The name of the compounds is to be given. Concept introduction: While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

Question

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Answer 1

Question

Chapter 2, Problem 83E

(a)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(a)

Expert Solution

Answer to Problem 83E

The naming of $CuI$ is copper (I) iodide.

Explanation of Solution

To determine: The naming of $CuI$

The name of $CuI$ is copper(I) iodide.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. Monatomic anions are named with the ending -ide.

Roman numerals are used to show the oxidation state because some elements exhibit more than one oxidation state. Copper exhibits $+1 and +2$ oxidation state.

Here copper $(Cu2+)$ ion is cation and iodine $(I−)$ ion is mono-atomic anion. The oxidation state of copper is $+1$ because total charge on anion is $−1$ . Since compound should be electrically neutral, the name of $CuI$ compound is copper (I) iodide.

Conclusion

The name of $CuI$ is copper(I) iodide.

(b)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(b)

Expert Solution

Answer to Problem 83E

The naming of $CuI2$ is copper (II) iodide.

Explanation of Solution

To determine: The naming of $CuI2$

The name of $CuI2$ is copper(II) iodide.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. Monatomic anions are named with the ending -ide.

Roman numerals are used to show the oxidation state because some elements exhibit more than one oxidation state. Copper exhibits $+1 and +2$ oxidation state.

Here copper $(Cu2+)$ ion is cation and iodine $(I−)$ ion is mono-atomic anion. The oxidation state of copper is $+2$ because total charge on anion is $−2$ . Since compound should be electrically neutral, the name of $CuI2$ compound is copper (II) iodide.

Conclusion

The name of $CuI2$ is copper(II) iodide.

(c)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(c)

Expert Solution

Answer to Problem 83E

The naming of $CuI2$ is cobalt (II) iodide.

Explanation of Solution

To determine: The naming of $CoI2$

The name of $CuI2$ is cobalt (II) iodide.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. Monatomic anions are named with the ending -ide.

Roman numerals are used to show the oxidation state because some elements exhibit more than one oxidation state. Cobalt exhibits $+2 and +3$ oxidation state.

Here cobalt $(Cu2+)$ ion is cation and iodine $(I-)$ ion is mono-atomic anion. The oxidation state of copper is $+2$ because total charge on anion is $−2$ . Since compound should be electrically neutral, the name of $CoI2$ compound is cobalt (II) iodide.

Conclusion

The name of $CuI2$ is cobalt (II) iodide.

(d)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(d)

Expert Solution

Answer to Problem 83E

The naming of $Na2CO3$ is sodium carbonate

Explanation of Solution

To determine: The naming of $Na2CO3$

The name of $Na2CO3$ is sodium carbonate

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. The anions are named with the ending -ide.

The compound $Na2CO3$ is a molecular compound. While naming a molecular compound, the number of respective atoms present is denoted by using prefixes like mono, di tri.

Here sodium ion $(Na+)$ is cation and carbonate ion $(CO32-)$ is anion. The total charge on cation is $+2$ and total charge on anion is $−2$ . Since compound should be electrically neutral, the name of $Na2CO3$ compound is sodium carbonate

Conclusion

The name of $Na2CO3$ is sodium carbonate

(e)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(e)

Expert Solution

Answer to Problem 83E

The naming of $NaHCO3$ is sodium bicarbonate.

Explanation of Solution

To determine: The naming of $NaHCO3$

The name of $NaHCO3$ is sodium bicarbonate

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. The anions are named with the ending -ide.

The compound $NaHCO3$ is a molecular compound. While naming a molecular compound, the number of respective atoms present is denoted by using prefixes like mono, di tri.

Here sodium ion $(Na+)$ is cation and bicarbonate ion $(HCO3−)$ is anion. The total charge on cation is $+1$ and total charge on anion is $−1$ . Since compound should be electrically neutral, the name of $NaHCO3$ compound is sodium bicarbonate

Conclusion

The name of $NaHCO3$ is sodium bicarbonate

(f)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(f)

Expert Solution

Answer to Problem 83E

The naming of $S4N4$ is tetra sulfur tetra nitride.

Explanation of Solution

To determine: The naming of $S4N4$

The name of $S4N4$ is tetra sulfur tetra nitride.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. The anions are named with the ending -ide. The prefix mono-, di-, tri- is used in prefix to indicate the number of atoms in each element.

Here sulfur ion $(S2−)$ is cation having four atoms and nitrogen ion $(N3-)$ is poly-atomic anion having four atoms. Hence the name of $S4N4$ compound is tetra sulfur tetra nitride.

Conclusion

The name of $S4N4$ is tetra sulfur tetra nitride.

(g)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(g)

Expert Solution

Answer to Problem 83E

The naming of $SF6$ is tetra sulfur tetra nitride.

Explanation of Solution

To determine: The naming of $SF6$

The name of $SF6$ is tetra sulfur tetra nitride.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. The anions are named with the ending -ide. The prefix mono-, di-, tri- is used in prefix to indicate the number of atoms in each element.

Here sulfur ion $(S2−)$ is cation having single atom and fluorine ion $(F-)$ is mono-atomic anion having six atoms. Hence the name of $SF6$ compound is sulfur hexafluoride.

Conclusion

The name of $SF6$ is tetra sulfur tetra nitride.

(h)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(h)

Expert Solution

Answer to Problem 83E

The naming of $NaOCl$ is sodium Hypochlorite.

Explanation of Solution

To determine: The naming of $NaOCl$

The name of $NaOCl$ is sodium Hypochlorite.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. When there are only two members in the same series, then the anion with the least number of oxygen’s ends in -ite, and the anion with the most ends in -ate.

If more than two oxyanion is present in compound then “hypo-” (less than) and “per-” (more than) is used in prefix.

Hence the name of $NaOCl$ compound is sodium Hypochlorite.

Conclusion

The name of $NaOCl$ is sodium Hypochlorite.

(i)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(i)

Expert Solution

Answer to Problem 83E

The naming of $BaCrO4$ is barium chromate.

Explanation of Solution

To determine: The naming of $BaCrO4$

The name of $BaCrO4$ is barium chromate.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. When there are only two members in the same series, then the anion with the least number of oxygen ends in -ite, and the anion with the most ends in -ate.

Here barium ion $(Ba2+)$ is cation and chromate ion $(CrO42-)$ is anion.

Hence the name of $BaCrO4$ is barium chromate.

Conclusion

The name of $BaCrO4$ is barium chromate.

(j)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(j)

Expert Solution

Answer to Problem 83E

The naming of $NH4NO3$ is ammonium nitrate.

Explanation of Solution

To determine: The naming of $NH4NO3$

The name of $NH4NO3$ is ammonium nitrate.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. When there are only two members in the same series, then the anion with the least number of oxygen ends in -ite, and the anion with the most ends in -ate.

Here ammonium $(NH4+)$ is cation and nitratee ion $(NO3-)$ is anion.

Hence the name of $NH4NO3$ is ammonium nitrate.

Conclusion

The name of $NH4NO3$ is ammonium nitrate.

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a. The change in the Gibbs energy of a certain constant pressure process is found to fit the expression: AG-85.1 J mol −1 +36.5 J mol ¹K-1 × T A. Calculate the value of AS for the process. B. Next, use the Gibbs-Helmholtz equation: (a(AG/T)) ΔΗ - T2 to calculate the value of AH for the process.

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Answer 2

Question

Chapter 2, Problem 83E

(a)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(a)

Expert Solution

Answer to Problem 83E

The naming of $CuI$ is copper (I) iodide.

Explanation of Solution

To determine: The naming of $CuI$

The name of $CuI$ is copper(I) iodide.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. Monatomic anions are named with the ending -ide.

Roman numerals are used to show the oxidation state because some elements exhibit more than one oxidation state. Copper exhibits $+1 and +2$ oxidation state.

Here copper $(Cu2+)$ ion is cation and iodine $(I−)$ ion is mono-atomic anion. The oxidation state of copper is $+1$ because total charge on anion is $−1$ . Since compound should be electrically neutral, the name of $CuI$ compound is copper (I) iodide.

Conclusion

The name of $CuI$ is copper(I) iodide.

(b)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(b)

Expert Solution

Answer to Problem 83E

The naming of $CuI2$ is copper (II) iodide.

Explanation of Solution

To determine: The naming of $CuI2$

The name of $CuI2$ is copper(II) iodide.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. Monatomic anions are named with the ending -ide.

Roman numerals are used to show the oxidation state because some elements exhibit more than one oxidation state. Copper exhibits $+1 and +2$ oxidation state.

Here copper $(Cu2+)$ ion is cation and iodine $(I−)$ ion is mono-atomic anion. The oxidation state of copper is $+2$ because total charge on anion is $−2$ . Since compound should be electrically neutral, the name of $CuI2$ compound is copper (II) iodide.

Conclusion

The name of $CuI2$ is copper(II) iodide.

(c)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(c)

Expert Solution

Answer to Problem 83E

The naming of $CuI2$ is cobalt (II) iodide.

Explanation of Solution

To determine: The naming of $CoI2$

The name of $CuI2$ is cobalt (II) iodide.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. Monatomic anions are named with the ending -ide.

Roman numerals are used to show the oxidation state because some elements exhibit more than one oxidation state. Cobalt exhibits $+2 and +3$ oxidation state.

Here cobalt $(Cu2+)$ ion is cation and iodine $(I-)$ ion is mono-atomic anion. The oxidation state of copper is $+2$ because total charge on anion is $−2$ . Since compound should be electrically neutral, the name of $CoI2$ compound is cobalt (II) iodide.

Conclusion

The name of $CuI2$ is cobalt (II) iodide.

(d)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(d)

Expert Solution

Answer to Problem 83E

The naming of $Na2CO3$ is sodium carbonate

Explanation of Solution

To determine: The naming of $Na2CO3$

The name of $Na2CO3$ is sodium carbonate

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. The anions are named with the ending -ide.

The compound $Na2CO3$ is a molecular compound. While naming a molecular compound, the number of respective atoms present is denoted by using prefixes like mono, di tri.

Here sodium ion $(Na+)$ is cation and carbonate ion $(CO32-)$ is anion. The total charge on cation is $+2$ and total charge on anion is $−2$ . Since compound should be electrically neutral, the name of $Na2CO3$ compound is sodium carbonate

Conclusion

The name of $Na2CO3$ is sodium carbonate

(e)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(e)

Expert Solution

Answer to Problem 83E

The naming of $NaHCO3$ is sodium bicarbonate.

Explanation of Solution

To determine: The naming of $NaHCO3$

The name of $NaHCO3$ is sodium bicarbonate

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. The anions are named with the ending -ide.

The compound $NaHCO3$ is a molecular compound. While naming a molecular compound, the number of respective atoms present is denoted by using prefixes like mono, di tri.

Here sodium ion $(Na+)$ is cation and bicarbonate ion $(HCO3−)$ is anion. The total charge on cation is $+1$ and total charge on anion is $−1$ . Since compound should be electrically neutral, the name of $NaHCO3$ compound is sodium bicarbonate

Conclusion

The name of $NaHCO3$ is sodium bicarbonate

(f)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(f)

Expert Solution

Answer to Problem 83E

The naming of $S4N4$ is tetra sulfur tetra nitride.

Explanation of Solution

To determine: The naming of $S4N4$

The name of $S4N4$ is tetra sulfur tetra nitride.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. The anions are named with the ending -ide. The prefix mono-, di-, tri- is used in prefix to indicate the number of atoms in each element.

Here sulfur ion $(S2−)$ is cation having four atoms and nitrogen ion $(N3-)$ is poly-atomic anion having four atoms. Hence the name of $S4N4$ compound is tetra sulfur tetra nitride.

Conclusion

The name of $S4N4$ is tetra sulfur tetra nitride.

(g)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(g)

Expert Solution

Answer to Problem 83E

The naming of $SF6$ is tetra sulfur tetra nitride.

Explanation of Solution

To determine: The naming of $SF6$

The name of $SF6$ is tetra sulfur tetra nitride.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. The anions are named with the ending -ide. The prefix mono-, di-, tri- is used in prefix to indicate the number of atoms in each element.

Here sulfur ion $(S2−)$ is cation having single atom and fluorine ion $(F-)$ is mono-atomic anion having six atoms. Hence the name of $SF6$ compound is sulfur hexafluoride.

Conclusion

The name of $SF6$ is tetra sulfur tetra nitride.

(h)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(h)

Expert Solution

Answer to Problem 83E

The naming of $NaOCl$ is sodium Hypochlorite.

Explanation of Solution

To determine: The naming of $NaOCl$

The name of $NaOCl$ is sodium Hypochlorite.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. When there are only two members in the same series, then the anion with the least number of oxygen’s ends in -ite, and the anion with the most ends in -ate.

If more than two oxyanion is present in compound then “hypo-” (less than) and “per-” (more than) is used in prefix.

Hence the name of $NaOCl$ compound is sodium Hypochlorite.

Conclusion

The name of $NaOCl$ is sodium Hypochlorite.

(i)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(i)

Expert Solution

Answer to Problem 83E

The naming of $BaCrO4$ is barium chromate.

Explanation of Solution

To determine: The naming of $BaCrO4$

The name of $BaCrO4$ is barium chromate.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. When there are only two members in the same series, then the anion with the least number of oxygen ends in -ite, and the anion with the most ends in -ate.

Here barium ion $(Ba2+)$ is cation and chromate ion $(CrO42-)$ is anion.

Hence the name of $BaCrO4$ is barium chromate.

Conclusion

The name of $BaCrO4$ is barium chromate.

(j)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(j)

Expert Solution

Answer to Problem 83E

The naming of $NH4NO3$ is ammonium nitrate.

Explanation of Solution

To determine: The naming of $NH4NO3$

The name of $NH4NO3$ is ammonium nitrate.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. When there are only two members in the same series, then the anion with the least number of oxygen ends in -ite, and the anion with the most ends in -ate.

Here ammonium $(NH4+)$ is cation and nitratee ion $(NO3-)$ is anion.

Hence the name of $NH4NO3$ is ammonium nitrate.

Conclusion

The name of $NH4NO3$ is ammonium nitrate.

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Answer 3

Question

Chapter 2, Problem 83E

(a)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(a)

Expert Solution

Answer to Problem 83E

The naming of $CuI$ is copper (I) iodide.

Explanation of Solution

To determine: The naming of $CuI$

The name of $CuI$ is copper(I) iodide.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. Monatomic anions are named with the ending -ide.

Roman numerals are used to show the oxidation state because some elements exhibit more than one oxidation state. Copper exhibits $+1 and +2$ oxidation state.

Here copper $(Cu2+)$ ion is cation and iodine $(I−)$ ion is mono-atomic anion. The oxidation state of copper is $+1$ because total charge on anion is $−1$ . Since compound should be electrically neutral, the name of $CuI$ compound is copper (I) iodide.

Conclusion

The name of $CuI$ is copper(I) iodide.

(b)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(b)

Expert Solution

Answer to Problem 83E

The naming of $CuI2$ is copper (II) iodide.

Explanation of Solution

To determine: The naming of $CuI2$

The name of $CuI2$ is copper(II) iodide.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. Monatomic anions are named with the ending -ide.

Roman numerals are used to show the oxidation state because some elements exhibit more than one oxidation state. Copper exhibits $+1 and +2$ oxidation state.

Here copper $(Cu2+)$ ion is cation and iodine $(I−)$ ion is mono-atomic anion. The oxidation state of copper is $+2$ because total charge on anion is $−2$ . Since compound should be electrically neutral, the name of $CuI2$ compound is copper (II) iodide.

Conclusion

The name of $CuI2$ is copper(II) iodide.

(c)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(c)

Expert Solution

Answer to Problem 83E

The naming of $CuI2$ is cobalt (II) iodide.

Explanation of Solution

To determine: The naming of $CoI2$

The name of $CuI2$ is cobalt (II) iodide.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. Monatomic anions are named with the ending -ide.

Roman numerals are used to show the oxidation state because some elements exhibit more than one oxidation state. Cobalt exhibits $+2 and +3$ oxidation state.

Here cobalt $(Cu2+)$ ion is cation and iodine $(I-)$ ion is mono-atomic anion. The oxidation state of copper is $+2$ because total charge on anion is $−2$ . Since compound should be electrically neutral, the name of $CoI2$ compound is cobalt (II) iodide.

Conclusion

The name of $CuI2$ is cobalt (II) iodide.

(d)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(d)

Expert Solution

Answer to Problem 83E

The naming of $Na2CO3$ is sodium carbonate

Explanation of Solution

To determine: The naming of $Na2CO3$

The name of $Na2CO3$ is sodium carbonate

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. The anions are named with the ending -ide.

The compound $Na2CO3$ is a molecular compound. While naming a molecular compound, the number of respective atoms present is denoted by using prefixes like mono, di tri.

Here sodium ion $(Na+)$ is cation and carbonate ion $(CO32-)$ is anion. The total charge on cation is $+2$ and total charge on anion is $−2$ . Since compound should be electrically neutral, the name of $Na2CO3$ compound is sodium carbonate

Conclusion

The name of $Na2CO3$ is sodium carbonate

(e)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(e)

Expert Solution

Answer to Problem 83E

The naming of $NaHCO3$ is sodium bicarbonate.

Explanation of Solution

To determine: The naming of $NaHCO3$

The name of $NaHCO3$ is sodium bicarbonate

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. The anions are named with the ending -ide.

The compound $NaHCO3$ is a molecular compound. While naming a molecular compound, the number of respective atoms present is denoted by using prefixes like mono, di tri.

Here sodium ion $(Na+)$ is cation and bicarbonate ion $(HCO3−)$ is anion. The total charge on cation is $+1$ and total charge on anion is $−1$ . Since compound should be electrically neutral, the name of $NaHCO3$ compound is sodium bicarbonate

Conclusion

The name of $NaHCO3$ is sodium bicarbonate

(f)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(f)

Expert Solution

Answer to Problem 83E

The naming of $S4N4$ is tetra sulfur tetra nitride.

Explanation of Solution

To determine: The naming of $S4N4$

The name of $S4N4$ is tetra sulfur tetra nitride.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. The anions are named with the ending -ide. The prefix mono-, di-, tri- is used in prefix to indicate the number of atoms in each element.

Here sulfur ion $(S2−)$ is cation having four atoms and nitrogen ion $(N3-)$ is poly-atomic anion having four atoms. Hence the name of $S4N4$ compound is tetra sulfur tetra nitride.

Conclusion

The name of $S4N4$ is tetra sulfur tetra nitride.

(g)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(g)

Expert Solution

Answer to Problem 83E

The naming of $SF6$ is tetra sulfur tetra nitride.

Explanation of Solution

To determine: The naming of $SF6$

The name of $SF6$ is tetra sulfur tetra nitride.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. The anions are named with the ending -ide. The prefix mono-, di-, tri- is used in prefix to indicate the number of atoms in each element.

Here sulfur ion $(S2−)$ is cation having single atom and fluorine ion $(F-)$ is mono-atomic anion having six atoms. Hence the name of $SF6$ compound is sulfur hexafluoride.

Conclusion

The name of $SF6$ is tetra sulfur tetra nitride.

(h)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(h)

Expert Solution

Answer to Problem 83E

The naming of $NaOCl$ is sodium Hypochlorite.

Explanation of Solution

To determine: The naming of $NaOCl$

The name of $NaOCl$ is sodium Hypochlorite.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. When there are only two members in the same series, then the anion with the least number of oxygen’s ends in -ite, and the anion with the most ends in -ate.

If more than two oxyanion is present in compound then “hypo-” (less than) and “per-” (more than) is used in prefix.

Hence the name of $NaOCl$ compound is sodium Hypochlorite.

Conclusion

The name of $NaOCl$ is sodium Hypochlorite.

(i)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(i)

Expert Solution

Answer to Problem 83E

The naming of $BaCrO4$ is barium chromate.

Explanation of Solution

To determine: The naming of $BaCrO4$

The name of $BaCrO4$ is barium chromate.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. When there are only two members in the same series, then the anion with the least number of oxygen ends in -ite, and the anion with the most ends in -ate.

Here barium ion $(Ba2+)$ is cation and chromate ion $(CrO42-)$ is anion.

Hence the name of $BaCrO4$ is barium chromate.

Conclusion

The name of $BaCrO4$ is barium chromate.

(j)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(j)

Expert Solution

Answer to Problem 83E

The naming of $NH4NO3$ is ammonium nitrate.

Explanation of Solution

To determine: The naming of $NH4NO3$

The name of $NH4NO3$ is ammonium nitrate.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. When there are only two members in the same series, then the anion with the least number of oxygen ends in -ite, and the anion with the most ends in -ate.

Here ammonium $(NH4+)$ is cation and nitratee ion $(NO3-)$ is anion.

Hence the name of $NH4NO3$ is ammonium nitrate.

Conclusion

The name of $NH4NO3$ is ammonium nitrate.

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Answer 4

Question

Chapter 2, Problem 83E

(a)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(a)

Expert Solution

Answer to Problem 83E

The naming of $CuI$ is copper (I) iodide.

Explanation of Solution

To determine: The naming of $CuI$

The name of $CuI$ is copper(I) iodide.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. Monatomic anions are named with the ending -ide.

Roman numerals are used to show the oxidation state because some elements exhibit more than one oxidation state. Copper exhibits $+1 and +2$ oxidation state.

Here copper $(Cu2+)$ ion is cation and iodine $(I−)$ ion is mono-atomic anion. The oxidation state of copper is $+1$ because total charge on anion is $−1$ . Since compound should be electrically neutral, the name of $CuI$ compound is copper (I) iodide.

Conclusion

The name of $CuI$ is copper(I) iodide.

(b)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(b)

Expert Solution

Answer to Problem 83E

The naming of $CuI2$ is copper (II) iodide.

Explanation of Solution

To determine: The naming of $CuI2$

The name of $CuI2$ is copper(II) iodide.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. Monatomic anions are named with the ending -ide.

Roman numerals are used to show the oxidation state because some elements exhibit more than one oxidation state. Copper exhibits $+1 and +2$ oxidation state.

Here copper $(Cu2+)$ ion is cation and iodine $(I−)$ ion is mono-atomic anion. The oxidation state of copper is $+2$ because total charge on anion is $−2$ . Since compound should be electrically neutral, the name of $CuI2$ compound is copper (II) iodide.

Conclusion

The name of $CuI2$ is copper(II) iodide.

(c)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(c)

Expert Solution

Answer to Problem 83E

The naming of $CuI2$ is cobalt (II) iodide.

Explanation of Solution

To determine: The naming of $CoI2$

The name of $CuI2$ is cobalt (II) iodide.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. Monatomic anions are named with the ending -ide.

Roman numerals are used to show the oxidation state because some elements exhibit more than one oxidation state. Cobalt exhibits $+2 and +3$ oxidation state.

Here cobalt $(Cu2+)$ ion is cation and iodine $(I-)$ ion is mono-atomic anion. The oxidation state of copper is $+2$ because total charge on anion is $−2$ . Since compound should be electrically neutral, the name of $CoI2$ compound is cobalt (II) iodide.

Conclusion

The name of $CuI2$ is cobalt (II) iodide.

(d)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(d)

Expert Solution

Answer to Problem 83E

The naming of $Na2CO3$ is sodium carbonate

Explanation of Solution

To determine: The naming of $Na2CO3$

The name of $Na2CO3$ is sodium carbonate

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. The anions are named with the ending -ide.

The compound $Na2CO3$ is a molecular compound. While naming a molecular compound, the number of respective atoms present is denoted by using prefixes like mono, di tri.

Here sodium ion $(Na+)$ is cation and carbonate ion $(CO32-)$ is anion. The total charge on cation is $+2$ and total charge on anion is $−2$ . Since compound should be electrically neutral, the name of $Na2CO3$ compound is sodium carbonate

Conclusion

The name of $Na2CO3$ is sodium carbonate

(e)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(e)

Expert Solution

Answer to Problem 83E

The naming of $NaHCO3$ is sodium bicarbonate.

Explanation of Solution

To determine: The naming of $NaHCO3$

The name of $NaHCO3$ is sodium bicarbonate

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. The anions are named with the ending -ide.

The compound $NaHCO3$ is a molecular compound. While naming a molecular compound, the number of respective atoms present is denoted by using prefixes like mono, di tri.

Here sodium ion $(Na+)$ is cation and bicarbonate ion $(HCO3−)$ is anion. The total charge on cation is $+1$ and total charge on anion is $−1$ . Since compound should be electrically neutral, the name of $NaHCO3$ compound is sodium bicarbonate

Conclusion

The name of $NaHCO3$ is sodium bicarbonate

(f)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(f)

Expert Solution

Answer to Problem 83E

The naming of $S4N4$ is tetra sulfur tetra nitride.

Explanation of Solution

To determine: The naming of $S4N4$

The name of $S4N4$ is tetra sulfur tetra nitride.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. The anions are named with the ending -ide. The prefix mono-, di-, tri- is used in prefix to indicate the number of atoms in each element.

Here sulfur ion $(S2−)$ is cation having four atoms and nitrogen ion $(N3-)$ is poly-atomic anion having four atoms. Hence the name of $S4N4$ compound is tetra sulfur tetra nitride.

Conclusion

The name of $S4N4$ is tetra sulfur tetra nitride.

(g)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(g)

Expert Solution

Answer to Problem 83E

The naming of $SF6$ is tetra sulfur tetra nitride.

Explanation of Solution

To determine: The naming of $SF6$

The name of $SF6$ is tetra sulfur tetra nitride.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. The anions are named with the ending -ide. The prefix mono-, di-, tri- is used in prefix to indicate the number of atoms in each element.

Here sulfur ion $(S2−)$ is cation having single atom and fluorine ion $(F-)$ is mono-atomic anion having six atoms. Hence the name of $SF6$ compound is sulfur hexafluoride.

Conclusion

The name of $SF6$ is tetra sulfur tetra nitride.

(h)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(h)

Expert Solution

Answer to Problem 83E

The naming of $NaOCl$ is sodium Hypochlorite.

Explanation of Solution

To determine: The naming of $NaOCl$

The name of $NaOCl$ is sodium Hypochlorite.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. When there are only two members in the same series, then the anion with the least number of oxygen’s ends in -ite, and the anion with the most ends in -ate.

If more than two oxyanion is present in compound then “hypo-” (less than) and “per-” (more than) is used in prefix.

Hence the name of $NaOCl$ compound is sodium Hypochlorite.

Conclusion

The name of $NaOCl$ is sodium Hypochlorite.

(i)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(i)

Expert Solution

Answer to Problem 83E

The naming of $BaCrO4$ is barium chromate.

Explanation of Solution

To determine: The naming of $BaCrO4$

The name of $BaCrO4$ is barium chromate.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. When there are only two members in the same series, then the anion with the least number of oxygen ends in -ite, and the anion with the most ends in -ate.

Here barium ion $(Ba2+)$ is cation and chromate ion $(CrO42-)$ is anion.

Hence the name of $BaCrO4$ is barium chromate.

Conclusion

The name of $BaCrO4$ is barium chromate.

(j)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(j)

Expert Solution

Answer to Problem 83E

The naming of $NH4NO3$ is ammonium nitrate.

Explanation of Solution

To determine: The naming of $NH4NO3$

The name of $NH4NO3$ is ammonium nitrate.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. When there are only two members in the same series, then the anion with the least number of oxygen ends in -ite, and the anion with the most ends in -ate.

Here ammonium $(NH4+)$ is cation and nitratee ion $(NO3-)$ is anion.

Hence the name of $NH4NO3$ is ammonium nitrate.

Conclusion

The name of $NH4NO3$ is ammonium nitrate.

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Answer 5

Chapter 2, Problem 83E

(a)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(a)

Expert Solution

Answer to Problem 83E

The naming of $CuI$ is copper (I) iodide.

Explanation of Solution

To determine: The naming of $CuI$

The name of $CuI$ is copper(I) iodide.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. Monatomic anions are named with the ending -ide.

Roman numerals are used to show the oxidation state because some elements exhibit more than one oxidation state. Copper exhibits $+1 and +2$ oxidation state.

Here copper $(Cu2+)$ ion is cation and iodine $(I−)$ ion is mono-atomic anion. The oxidation state of copper is $+1$ because total charge on anion is $−1$ . Since compound should be electrically neutral, the name of $CuI$ compound is copper (I) iodide.

Conclusion

The name of $CuI$ is copper(I) iodide.

(b)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(b)

Expert Solution

Answer to Problem 83E

The naming of $CuI2$ is copper (II) iodide.

Explanation of Solution

To determine: The naming of $CuI2$

The name of $CuI2$ is copper(II) iodide.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. Monatomic anions are named with the ending -ide.

Roman numerals are used to show the oxidation state because some elements exhibit more than one oxidation state. Copper exhibits $+1 and +2$ oxidation state.

Here copper $(Cu2+)$ ion is cation and iodine $(I−)$ ion is mono-atomic anion. The oxidation state of copper is $+2$ because total charge on anion is $−2$ . Since compound should be electrically neutral, the name of $CuI2$ compound is copper (II) iodide.

Conclusion

The name of $CuI2$ is copper(II) iodide.

(c)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(c)

Expert Solution

Answer to Problem 83E

The naming of $CuI2$ is cobalt (II) iodide.

Explanation of Solution

To determine: The naming of $CoI2$

The name of $CuI2$ is cobalt (II) iodide.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. Monatomic anions are named with the ending -ide.

Roman numerals are used to show the oxidation state because some elements exhibit more than one oxidation state. Cobalt exhibits $+2 and +3$ oxidation state.

Here cobalt $(Cu2+)$ ion is cation and iodine $(I-)$ ion is mono-atomic anion. The oxidation state of copper is $+2$ because total charge on anion is $−2$ . Since compound should be electrically neutral, the name of $CoI2$ compound is cobalt (II) iodide.

Conclusion

The name of $CuI2$ is cobalt (II) iodide.

(d)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(d)

Expert Solution

Answer to Problem 83E

The naming of $Na2CO3$ is sodium carbonate

Explanation of Solution

To determine: The naming of $Na2CO3$

The name of $Na2CO3$ is sodium carbonate

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. The anions are named with the ending -ide.

The compound $Na2CO3$ is a molecular compound. While naming a molecular compound, the number of respective atoms present is denoted by using prefixes like mono, di tri.

Here sodium ion $(Na+)$ is cation and carbonate ion $(CO32-)$ is anion. The total charge on cation is $+2$ and total charge on anion is $−2$ . Since compound should be electrically neutral, the name of $Na2CO3$ compound is sodium carbonate

Conclusion

The name of $Na2CO3$ is sodium carbonate

(e)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(e)

Expert Solution

Answer to Problem 83E

The naming of $NaHCO3$ is sodium bicarbonate.

Explanation of Solution

To determine: The naming of $NaHCO3$

The name of $NaHCO3$ is sodium bicarbonate

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. The anions are named with the ending -ide.

The compound $NaHCO3$ is a molecular compound. While naming a molecular compound, the number of respective atoms present is denoted by using prefixes like mono, di tri.

Here sodium ion $(Na+)$ is cation and bicarbonate ion $(HCO3−)$ is anion. The total charge on cation is $+1$ and total charge on anion is $−1$ . Since compound should be electrically neutral, the name of $NaHCO3$ compound is sodium bicarbonate

Conclusion

The name of $NaHCO3$ is sodium bicarbonate

(f)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(f)

Expert Solution

Answer to Problem 83E

The naming of $S4N4$ is tetra sulfur tetra nitride.

Explanation of Solution

To determine: The naming of $S4N4$

The name of $S4N4$ is tetra sulfur tetra nitride.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. The anions are named with the ending -ide. The prefix mono-, di-, tri- is used in prefix to indicate the number of atoms in each element.

Here sulfur ion $(S2−)$ is cation having four atoms and nitrogen ion $(N3-)$ is poly-atomic anion having four atoms. Hence the name of $S4N4$ compound is tetra sulfur tetra nitride.

Conclusion

The name of $S4N4$ is tetra sulfur tetra nitride.

(g)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(g)

Expert Solution

Answer to Problem 83E

The naming of $SF6$ is tetra sulfur tetra nitride.

Explanation of Solution

To determine: The naming of $SF6$

The name of $SF6$ is tetra sulfur tetra nitride.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. The anions are named with the ending -ide. The prefix mono-, di-, tri- is used in prefix to indicate the number of atoms in each element.

Here sulfur ion $(S2−)$ is cation having single atom and fluorine ion $(F-)$ is mono-atomic anion having six atoms. Hence the name of $SF6$ compound is sulfur hexafluoride.

Conclusion

The name of $SF6$ is tetra sulfur tetra nitride.

(h)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(h)

Expert Solution

Answer to Problem 83E

The naming of $NaOCl$ is sodium Hypochlorite.

Explanation of Solution

To determine: The naming of $NaOCl$

The name of $NaOCl$ is sodium Hypochlorite.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. When there are only two members in the same series, then the anion with the least number of oxygen’s ends in -ite, and the anion with the most ends in -ate.

If more than two oxyanion is present in compound then “hypo-” (less than) and “per-” (more than) is used in prefix.

Hence the name of $NaOCl$ compound is sodium Hypochlorite.

Conclusion

The name of $NaOCl$ is sodium Hypochlorite.

(i)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(i)

Expert Solution

Answer to Problem 83E

The naming of $BaCrO4$ is barium chromate.

Explanation of Solution

To determine: The naming of $BaCrO4$

The name of $BaCrO4$ is barium chromate.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. When there are only two members in the same series, then the anion with the least number of oxygen ends in -ite, and the anion with the most ends in -ate.

Here barium ion $(Ba2+)$ is cation and chromate ion $(CrO42-)$ is anion.

Hence the name of $BaCrO4$ is barium chromate.

Conclusion

The name of $BaCrO4$ is barium chromate.

(j)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(j)

Expert Solution

Answer to Problem 83E

The naming of $NH4NO3$ is ammonium nitrate.

Explanation of Solution

To determine: The naming of $NH4NO3$

The name of $NH4NO3$ is ammonium nitrate.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. When there are only two members in the same series, then the anion with the least number of oxygen ends in -ite, and the anion with the most ends in -ate.

Here ammonium $(NH4+)$ is cation and nitratee ion $(NO3-)$ is anion.

Hence the name of $NH4NO3$ is ammonium nitrate.

Conclusion

The name of $NH4NO3$ is ammonium nitrate.

Answer 6

Chapter 2, Problem 83E

(a)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(a)

Expert Solution

Answer to Problem 83E

The naming of $CuI$ is copper (I) iodide.

Explanation of Solution

To determine: The naming of $CuI$

The name of $CuI$ is copper(I) iodide.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. Monatomic anions are named with the ending -ide.

Roman numerals are used to show the oxidation state because some elements exhibit more than one oxidation state. Copper exhibits $+1 and +2$ oxidation state.

Here copper $(Cu2+)$ ion is cation and iodine $(I−)$ ion is mono-atomic anion. The oxidation state of copper is $+1$ because total charge on anion is $−1$ . Since compound should be electrically neutral, the name of $CuI$ compound is copper (I) iodide.

Conclusion

The name of $CuI$ is copper(I) iodide.

Answer 7

Chapter 2, Problem 83E

(a)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

Answer 8

(a)

Expert Solution

Answer to Problem 83E

The naming of $CuI$ is copper (I) iodide.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

To determine: The naming of $CuI$

The name of $CuI$ is copper(I) iodide.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. Monatomic anions are named with the ending -ide.

Roman numerals are used to show the oxidation state because some elements exhibit more than one oxidation state. Copper exhibits $+1 and +2$ oxidation state.

Here copper $(Cu2+)$ ion is cation and iodine $(I−)$ ion is mono-atomic anion. The oxidation state of copper is $+1$ because total charge on anion is $−1$ . Since compound should be electrically neutral, the name of $CuI$ compound is copper (I) iodide.

Answer 13

Conclusion

The name of $CuI$ is copper(I) iodide.

Answer 14

(b)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(b)

Expert Solution

Answer to Problem 83E

The naming of $CuI2$ is copper (II) iodide.

Explanation of Solution

To determine: The naming of $CuI2$

The name of $CuI2$ is copper(II) iodide.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. Monatomic anions are named with the ending -ide.

Roman numerals are used to show the oxidation state because some elements exhibit more than one oxidation state. Copper exhibits $+1 and +2$ oxidation state.

Here copper $(Cu2+)$ ion is cation and iodine $(I−)$ ion is mono-atomic anion. The oxidation state of copper is $+2$ because total charge on anion is $−2$ . Since compound should be electrically neutral, the name of $CuI2$ compound is copper (II) iodide.

Conclusion

The name of $CuI2$ is copper(II) iodide.

Answer 15

(b)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

Answer 16

(b)

Expert Solution

Answer to Problem 83E

The naming of $CuI2$ is copper (II) iodide.

Answer 17

(b)

Expert Solution

Answer 18

(b)

Expert Solution

Answer 19

Expert Solution

Answer 20

Explanation of Solution

To determine: The naming of $CuI2$

The name of $CuI2$ is copper(II) iodide.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. Monatomic anions are named with the ending -ide.

Roman numerals are used to show the oxidation state because some elements exhibit more than one oxidation state. Copper exhibits $+1 and +2$ oxidation state.

Here copper $(Cu2+)$ ion is cation and iodine $(I−)$ ion is mono-atomic anion. The oxidation state of copper is $+2$ because total charge on anion is $−2$ . Since compound should be electrically neutral, the name of $CuI2$ compound is copper (II) iodide.

Answer 21

Conclusion

The name of $CuI2$ is copper(II) iodide.

Answer 22

(c)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(c)

Expert Solution

Answer to Problem 83E

The naming of $CuI2$ is cobalt (II) iodide.

Explanation of Solution

To determine: The naming of $CoI2$

The name of $CuI2$ is cobalt (II) iodide.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. Monatomic anions are named with the ending -ide.

Roman numerals are used to show the oxidation state because some elements exhibit more than one oxidation state. Cobalt exhibits $+2 and +3$ oxidation state.

Here cobalt $(Cu2+)$ ion is cation and iodine $(I-)$ ion is mono-atomic anion. The oxidation state of copper is $+2$ because total charge on anion is $−2$ . Since compound should be electrically neutral, the name of $CoI2$ compound is cobalt (II) iodide.

Conclusion

The name of $CuI2$ is cobalt (II) iodide.

Answer 23

(c)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

Answer 24

(c)

Expert Solution

Answer to Problem 83E

The naming of $CuI2$ is cobalt (II) iodide.

Answer 25

(c)

Expert Solution

Answer 26

(c)

Expert Solution

Answer 27

Expert Solution

Answer 28

Explanation of Solution

To determine: The naming of $CoI2$

The name of $CuI2$ is cobalt (II) iodide.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. Monatomic anions are named with the ending -ide.

Roman numerals are used to show the oxidation state because some elements exhibit more than one oxidation state. Cobalt exhibits $+2 and +3$ oxidation state.

Here cobalt $(Cu2+)$ ion is cation and iodine $(I-)$ ion is mono-atomic anion. The oxidation state of copper is $+2$ because total charge on anion is $−2$ . Since compound should be electrically neutral, the name of $CoI2$ compound is cobalt (II) iodide.

Answer 29

Conclusion

The name of $CuI2$ is cobalt (II) iodide.

Answer 30

(d)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(d)

Expert Solution

Answer to Problem 83E

The naming of $Na2CO3$ is sodium carbonate

Explanation of Solution

To determine: The naming of $Na2CO3$

The name of $Na2CO3$ is sodium carbonate

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. The anions are named with the ending -ide.

The compound $Na2CO3$ is a molecular compound. While naming a molecular compound, the number of respective atoms present is denoted by using prefixes like mono, di tri.

Here sodium ion $(Na+)$ is cation and carbonate ion $(CO32-)$ is anion. The total charge on cation is $+2$ and total charge on anion is $−2$ . Since compound should be electrically neutral, the name of $Na2CO3$ compound is sodium carbonate

Conclusion

The name of $Na2CO3$ is sodium carbonate

Answer 31

(d)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

Answer 32

(d)

Expert Solution

Answer to Problem 83E

The naming of $Na2CO3$ is sodium carbonate

Answer 33

(d)

Expert Solution

Answer 34

(d)

Expert Solution

Answer 35

Expert Solution

Answer 36

Explanation of Solution

To determine: The naming of $Na2CO3$

The name of $Na2CO3$ is sodium carbonate

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. The anions are named with the ending -ide.

The compound $Na2CO3$ is a molecular compound. While naming a molecular compound, the number of respective atoms present is denoted by using prefixes like mono, di tri.

Here sodium ion $(Na+)$ is cation and carbonate ion $(CO32-)$ is anion. The total charge on cation is $+2$ and total charge on anion is $−2$ . Since compound should be electrically neutral, the name of $Na2CO3$ compound is sodium carbonate

Answer 37

Conclusion

The name of $Na2CO3$ is sodium carbonate

Answer 38

(e)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(e)

Expert Solution

Answer to Problem 83E

The naming of $NaHCO3$ is sodium bicarbonate.

Explanation of Solution

To determine: The naming of $NaHCO3$

The name of $NaHCO3$ is sodium bicarbonate

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. The anions are named with the ending -ide.

The compound $NaHCO3$ is a molecular compound. While naming a molecular compound, the number of respective atoms present is denoted by using prefixes like mono, di tri.

Here sodium ion $(Na+)$ is cation and bicarbonate ion $(HCO3−)$ is anion. The total charge on cation is $+1$ and total charge on anion is $−1$ . Since compound should be electrically neutral, the name of $NaHCO3$ compound is sodium bicarbonate

Conclusion

The name of $NaHCO3$ is sodium bicarbonate

Answer 39

(e)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

Answer 40

(e)

Expert Solution

Answer to Problem 83E

The naming of $NaHCO3$ is sodium bicarbonate.

Answer 41

(e)

Expert Solution

Answer 42

(e)

Expert Solution

Answer 43

Expert Solution

Answer 44

Explanation of Solution

To determine: The naming of $NaHCO3$

The name of $NaHCO3$ is sodium bicarbonate

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. The anions are named with the ending -ide.

The compound $NaHCO3$ is a molecular compound. While naming a molecular compound, the number of respective atoms present is denoted by using prefixes like mono, di tri.

Here sodium ion $(Na+)$ is cation and bicarbonate ion $(HCO3−)$ is anion. The total charge on cation is $+1$ and total charge on anion is $−1$ . Since compound should be electrically neutral, the name of $NaHCO3$ compound is sodium bicarbonate

Answer 45

Conclusion

The name of $NaHCO3$ is sodium bicarbonate

Answer 46

(f)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(f)

Expert Solution

Answer to Problem 83E

The naming of $S4N4$ is tetra sulfur tetra nitride.

Explanation of Solution

To determine: The naming of $S4N4$

The name of $S4N4$ is tetra sulfur tetra nitride.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. The anions are named with the ending -ide. The prefix mono-, di-, tri- is used in prefix to indicate the number of atoms in each element.

Here sulfur ion $(S2−)$ is cation having four atoms and nitrogen ion $(N3-)$ is poly-atomic anion having four atoms. Hence the name of $S4N4$ compound is tetra sulfur tetra nitride.

Conclusion

The name of $S4N4$ is tetra sulfur tetra nitride.

Answer 47

(f)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

Answer 48

(f)

Expert Solution

Answer to Problem 83E

The naming of $S4N4$ is tetra sulfur tetra nitride.

Answer 49

(f)

Expert Solution

Answer 50

(f)

Expert Solution

Answer 51

Expert Solution

Answer 52

Explanation of Solution

To determine: The naming of $S4N4$

The name of $S4N4$ is tetra sulfur tetra nitride.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. The anions are named with the ending -ide. The prefix mono-, di-, tri- is used in prefix to indicate the number of atoms in each element.

Here sulfur ion $(S2−)$ is cation having four atoms and nitrogen ion $(N3-)$ is poly-atomic anion having four atoms. Hence the name of $S4N4$ compound is tetra sulfur tetra nitride.

Answer 53

Conclusion

The name of $S4N4$ is tetra sulfur tetra nitride.

Answer 54

(g)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(g)

Expert Solution

Answer to Problem 83E

The naming of $SF6$ is tetra sulfur tetra nitride.

Explanation of Solution

To determine: The naming of $SF6$

The name of $SF6$ is tetra sulfur tetra nitride.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. The anions are named with the ending -ide. The prefix mono-, di-, tri- is used in prefix to indicate the number of atoms in each element.

Here sulfur ion $(S2−)$ is cation having single atom and fluorine ion $(F-)$ is mono-atomic anion having six atoms. Hence the name of $SF6$ compound is sulfur hexafluoride.

Conclusion

The name of $SF6$ is tetra sulfur tetra nitride.

Answer 55

(g)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

Answer 56

(g)

Expert Solution

Answer to Problem 83E

The naming of $SF6$ is tetra sulfur tetra nitride.

Answer 57

(g)

Expert Solution

Answer 58

(g)

Expert Solution

Answer 59

Expert Solution

Answer 60

Explanation of Solution

To determine: The naming of $SF6$

The name of $SF6$ is tetra sulfur tetra nitride.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. The anions are named with the ending -ide. The prefix mono-, di-, tri- is used in prefix to indicate the number of atoms in each element.

Here sulfur ion $(S2−)$ is cation having single atom and fluorine ion $(F-)$ is mono-atomic anion having six atoms. Hence the name of $SF6$ compound is sulfur hexafluoride.

Answer 61

Conclusion

The name of $SF6$ is tetra sulfur tetra nitride.

Answer 62

(h)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(h)

Expert Solution

Answer to Problem 83E

The naming of $NaOCl$ is sodium Hypochlorite.

Explanation of Solution

To determine: The naming of $NaOCl$

The name of $NaOCl$ is sodium Hypochlorite.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. When there are only two members in the same series, then the anion with the least number of oxygen’s ends in -ite, and the anion with the most ends in -ate.

If more than two oxyanion is present in compound then “hypo-” (less than) and “per-” (more than) is used in prefix.

Hence the name of $NaOCl$ compound is sodium Hypochlorite.

Conclusion

The name of $NaOCl$ is sodium Hypochlorite.

Answer 63

(h)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

Answer 64

(h)

Expert Solution

Answer to Problem 83E

The naming of $NaOCl$ is sodium Hypochlorite.

Answer 65

(h)

Expert Solution

Answer 66

(h)

Expert Solution

Answer 67

Expert Solution

Answer 68

Explanation of Solution

To determine: The naming of $NaOCl$

The name of $NaOCl$ is sodium Hypochlorite.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. When there are only two members in the same series, then the anion with the least number of oxygen’s ends in -ite, and the anion with the most ends in -ate.

If more than two oxyanion is present in compound then “hypo-” (less than) and “per-” (more than) is used in prefix.

Hence the name of $NaOCl$ compound is sodium Hypochlorite.

Answer 69

Conclusion

The name of $NaOCl$ is sodium Hypochlorite.

Answer 70

(i)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(i)

Expert Solution

Answer to Problem 83E

The naming of $BaCrO4$ is barium chromate.

Explanation of Solution

To determine: The naming of $BaCrO4$

The name of $BaCrO4$ is barium chromate.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. When there are only two members in the same series, then the anion with the least number of oxygen ends in -ite, and the anion with the most ends in -ate.

Here barium ion $(Ba2+)$ is cation and chromate ion $(CrO42-)$ is anion.

Hence the name of $BaCrO4$ is barium chromate.

Conclusion

The name of $BaCrO4$ is barium chromate.

Answer 71

(i)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

Answer 72

(i)

Expert Solution

Answer to Problem 83E

The naming of $BaCrO4$ is barium chromate.

Answer 73

(i)

Expert Solution

Answer 74

(i)

Expert Solution

Answer 75

Expert Solution

Answer 76

Explanation of Solution

To determine: The naming of $BaCrO4$

The name of $BaCrO4$ is barium chromate.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. When there are only two members in the same series, then the anion with the least number of oxygen ends in -ite, and the anion with the most ends in -ate.

Here barium ion $(Ba2+)$ is cation and chromate ion $(CrO42-)$ is anion.

Hence the name of $BaCrO4$ is barium chromate.

Answer 77

Conclusion

The name of $BaCrO4$ is barium chromate.

Answer 78

(j)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

(j)

Expert Solution

Answer to Problem 83E

The naming of $NH4NO3$ is ammonium nitrate.

Explanation of Solution

To determine: The naming of $NH4NO3$

The name of $NH4NO3$ is ammonium nitrate.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. When there are only two members in the same series, then the anion with the least number of oxygen ends in -ite, and the anion with the most ends in -ate.

Here ammonium $(NH4+)$ is cation and nitratee ion $(NO3-)$ is anion.

Hence the name of $NH4NO3$ is ammonium nitrate.

Conclusion

The name of $NH4NO3$ is ammonium nitrate.

Answer 79

(j)

Interpretation Introduction

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

Answer 80

(j)

Expert Solution

Answer to Problem 83E

The naming of $NH4NO3$ is ammonium nitrate.

Answer 81

(j)

Expert Solution

Answer 82

(j)

Expert Solution

Answer 83

Expert Solution

Answer 84

Explanation of Solution

To determine: The naming of $NH4NO3$

The name of $NH4NO3$ is ammonium nitrate.

In naming of covalent compound, the name of cation comes first which is the name of element. Anion is named second which is also the name of element. When there are only two members in the same series, then the anion with the least number of oxygen ends in -ite, and the anion with the most ends in -ate.

Here ammonium $(NH4+)$ is cation and nitratee ion $(NO3-)$ is anion.

Hence the name of $NH4NO3$ is ammonium nitrate.

Answer 85

Conclusion

The name of $NH4NO3$ is ammonium nitrate.

Answer 86

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The name of the compounds is to be given. Concept introduction: While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

Interpretation: The name of the compounds is to be given. Concept introduction: While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

Answer to Problem 83E

Explanation of Solution

Interpretation: The name of the compounds is to be given. Concept introduction: While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

Answer to Problem 83E

Explanation of Solution

Interpretation: The name of the compounds is to be given. Concept introduction: While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

Answer to Problem 83E

Explanation of Solution

Interpretation: The name of the compounds is to be given. Concept introduction: While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

Answer to Problem 83E

Explanation of Solution

Interpretation: The name of the compounds is to be given. Concept introduction: While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

Answer to Problem 83E

Explanation of Solution

Interpretation: The name of the compounds is to be given. Concept introduction: While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

Answer to Problem 83E

Explanation of Solution

Interpretation: The name of the compounds is to be given. Concept introduction: While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

Answer to Problem 83E

Explanation of Solution

Interpretation: The name of the compounds is to be given. Concept introduction: While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

Answer to Problem 83E

Explanation of Solution

Interpretation: The name of the compounds is to be given. Concept introduction: While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

Answer to Problem 83E

Explanation of Solution

Interpretation: The name of the compounds is to be given. Concept introduction: While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

Answer to Problem 83E

Explanation of Solution

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Chapter 2 Solutions

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.

Interpretation:

The name of the compounds is to be given.

Concept introduction:

While naming covalent compound, the name of the cation is written first, followed by the name of the anion. Generally, the cation present is a metal or a polyatomic cation, whereas the anion is a non-metal.