Explain the reasons for the observed changes in the mobility of sulfur at different temperatures and the reactivity of lead(II) carbonate with dilute hydrochloric acid.

Question

Asked on April 27, 2026Chemistry

This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.

Accepted Answer · 2026-04-27T09:45:18.128Z

Step 1: Answer Question 6. It is not suitable to prepare carbon dioxide by reacting dilute hydrochloric acid with lead(II) carbonate because the lead(II) chloride (PbCl_2) formed is insoluble. This insoluble layer coats the surface of the lead(II) carbonate, preventing further reaction and stopping the production of carbon dioxide. Step 2: Answer Question 7. a) X is very mobile* because at 113^, sulfur exists as discrete S_8 rings. These rings can slide past each other easily due to weak intermolecular forces. Y is viscous* because at 116^, the S_8 rings break open and polymerize into long, tangled chains. These long chains become entangled, increasing the resistance to flow. Z is very mobile* because at 400^, the long polymeric chains break down into shorter chains. These shorter chains can move past each other more easily, reducing the viscosity. b) At 30^, sulfur is a solid composed of S_8 rings. At 113^ (X), sulfur melts to form a liquid of S_8 rings, which are mobile. At 116^ (Y), the S_8 rings break and link to form long, tangled polymeric chains, making the liquid viscous. At 400^ (Z), the long polymeric chains break into shorter chains, which are more mobile. Step 3: Answer Question 8. Carbon(IV) oxide (CO_2) is preferred to water for extinguishing kerosene fires because CO_2 is denser than air and forms a blanket over the fire, cutting off the oxygen supply. Water is unsuitable as kerosene is less dense than water and immiscible, causing the fire to spread. Step 4: Answer Question 9. The term is electron affinity. Step 5: Answer Question 10. The presence of nitrate ions can be determined using the brown ring test: 1. Dissolve the fertilizer in water and acidify with dilute sulfuric acid. 2. Add freshly prepared iron(II) sulfate solution. 3. Carefully add concentrated sulfuric acid down the side of the test tube to form a separate layer. 4. A brown ring at the junction of the two layers indicates the presence of nitrate ions. Step 6: Answer Question 11. a) To show that graphite and diamond are both allotropes of carbon: Burn samples of both in excess oxygen. Both will produce only carbon dioxide* (CO_2) as a product, with no residue, indicating they are composed solely of carbon. b) Graphite is used as a lubricant because it has a layered structure* where carbon atoms are arranged in hexagonal rings within layers. The weak intermolecular forces between these layers allow them to slide past each other easily. Diamond is used as an abrasive because it has a giant covalent structure* where each carbon atom is strongly covalently bonded to four other carbon atoms in a rigid tetrahedral arrangement. This extensive network of strong covalent bonds makes it extremely hard. Step 7: Answer Question 12. i) The equation for the burning of phosphorus in excess air is: P_4(s) + 5O_2(g) P_4O_10(s) ii) The solution obtained had a pH of 2.0 because phosphorus(V) oxide (P_4O_10) is an acidic oxide. When it dissolves in water, it reacts to form phosphoric acid (H_3PO_4), which is a strong acid, resulting in a low pH. P_4O_10(s) + 6H_2O(l) 4H_3PO_4(aq) Step 8: Answer Question 13. a) The equation for the dehydration of ethenedioic acid is: (COOH)_2(l) conc. H_2SO_4 CO(g) + CO_2(g) + H_2O(l) b) To obtain dry carbon(II) oxide (CO) from the mixture of gases: 1. Pass the gas mixture through a drying agent (e.g., concentrated sulfuric acid) to remove water vapor. 2. Then, pass the remaining gas through a strong alkali solution (e.g., concentrated potassium hydroxide solution) to absorb the acidic carbon(IV) oxide (CO_2). Carbon(II) oxide (CO) is a neutral gas and will not react. The remaining gas will be dry CO. Step 9: Answer Question 14. Silicon(IV) oxide (SiO_2) exists as a solid with a high melting point because it has a giant covalent structure*. Each silicon atom is covalently bonded to four oxygen atoms, forming a continuous network of strong covalent bonds. A large amount of energy is required to break these numerous strong bonds. Sulfur(IV) oxide (SO_2) exists as a gas because it has a simple molecular structure. Within each SO_2 molecule, there are strong covalent bonds, but between* the molecules, there are only weak intermolecular forces. Little energy is required to

Accepted Answer · 2026-04-27T09:45:18.128Z

Step 1: Answer Question 6. It is not suitable to prepare carbon dioxide by reacting dilute hydrochloric acid with lead(II) carbonate because the lead(II) chloride (PbCl_2) formed is insoluble. This insoluble layer coats the surface of the lead(II) carbonate, preventing further reaction and stopping the production of carbon dioxide. Step 2: Answer Question 7. a) X is very mobile* because at 113^, sulfur exists as discrete S_8 rings. These rings can slide past each other easily due to weak intermolecular forces. Y is viscous* because at 116^, the S_8 rings break open and polymerize into long, tangled chains. These long chains become entangled, increasing the resistance to flow. Z is very mobile* because at 400^, the long polymeric chains break down into shorter chains. These shorter chains can move past each other more easily, reducing the viscosity. b) At 30^, sulfur is a solid composed of S_8 rings. At 113^ (X), sulfur melts to form a liquid of S_8 rings, which are mobile. At 116^ (Y), the S_8 rings break and link to form long, tangled polymeric chains, making the liquid viscous. At 400^ (Z), the long polymeric chains break into shorter chains, which are more mobile. Step 3: Answer Question 8. Carbon(IV) oxide (CO_2) is preferred to water for extinguishing kerosene fires because CO_2 is denser than air and forms a blanket over the fire, cutting off the oxygen supply. Water is unsuitable as kerosene is less dense than water and immiscible, causing the fire to spread. Step 4: Answer Question 9. The term is electron affinity. Step 5: Answer Question 10. The presence of nitrate ions can be determined using the brown ring test: 1. Dissolve the fertilizer in water and acidify with dilute sulfuric acid. 2. Add freshly prepared iron(II) sulfate solution. 3. Carefully add concentrated sulfuric acid down the side of the test tube to form a separate layer. 4. A brown ring at the junction of the two layers indicates the presence of nitrate ions. Step 6: Answer Question 11. a) To show that graphite and diamond are both allotropes of carbon: Burn samples of both in excess oxygen. Both will produce only carbon dioxide* (CO_2) as a product, with no residue, indicating they are composed solely of carbon. b) Graphite is used as a lubricant because it has a layered structure* where carbon atoms are arranged in hexagonal rings within layers. The weak intermolecular forces between these layers allow them to slide past each other easily. Diamond is used as an abrasive because it has a giant covalent structure* where each carbon atom is strongly covalently bonded to four other carbon atoms in a rigid tetrahedral arrangement. This extensive network of strong covalent bonds makes it extremely hard. Step 7: Answer Question 12. i) The equation for the burning of phosphorus in excess air is: P_4(s) + 5O_2(g) P_4O_10(s) ii) The solution obtained had a pH of 2.0 because phosphorus(V) oxide (P_4O_10) is an acidic oxide. When it dissolves in water, it reacts to form phosphoric acid (H_3PO_4), which is a strong acid, resulting in a low pH. P_4O_10(s) + 6H_2O(l) 4H_3PO_4(aq) Step 8: Answer Question 13. a) The equation for the dehydration of ethenedioic acid is: (COOH)_2(l) conc. H_2SO_4 CO(g) + CO_2(g) + H_2O(l) b) To obtain dry carbon(II) oxide (CO) from the mixture of gases: 1. Pass the gas mixture through a drying agent (e.g., concentrated sulfuric acid) to remove water vapor. 2. Then, pass the remaining gas through a strong alkali solution (e.g., concentrated potassium hydroxide solution) to absorb the acidic carbon(IV) oxide (CO_2). Carbon(II) oxide (CO) is a neutral gas and will not react. The remaining gas will be dry CO. Step 9: Answer Question 14. Silicon(IV) oxide (SiO_2) exists as a solid with a high melting point because it has a giant covalent structure*. Each silicon atom is covalently bonded to four oxygen atoms, forming a continuous network of strong covalent bonds. A large amount of energy is required to break these numerous strong bonds. Sulfur(IV) oxide (SO_2) exists as a gas because it has a simple molecular structure. Within each SO_2 molecule, there are strong covalent bonds, but between* the molecules, there are only weak intermolecular forces. Little energy is required to

Explain the reasons for the observed changes in the mobility of sulfur at different temperatures and the reactivity of lead(II) carbonate with dilute hydrochloric acid.

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Explain the reasons for the observed changes in the mobility of sulfur at different temperatures and the reactivity of lead(II) carbonate with dilute hydrochloric acid.

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