(a)
Interpretation: The formula for simple binary ionic compound calcium chloride is to be named.
Concept introduction: Binary ionic compounds contain a metal that can form more than one type of cation.
(a)
Answer to Problem 49A
Explanation of Solution
The given compound is calcium chloride. Here, Ca has +2 charge and Cl has -1 charge. Thus, Ca and Cl must be combined in 1:2 ratio to form CaCl2.
Name | Formula | Comments |
Calcium chloride | Contains two |
(b)
Interpretation: The formula for simple binary ionic compound silver (I) oxide is to be named.
Concept introduction: Binary ionic compounds contain a metal that can form more than one type of cation.
(b)
Answer to Problem 49A
Explanation of Solution
Given ionic compound is silver (I) oxide. Here, silver has +1 charge and O has -2 charge. Thus, Ag and O must be combined in 2:1 ratio.
Name | Formula |
Silver(I) oxide |
(c)
Interpretation: The formula for simple binary ionic compound aluminium sulfide is to be named.
Concept introduction: Binary ionic compounds contain a metal that can form more than one type of cation.
(c)
Answer to Problem 49A
Explanation of Solution
The given compound is aluminium sulfide. Here, Al has +3 charge and sulfur has -2 charge thus, Al and S must be combined in 2:3 ratio.
Name | Formula |
Aluminum sulfide |
(d)
Interpretation: The formula for simple binary ionic compound beryllium bromide is to be named.
Concept introduction: Binary ionic compounds contain a metal that can form more than one type of cation.
(d)
Answer to Problem 49A
Explanation of Solution
The given compound is beryllium bromide. The charge on Be is 2+ and that on Br is -1 thus, Be and Br must be combined in 1:2 ratio.
Name | Formula |
Beryllium bromide |
(e)
Interpretation: The formula for simple binary ionic compounds is to be named.
Concept introduction: Binary ionic compounds contain a metal that can form more than one type of cation.
(e)
Answer to Problem 49A
Explanation of Solution
The given compound is hydrogen sulfide or hydrosulfuric acid. Here, H has +1 charge and S has -2 charge thus, H and S must be combined in 2:1 ratio.
Name | Formula | Comments |
Hydro sulfuric acid | 2 |
(f)
Interpretation: The formula for simple binary ionic compound potassium hydride is to be named.
Concept introduction: Binary ionic compounds contain a metal that can form more than one type of cation.
(f)
Answer to Problem 49A
KH.
Explanation of Solution
The given compound is potassium hydride. The charge on K is +1 and that on H is -1 thus, they ,must be combined in 1:1 ratio.
Name | Formula |
Potassium hydride | KH |
(g)
Interpretation: The formula for simple binary ionic compound magnesium iodide is to be named.
Concept introduction: Binary ionic compounds contain a metal that can form more than one type of cation.
(g)
Answer to Problem 49A
Explanation of Solution
The given compound is magnesium iodide. Here, Mg has +2 charge and I has -1 charge. In the formula, Mg and I must be combined in 1:2 ratio.
Name | Formula |
Magnesium iodide |
(h)
Interpretation: The formula for simple binary ionic compound cesium fluoride is to be named.
Concept introduction: Binary ionic compounds contain a metal that can form more than one type of cation.
(h)
Answer to Problem 49A
Explanation of Solution
The given compound is cesium fluoride. Here, Cs has +1 charge and F has -1 charge thus, Cs and F must be combined in 1:1 ratio.
Name | Formula |
Cesium fluoride |
Chapter 4 Solutions
World of Chemistry, 3rd edition
- 6. Consider the following exothermic reaction below. 2Cu2+(aq) +41 (aq)2Cul(s) + 12(aq) a. If Cul is added, there will be a shift left/shift right/no shift (circle one). b. If Cu2+ is added, there will be a shift left/shift right/no shift (circle one). c. If a solution of AgNO3 is added, there will be a shift left/shift right/no shift (circle one). d. If the solvent hexane (C6H14) is added, there will be a shift left/shift right/no shift (circle one). Hint: one of the reaction species is more soluble in hexane than in water. e. If the reaction is cooled, there will be a shift left/shift right/no shift (circle one). f. Which of the changes above will change the equilibrium constant, K?arrow_forwardShow work. don't give Aiarrow_forwardShow work with explanation needed. don't give Ai generated solutionarrow_forward
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- 3. Provide the missing compounds or reagents. 1. H,NNH КОН 4 EN MN. 1. HBUCK = 8 хно Panely prowseful kanti-chuprccant fad, winddively, can lead to the crading of deduc din-willed, tica, The that chemooices in redimi Грин. " like (for alongan Ridovi MN نيا . 2. Cl -BuO 1. NUH 2.A A -BuOK THE CF,00,H Ex 5)arrow_forward2. Write a complete mechanism for the reaction shown below. NaOCH LOCH₁ O₂N NO2 CH₂OH, 20 °C O₂N NO2arrow_forward4. Propose a synthesis of the target molecules from the respective starting materials. a) b) LUCH C Br OHarrow_forward
- The following mechanism for the gas phase reaction of H2 and ICI that is consistent with the observed rate law is: step 1 step 2 slow: H2(g) +ICI(g) → HCl(g) + HI(g) fast: ICI(g) + HI(g) → HCl(g) + |2(g) (1) What is the equation for the overall reaction? Use the smallest integer coefficients possible. If a box is not needed, leave it blank. + → + (2) Which species acts as a catalyst? Enter formula. If none, leave box blank: (3) Which species acts as a reaction intermediate? Enter formula. If none, leave box blank: (4) Complete the rate law for the overall reaction that is consistent with this mechanism. (Use the form k[A][B]"..., where '1' is understood (so don't write it) for m, n etc.) Rate =arrow_forwardPlease correct answer and don't use hand rating and don't use Ai solutionarrow_forward1. For each of the following statements, indicate whether they are true of false. ⚫ the terms primary, secondary and tertiary have different meanings when applied to amines than they do when applied to alcohols. • a tertiary amine is one that is bonded to a tertiary carbon atom (one with three C atoms bonded to it). • simple five-membered heteroaromatic compounds (e.g. pyrrole) are typically more electron rich than benzene. ⚫ simple six-membered heteroaromatic compounds (e.g. pyridine) are typically more electron rich than benzene. • pyrrole is very weakly basic because protonation anywhere on the ring disrupts the aromaticity. • thiophene is more reactive than benzene toward electrophilic aromatic substitution. • pyridine is more reactive than nitrobenzene toward electrophilic aromatic substitution. • the lone pair on the nitrogen atom of pyridine is part of the pi system.arrow_forward
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