Answer:
(a): Precipitate of calcium carbonate will form.
(b): Precipitate of barium sulfate will form.
(c): Precipitate of silver iodide will form.
(d): Precipitate of sodium chromate will form.
Explanation:
Complete ionic equation is defined as the equation in which all the substances that are strong electrolytes present in an aqueous state and are represented in the form of ions.
Net ionic equation is defined as the equations in which spectator ions are not included.
Spectator ions are the ones that are present equally on the reactant and product sides. They do not participate in the reaction.
(a):
The balanced molecular equation is:
[tex]CaCl_2(aq)+K_2CO_3(aq)\rightarrow 2KCl(aq)+CaCO_3(s)[/tex]
The complete ionic equation follows:
[tex]Ca^{2+}(aq)+2Cl^-(aq)+2K^+(aq)+CO_3^{2-}(aq)\rightarrow 2K^+(aq)+2Cl^-(aq)+CaCO_3(s)[/tex]
As potassium and chloride ions are present on both sides of the reaction. Thus, they are considered spectator ions.
The net ionic equation follows:
[tex]Ca^{2+}(aq)+CO_3^{2-}(aq)\rightarrow CaCO_3(s)[/tex]
Precipitate of calcium carbonate will form.
(b)
The balanced molecular equation is:
[tex]BaCl_2(aq)+MgSO_4(aq)\rightarrow MgCl_2(aq)+BaSO_4(s)[/tex]
The complete ionic equation follows:
[tex]Ba^{2+}(aq)+2Cl^-(aq)+Mg^{2+}(aq)+SO_4^{2-}(aq)\rightarrow Mg^{2+}(aq)+2Cl^-(aq)+BaSO_4(s)[/tex]
As magnesium and chloride ions are present on both sides of the reaction. Thus, they are considered spectator ions.
The net ionic equation follows:
[tex]Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)[/tex]
Precipitate of barium sulfate will form.
(c):
The balanced molecular equation is:
[tex]AgNO_3(aq)+KI(aq)\rightarrow KNO_3(aq)+AgI(s)[/tex]
The complete ionic equation follows:
[tex]Ag^{+}(aq)+NO_3^-(aq)+K^+(aq)+I^{-}(aq)\rightarrow K^+(aq)+NO_3^-(aq)+AgI(s)[/tex]
As potassium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.
The net ionic equation follows:
[tex]Ag^{+}(aq)+I^{-}(aq)\rightarrow AgI(s)[/tex]
Precipitate of silver iodide will form.
(d):
The balanced molecular equation is:
[tex]2NaCl(aq)+(NH_4)_2CrO_4(aq)\rightarrow 2NH_4Cl(aq)+Na_2CrO_4(s)[/tex]
The complete ionic equation follows:
[tex]2Na^{+}(aq)+2Cl^-(aq)+2NH_4^+(aq)+CrO_4^{2-}(aq)\rightarrow 2NH_4^+(aq)+2Cl^-(aq)+Na_2CrO_4(s)[/tex]
As ammonium and chloride ions are present on both sides of the reaction. Thus, they are considered spectator ions.
The net ionic equation follows:
[tex]2Na^{+}(aq)+CrO_4^{2-}(aq)\rightarrow Na_2CrO_4(s)[/tex]
Precipitate of sodium chromate will form.
A pure copper cube has an edge length of 1.76 cm. How many copper atoms does it contain? (volume of a cube = (edge length)^3; density of copper = 8.96 g/cm^3 )
The Fischer esterification mechanism is examined in this question. The overall reaction is: Benzoic acid, C H 3 O H and H C l react to form a methyl ester, H 2 O and H C l. Benzoic acid is a carboxylic acid bonded to a benzene ring. Identify the results or mechanism of each step.
Answer:
See explanation and image attached
Explanation:
Fischer esterification is a type of reaction used to convert carboxylic acids to ester in the presence of excess alcohol and a strong acid which acts as a catalyst. Another final product formed in the reaction is water.
The mechanism for the fischer esterification of Benzoic acid and C H 3 O H in the presence of HCl as the catalyst is shown in the image attached to this answer.
The final products of the reaction are methyl benzoate, water and H^+ as shown in the image attached.
The methyl ester, water, and the acid catalyst (HCl) are byproducts of the Fischer esterification process, which involves protonation, nucleophilic attack, elimination, and deprotonation processes.
Carbonyl oxygen protonation: The carbonyl oxygen of the carboxylic acid (benzoic acid) is protonated by the acid catalyst (HCl) in the first step. The protonation of the carbonyl carbon increases its electrophilicity and promotes the alcohol's nucleophilic assault. Attack by the alcohol's nucleophilic oxygen (methanol, CH3OH) on the protonated carboxylic acid's carbonyl carbon results in the formation of a tetrahedral intermediate. The acid catalyst also helps with this phase. Elimination of water: In the following step, the water molecule must be removed from the tetrahedral intermediate. The hydroxyl group (-OH) from the carboxylic acid and a hydrogen from the hydroxyl group of the alcohol are removed to create this water molecule. Deprotonation: A deprotonation occurs after the removal of water.
To know more about Fischer esterification process, here:
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