(NO3) in the compounds HNO3 or KNO3; (SO4) in H2SO4. In HNO3 the radical has a valence of 1, to balance that of H, H-NO3). In H2SO4, what is the valence of (SO4)? Give it in each of these radicals, noting first that of the first element: K(NO3), Na2(SO4), Na2(CO3), K(ClO3), H3(PO4), Ca3(PO4)2, Na4(SiO4).
Suppose we wish to know the symbol for calcium phosphate. Ca and PO4 are the two parts. In H3(PO4) the radical is a triad, to balance H3. Ca is a dyad, Ca==(P04). The least common multiple of the bonds (2 and 3) is 6, which, divided by 2 (no. Ca bonds), gives 3 (no. Ca atoms to be taken). 6 / 3 (no. (PO4) bonds) gives 2 (no. PO4 radicals to be taken). Hence the symbol Ca3(P04)2.
Verify this by writing graphically.
Write symbols for the union of Mg and (SO4), Na and (PO4), Zn and (NO3), K and (NO3), K and (SO4), Mg and (PO4), Fe and (SO4) (both valences of Fe), Fe and (NO3), taking the valences of the radicals from HNO3, H2SO4, H3PO4.
Chapter XII.
ELECTRO-CHEMICAL RELATION OF ELEMENTS.
58. Examine untarnished pieces of iron, silver, nickel, lead, etc.; also quartz, resin, silk, wood, paper. Notice that from the first four light is reflected in a different way from that of the others. This property of reflecting light is known as l.u.s.ter.
Metals have a metallic l.u.s.ter which is peculiar to themselves; and this, for the present, may be regarded as their chief characteristic. Are they at the positive or negative end of the list? See page 43. How is it with the non-metals? This arrangement has a significance in chemistry which we must now examine. The three appended experiments show how one metal can be withdrawn from solution by a second, this second by a third, the third by a fourth, and so on. For expedition, three pupils can work together for the three following experiments, each doing one, and examining the results of the others.
59. Deposition of Silver.
Experiment 35.--Put a ten-cent Ag coin into an evaporating-dish, and pour over it a mixture of 5 cc. HNO3 and 10 cc. H2O. Warm till all, or nearly all, the Ag dissolves. Remove the lamp. 3 Ag + 4 HNO3 = 3 AgNO3 + 2 H2O + NO. Then add 10 cc. H2O, and at once put in a short piece of Cu wire, or a cent. Leave till quite a deposit appears, then pour off the liquid, wash the deposit thoroughly, and remove it from the coin. See whether the metal resembles Ag. 2 AgNO3 + Cu =?60. Deposition of Copper.
Experiment 36.--Dissolve a cent or some Cu turnings in dilute HNO3, as in Experiment 35, and dilute the solution. 3 Cu + 8 HN09 - 3 Cu (NOA+4 H2O+2 NO.)
Then put in a clean strip of Pb, and set aside as before, examining the deposit finally. Cu(NO3), + Pb - ?
61. Deposition of Lead.
Experiment 37.--Perform this experiment in the same manner as the two previous ones, dissolving a small piece of Pb, and using a strip of Zn to precipitate the Pb. 3 Pb + 8 HNO3 - 3 Pb (NO4)2 + 4 Ha0 + 2 NO. Pb (NO3) 2 + Zn = ? h.
62. Explanation. -These experiments show that Cu will replace Ag in a solution of AgNO3, that Pb will replace and deposit Cu from a similar compound, and that Zn will deposit Pb in the same way.
They show that the affinity of Zn for (NO3) is stronger than either Ag, Cu, or Pb. We. express this affinity by saying that Zn is the most positive of the four metals, while Ag is the most nega- tive. Cu is positive to Ag, but negative to Pb and Zn.
Which of the four elements are positive to Pb, and which negative? Mg would withdraw Zn from a similar solution, and be in its turn withdrawn by Na. The table on page 43 is founded on this relation. A given element is positive to every element above it in the list, and negative to all below it.
Metals are usually cla.s.sed as positive, non-metals as negative.
Each in union with O and 1=I gives rise to a very important cla.s.s of compounds,=--the negative to acids, the positive to bases.
In the following, note whether the positive or the negative element is written first:--HCl, Na20,-As2S3, -MgBr2, Ag2S. Na2SO4 is made up of two parts, Na2 being positive, the radical SO4 negative. Like elements, radicals are either positive or negative. In the following, separate the positive element from the negative radical by a vertical line: Na2CO3, NaNO3, ZnSO4, KClO3.
The most common positive radical is NH4, ammonium, as in NH4Cl.
It always deports itself as a metal. The commonest radical is the negative OH, called hydroxyl, from hydrogen- oxygen. Take away H from the symbol of water, H-O-H, and hydroxyl --(OH) with one free bond is left. If an element takes the place of H, i.e.
unites with OH, the compound is called a hydrate. KOH is pota.s.sium hydrate. Name NaOH, Ca(OH)2, NH4OH, Zn(OH)2, Al2(OH)6.
Is the first part of each symbol above positive or negative?
H has an intermediate place in the list. It is a const.i.tuent of both acids and bases, and of the neutral substance, water.
ORDER.
Negative or Non-Metallic Elements.
Acid-forming with H(usually OH).
Oxygen Sulphur Nitrogen Fluorine Chlorine Bromine Iodine Phosphorus a.r.s.enic Carbon Silicon Hydrogen
Positive or Metallic Elements.
Base-forming with OH.
Gold Platinum Mercury Silver Copper Tin Lead Iron Zinc Aluminium Magnesium Calcium Sodium Pota.s.sium
CHAPTER XIII.
ELECTROLYSIS.
The following experiment is to be performed only by the teacher, but pupils should make drawings and explain.
63. Decomposition of Water.
Experiment 38.--Arrange "in series" two or more cells of a Bunsen battery (Physics, page 164), [References are made in this book to Gage"s Introduction to Physical Science.] and attach the terminal wires to an electrolytic apparatus (Fig. 19) filled with water made slightly acid with H2SO4. Construct a diagram of the apparatus, marking the Zn in the liquid +, since it is positive, and the C, or other element, -. Mark the electrode attached to the Zn -, and that attached to the C +; positive electricity at one end of a body commonly implies negative at the other.
Opposites attract, while like electricities repel each other.
These a.n.a.logies will aid the memory. At the + electrode is the - element of H2O, and at the - electrode the + element. Note, page 43, whether H or O is positive with reference to the other, and write the symbol for each at the proper electrode. Compare the diagram with the apparatus, to verify your conclusion. Why does gas collect twice as fast at one electrode as at the other? What does this prove of the composition of water? When filled, test the gases in each tube, for O and H, with a burning stick.
Electrical a.n.a.lysis is called electrolysis.
If a solution of NaCl be electrolyzed, which element will go to the + pole? Which, if the salt were K2SO4? Explain these reactions in the electrolysis of that salt. K2SO4 = K2 + S03 + O.
SO4 is unstable, and breaks up into SO3 and O. Both K and SO3 have great affinity for water. K2 + 2 H2O = 2 KOH + H2. S03 + H2O = H2SO4.
The base KOH would be found at the - electrode, and the acid H2SO4 at the + electrode.
The positive portion, K, uniting with H2O forms a base; the negative part, S03, with H2O forms an acid. Of what does this show a salt to be composed?
64. Conclusions.--These experiments show (1) that at the + electrode there always appears the negative element, or radical, of the compound, and at the - electrode the positive element; (2) that these elements unite with those of water, to make, in the former case, acids, in the latter, bases; (3) that acids and bases differ as negative and positive elements differ, each being united with O and H, and yet producing compounds of a directly opposite character; (4) that salts are really compounded of acids and bases. This explains why salts are usually inactive and neutral in character, while acids and bases are active agents.
Thus we see why the most positive or the most negative elements in general have the strongest affinities, while those intermediate in the list are inactive, and have weak affinities; why alloys of the metals are weak compounds; why a neutral substance, like water, has such a weak affinity for the salts which it holds in solution; and why an aqueous solution is regarded as a mechanical mixture rather than a chemical compound.
In this view, the division line between chemistry and physics is not a distinct one. These will be better understood after studying the chapters on acids, bases and salts.
Chapter XIV.
UNION BY VOLUME.
66. Avogadro"s Law of Gases.--Equal volumes of all gases, the temperature and pressure being the same, have the same number of molecules. This law is the foundation of modern chemistry. A cubic centimeter of O has as many molecules as a cubic centimeter of H, a liter of N the same number as a liter of steam, under similar conditions. Compare the number of molecules in 5 l. of N2O with that in 10 l. Cl. 7 cc. vapor of I to 6 cc. vapor of S.
The half-molecules of two gases have, of course, the same relation to each other, and in elements the half-molecule is usually the atom.
The molecular volumes--molecules and the surrounding s.p.a.ce--of all gases must therefore be equal, as must the half-volumes.
Notice that this law applies only to gases, not to liquids or solids. Let us apply it to the experiment for the electrolysis of water. In this we found twice as much H by volume as O.
Evidently, then, steam has twice as many molecules of H as of O, and twice as many half-molecules, or atoms. If the molecule has one atom of O, it must have two of H, and the formula will be H2O.
Suppose we reverse the process and synthesize steam, which can be done by pa.s.sing an electric spark through a mixture of H and O in a eudiometer over mercury; we should need to take twice as much H as O. Now when 2 cc. of H combine thus with 1 cc. of O, only 2 cc.of steam are produced. Three volumes are condensed into two volumes, and of course three molecular volumes into two, three atomic volumes into two. This may be written as follows:--
H + H + O = H2O.
This is a condensation of one-third.
If 2 l. of chlorhydric acid gas be a.n.a.lyzed, there will result 1 l. of H and 1 l. of Cl. The same relation exists between the molecules and the atoms, and the reaction is:--