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(a)
Interpretation:
The cations & anions derived from Barium & Bromine should be combined in order to get an electrically neutralized ionic compound.
Concept introduction:
An ionic compound forms by the mutual attraction of 2 oppositely charged ions called cation & anion. Cationis derived from Barium (Ba), while anion is derived from Bromine (Br).
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Answer to Problem 48P
BaBr2
Explanation of Solution
Cations are derived from metals. A cation is formed by removing 1 or few electrons from a metal atom. As electrons are negatively charged, the removal of electrons (negative charges) gives a positive charge to the rest of the atom.
Oppositely anions are derived from non-metal elements. An anion is formed by gaining 1 or few electrons. As electrons are negatively charged, the gaining of electrons (negative charges) gives a negative charge to the rest of the atom.
So, in the given two elements metal is Barium (Ba). So cation is derived from Ba. The only cation derived from Ba is Ba2+. The cation is derived by removing 2 electrons from Ba.
Ba → Ba2+ + 2e (electrons are indicated as e)
Bromine is the non-metal which is forming anion, Br- / Bromide, by gaining 1 electron to a Br atom.
Br + e → Br-
From this cation & anion an ionic compound is formed. Usually a simple ionic compound is electrically neutral. So cation & anion combined in a way to neutralize opposite charges.
As Ba2+ is having +2 positive charge, 2 Br- ions should becombined.
So the ionic compound form from Barium & Bromine is BaBr2.
(b)
Interpretation:
The cations & anions derived from Aluminum&Sulfur should be combined in order to get an electrically neutralized ionic compound.
Concept introduction:
An ionic compound forms by the mutual attraction of 2 oppositely charged ions called cation & anion. Cation is derived from Aluminum (Al), while anion is derived from Sulfur(S).
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Answer to Problem 48P
Al2S3
Explanation of Solution
Cations are derived from metals. A cation is formed by removing 1 or few electrons from a metal atom. As electrons are negatively charged, the removal of electrons (negative charges) gives a positive charge to the rest of the atom.
Oppositely anions are derived from non-metal elements. An anion is formed by gaining 1 or few electrons. As electrons are negatively charged, the gaining of electrons (negative charges) gives a negative charge to the rest of the atom.
In the given two elements metal is Aluminum (Al). So cation is derived from Al. The only cation derived from Al is Al3+. The cation is derived by removing 3 electrons from Al.
Al → Al3++ 3e (electrons are indicated as e)
Sulfur is the non-metal which is forming anion, S2-/Sulfide, by gaining 2 electrons.
S + 2e → S2-
From this cation & anion an ionic compound is formed. Usually a simple ionic compound is electrically neutral. So cation & anion combined in a way to neutralize opposite charges.
Al3+ has a +3 charge. S2-/ Sulfide ion has a -2 charge. So S2-: Al3+ should be combined in a ratio of 3: 2, to maintain the charge neutrality. This is the first ratio that the both values are integers.
The above ratio is proven correct by the following simple calculation to balance charges.
(Equals zero because the ionic compound should have no net charge)
So in the ionic compound form from Aluminum & Sulfur there should be 2 Al3+s & 3 S2- s. So the compound is Al2S3
(c)
Interpretation:
The cations & anions derived from Manganese&Chlorine should be combined in order to get an electrically neutralized ionic compound.
Concept introduction:
An ionic compound forms by the mutual attraction of 2 oppositely charged ions called cation & anion. Cation is derived from Manganese (Mn), while anion is derived from Chlorine (Cl).
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Answer to Problem 48P
MnCl2
Explanation of Solution
Cations are derived from metals. A cation is formed by removing 1 or few electrons from a metal atom. As electrons are negatively charged, the removal of electrons (negative charges) gives a positive charge to the rest of the atom.
Oppositely anions are derived from non-metal elements. An anion is formed by gaining 1 or few electrons. As electrons are negatively charged, the gaining of electrons (negative charges) gives a negative charge to the rest of the atom.
So in the given two elements metal is Manganese (Mn). So cation is derived from Mn. Manganese has the ability to form several cations like Mn2+, Mn3+&Mn4+. Out of them the most stable cation is Mn2+. This cation is derived by removing 2 electrons from Mn.
Mn → Mn2+ + 2e (electrons are indicated as e)
Chlorine is the non-metal which is forming anion, Cl-/Chloride, by gaining 1 electron to a Cl atom.
Cl + e → Cl-
From this cation & anion an ionic compound is formed. Usually a simple ionic compound is electrically neutral. So cation & anion combined in a way to neutralize opposite charges.
As Mn2+ is having +2 positive charge, 2 Cl- ions should becombined.
So, the ionic compound form from Manganese&Chlorine is MnCl2.
(d)
Interpretation:
The cations & anions derived from Zinc&Sulfur should be combined in order to get an electrically neutralized ionic compound.
Concept introduction:
An ionic compound forms by the mutual attraction of 2 oppositely charged ions called cation & anion. Cation is derived from Zinc (Zn), while anion is derived from Sulfur (S).
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Answer to Problem 48P
ZnS
Explanation of Solution
Cations are derived from metals. A cation is formed by removing 1 or few electrons from a metal atom. As electrons are negatively charged, the removal of electrons (negative charges) gives a positive charge to the rest of the atom.
Oppositely anions are derived from non-metal elements. An anion is formed by gaining 1 or few electrons. As electrons are negatively charged, the gaining of electrons (negative charges) gives a negative charge to the rest of the atom.
So in the given two elements metal is Zinc (Zn). So cation is derived from Zn. The most common & stable cation derived from Zn is Zn2+. The cation is derived by removing 2 electrons from Zn.
Zn → Zn2+ + 2e (electrons are indicated as e)
Sulfur is the non-metal which is forming anion, S2- / Sulfide, by gaining 2 electrons.
S + 2e → S2-
From this cation & anion an ionic compound is formed. Usually a simple ionic compound is electrically neutral. So cation & anion combined in a way to neutralize opposite charges.
As both Zn2+& S2-is having +2 & -2 charge, one cation & one anion are combined to form the Compound ZnS.
So, the ionic compound form from Zinc&Sulfur is ZnS.
(e)
Interpretation:
The cations & anions derived from Magnesium&Fluorine should be combined in order to get an electrically neutralized ionic compound.
Concept introduction:
An ionic compound forms by the mutual attraction of 2 oppositely charged ions called cation & anion. Cation is derived from Magnesium (Mg), while anion is derived from Fluorine (F).
![Check Mark](/static/check-mark.png)
Answer to Problem 48P
MgF2
Explanation of Solution
Cations are derived from metals. A cation is formed by removing 1 or few electrons from a metal atom. As electrons are negatively charged, the removal of electrons (negative charges) gives a positive charge to the rest of the atom.
Oppositely anions are derived from non-metal elements. An anion is formed by gaining 1 or few electrons. As electrons are negatively charged, the gaining of electrons (negative charges) gives a negative charge to the rest of the atom.
So in the given two elements metal is Magnesium (Mg). So cation is derived from Mg. The only cation derived from Mg is Mg2+. The cation is derived by removing 2 electrons from Mg.
Mg → Mg2+ + 2e (electrons are indicated as e)
Fluorine is the non-metal, which is forming anion, F- / Fluoride, by gaining 1 electron to a F atom.
F + e → F-
From this cation & anion an ionic compound is formed. Usually a simple ionic compound is electrically neutral. So cation & anion combined in a way to neutralize opposite charges.
As Mg2+ is having +2 positive charge, 2 F- ions should becombined.
So the ionic compound form from Magnesium&Fluorine is MgF2.
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Chapter 3 Solutions
General, Organic, and Biological Chemistry - 4th edition
- Use the References to access important values if needed for this question. What is the IUPAC name of each of the the following? 0 CH3CHCNH₂ CH3 CH3CHCNHCH2CH3 CH3arrow_forwardYou have now performed a liquid-liquid extraction protocol in Experiment 4. In doing so, you manipulated and exploited the acid-base chemistry of one or more of the compounds in your mixture to facilitate their separation into different phases. The key to understanding how liquid- liquid extractions work is by knowing which layer a compound is in, and in what protonation state. The following liquid-liquid extraction is different from the one you performed in Experiment 4, but it uses the same type of logic. Your task is to show how to separate apart Compound A and Compound B. . Complete the following flowchart of a liquid-liquid extraction. Handwritten work is encouraged. • Draw by hand (neatly) only the appropriate organic compound(s) in the boxes. . Specify the reagent(s)/chemicals (name is fine) and concentration as required in Boxes 4 and 5. • Box 7a requires the solvent (name is fine). • Box 7b requires one inorganic compound. • You can neatly complete this assignment by hand and…arrow_forwardb) Elucidate compound D w) mt at 170 nd shows c-1 stretch at 550cm;' The compound has the ff electronic transitions: 0%o* and no a* 1H NMR Spectrum (CDCl3, 400 MHz) 3.5 3.0 2.5 2.0 1.5 1.0 0.5 ppm 13C{H} NMR Spectrum (CDCl3, 100 MHz) Solvent 80 70 60 50 40 30 20 10 0 ppm ppm ¹H-13C me-HSQC Spectrum ppm (CDCl3, 400 MHz) 5 ¹H-¹H COSY Spectrum (CDCl3, 400 MHz) 0.5 10 3.5 3.0 2.5 2.0 1.5 1.0 10 15 20 20 25 30 30 -35 -1.0 1.5 -2.0 -2.5 3.0 -3.5 0.5 ppm 3.5 3.0 2.5 2.0 1.5 1.0 0.5 ppmarrow_forward
- Part I. a) Elucidate the structure of compound A using the following information. • mass spectrum: m+ = 102, m/2=57 312=29 • IR spectrum: 1002.5 % TRANSMITTANCE Ngg 50 40 30 20 90 80 70 60 MICRONS 5 8 9 10 12 13 14 15 16 19 1740 cm M 10 0 4000 3600 3200 2800 2400 2000 1800 1600 13 • CNMR 'H -NMR Peak 8 ppm (H) Integration multiplicity a 1.5 (3H) triplet b 1.3 1.5 (3H) triplet C 2.3 1 (2H) quartet d 4.1 1 (2H) quartet & ppm (c) 10 15 28 60 177 (C=0) b) Elucidate the structure of compound B using the following information 13C/DEPT NMR 150.9 MHz IIL 1400 WAVENUMBERS (CM-1) DEPT-90 DEPT-135 85 80 75 70 65 60 55 50 45 40 35 30 25 20 ppm 1200 1000 800 600 400arrow_forward• Part II. a) Elucidate The structure of compound c w/ molecular formula C10 11202 and the following data below: • IR spectra % TRANSMITTANCE 1002.5 90 80 70 60 50 40 30 20 10 0 4000 3600 3200 2800 2400 2000 1800 1600 • Information from 'HAMR MICRONS 8 9 10 11 14 15 16 19 25 1400 WAVENUMBERS (CM-1) 1200 1000 800 600 400 peak 8 ppm Integration multiplicity a 2.1 1.5 (3H) Singlet b 3.6 1 (2H) singlet с 3.8 1.5 (3H) Singlet d 6.8 1(2H) doublet 7.1 1(2H) doublet Information from 13C-nmR Normal carbon 29ppm Dept 135 Dept -90 + NO peak NO peak 50 ppm 55 ppm + NO peak 114 ppm t 126 ppm No peak NO peak 130 ppm t + 159 ppm No peak NO peak 207 ppm по реак NO peakarrow_forwardCould you redraw these and also explain how to solve them for me pleasarrow_forward
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