1.3.3-1. Simple substance, of course. Like any other gas, chlorine gas (Cl₂) can be put in a gas container, transported, and used.

1.3.3-2. Chemical element iron (Fe). As a simple substance, iron is a metal that iron nails and other items are made of. There is no iron metal in spinach, tofu, broccoli, and other foods that are high in iron.

1.3.3-3. Simple substance, Br₂. The element bromine cannot be a liquid or a solid or a gas because a chemical element is not a substance composed of atoms or molecules, but is a type of atom.

1.3.3-4. Simple substance. Since a chemical element is a type of atom, how can it have a melting or boiling point?

1.3.3-5. Chemical element phosphorus. As a simple substance, phosphorus (white phosphorus) is highly hazardous, being both toxic and pyrophoric.

1.4.8-1. Homogeneous. Mixtures of gases are always homogeneous.

1.4.8-2. heterogeneous; homogeneous; heterogeneous.

1.4.8-3. (a) is correct and (b) is not. The bubbles are the air that, like any other gas, has a higher solubility in water and other liquids at lower temperatures than at higher temperatures. If air were insoluble in water, there would be no life in water because fish and other aquatic animals would not be able to breathe. There is no such thing as "slow boiling" below the boiling point of a liquid.

1.4.8-4. Adding water and filtering the mixture was a good idea because sand is insoluble in water and could be separated by filtration. However, evaporating the filtrate to dryness was a bad idea. After all of the water had been gone, the NaCl was still contaminated with KNO₃ because KNO₃ is also soluble in water. The filtrate should have been evaporated only partially, so that NaCl crystallized out whereas KNO₃ stayed in the solution.

1.4.8-5. Both contain no or very little mineral salts because both are produced as a result of condensation of water vapors. In general, however, distilled water is more pure because as rainwater passes through the atmosphere it picks up tiny particles of dust and some gaseous contaminants present there, such as nitrogen oxides and sulfur oxides. Those oxides react with water to give acids. That is why areas where air pollution is severe suffer from so-called "acid rains".

1.4.8-6. See section 4.7 and Figure 1-19. Like sulfur, copper is not attracted by magnets, whereas iron is. Just use a magnet to pick up the iron filings from their mixture with their copper powder.

1.4.8-7. (c).

1.4.8-8. A filtrate is a solid-free solution after filtration. A mother liquor is the part of a solution that is left over after crystallization.

1.5.3-1. (a) physical; (b) chemical; (c) physical; (d) chemical (e) chemical.

1.6.5-1. (A) Zn + S = ZnS. (B) 32.5 g of Zn; 48.5 g of ZnS. (C) 44.8 g; Zn.

1.6.5-2. Iron metal is attracted by a magnet, whereas FeS is not. Use the magnet separation method (section 4.7 and Figure 1-19). However, before the FeS and Fe can be separated with a magnet, we have to thoroughly grind up the mixture, which is produced in the form of a hard solid cake in the reaction.

1.7.3-1. (a) Ca 71.5%; O 28.5%; (b) Al 52.9%; O 47.1%; (c) Mg 13.2%; Br 86.8%; (d) Cu 39.8%; S 20.1%; O 40.1%; (e) Cu 25.5%; S 12.8%; O 57.7%; H 4.0%.

1.7.3-2. First, we calculate the content of pure gold in the two compounds using the atomic masses. For AuCl₃, the mass percentages are: Au 64.9%; Cl 35.1%. For AuBr₃, the mass percentages are: Au 45.1%; Br 54.9%. Therefore, 1 g of AuCl₃ contains 0.649 g of pure gold and 1 g of AuBr₃ 0.451 g of pure gold. You would pay $140.00 for 0.649 g of pure gold if you bought AuCl₃ or $100.00 for 0.451 g of pure gold if you bought AuBr₃. To calculate the cost of pure gold per g for AuCl₃, we divide $140.00 by 0.649 = $215.72 per gram. For AuBr₃, the cost of 1 g of pure gold is $100.00 divided by 0.451 = $221.73, roughly $6.00 more. You would get more value for your money if you bought AuCl₃.

1.8.8-1. Hint: The last reaction (d) might be a bit trickier to balance. Just pay attention to the fact that the number of oxygen atoms on the right is even and will always remain even no matter what coefficient is placed before the BaO₂. In contrast, the number of oxygen atoms on the left is currently odd because in O₂ it is even but in BaO it is odd. Therefore, an even coefficient number should be used for BaO.

1.8.8-3. (a) decomposition; (b) combination; (c) substitution; (d) combination; (e) substitution; (f) decomposition; (g) exchange; (h) combination.

1.8.8-6. First we need a balanced equation for the reaction. Here it is, CaCO₃ = CaO + CO₂, showing that the decomposition of one molecule of CaCO₃ gives rise to one molecule of CaO and one molecule of CO₂. Next we calculate molecular weights of CaCO₃ and of CaO using the atomic masses of Ca (40), C (12), and O (16) from the periodic table. For CaCO₃, the molecular weight is 100, for CaO it is 56, and for CO₂ 44. Let us write these numbers right below the corresponding substances in the equation.

The equation with the numbers shows that the decomposition of 100 g (or pounds or kilos or tons, etc.) of CaCO₃ would produce 56 g (or pounds or kilos or tons, etc.) of CaO. In other words, if we needed to make 56 tons of CaO, we would have to use 100 tons of CaCO₃. However, we need to make 112 tons of CaO. For that, we need to decompose x tons of CaCO₃. Let us write this figure and the unknown above the corresponding formulas in the equation. It is a good idea to delete the molecular weight of CO₂ (44) because this number is not needed for our calculation.

Now we can solve the proportion for x by multiplying 112 by 100 and dividing the result by 56. x = (112 x 100)/56 = 200. To make 112 tons of CaO, we need 200 tons of CaCO₃.

1.9.2-1. (a) x = 3; (b) y = 2; z = 3.

1.9.2-2. K₂O, BaO, Al₂O₃, CO, CO₂, P₂O₃, P₂O₅, CrO, Cr₂O₃, CrO₃.

1.10.7-2. Yes, although not that many. Examples include gold (Au) and platinum (Pt).

1.10.7-3. The burning of magnesium (Mg) in air. You are encouraged to do research on the Internet to learn more about this application.

1.10.7-4. Absolutely not. A substance of a particular formula (composition) and structure displays exactly the same properties regardless of its source.

1.10.7-5. 2 C₂H₆ + 7 O₂ = 4 CO₂ + 6 H₂O.

1.10.7-6. Light blue.

1.10.7-7. Sb₂O₃. The equation: 4 Sb + 3 O₂ = 2 Sb₂O₃.

1.10.7-8. (b).

1.11.6-1. AlH₃; 2 Al + 3 H₂ = 2 AlH₃

1.11.6-3. Make H₂ using the nails and sulfuric acid (Figure 1-36). Do the reduction of CuO and WO₃ with the H₂ produced (Figures 1-38 and 1-39).

1.11.6-4. It is clear from the chemical equations that one molecule of H₂ can be formed from either one atom of Fe or one atom of Zn. Atomic masses of Zn and Fe are approximately 65 and 56, respectively. Therefore, one atom of Zn is heavier than one atom of Fe. Accordingly, 1 g of iron contains more Fe atoms than 1 g of zinc contains Zn atoms. Consequently, 1 g of Fe will produce more H₂ in the reaction with sulfuric acid than 1 g of Zn.

1.12.13-1. Nitric acid, potassium hydroxide, phosphoric acid, calcium hydroxide, sulfuric acid, hydrochloric acid (hydrogen chloride if gas), sodium hydroxide, aluminum hydroxide, iron (II) hydroxide, iron (III) hydroxide.

1.12.13-2. Mg(OH)₂, HBr, LiOH, H₂CO₃, Co(OH)₂, H₂SiO₃.

1.12.13-3. Not necessarily. We should remember that samples of concentrated HNO₃ most often contain orange-brown NO₂, a product of its decomposition.

1.12.13-4. Look for another supervisor. Pure carbonic acid does not exist.

1.12.13-5. The technician made at least five very serious mistakes. Making any one of these five is enough to fire the technician on the spot. Mistake 1: To prepare 20% sulfuric acid one needs 20 g of H₂SO₄ and 80 g (not 100 g) of water (see the definition of mass percentage concentration above). Mistake 2: Indeed, the density of water is 1 g/mL, so 100 g of water = 100 mL of water. However, the density of sulfuric acid is 1.84 g/mL. Therefore, the volume taken by 20 g of H₂SO₄ is 20 g divided by 1.84 g/mL ≈ 11 mL, not 20 mL. Mistake 3: When diluting sulfuric acid with water one must add acid to water, not the other way around. The technician was lucky to avoid chemical and thermal burns. Mistake 4: Phenolphthalein turns raspberry red in the presence of a base; in the presence of acids, it remains colorless. Therefore, colorless solutions of phenolphthalein are useless for testing samples for the presence of an acid. Mistake 5: Sulfuric acid does not decompose upon dilution with water, no matter if it is diluted properly (ATW: acid to water) or not (WTA: water to acid).

1.12.13-6. (d) and (f).

1.12.13-7. Both are wrong. The two solutions contain the same amount of H₂O because the mass percentage concentration is the same. By mass percentage concentration definition, 100 g of a 10% solution of any substance comprises 90 g of water and 10 g of the solute. What the solute is does not matter. 50 g of any 10% solution contains half the quantity of water, 45 g.

1.12.13-8. KOH and NaOH are highly hygroscopic, meaning that they avidly attract and hold water vapors from the atmosphere. Both absorb so much moisture from the air that eventually they can dissolve in the water absorbed (watch this video). The process by which a hygroscopic substance self-dissolves in the water it has absorbed from the atmosphere is called deliquescence. The absorption of water vapors from the air commences on the surface of a hygroscopic material. The larger the surface area, the faster the rate of the moisture uptake will be. A powdered solid substance always has a larger surface area than a pellet or block made of the same quantity of the same substance. It is for this reason that pellets of KOH and NaOH are less hygroscopic and more convenient to work with than KOH and NaOH powders.

1.13.3-2. First, we need to write a balanced chemical equation for this reaction: NaOH + HBr = NaBr + H₂O. Next, we calculate the molecular weights of NaOH and HBr using the atomic masses from the periodic table. The molecular weight of NaOH is: 23 + 16 + 1 = 40. The molecular weight of HBr is: 1 + 80 = 81. Now we write down these numbers below the corresponding formulas in the equation:

The equation now shows that to neutralize 40 g of NaOH, we would need 81 g of HBr. We need to neutralize 60 g of NaOH. The amount of HBr needed for that (x) should be proportional:

The proportion is easily solved for x: x = (60 x 81)/40 = 121.5. To neutralize 60 g of NaOH, 121.5 g of HBr is needed.

1.13.3-3. First we need to calculate how much pure HCl and pure KOH the solutions contained. Once we know that, we will determine which of the two reagents was limiting and which was in excess. If it is the HCl that was in excess, the litmus paper color must have changed to red. If it is the KOH that was in excess (or, neither reagent was in excess, i.e., the HCl to KOH ratio was exactly 1:1), the blue color of the indicator remained the same.

We had 50 g of a 7% solution of HCl, which means that 100 g of this solution would contain 7 g of HCl. The proportion is:

100 g of 7% HCl contains ------ 7 g of pure HCl
50 g of 7% HCl contains ------ x g of pure HCl

Solving for x: x = (7 x 50)/100 = 3.5 g. This is the amount of pure HCl in our solution.

The amount of pure KOH in 30 g of a 9% solution of KOH is calculated similarly:

100 g of 9% KOH solution contains ------ 9 g of pure KOH
30 g of 9% KOH solution contains ------ y g of pure KOH

Solving for y: y = (9 x 30)/100 = 2.7 g. This is the amount of pure KOH in our solution.

Now that we know that there were 3.5 g of HCl and 2.7 g of KOH used for the reaction, we can calculate which of the two was in excess. First we write the equation:

KOH + HCl = KCl + H₂O

Next, we calculate the molecular weights for HCl and for KOH. These are 36.5 and 56, respectively. We now place these numbers below the formulas in the equation to show that to neutralize 56 g of pure KOH, one would need 36.5 g of pure HCl.

The amount of KOH used for the reaction was 2.7 g. Let us calculate how much HCl (z) is needed to neutralize 2.7 g of pure KOH.

z = (2.7 x 36.5)/56 = 1.76 g. This is the quantity of pure HCl needed to neutralize 2.7 g of KOH. Our HCl solution contained 3.5 g of HCl, nearly twice as much. Therefore, the limiting reagent was KOH and HCl was used in excess in the reaction. Consequently, the resultant solution was acidic and the litmus paper must have turned red.

1.14.6-1. True.

1.14.6-2. (a) potassium oxide; (b) iron (III) oxide; (c) iron (II) oxide; (d) manganese (VII) oxide; (e) copper (I) oxide; (f) copper (II) oxide; (g) aluminum oxide; (h) carbon (IV) oxide or carbon dioxide; (i) carbon (II) oxide or carbon monoxide; (j) sulfur (VI) oxide or sulfur trioxide; (k) sulfur (IV) oxide or sulfur dioxide; (l) magnesium oxide; (m) water; (n) silicon (IV) oxide or silicon dioxide; (o) calcium oxide; (p) zinc oxide.

1.14.6-3. (a) Na₂O; (b) P₂O₅; (c) CO₂; (d) CaO; (e) TiO₂; (f) CO; (g) Ag₂O; (h) NO₂; (i) N₂O; (j) WO₃.

1.14.6-4. (a).

1.14.6-5. Heat limestone to decompose it to quicklime, CaCO₃ = CaO + CO₂, then treat the quicklime produced with water, CaO + H₂O = Ca(OH)₂.

1.14.6-6. See 1.14 for the balanced equations for all of these reactions, except for (d), (f) and (i). For (d), S + O₂ = SO₂. Being a basic oxide, Ag₂O reacts with acids but does not react with bases (f). Oxides efficiently react with water only if the resultant acid or base is water-soluble (i), see Figure 1-62.