration with the common ion effect, and thereby shifts the equilibrium condition.

The ions liberated by the following compounds has been:

\rm CCl_4\;=\text{not dissolved}\\CO_2\;+\;H_2O\leftrightharpoons CO_3^-\;+\;H^+\\CuSO_4\leftrightharpoons Cu^2^+\;+\;SO_4^2^-\\Na_2CO_3\leftrightharpoons 2\;Na^+\;+\;CO_3^-

The compound, that has been capable of liberating the carbonate ions in the medium has been \rm \bold{Na_2CO_3}. Thus, it results in the shift of the equilibrium due to common ion effect. Hence, option D is correct.

For more information about common ion effect, refer to the link:

brainly.com/question/4090548

6 0
2 years ago
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