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 assay 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; y = 2; (b) z = 3.
1.9.2-2. K2O, BaO, Al2O3, CO2, 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. Answer: 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 was 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 and should be given an F for introductory chemistry. 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 balancedchemical 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, 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) CO2; (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 (f) and (i). 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.
1.15.4-1. NaOH is a strong base. Therefore, it will react with acids and acidic and amphoteric oxides and hydroxides.

(a) NaOH + HCl = NaCl + H2O. HCl is a strong acid.

(b) No reaction with other bases, including Ca(OH)2.

(c) No reaction with basic oxides, including MgO.

(d) 2 NaOH + CO2 = Na2CO3 + H2O (NaOH in excess); NaOH + CO2 = NaHCO3 (CO2 in excess). CO2 is an acidic oxide.

(e) 2 NaOH + Zn(OH)2 = Na2ZnO2 + 2 H2O. Zn(OH)2 is amphoteric.

(f) 2 NaOH + H2SiO3 = Na2SiO3 + 2 H2O. H2SiO3 is an acid.
1.15.4-2. SO3 is an acidic oxide. Therefore, it will react with bases and basic species.

(a) No reaction. HCl is an acid.

(b) SO3 + Mg(OH)2 = MgSO4 + H2O. Mg(OH)2 is a base.

(c) SO3 + CaO = CaSO4. CaO is a basic oxide.

(d) No reaction. CO2 is an acidic oxide.

(e) SO3 + Ag2O = Ag2SO4. Ag2O is a basic oxide.
1.15.4-3. (a), (b), (d), (e), (g), (h), (i), (k), (l), (m), (n), (o), and (p).
1.16.4-1. False. Organic acids are, whereas mineral acids are not.
1.16.4-2. False. The reaction of H2SO4 with NaCl is a laboratory method for making HCl in small quantities. In industry, HCl is obtained as a co-product of chlorination reactions of some organic compounds.
(a) H2SO4 + Mg = MgSO4 + H2
(b) SO3 + H2O = H2SO4
(c) SO3 + 2 KOH = K2SO4
(d) CaCO3 + 2 HCl = CaCl2 + H2O + CO2
(e) H3PO4 + 3 NaOH = Na3PO4 + 3 H2O
(f) 2 HNO3 + Cu(OH)2 = Cu(NO3)2 + 2 H2O
(g) H2SO4 + BaCl2 = BaSO4↓ + 2 HCl
(h) P2O5 + 3 H2O = 2 H3PO4
1.17.4-1. False. NaOH and KOH are too reactive toward many substances to exist in nature, especially pure. Furthermore, NaOH and KOH are hygroscopic and deliquescent and, consequently, remain solid in humid air only for a very short period of time (see 1.12.9 and 1.12.13, problem 8).
1.17.4-2. True.

(a) NaOH + HNO3 = NaNO3 + H2O
(b) 2 KOH + CuCl2 = Cu(OH)2↓ + 2 KCl
(c) Ca(OH)2 + CO2 = CaCO3↓ + H2O
(d) Fe(OH)3 + 3 HCl = FeCl3 + 3 H2O
(e) AlCl3 + 3 NaOH = Al(OH)3↓ + 3 NaCl
(f) AlCl3 + 4 NaOH = NaAlO2 + 3 NaCl + 2 H2O
(g) Cu(OH)2 = CuO + H2O
(h) Zn(OH)2 + 2 KOH = K2ZnO2 + 2 H2O
(i) MgCl2 + 2 NaOH = Mg(OH)2↓ + 2 NaCl
(j) Na2O + H2O = 2 NaOH
1.18.3-1. (b) because Hg is more reactive than Ag.
(a) 2 K + 2 HCl = 2 KCl + H2
(b) Ag2SO4 + BaCl2 = BaSO4↓ + 2 AgCl↓
(c) Zn(OH)2 + 2 HCl = ZnCl2 + 2 H2O
(d) Cu(NO3)2 + Ca(OH)2 = Ca(NO3)2 + Cu(OH)2
(e) CuCl2 + Zn = ZnCl2 + Cu
(f) Cu + HCl = no reaction
(g) 2 Al(OH)3 + 3 H2SO4 = Al2(SO4)3 + 6 H2O
(h) Al(OH)3 + KOH = KAlO2 + 2 H2O
(i) Al2O3 + 2 KOH = 2 KAlO2 + H2O
(j) CaBr2 + Na2CO3 = 2 NaBr + CaCO3
(k) SiO2 + Na2O = Na2SiO3
(l) K2CO3 + 2 HCl = 2 KCl + H2O + CO2
(m) Zn + 2 HCl = ZnCl2 + H2
(n) Fe + H2SO4 = FeSO4 + H2
1.19.1-1. hydrogen (simple substance); oxygen (simple substance); copper (II) oxide (oxide); copper (I) oxide (oxide); potassium oxide (oxide); aluminum oxide (oxide); zinc oxide (oxide); barium hydroxide (hydroxide; base); iron (II) hydroxide (hydroxide; base); iron (III) hydroxide (hydroxide; base); sodium carbonate (salt); magnesium sulfate (salt); calcium chloride (salt); phosphoric acid (acid); hydrogen fluoride or hydrofluoric acid (acid); ozone (simple substance); potassium hydrosulfate or potassium bisulfate (hydro salt); strontium nitrate (salt); tin (II) chloride (salt); aluminum sulfate (salt).
1.19.1-2. H2O; HCl; CuO; Cu2O; CuCl; CuCl2; CuCl; CuCl2; H2SO4; H3PO4; HNO3; SnCl2; FeCl3; FeCl2; HBr; KOH; NaHCO3.
1.19.1-3. No. Allotropes are different simple substances formed by the same element.
1.19.1-4. False. See 1.2.2.
1.19.1-5. Amalgams. Homogeneous.
1.19.1-6. False. Carbon, phosphorus, sulfur, iodine, and some other nonmetals are solids.
1.19.1-7. Heterogeneous.
1.19.1-8. NaHCO3 + HCl = NaCl + H2O + CO2
1.19.1-9. There are many ways to solve problems like this one. I prefer to follow the simplest logic. What is mass percentage concentration? By definition, it is the number of grams of a solute in 100 g of solution. As the concentration of HCl is 36%, 100 g of that solution contains 36 g of pure HCl. The rest is 64 g of water: 100 g - 36 g = 64 g.

We need a 20% solution of HCl, which means that 100 g of the desired solution should contain 20 g of pure HCl, with the rest being water (80 g). However, the amount of HCl that we have is not 20 g but 36 g. How much water (x) do we need to add to 36 g of HCl to prepare a 20% solution? Let us solve the simple proportion:

For 20 g HCl – 80 g H2O is needed
For 36 g HCl – x g H2O is needed

x = (80 x 36)/20 = 144 g. This is the total amount of water that is needed to prepare 20% hydrochloric acid from 36 g of pure HCl. However, the solution in the bottle already contains 64 g of water. Therefore, we need to add to this solution 144 – 64 = 80 g of water.

There is another way to solve this problem. Our unknown, x, is the amount of water in grams that we need to dilute our given solution with in order to lower its concentration from 36% to 20%. The desired solution should contain 20 g of pure HCl per each 100 g of the solution. The proportion to be solved for x is:

100 g of desired solution contains – 20 g HCl
(100 g + x g) of desired solution contains – 36 g HCl

100 x 36 = (100 + x) x 20

3600 = 2000 + 20x

20x = 3600 – 2000 = 1600

x = 1600/20 = 80 g

The result is the same: we need to add 80 g of water to lower the original concentration of 36% to 20%.
1.19.1-10. False. Mixtures of gases are always homogeneous.
1.19.1-11. First, heat CaCO3 to decompose it to CaO and CO2: CaCO3 = CaO + CO2. Then treat the CaO produced with water: CaO + H2O = Ca(OH)2.
1.19.1-12. Reaction (a) will occur and reaction (b) will not. See the metal reactivity series (Figure 1-88).
1.19.1-13. First, we need to calculate how much KNO3 and water the starting solution contains. The mass percentage concentration of this solution is 10%, which means that 100 g of an identical solution would contain 90 g of water and 10 g of KNO3. To calculate the quantities of the salt and water in 50 g of our solution, we solve the proportion below, where x is the quantity of KNO3.

100 g of solution contains - 10 g KNO3
50 g of solution contains - x g KNO3

x = (10 x 50)/100 = 5 g

Therefore, 50 g of 10% solution of KNO3 contains 5 g of pure KNO3. The rest is water, 50 – 5 = 45 g.

Our unknown y is the needed quantity of solid KNO3 to be added. The total mass of the desired resultant 20% solution will therefore be (50 + y) g, of which (5 + y) g is KNO3. As the needed concentration is 20%, 100 g of such solution should contain 20 g of KNO3.

100 g of solution contains - 20 g KNO3
(50 + y) g of solution contains - (5 + y) g KNO3

100 x (5 + y) = 20 x (50 + y)

Solving this equation for y, we calculate y = 6.25 g.
1.19.1-14. Add water to dissolve NaCl and sugar. Filter off the insoluble SiO2 and BaSO4. Partially evaporate the filtrate until less soluble NaCl crystallizes out. Separate the crystals of NaCl from the mother liquor containing the dissolved sugar and remaining NaCl.
1.19.1-15. Chemical equations for all of these reactions can be found in Volume 1, except reaction (m): copper metal does not react with acids, see 1.15.3.
1.19.1-16. (b).
1.19.1-17. (a).
1.19.1-18. False. Although this reaction does produce H2, it is too violent to be used safely.
1.19.1-19. False. See 1.12.11.
1.19.1-20. Oxygen gas cannot be called a compound because it is a simple substance, a substance made up of atoms of only one element. Compounds are complex substances whose molecules are composed of atoms of more than one element.