Question

Написать молекулярные и молекулярно-ионные уравнения следующих превращений; ALCL3; AL(OH)3; K3{AL(OH)6}->AL(NO3)3; AL2O3 Осуществить превращения,назвать соединения хрома: Cr->Cr2O3->Cr(OH)3->Na3{Cr(OH)6} Cr(SO4)3 Составить уравнения реакций растворения меди и серебра в концентрированной серной кислоте при нагревании.

Answer 1

Ответ:

Объяснение:

AICI₃→ AI(OH)₃→ K₃[AI(OH)₆]→AI(NO₃)₃→AI(OH)₃→ AI₂O₃

AICI₃+3NaOH= AI(OH)₃↓+3NaCI

AI³⁺+3CI⁻+3Na⁺+3OH⁻= AI(OH)₃↓+3Na⁺+3CI⁻

AI³⁺+3OH⁻= AI(OH)₃↓

AI(OH)₃+3KOH= K₃[AI(OH)₆]

AI(OH)₃+3K⁺+OH⁻= 3K⁺+[AI(OH)₆]³⁻

AI(OH)₃+OH⁻= [AI(OH)₆]³⁻

K₃[AI(OH)₆]+6HNO₃=AI(NO₃)₃+ 3KNO₃+6H₂O

3K⁺+[AI(OH)₆]³⁻+6H⁺+6NO₃⁻=AI³⁺+3NO₃⁻+ 3K⁺+3NO₃⁻+6H₂O

[AI(OH)₆]³⁻+6H⁺=AI³⁺+6H₂O

AI(NO₃)₃+3NaOH=AI(OH)₃+3NaCI

AI³⁺+3NO₃⁻+3Na⁺+3OH⁻= AI(OH)₃↓+3Na⁺+3NO₃⁻

AI³⁺+3OH⁻= AI(OH)₃↓

2AI(OH)₃+t°C= AI₂O₃+3H₂O

Cr→Cr₂O₃→CrCI₃→Cr(OH)₃→Na₃{Cr(OH)₆}

→Cr₂(SO₄)₃

4Cr+3O₂=2Cr₂O₃ оксид хрома(lll)

Cr₂O₃+6HCI=2CrCI₃+3H₂O

                  хлорид хрома(lll)

CrCI₃+3NaOH=Cr(OH)₃+3NaCI

                  гидроксид хрома(lll)              

Cr(OH)₃+3NaOH=Na₃[Cr(OH)₆] гексагидроксохромат натрия

Na₃{Cr(OH)₆}

+H₂SO₄=Cr₂(SO₄)₃+ Na₂SO₄+H₂O

                              сульфат хрома(lll)

Cu+2H₂SO₄(конц)+t°=CuSO₄+SO₂↑+2H₂O

2Ag+2H₂SO₄(конц)+t°=Ag₂SO₄+SO₂↑+2H₂O

Answer 2

#### Molecular and Molecular-Ionic Equations for the Given Transformations 1. Transformation: ALCl3 to AL(NO3)3 The molecular equation for the transformation of ALCl3 to AL(NO3)3 can be represented as follows: **2AlCl3 + 3H2SO4 → Al2(SO4)3 + 6HCl** The molecular-ionic equation for this transformation can be represented as follows: **2Al^3+ + 6Cl^- + 3H2SO4 → Al2(SO4)3 + 6H^+ + 6Cl^-** Please note that the molecular-ionic equation represents the dissociation of the compounds into their respective ions. 2. Transformation: AL(OH)3 to AL2O3 The molecular equation for the transformation of AL(OH)3 to AL2O3 can be represented as follows: **2Al(OH)3 → Al2O3 + 3H2O** The molecular-ionic equation for this transformation can be represented as follows: **2Al(OH)3 → Al2O3 + 3H2O** In this case, there is no dissociation of compounds into ions. 3. Transformation: K3{AL(OH)6} to Cr(SO4)3 The molecular equation for the transformation of K3{AL(OH)6} to Cr(SO4)3 can be represented as follows: **2K3{AL(OH)6} + 3Cr2(SO4)3 → 6KAl(SO4)2 + 3Cr(OH)3** The molecular-ionic equation for this transformation can be represented as follows: **2K^+ + 6Al(OH)6^3- + 3Cr^3+ + 6SO4^2- → 6K^+ + 6Al(SO4)2^3- + 3Cr(OH)3** Again, please note that the molecular-ionic equation represents the dissociation of the compounds into their respective ions. #### Transformations of Chromium (Cr) and Naming of Chromium Compounds The transformations and names of chromium compounds are as follows: 1. Transformation: Cr to Cr2O3 The transformation of Cr to Cr2O3 involves the oxidation of chromium. The molecular equation for this transformation can be represented as follows: **4Cr + 3O2 → 2Cr2O3** The resulting compound is chromium(III) oxide. 2. Transformation: Cr2O3 to Cr(OH)3 The transformation of Cr2O3 to Cr(OH)3 involves the reaction with water. The molecular equation for this transformation can be represented as follows: **Cr2O3 + 3H2O → 2Cr(OH)3** The resulting compound is chromium(III) hydroxide. 3. Transformation: Cr(OH)3 to Na3{Cr(OH)6} The transformation of Cr(OH)3 to Na3{Cr(OH)6} involves the reaction with sodium hydroxide. The molecular equation for this transformation can be represented as follows: **Cr(OH)3 + 6NaOH → Na3{Cr(OH)6} + 3H2O** The resulting compound is sodium hexahydroxochromate(III). 4. Transformation: Na3{Cr(OH)6} to Cr(SO4)3 The transformation of Na3{Cr(OH)6} to Cr(SO4)3 involves the reaction with sulfuric acid. The molecular equation for this transformation can be represented as follows: **2Na3{Cr(OH)6} + 3H2SO4 → Cr(SO4)3 + 6NaOH + 6H2O** The resulting compound is chromium(III) sulfate. #### Equations for Dissolution Reactions of Copper and Silver in Concentrated Sulfuric Acid 1. Dissolution of Copper (Cu) in Concentrated Sulfuric Acid The equation for the dissolution of copper in concentrated sulfuric acid can be represented as follows: **Cu + 2H2SO4 → CuSO4 + SO2 + 2H2O** The resulting compound is copper(II) sulfate. 2. Dissolution of Silver (Ag) in Concentrated Sulfuric Acid The equation for the dissolution of silver in concentrated sulfuric acid can be represented as follows: **Ag + 2H2SO4 → Ag2SO4 + SO2 + 2H2O** The resulting compound is silver sulfate. Please note that these equations represent simplified representations of the reactions and may not account for all possible side reactions or conditions.