
(a)
Interpretation:
The Latin system name for ionic compound
Concept introduction:
The Latin system is an older method for naming metal cations which have two common ionic charges. The Latin system is also known as the suffix system because this system uses the Latin name of the metal and applies the suffix -ous or -ic. The lower ionic charge applies an -ous suffix and the higher ionic charge applies an -ic suffix.

Answer to Problem 46E
The Latin system name for ionic compound
Explanation of Solution
The Latin name for the copper is cuprum. In ionic compound
Cuprous sulfide is the Latin name of
(b)
Interpretation:
The Latin system name for ionic compound
Concept introduction:
The Latin system is an older method for naming metal cations which have two common ionic charges. The Latin system is also known as the suffix system because this system uses the Latin name of the metal and applies the suffix -ous or -ic. The lower ionic charge applies an -ous suffix and the higher ionic charge applies an -ic suffix.

Answer to Problem 46E
The Latin system name for ionic compound
Explanation of Solution
The Latin name of iron is ferrum. In the ionic compound
Ferric oxide is the Latin name of
(c)
Interpretation:
The Latin system name for ionic compound
Concept introduction:
The Latin system is an older method for naming metal cations which have two common ionic charges. The Latin system is also known as the suffix system because this system uses the Latin name of the metal and applies the suffix -ous or -ic. The lower ionic charge applies an -ous suffix and the higher ionic charge applies an -ic suffix.

Answer to Problem 46E
The Latin system name for ionic compound
Explanation of Solution
The Latin name of mercury is hydrargyrum or mercurum. In the ionic compound
Mercurous nitrite is the Latin name of
(d)
Interpretation:
The Latin system name for ionic compound
Concept introduction:
The Latin system is an older method for naming metal cations which have two common ionic charges. The Latin system is also known as the suffix system because this system uses the Latin name of the metal and applies the suffix -ous or -ic. The lower ionic charge applies an -ous suffix and the higher ionic charge applies an -ic suffix.

Answer to Problem 46E
The Latin system name for ionic compound
Explanation of Solution
The Latin name of the lead is plumbum. In the ionic compound
Plumbic acetate is the Latin name of
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Chapter 6 Solutions
EP INTRODUCTORY CHEM.-MOD.MASTERINGCHEM
- Provide an IUPAC name for each of the compounds shown. (Specify (E)/(Z) stereochemistry, if relevant, for straight chain alkenes only. Pay attention to commas, dashes, etc.) H₁₂C C(CH3)3 C=C H3C CH3 CH3CH2CH CI CH3 Submit Answer Retry Entire Group 2 more group attempts remaining Previous Nextarrow_forwardArrange the following compounds / ions in increasing nucleophilicity (least to most nucleophilic) CH3NH2 CH3C=C: CH3COO 1 2 3 5 Multiple Choice 1 point 1, 2, 3 2, 1, 3 3, 1, 2 2, 3, 1 The other answers are not correct 0000arrow_forwardcurved arrows are used to illustrate the flow of electrons. using the provided starting and product structures, draw the cured electron-pushing arrows for thw following reaction or mechanistic steps. be sure to account for all bond-breaking and bond making stepsarrow_forward
- Using the graphs could you help me explain the answers. I assumed that both graphs are proportional to the inverse of time, I think. Could you please help me.arrow_forwardSynthesis of Dibenzalacetone [References] Draw structures for the carbonyl electrophile and enolate nucleophile that react to give the enone below. Question 1 1 pt Question 2 1 pt Question 3 1 pt H Question 4 1 pt Question 5 1 pt Question 6 1 pt Question 7 1pt Question 8 1 pt Progress: 7/8 items Que Feb 24 at You do not have to consider stereochemistry. . Draw the enolate ion in its carbanion form. • Draw one structure per sketcher. Add additional sketchers using the drop-down menu in the bottom right corner. ⚫ Separate multiple reactants using the + sign from the drop-down menu. ? 4arrow_forwardShown below is the mechanism presented for the formation of biasplatin in reference 1 from the Background and Experiment document. The amounts used of each reactant are shown. Either draw or describe a better alternative to this mechanism. (Note that the first step represents two steps combined and the proton loss is not even shown; fixing these is not the desired improvement.) (Hints: The first step is correct, the second step is not; and the amount of the anhydride is in large excess to serve a purpose.)arrow_forward
- Hi I need help on the question provided in the image.arrow_forwardDraw a reasonable mechanism for the following reaction:arrow_forwardDraw the mechanism for the following reaction: CH3 CH3 Et-OH Et Edit the reaction by drawing all steps in the appropriate boxes and connecting them with reaction arrows. Add charges where needed. Electron-flow arrows should start on the electron(s) of an atom or a bond and should end on an atom, bond, or location where a new bond should be created. H± EXP. L CONT. י Α [1] осн CH3 а CH3 :Ö Et H 0 N о S 0 Br Et-ÖH | P LL Farrow_forward
- 20.00 mL of 0.150 M NaOH is titrated with 37.75 mL of HCl. What is the molarity of the HCl?arrow_forward20.00 mL of 0.025 M HCl is titrated with 0.035 M KOH. What volume of KOH is needed?arrow_forward20.00 mL of 0.150 M NaOH is titrated with 37.75 mL of HCl. What is the molarity of the HCl?arrow_forward
- Chemistry & Chemical ReactivityChemistryISBN:9781133949640Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage Learning
