Question 1

Question 4On adding dilute ammonia solution to a colourless solution of a salt, a white gelatinous precipitate appears. This precipitate however dissolves on addition of excess of ammonia solution.(a) From the following list, identify which metal salt solution was used above?    Na, Al, Zn, Pb, Fe(b) What is the formula of the white gelatinous precipitate obtained?(c) Give the balanced equation(s) when sulphate of this metal reacts with ammonia solution in excess.

Accepted Answer

(a) Zinc (Zn) metal salt solution was used(b) Zn(OH)2(c) The balanced equations are given below:ZnSO4colourless solution+2NH4OH⟶Zn(OH)2↓white gelatinous ppt+(NH4)2SO4colourless in solution\underset{	ext{colourless solution}}{{	ext{ZnSO}_4}} + 2	ext{NH}_4	ext{OH} \longrightarrow \underset{	ext{white gelatinous ppt}}{	ext{Zn(OH)}_2↓} + \underset{	ext{colourless in solution}}{(	ext{NH}_4)_2	ext{SO}_4}colourless solutionZnSO4​​+2NH4​OH⟶white gelatinous pptZn(OH)2​↓​+colourless in solution(NH4​)2​SO4​​With excess of NH4OH ppt. dissolvesZn(OH)2+(NH4)2SO4+2NH4OH⟶[(Zn(NH3)4]SO4Tetraamminezinc (II) sulphate+4H2O	ext{Zn(OH)}_2 + (	ext{NH}_4)_2	ext{SO}_4 + 2	ext{NH}_4	ext{OH} \longrightarrow \underset{\underset{	ext{zinc (II) sulphate}}{	ext{Tetraammine}}}{[(	ext{Zn(NH}_3)_4]	ext{SO}_4} + 4	ext{H}_2	ext{O}Zn(OH)2​+(NH4​)2​SO4​+2NH4​OH⟶zinc (II) sulphateTetraammine​[(Zn(NH3​)4​]SO4​​+4H2​O

Question 2

Question 9Assertion (A): Oxides of most of the metals are basic in nature.Reason (R): All metal oxides dissolve in water forming alkalis.Both A and R are true and R is the correct explanation of A.Both A and R are true but R is not the correct explanation of A.A is true but R is false.A is false but R is true.

Accepted Answer

A is true but R is false.Explanation — Oxides of most of the metals are basic in nature. They dissolve in water forming hydroxides (or alkalis).For example:Na2Osodium oxide+H2O⟶2NaOHsodium hydroxide\underset{	ext{sodium oxide}}{{	ext{Na}_2	ext{O}}} + 	ext{H}_2	ext{O} \longrightarrow \underset{	ext{sodium hydroxide}}{2	ext{NaOH}}sodium oxideNa2​O​+H2​O⟶sodium hydroxide2NaOH​Hence the assertion (A) is true.Reason (R) is false because, not all metal oxides dissolve in water to form alkalis. A few metallic oxides and hydroxides exhibit dual character, i.e., they show acidic as well as basic character. They are said to be amphoteric in nature.For example : Copper(II) oxide (CuO)CuO + H2O ⟶ No reaction

Question 3

Question 10Assertion (A): Zinc oxide reacts with acids as well as bases to form salt and water.Reason (R): Zinc oxide is amphoteric in nature.Both A and R are true and R is the correct explanation of A.Both A and R are true but R is not the correct explanation of A.A is true but R is false.A is false but R is true.

Accepted Answer

Both A and R are true and R is the correct explanation of A.Explanation — Zinc oxide reacts with both acids and concentrated alkalis (NaOH and KOH) forming salt and water.ZnO + 2HCl ⟶ ZnCl2 + H2OZnO + 2NaOH ⟶ Na2ZnO2 + H2OHence the assertion (a) is true.Reason (R) is true because zinc oxide is amphoteric in nature, reacts with both acids and concentrated alkalis forming salt and water. Hence Both A and R are true and reason (R) is the correct explanation of assertion(A).

Question 4

Question 1Name the chloride of a metal which is soluble in excess of ammonium hydroxide. Write equation for the same.

Accepted Answer

Zinc chloride (ZnCl2) is soluble in excess of ammonium hydroxide.When ammonia solution is added dropwise to zinc chloride solution, a white gelatinous ppt of zinc hydroxide is obtained.ZnCl2colourless solution+2NH4OH⟶Zn(OH)2↓white gelatinous ppt+2NH4Cl\underset{	ext{colourless solution}}{{	ext{ZnCl}_2}} + 2	ext{NH}_4	ext{OH} \longrightarrow \underset{	ext{white gelatinous ppt}}{	ext{Zn(OH)}_2↓} + 2	ext{NH}_4	ext{Cl}colourless solutionZnCl2​​+2NH4​OH⟶white gelatinous pptZn(OH)2​↓​+2NH4​ClOn adding excess of ammonia solution, the ppt dissolves and a colourless solution is obtained.Zn(OH)2+2NH4Cl+2NH4OH⟶[(Zn(NH3)4]Cl2Tetraamminezinc (II) sulphate+4H2O	ext{Zn(OH)}_2 + 2	ext{NH}_4	ext{Cl} + 2	ext{NH}_4	ext{OH} \longrightarrow \underset{\underset{	ext{zinc (II) sulphate}}{	ext{Tetraammine}}}{[(	ext{Zn(NH}_3)_4]	ext{Cl}_2} + 4	ext{H}_2	ext{O}Zn(OH)2​+2NH4​Cl+2NH4​OH⟶zinc (II) sulphateTetraammine​[(Zn(NH3​)4​]Cl2​​+4H2​O

Question 5

Question 2What happens when ammonia solution is added first dropwise and then in excess to the following solutions:(i) CuSO4(ii) ZnSO4(iii) FeCl3Write balanced equations for these reactions.

Accepted Answer

(i) When ammonia solution is added dropwise to cupper sulphate, a pale blue ppt of copper hydroxide is obtained.CuSO4blue+2NH4OH⟶Cu(OH)2pale blue ppt↓+(NH4)2SO4colourless in solution\underset{	ext{blue}}{{	ext{CuSO}_4}} + 2	ext{NH}_4	ext{OH} \longrightarrow \underset{	ext{pale blue ppt}}{	ext{Cu(OH)}_2} ↓ + \underset{	ext{colourless in solution}}{(	ext{NH}_4)_2	ext{SO}_4}blueCuSO4​​+2NH4​OH⟶pale blue pptCu(OH)2​​↓+colourless in solution(NH4​)2​SO4​​On adding excess of ammonia solution, the ppt dissolves and a deep blue solution is obtained.Cu(OH)2+(NH4)2SO4+2NH4OH⟶[(Cu(NH3)4]SO4Tetraamminecopper (II) sulphate+4H2O	ext{Cu(OH)}_2 + (	ext{NH}_4)_2	ext{SO}_4 + 2	ext{NH}_4	ext{OH} \longrightarrow \underset{\underset{	ext{copper (II) sulphate}}{	ext{Tetraammine}}}{[(	ext{Cu(NH}_3)_4]	ext{SO}_4} + 4	ext{H}_2	ext{O}Cu(OH)2​+(NH4​)2​SO4​+2NH4​OH⟶copper (II) sulphateTetraammine​[(Cu(NH3​)4​]SO4​​+4H2​O(ii) When ammonia solution is added dropwise to zinc sulphate, a white gelatinous ppt of zinc hydroxide is obtained.ZnSO4colourless solution+2NH4OH⟶Zn(OH)2↓white gelatinous ppt+(NH4)2SO4colourless in solution\underset{	ext{colourless solution}}{{	ext{ZnSO}_4}} + 2	ext{NH}_4	ext{OH} \longrightarrow \underset{	ext{white gelatinous ppt}}{	ext{Zn(OH)}_2↓} + \underset{	ext{colourless in solution}}{(	ext{NH}_4)_2	ext{SO}_4}colourless solutionZnSO4​​+2NH4​OH⟶white gelatinous pptZn(OH)2​↓​+colourless in solution(NH4​)2​SO4​​On adding excess of ammonia solution, the ppt dissolves and a colourless solution is obtained.Zn(OH)2+(NH4)2SO4+2NH4OH⟶[(Zn(NH3)4]SO4Tetraamminezinc (II) sulphate+4H2O	ext{Zn(OH)}_2 + (	ext{NH}_4)_2	ext{SO}_4 + 2	ext{NH}_4	ext{OH} \longrightarrow \underset{\underset{	ext{zinc (II) sulphate}}{	ext{Tetraammine}}}{[(	ext{Zn(NH}_3)_4]	ext{SO}_4} + 4	ext{H}_2	ext{O}Zn(OH)2​+(NH4​)2​SO4​+2NH4​OH⟶zinc (II) sulphateTetraammine​[(Zn(NH3​)4​]SO4​​+4H2​O(iii) When ammonia solution is added dropwise to iron (III) chloride, a reddish brown ppt. of Fe(OH)3 is obtained.FeCl3yellow solution+3NH4OH⟶Fe(OH)3↓reddish brown ppt+3NH4Clcolourless in solution\underset{	ext{yellow solution}}{{	ext{FeCl}_3}} + 3	ext{NH}_4	ext{OH} \longrightarrow \underset{	ext{reddish brown ppt}}{	ext{Fe(OH)}_3↓} + \underset{	ext{colourless in solution}}{3	ext{NH}_4	ext{Cl}}yellow solutionFeCl3​​+3NH4​OH⟶reddish brown pptFe(OH)3​↓​+colourless in solution3NH4​Cl​Excess of ammonia solution addition doesn't dissolve ppt.

Question 6

Question 3What do you observe when caustic soda solution is added to the following solution, first a little and then in excess :(a) FeCl3(b) ZnSO4(c) Pb(NO3)2(d) CuSO4Write balanced equations for these reactions.

Accepted Answer

(a) When caustic soda solution is added to FeCl3 dropwise, a reddish brown ppt is obtained, which is insoluble in excess of NaOH:FeCl3yellow+3NaOHcolourless⟶Fe(OH)3↓reddish brown ppt+3NaClcolourless\underset{	ext{yellow}}{{	ext{FeCl}_3}} + \underset{	ext{colourless}}{3	ext{NaOH}} \longrightarrow \underset{	ext{reddish brown ppt}}{	ext{Fe(OH)}_3↓} + \underset{	ext{colourless}}{3	ext{NaCl}}yellowFeCl3​​+colourless3NaOH​⟶reddish brown pptFe(OH)3​↓​+colourless3NaCl​(b) When caustic soda solution is added to Zinc sulphate dropwise, a white gelatinous ppt is obtained, which dissolves in excess of NaOH:ZnSO4colourless+2NaOHcolourless⟶Zn(OH)2↓white gelatinous ppt+Na2SO4 colourless\underset{	ext{colourless}}{{	ext{ZnSO}_4}} + \underset{	ext{colourless}}{2	ext{NaOH}} \longrightarrow \underset{	ext{white gelatinous ppt}}{	ext{Zn(OH)}_2↓} + \underset{	ext{ colourless}}{	ext{Na}_2	ext{SO}_4}colourlessZnSO4​​+colourless2NaOH​⟶white gelatinous pptZn(OH)2​↓​+ colourlessNa2​SO4​​Zn(OH)2+2NaOH excess⟶Na2ZnO2↓colourless+2H2O{	ext{Zn(OH)}_2} + \underset{	ext{ excess}}{2	ext{NaOH}} \longrightarrow \underset{	ext{colourless}}{	ext{Na}_2	ext{ZnO}_2↓} + 2	ext{H}_2	ext{O}Zn(OH)2​+ excess2NaOH​⟶colourlessNa2​ZnO2​↓​+2H2​O(c) When caustic soda solution is added to Pb(NO3)2 dropwise, a chalky white ppt is obtained, which dissolves in excess of NaOH:Pb(NO3)2colourless+2NaOHcolourless⟶Pb(OH)2↓white ppt+2NaNO3 colourless\underset{	ext{colourless}}{{	ext{Pb(NO}_3)_2}} + \underset{	ext{colourless}}{2	ext{NaOH}} \longrightarrow \underset{	ext{white ppt}}{	ext{Pb(OH)}_2↓} + \underset{	ext{ colourless}}{2	ext{NaNO}_3}colourlessPb(NO3​)2​​+colourless2NaOH​⟶white pptPb(OH)2​↓​+ colourless2NaNO3​​Pb(OH)2+2NaOHexcess⟶Na2PbO2↓sodium plumbite - colourless+2H2O{	ext{Pb(OH)}_2} + \underset{	ext{excess}}{2	ext{NaOH}} \longrightarrow \underset{	ext{sodium plumbite - colourless}}{	ext{Na}_2	ext{PbO}_2↓} + 2	ext{H}_2	ext{O}Pb(OH)2​+excess2NaOH​⟶sodium plumbite - colourlessNa2​PbO2​↓​+2H2​O(d) When caustic soda solution is added to CuSO4 dropwise, a pale blue ppt is obtained, which is insoluble in excess of NaOH:CuSO4blue+2NaOHcolourless⟶Cu(OH)2↓pale blue ppt+Na2SO4colourless\underset{	ext{blue}}{{	ext{CuSO}_4}} + \underset{	ext{colourless}}{2	ext{NaOH}} \longrightarrow \underset{	ext{pale blue ppt}}{	ext{Cu(OH)}_2↓} + \underset{	ext{colourless}}{	ext{Na}_2	ext{SO}_4}blueCuSO4​​+colourless2NaOH​⟶pale blue pptCu(OH)2​↓​+colourlessNa2​SO4​​

Question 7

Question 5What is observed when hot concentrated caustic soda solution is added to(a) Zinc(b) Aluminium?Write balanced equations.

Accepted Answer

(a) When hot concentrated caustic soda solution is added to zinc, soluble salt of sodium zincate [Na2ZnO2] is formed and hydrogen gas is liberated.The balanced equation is:Zn+2NaOHhot and conc.⟶Na2ZnO2sodium zincatecolourless+H2↑	ext{Zn} + \underset{	ext{hot and conc.}}{2	ext{NaOH}} \longrightarrow \underset{\underset{	ext{colourless}}{	ext{sodium zincate}}}{	ext{Na}_2	ext{ZnO}_2} + 	ext{H}_2↑Zn+hot and conc.2NaOH​⟶colourlesssodium zincate​Na2​ZnO2​​+H2​↑(b) When hot concentrated caustic soda solution is added to aluminium, soluble salt of sodium meta aluminate [NaAlO2] is formed and hydrogen gas is liberated.2Al+2NaOH+2H2O⟶2NaAlO2sodium meta aluminatecolourless+3H2↑2	ext{Al} + 2	ext{NaOH} + 2	ext{H}_2	ext{O} \longrightarrow \underset{\underset{	ext{colourless}}{	ext{sodium meta aluminate}}}{2	ext{NaAlO}_2} + 3	ext{H}_2↑2Al+2NaOH+2H2​O⟶colourlesssodium meta aluminate​2NaAlO2​​+3H2​↑

Question 8

Question 6Distinguish by adding: Sodium hydroxide solution or ammonium hydroxide solution to(a) Calcium salt solution and lead salt solution(b) Lead nitrate solution and zinc nitrate solution(c) Copper salt solution and ferrous salt solution(d) Fe(II) salt solution and Fe(III) salt solution(e) Ferrous nitrate and lead nitrate

Accepted Answer

(a) Ammonium hydroxide on reaction with lead salt solution gives chalky white precipitate of Pb(OH)2. No precipitation occurs on adding Ammonium hydroxide to Calcium salt solution even when it is added in excess.Pb(NO3)2colourless+2NH4OH⟶Pb(OH)2white ppt↓+2NH4NO3\underset{	ext{colourless}}{{	ext{Pb(NO}_3)_2}} + 2	ext{NH}_4	ext{OH} \longrightarrow \underset{	ext{white ppt}}{	ext{Pb(OH)}_2} ↓ + 2	ext{NH}_4	ext{NO}_3colourlessPb(NO3​)2​​+2NH4​OH⟶white pptPb(OH)2​​↓+2NH4​NO3​(b) When ammonium hydroxide solution is added to each of the compounds, lead nitrate forms a chalky white precipitate of lead hydroxide [Pb(OH)2] which is insoluble in excess of ammonium hydroxide.Pb(NO3)2colourless+2NH4OH⟶Pb(OH)2white ppt↓+2NH4NO3\underset{	ext{colourless}}{{	ext{Pb(NO}_3)_2}} + 2	ext{NH}_4	ext{OH} \longrightarrow \underset{	ext{white ppt}}{	ext{Pb(OH)}_2} ↓ + 2	ext{NH}_4	ext{NO}_3colourlessPb(NO3​)2​​+2NH4​OH⟶white pptPb(OH)2​​↓+2NH4​NO3​Whereas a gelatinous white precipitate of zinc hydroxide [Zn(OH)2] is formed in case of zinc nitrate, which is soluble in excess of ammonium hydroxide.Zn(NO3)2 + 2NH4OH ⟶ 2NH4NO3 + Zn(OH)2 ↓(c) On adding Sodium hydroxide to Copper salt pale blue coloured precipitate is obtained which is insoluble in excess of Sodium hydroxide. Ferrous salt solution gives a dirty green coloured precipitate with Sodium hydroxide which is insoluble in excess of NaOH.CuSO4blue+2NaOHcolourless⟶Cu(OH)2↓pale blue ppt+ Na2SO4colourless\underset{	ext{blue}}{{	ext{CuSO}_4}} + \underset{	ext{colourless}}{2	ext{NaOH}} \longrightarrow \underset{	ext{pale blue ppt}}{	ext{Cu(OH)}_2↓} + \underset{	ext{colourless}}{	ext{ Na}_2	ext{SO}_4}blueCuSO4​​+colourless2NaOH​⟶pale blue pptCu(OH)2​↓​+colourless Na2​SO4​​FeSO4pale green+2NaOHcolourless⟶Fe(OH)2↓dirty green ppt+Na2SO4colourless\underset{	ext{pale green}}{{	ext{FeSO}_4}} + \underset{	ext{colourless}}{2	ext{NaOH}} \longrightarrow \underset{	ext{dirty green ppt}}{	ext{Fe(OH)}_2↓} + \underset{	ext{colourless}}{	ext{Na}_2	ext{SO}_4}pale greenFeSO4​​+colourless2NaOH​⟶dirty green pptFe(OH)2​↓​+colourlessNa2​SO4​​(d) Sodium hydroxide on reaction with Fe(II) salt gives dirty green coloured precipitate, while with Fe(III) salt solution it forms reddish brown precipitate. Both precipitates are insoluble in excess NaOH.Fe(II) salt:FeSO4pale green solution+2NaOH⟶Fe(OH)2↓dirty green ppt+Na2SO4colourless in solution\underset{	ext{pale green solution}}{{	ext{FeSO}_4}} + 2	ext{NaOH} \longrightarrow \underset{	ext{dirty green ppt}}{	ext{Fe(OH)}_2↓} + \underset{	ext{colourless in solution}}{	ext{Na}_2	ext{SO}_4}pale green solutionFeSO4​​+2NaOH⟶dirty green pptFe(OH)2​↓​+colourless in solutionNa2​SO4​​Fe(III) salt :FeCl3yellow+3NaOHcolourless⟶Fe(OH)3↓reddish brown ppt+3NaClcolourless\underset{	ext{yellow}}{{	ext{FeCl}_3}} + \underset{	ext{colourless}}{3	ext{NaOH}} \longrightarrow \underset{	ext{reddish brown ppt}}{	ext{Fe(OH)}_3↓} + \underset{	ext{colourless}}{3	ext{NaCl}}yellowFeCl3​​+colourless3NaOH​⟶reddish brown pptFe(OH)3​↓​+colourless3NaCl​(e) Ammonium hydroxide on reaction with lead nitrate gives a chalky white insoluble precipitate, and with ferrous nitrate forms a dirty green ppt.Pb(NO3)2colourless+2NH4OH⟶Pb(OH)2↓white ppt+2NH4NO3\underset{	ext{colourless}}{{	ext{Pb(NO}_3)_2}} + 2	ext{NH}_4	ext{OH} \longrightarrow \underset{	ext{white ppt}}{	ext{Pb(OH)}_2↓} + 2	ext{NH}_4	ext{NO}_3colourlessPb(NO3​)2​​+2NH4​OH⟶white pptPb(OH)2​↓​+2NH4​NO3​Fe(NO3)2colourless+2NH4OH⟶Fe(OH)2↓dirty green ppt+2NH4NO3 colourless\underset{	ext{colourless}}{{	ext{Fe(NO}_3)_2}} + 2	ext{NH}_4	ext{OH} \longrightarrow \underset{	ext{dirty green ppt}}{	ext{Fe(OH)}_2↓} + \underset{	ext{ colourless}}{2	ext{NH}_4	ext{NO}_3}colourlessFe(NO3​)2​​+2NH4​OH⟶dirty green pptFe(OH)2​↓​+ colourless2NH4​NO3​​

Question 9

Question 7How will you distinguish calcium nitrate and zinc nitrate solution?

Accepted Answer

When ammonium hydroxide (NH4OH) is added to zinc nitrate solution [Zn(NO3)2], a gelatinous white ppt of zinc hydroxide [Zn(OH)2] is obtained which is soluble in excess of NH4OH.ZnSO4colourless solution+2NH4OH⟶Zn(OH)2↓white gelatinous ppt+(NH4)2SO4colourless in solution\underset{	ext{colourless solution}}{{	ext{ZnSO}_4}} + 2	ext{NH}_4	ext{OH} \longrightarrow \underset{	ext{white gelatinous ppt}}{	ext{Zn(OH)}_2↓} + \underset{	ext{colourless in solution}}{(	ext{NH}_4)_2	ext{SO}_4}colourless solutionZnSO4​​+2NH4​OH⟶white gelatinous pptZn(OH)2​↓​+colourless in solution(NH4​)2​SO4​​(With excess NH4OH ppt. dissolves)Zn(OH)2+(NH4)2SO4+2NH4OH⟶[(Zn(NH3)4]SO4Tetraamminezinc (II) sulphate+4H2O	ext{Zn(OH)}_2 + (	ext{NH}_4)_2	ext{SO}_4 + 2	ext{NH}_4	ext{OH} \longrightarrow \underset{\underset{	ext{zinc (II) sulphate}}{	ext{Tetraammine}}}{[(	ext{Zn(NH}_3)_4]	ext{SO}_4} + 4	ext{H}_2	ext{O}Zn(OH)2​+(NH4​)2​SO4​+2NH4​OH⟶zinc (II) sulphateTetraammine​[(Zn(NH3​)4​]SO4​​+4H2​OOn the other hand, calcium nitrate solution [Ca(NO3)2] does not give any ppt. even when excess of ammonium hydroxide is added.

Question 10

Question 8Write balanced equation:(a) Reaction of sodium hydroxide solution with Iron (III) chloride solution(b) Copper sulphate solution with ammonium hydroxide solution

Accepted Answer

(a) When sodium hydroxide solution is added to FeCl3 dropwise, a reddish brown ppt is obtained, which is insoluble in excess of NaOH:FeCl3Yellow+3NaOHColourless⟶Fe(OH)3↓Reddish brown ppt+3NaClColourless\underset{	ext{Yellow}}{{	ext{FeCl}_3}} + \underset{	ext{Colourless}}{3	ext{NaOH}} \longrightarrow \underset{	ext{Reddish brown ppt}}{	ext{Fe(OH)}_3↓} + \underset{	ext{Colourless}}{3	ext{NaCl}}YellowFeCl3​​+Colourless3NaOH​⟶Reddish brown pptFe(OH)3​↓​+Colourless3NaCl​(b) When ammonia solution is added dropwise to cupper sulphate, a pale blue ppt of copper hydroxide is obtained.CuSO4Blue+2NH4OH⟶Cu(OH)2Pale Blue ppt↓+(NH4)2SO4Colourless in solution\underset{	ext{Blue}}{{	ext{CuSO}_4}} + 2	ext{NH}_4	ext{OH} \longrightarrow \underset{	ext{Pale Blue ppt}}{	ext{Cu(OH)}_2} ↓ + \underset{	ext{Colourless in solution}}{(	ext{NH}_4)_2	ext{SO}_4}BlueCuSO4​​+2NH4​OH⟶Pale Blue pptCu(OH)2​​↓+Colourless in solution(NH4​)2​SO4​​On adding excess of ammonia solution, the ppt dissolves and a deep blue solution is obtained.Cu(OH)2+(NH4)2SO4+2NH4OH⟶[(Cu(NH3)4]SO4Tetraammine copper (II) sulphate+4H2O	ext{Cu(OH)}_2 + (	ext{NH}_4)_2	ext{SO}_4 + 2	ext{NH}_4	ext{OH} \longrightarrow \underset{	ext{Tetraammine copper (II) sulphate}}{[(	ext{Cu(NH}_3)_4]	ext{SO}_4} + 4	ext{H}_2	ext{O}Cu(OH)2​+(NH4​)2​SO4​+2NH4​OH⟶Tetraammine copper (II) sulphate[(Cu(NH3​)4​]SO4​​+4H2​O

Analytical Chemistry

Exercise 4 Long Answer Type

Exercise 4 Multiple Choice Type

Exercise 4 Short Answer Type

Intext Questions 1