Whether FeS will precipitate or not when Fe ( NO 3 ) 2 and HCl should be determined. Also, whether FeS will precipitate or not when pH is adjusted to 8 should be determined. Concept introduction: Solubility product in acid is equilibrium constant for reaction that occurs in acid solution when an ionic compound is dissolved to produce ions. It is represented by K spa . Consider MS an ionic compound dissolved in acidic solution. Its dissociation occurs as: MS ( s ) + 2H 3 O + ( a q ) ⇌ M 2 + ( a q ) + H 2 S ( a q ) + 2H 2 O ( l ) The expression for its K spa is as follows: K spa = [ M 2 + ] [ H 2 S ] [ H 3 O + ] A precipitate of an ionic compound will form when solutions that contain respective ions are mixed. The precipitation depends on value of reaction quotient Q c . Q c is defined in same way as K spa . Also, concentrations in the expression for Q c are concentration at time t and not equilibrium concentrations. Consider MS to be an ionic compound. Its dissociation occurs as: MS ( s ) + 2H 3 O + ( a q ) ⇌ M 2 + ( a q ) + H 2 S ( a q ) + 2H 2 O ( l ) The expression for Q c is as follows: Q c = [ M 2 + ] [ H 2 S ] [ H 3 O + ] Metal cations can be separated into two groups by the precipitation of metal sulfide. The cations which form very insoluble sulfides can be separated from cations which form soluble sulfides. The separation takes place in an acidic solution and use solubility equilibrium. The separation depends on the H 3 O + concentration so that reaction quotient Q c exceeds K spa for insoluble sulfides but not for soluble sulfide. As a result, insoluble sulfides precipitate under acidic condition but soluble sulfides remain in solution. The negative logarithm of molar concentration of hydronium ion is called pH . The expression for pH is as follows: pH = − log 10 [ H 3 O + ]
Whether FeS will precipitate or not when Fe ( NO 3 ) 2 and HCl should be determined. Also, whether FeS will precipitate or not when pH is adjusted to 8 should be determined. Concept introduction: Solubility product in acid is equilibrium constant for reaction that occurs in acid solution when an ionic compound is dissolved to produce ions. It is represented by K spa . Consider MS an ionic compound dissolved in acidic solution. Its dissociation occurs as: MS ( s ) + 2H 3 O + ( a q ) ⇌ M 2 + ( a q ) + H 2 S ( a q ) + 2H 2 O ( l ) The expression for its K spa is as follows: K spa = [ M 2 + ] [ H 2 S ] [ H 3 O + ] A precipitate of an ionic compound will form when solutions that contain respective ions are mixed. The precipitation depends on value of reaction quotient Q c . Q c is defined in same way as K spa . Also, concentrations in the expression for Q c are concentration at time t and not equilibrium concentrations. Consider MS to be an ionic compound. Its dissociation occurs as: MS ( s ) + 2H 3 O + ( a q ) ⇌ M 2 + ( a q ) + H 2 S ( a q ) + 2H 2 O ( l ) The expression for Q c is as follows: Q c = [ M 2 + ] [ H 2 S ] [ H 3 O + ] Metal cations can be separated into two groups by the precipitation of metal sulfide. The cations which form very insoluble sulfides can be separated from cations which form soluble sulfides. The separation takes place in an acidic solution and use solubility equilibrium. The separation depends on the H 3 O + concentration so that reaction quotient Q c exceeds K spa for insoluble sulfides but not for soluble sulfide. As a result, insoluble sulfides precipitate under acidic condition but soluble sulfides remain in solution. The negative logarithm of molar concentration of hydronium ion is called pH . The expression for pH is as follows: pH = − log 10 [ H 3 O + ]
Solution Summary: The author explains the equilibrium constant for reaction that occurs in acidic solution when an ionic compound is dissolved to produce ions.
Whether FeS will precipitate or not when Fe(NO3)2 and HCl should be determined. Also, whether FeS will precipitate or not when pH is adjusted to 8 should be determined.
Concept introduction:
Solubility product in acid is equilibrium constant for reaction that occurs in acid solution when an ionic compound is dissolved to produce ions. It is represented by Kspa. Consider MS an ionic compound dissolved in acidic solution. Its dissociation occurs as:
MS(s)+2H3O+(aq)⇌M2+(aq)+H2S(aq)+2H2O(l)
The expression for its Kspa is as follows:
Kspa=[M2+][H2S][H3O+]
A precipitate of an ionic compound will form when solutions that contain respective ions are mixed. The precipitation depends on value of reaction quotient Qc. Qc is defined in same way as Kspa. Also, concentrations in the expression for Qc are concentration at time t and not equilibrium concentrations. Consider MS to be an ionic compound. Its dissociation occurs as:
MS(s)+2H3O+(aq)⇌M2+(aq)+H2S(aq)+2H2O(l)
The expression for Qc is as follows:
Qc=[M2+][H2S][H3O+]
Metal cations can be separated into two groups by the precipitation of metal sulfide. The cations which form very insoluble sulfides can be separated from cations which form soluble sulfides. The separation takes place in an acidic solution and use solubility equilibrium.
The separation depends on the H3O+ concentration so that reaction quotient Qc exceeds Kspa for insoluble sulfides but not for soluble sulfide. As a result, insoluble sulfides precipitate under acidic condition but soluble sulfides remain in solution.
The negative logarithm of molar concentration of hydronium ion is called pH. The expression for pH is as follows:
Please answer the question and provide a detailed drawing of the structure. If there will not be a new C – C bond, then the box under the drawing area will be checked.
Will the following reaction make a molecule with a new C – C bond as its major product:
Draw the major organic product or products, if the reaction will work. Be sure you use wedge and dash bonds if necessary, for example to distinguish between major products with different stereochemistry.
Please do not use AI. AI cannot "see" the molecules properly, and it therefore gives the wrong answer while giving incorrect descriptions of the visual images we're looking at. All of these compounds would be produced (I think). In my book, I don't see any rules about yield in this case, like explaining that one product would be present in less yield for this reason or that reason. Please explain why some of these produce less yield than others.
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Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell