Nelson Science Perspectives 10
Nelson Science Perspectives 10
1st Edition
Christy C. Hayhoe, Doug Hayhoe, Jeff Major, Maurice DiGiuseppe
ISBN: 9780176355289
Table of contents
Textbook solutions

All Solutions

Section 5-7: Names and Formulas of Ionic Compounds

Exercise 1
Step 1
1 of 3
In this question, we are going to describe the skill that I think needs practicing, and how we are planning on improving this skill.
Step 2
2 of 3
The skill that is needed practicing is the writing of chemical formulas of ionic compounds, and the reason that the writing of chemical formulas can sometimes be tricky so you need to practice so you can master it.
Step 3
3 of 3
To improve this skill you need to understand the rules of writing the chemical formula, and then keep practicing on writing so you can develop this skill.
Exercise 2
Step 1
1 of 6
a.

CaF$_2$ is calcium fluoride.

Step 2
2 of 6
b.

K$_2$S is potassium sulfide.

Step 3
3 of 6
c.

Al$_2$O$_3$ is aluminum oxide.

Step 4
4 of 6
d.

LiBr is lithium bromide.

Step 5
5 of 6
e.

Ca$_3$P$_2$ is calcium phosphide.

Result
6 of 6
a. Calcium fluoride.

b. Potassium sulfide.

c. Aluminum oxide.

d. Lithium bromide.

e. Calcium phosphide.

Exercise 3
Step 1
1 of 4
a.

The ionic charge on K ion is $+1$ while that on Br ion is $-1$, therefore the compound potassium bromide is KBr.

Step 2
2 of 4
b.

The ionic charge on Ca ion is $+2$ while that on O ion is $-2$, therefore the compound calcium oxide is CaO.

Step 3
3 of 4
c.

The ionic charge on Na ion is $+1$ while that on S ion is $-2$. This implies that 2 sodium atoms will need to lose each of their valence electrons to 1 atom of sulfur to form the compound, therefore the compound sodium sulfide is Na$_2$S.

Result
4 of 4
a. KBr.

b. CaO.

c. Na$_2$S.

Exercise 4
Step 1
1 of 2
The correct formula of tin(IV) oxide is SnO$_2$. This is because 1 atom of tin loses 4 of its valence electrons to 2 atoms of oxygen. Each atom of oxygen gains 2 electrons to form the oxide.
Result
2 of 2
SnO$_2$.
Exercise 5
Step 1
1 of 3
The ionic charge on the bromide ion $-1$. This implies that if its compound contains 1 ion of copper per bromide ion, then their ratio is $1:1$ in the compound and therefore its chemical formula must be CuBr. This implies that the ionic charge on copper in this compound is $+1$ and therefore the name of this compound is copper(I) bromide.
Step 2
2 of 3
The ionic charge on the bromide ion $-1$. This implies that if its compound contains 1 ion of copper for every 2 bromide ions, then their ratio is $1:2$ in the compound and therefore its chemical formula must be CuBr$_2$. This implies that the ionic charge on copper in this compound is $+2$ and therefore the name of this compound is copper(II) bromide.
Result
3 of 3
Copper(I) bromide – CuBr.

Copper(II) bromide – CuBr$_2$.

Exercise 6
Result
1 of 1
An ionic compound is made by the transfer of electrons making the compound an electrically neutral entity and therefore the net charge on it must always be $0$.
Exercise 7
Step 1
1 of 6
a.

The ionic charge on calcium ion is $+2$ while that on chlorine ion is $-1$. This implies that 2 chloride atoms will be needed to gain electrons lost from 1 calcium atom to form the compound, therefore the compound calcium chloride is CaCl$_2$.

Step 2
2 of 6
b.

The ionic charge on aluminum ion is $+3$ while that on bromide ion is $-1$. This implies that 3 bromine atoms will be needed to gain electrons lost from 1 aluminum atom to form the compound, therefore the compound aluminum bromide is AlBr$_3$.

Step 3
3 of 6
c.

The ionic charge on magnesium ion is $+2$ while that on sulfide ion is $-2$, therefore the compound magnesium sulfide is MgS.

Step 4
4 of 6
d.

The ionic charge on lithium ion is $+1$ while that on nitride ion is $-3$. This implies that 3 lithium atoms will need to lose each of their valence electrons to 1 atom of nitrogen to form the compound, therefore the compound lithium nitride is Li$_3$N.

Step 5
5 of 6
e.

The ionic charge on calcium ion is $+2$ while that on nitride ion is $-3$. This implies that 3 calcium atoms will lose $3times2=6$ of their valence electrons to $dfrac{6}{3}=2$ atoms of nitrogen to form the compound, therefore the compound calcium nitride is Ca$_3$N$_2$.

Result
6 of 6
a. CaCl$_2$.

b. AlBr$_3$.

c. Mg$_2$S.

d. Li$_3$N.

e. Ca$_3$N$_2$.

Exercise 8
Step 1
1 of 2
The systematic naming convention is a more convenient naming convention because it tells the chemical composition of the minerals.
Result
2 of 2
Click to see answer.
Exercise 9
Step 1
1 of 16
Name
Formula
Step 2
2 of 16
iron(II) bromide
FeBr$_2$
Step 3
3 of 16
manganese(IV) oxide
MnO$_2$
Step 4
4 of 16
tin(IV) chloride
SnCl$_4$
Step 5
5 of 16
copper(I) sulfide
Cu$_2$S
Step 6
6 of 16
iron(III) nitride
FeN
Step 7
7 of 16
copper(II) oxide
CuO
Step 8
8 of 16
lead chloride
PbCl$_2$
Step 9
9 of 16
iron(III) oxide
Fe$_2$O$_3$
Step 10
10 of 16
tin(II) sulfide
SnS
Step 11
11 of 16
copper(II) phosphide
Cu$_3$P$_2$
Step 12
12 of 16
calcium bromide
CaBr$_2$
Step 13
13 of 16
copper(II) fluoride
CuF$_2$
Step 14
14 of 16
potassium phosphide
K$_3$P
Step 15
15 of 16
copper(I) phosphide
Cu$_3$P
Result
16 of 16
Click to see table
Exercise 10
Result
1 of 1
Two charges of $Fe^{3+}$ and one charge of $Fe^{2+}$ with four oxygens will give the desired formula
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Section 4-1: Systems in Plants
Section 4-2: Plant Tissue Systems
Section 4-4: Tissues Working Together
Section 4-6: Plant Growth
Page 150: Review
Page 152: Self-Quiz
Page 159: Unit Review
Page 164: Self-Quiz
Chapter 5: Chemicals and Their Properties
Section 5-1: Properties and Changes
Section 5-3: Hazardous Products and Workplace Safety
Section 5-4: Patterns and the Periodic Table
Section 5-5: Atoms and Ions
Section 5-6: Ionic Compounds
Section 5-7: Names and Formulas of Ionic Compounds
Section 5-9: Polyatomic Ions
Section 5-10: Molecules and Covalent Bonding
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Page 218: Self-Quiz
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