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Concept explainers
(a)
Interpretation:
The chemical formula for the following binary ionic compound–Cobalt(II) sulphide is to be written.
Concept Introduction:
An ionic compound has two elements one is metal and another one is non-metal. The metal ion always has positive charge and the nonmetal ion always has negative ion in binary compounds.
Example:
The following rule can be used for the naming of binary ionic compounds.
The full name of the metallic element is given first, followed by a separate word containing the stem of the metallic element name and the suffix –ide.
The positive charge on metal ions from IA, IIA and IIIA is equal to the group number, while the negative charge on non-metal ions from VA, VIA and VIIA is equal to the group number eight.
(b)
Interpretation:
The chemical formula for the following binary ionic compound–Cobalt(III) sulphide is to be written.
Concept Introduction:
An ionic compound has two elements one is metal and another one is non-metal. The metal ion always has positive charge and the nonmetal ion always has negative ion in binary compounds.
Example:
The following rule can be used for the naming of binary ionic compounds.
The full name of the metallic element is given first, followed by a separate word containing the stem of the metallic element name and the suffix –ide.
The positive charge on metal ions from IA, IIA and IIIA is equal to the group number, while the negative charge on non-metal ions from VA, VIA and VIIA is equal to the group number eight.
(c)
Interpretation:
The chemical formula for the following binary ionic compound–Tin(IV) iodide is to be written.
Concept Introduction:
An ionic compound has two elements one is metal and another one is non-metal. The metal ion always has positive charge and the nonmetal ion always has negative ion in binary compounds.
Example:
The following rule can be used for the naming of binary ionic compounds.
The full name of the metallic element is given first, followed by a separate word containing the stem of the metallic element name and the suffix –ide.
The positive charge on metal ions from IA, IIA and IIIA is equal to the group number, while the negative charge on non-metal ions from VA, VIA and VIIA is equal to the group number eight.
(d)
Interpretation:
The chemical formula for the following binary ionic compound –Lead(II) nitride is to be written.
Concept Introduction:
An ionic compound has two elements one is metal and another one is non-metal. The metal ion always has positive charge and the nonmetal ion always has negative ion in binary compounds.
Example:
The following rule can be used for the naming of binary ionic compounds.
The full name of the metallic element is given first, followed by a separate word containing the stem of the metallic element name and the suffix –ide.
The positive charge on metal ions from IA, IIA and IIIA is equal to the group number, while the negative charge on non-metal ions from VA, VIA and VIIA is equal to the group number eight.
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Chapter 4 Solutions
General, Organic, and Biological Chemistry
- Q1: For each molecule, assign each stereocenter as R or S. Circle the meso compounds. Label each compound as chiral or achiral. OH HO CI Br H CI CI Br CI CI Xf x f g Br D OH Br Br H₂N R. IN Ill I -N S OMe D II H CO₂H 1/111 DuckDuckGarrow_forwardThese are synthesis questions. You need to show how the starting material can be converted into the product(s) shown. You may use any reactions we have learned. Show all the reagents you need. Show each molecule synthesized along the way and be sure to pay attention to the regiochemistry and stereochemistry preferences for each reaction. If a racemic molecule is made along the way, you need to draw both enantiomers and label the mixture as "racemic". All of the carbon atoms of the products must come from the starting material! ? H Harrow_forwardQ5: Draw every stereoisomer for 1-bromo-2-chloro-1,2-difluorocyclopentane. Clearly show stereochemistry by drawing the wedge-and-dashed bonds. Describe the relationship between each pair of the stereoisomers you have drawn.arrow_forward
- Classify each pair of molecules according to whether or not they can participate in hydrogen bonding with one another. Participate in hydrogen bonding CH3COCH3 and CH3COCH2CH3 H2O and (CH3CH2)2CO CH3COCH3 and CH₂ CHO Answer Bank Do not participate in hydrogen bonding CH3CH2OH and HCHO CH3COCH2CH3 and CH3OHarrow_forwardNonearrow_forwardQ4: Comparing (3S,4S)-3,4-dimethylhexane and (3R,4S)-3,4-dimethylhexane, which one is optically active? Briefly explain.arrow_forward
- Nonearrow_forwardNonearrow_forwardGiven the standard enthalpies of formation for the following substances, determine the reaction enthalpy for the following reaction. 4A (g) + 2B (g) → 2C (g) + 7D (g) AHrxn =?kJ Substance AH in kJ/mol A (g) - 20.42 B (g) + 32.18 C (g) - 72.51 D (g) - 17.87arrow_forward
- Determine ASran for Zn(s) + 2HCl(aq) = ZnCl2(aq) + H2(aq) given the following information: Standard Entropy Values of Various Substance Substance So (J/mol • K) 60.9 Zn(s) HCl(aq) 56.5 130.58 H2(g) Zn2+(aq) -106.5 55.10 CI (aq)arrow_forward3) Catalytic hydrogenation of the compound below produced the expected product. However, a byproduct with molecular formula C10H12O is also formed in small quantities. What is the by product?arrow_forwardWhat is the ΔHorxn of the reaction? NaOH(aq) + HCl(aq) → H2O(l) + NaCl(aq) ΔHorxn 1= ________ kJ/molarrow_forward
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