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An Introduction to Chemical Science by R.P. Williams

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Opposites attract, while like electricities repel each other.
These analogies 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 analysis 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 space--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 passing 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 analyzed, 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:--

HCl = H + Cl.

Reverse the process, and 1 l. of H unites with 1 l. of Cl to
produce 2 l. of the acid gas; there is no condensation, and the
symbol is HCl. In seven volumes HCl how many of each constituent?

The combination of two volumes of H with one volume of S is found
to produce two volumes of hydrogen sulphide. Therefore two atoms
of H combine with one of S to form a molecule whose symbol is
H2S.

H + H + S = H2S.

What is the condensation in this case?

PROBLEMS.

(1) How many liters of S will it take to unite with 4 l. of H?
How much H2S will be formed?

(2) How many liters of H will it take to combine with 5 l. of S?
How much H2S results?

(3) In 6 l. H2S how many liters H, and how much S? Prove.

(4) In four volumes H2S how many volumes of each constituent?

(5) If three volumes of H be mixed with two volumes of S, so as
to make H2S, how much will be formed? How much of either element
will be left? An analysis of 2 cc. of ammonia gives 1 cc. N and 3
cc. H. The symbol must then be NH3, the reaction,--

NH3 = N + H + H + H.

What condensation in the synthesis of NH3?

In 12 cc. NH3 how many cubic centimeters of each element? In 2
1/2 cc? How much H by volume is required to combine with nine
volumes of N? How many volumes of NH3 are produced?

In elements that have not been weighed in the gaseous state, as
C, the evidence of atomic volume is not direct, but we will
assume it. Thus two volumes of marsh gas would separate into one
of C and four of H. What is its symbol and supposed condensation?
Two volumes of alcohol vapor resolve into two of C, six of H, and
one of O. What is its symbol? its condensation?

The symbol itself of a compound will usually show what its
condensation is; e.g. HCl, HBr, HF, etc., have two atoms; hence
there will be no shrinkage. In H2O, SO2, CO2, the molecule has
three atoms condensed into the space of two, or one-third
shrinkage. In NH3 four volumes are crowded into the space of two,
a condensation of one-half.

P, As, Hg, Zn, have exceptional atomic volumes.

Chapter XV.

ACIDS AND BASES.

66. What Acids Are.

Experiment 39.--Pour a few drops of chlorhydric acid, HCl, into a
clean evaporating-dish. Add 5 cc. H2O, and stir. Touch a drop to
the tongue, noting the taste. Dip into it the end of a piece of
blue litmus paper, and record the result. Thoroughly wash the
dish, then pour in a few drops of nitric acid, HNO3, and 5 cc.
H2O, and stir. Taste, and test with blue litmus. Test in the same
way sulphuric acid, H2SO4. Name two characteristics of an acid.
In a vertical line write the formulae of the acids above. What
element is common to them all? Is the rest of the formula
positive or negative?

67. An Acid is a substance composed of H and a negative element
or radical. It has usually a sour taste, and turns blue litmus
red. Litmus is a vegetable extract obtained from lichens in
Southern Europe. Acids have the same action on many other
vegetable pigments. Are the following acid formulae, and why?
H2SO3, HBr, HNO2, H3PO3, H4SiO4. Most acids have O as well as H.
Complete the symbols for acids in the following list, and name
them, from the type given:--

HCl, chlorhydric acid. HN03, nitric acid.
?Br, ? ?Cl? ?
?I, ? ?Br? ?
?F, ? ?I? ?
H3PO4, phosphoric acid. H3PO3, phosphorous acid.
?As? ? ?As? ?

Complete these equations:--

H2SO3 - H2O = ? | 2 HN03 - H2O = ?
H2SO4 - H2O = ? | 2 HNO2 - H2O = ?
H2CO3 - H2O = ? | 2 H3AsO4 - 3 H2O = ?

Are the products in each case metallic or non-metallic oxides?
They are called anhydrides. Notice that each is formed by the
withdrawal of water from an acid. Reverse the equations; as, SO3
+ H2O = ?

68. An Anhydride is what remains after water has been removed
from an acid; or, it is the oxide of a non- metallic element,
which, united with water, forms an acid. SO2 is sulphurous
anhydride, SO2 sulphuric anhydride, the ending ic meaning more O,
or negative element, than ous. Name the others above.

Anhydrides were formerly called acids,--anhydrous acids, in
distinction from hydrated ones, as CO2 even now is often called
carbonic acid.

Experiment 40.--Hold a piece of wet blue litmus paper in the
fumes of SO2, and note the acid test. Try the same with dry
litmus paper.

Experiment 41.--Burn a little S in a receiver of air containing
10 cc. H2O, and loosely covered, as in the O experiment. Then
shake to dissolve the SO2. H2O + SO2 = H2SO3. Apply test paper.

69. Naming Acids.--Compare formulae H2SO3 and H2SO4. Of two acids
having the same elements, the name of the one with least O, or
negative element, ends in ous, the other in ic. H2SO3 is
sulphurous acid, H2SO4, sulphuric acid. Name H3PO4 and H3PO3;
H3AsO3 and H3ASO4; HNO2 and HNO3.

If there are more than two acids in a series, the prefixes hypo,
less, and per, more, are used. The following is such a series:
HClO, HClO2, HClO3, HClO4.

HClO3 is chloric acid; HClO2, chlorous; HClO, hypochlorous; HClO4
perchloric. Hypo means less of the negative element than ous; per
means more of the negative element than ic. Name: H3PO4 (ic),
H3PO3, H3PO2. Also HBrO (HBrO2 does not exist), HBrO3 (ic),
HBrO4.

What are the three most negative elements? Note their occurrence
in the three strongest and most common acids. Hereafter note the
names and symbols of all the acids you see.

70. What Bases Are.

Experiment 42.--Put a few drops of NH4OH into an evaporating-
dish. Add 5 cc. H2O, and stir. Taste a drop. Dip into it a piece
of red litmus paper, noting the effect. Cleanse the dish, and
treat in the same way a few drops NaOH solution, recording the
result. Do the same with KOH. Acid stains on the clothing, with
the exception of those made by HNO3, maybe removed by NH4OH.
H2SO4, however, rapidly destroys the fiber of the cloth.

Name two characteristics of a base. In the formulae of those
bases, what two common elements? Name the radical. Compare those
symbols with the symbol for water, HOH. Is (OH) positive or
negative? Is the other part of each formula positive or negative?
What are two constituents, then, of a base? Bases are called
hydrates. Write in a vertical line five positive elements. Note
the valence of each, and complete the formula for its base. Affix
the names. Can you see any reason why the three bases above given
are the strongest?

Taking the valences of Cr and Fe, write symbols for two sets of
hydrates, and name them. Try to recognize and name every base
hereafter met with.

A Base is a substance which is composed of a metal, or positive
radical, and OH. It generally turns red litmus blue, and often
has an acrid taste.

An Alkali is a base which is readily soluble in water. The three
principal alkalies are NH4OH, KOH, and NaOH.

Alkali Metals are those which form alkalies. Name three.

An Alkaline Reaction is the turning of red litmus blue.

An Acid Reaction is the turning of blue litmus red.

Experiment 43.--Pour 5 cc. of a solution of litmus in water, into
a clean t.t. or small beaker. Pour 2 or 3 cc. of HCl into an
evaporating-dish, and the same quantity of NH4OH into another
dish. Take a drop of the HCl on a stirring-rod and stir the
litmus solution with it. Note the acid reaction. Clean the rod,
and with it take a drop (or more if necessary) of NH4OH, and add
this to the red litmus solution, noting the alkaline reaction.
Experiment in the same way with the two other principal acids and
the two other alkalies.

Litmus paper is commonly used to test these reactions, and
hereafter whenever the term LITMUS is employed in that sense, the
test-paper should be understood. This paper can be prepared by
dipping unglazed paper into a strong aqueous solution of
litmus.

CHAPTER XVI.

SALTS.

71. Acids and Bases are usually Opposite in Character.--When two
forces act in opposition they tend to neutralize each other. We
may see an analogy to this in the union of the two opposite
classes of compounds, acids and bases, to form salts.

72. Neutralization.

Experiment 44.--Put into an evaporating-dish 5 cc. of NaOH
solution. Add HCl to this from a t.t., a few drops at a time,
stirring the mixture with a glass rod (Fig. 20), and testing it
with litmus paper, until the liquid is neutral, i.e. will not
turn the test paper from blue to red, or red to blue. Test with
both colors. If it turns blue to red, too much acid has been
added; if red to blue, too much base. When it is very nearly
neutral, add the reagent, HCl or NaOH, a drop at a time with the
stirring-rod. It must be absolutely neutral to both colors.
Evaporate the water by heating the dish over asbestus paper, wire
gauze, or sand, in an iron plate (Fig. 21) till the residue
becomes dry and white. Cool the residue, taste, and name it. The
equation is: HCl + NaOH = NaCl + HOH or H2O. Note which elements,
positive or negative, change places. Why was the liquid boiled?
The residue is a type of a large class of compounds, called
salts.

(Fig. 20) (Fig. 21)

Experiment 45. -- Experiment in the same way with KOH solution
and H2SO4, applying the same tests. H2SO4 + 2 KOH = K2SO4 + 2
HOH. What is the solid product?

Experiment 46.--Neutralize NH4OH with HNO3, evaporate, apply the
tests, and write the equation. Write equations for the
combination of NaOH and H2SO4; NaOH and HNO3; KOH and HCl; KOH
and HNO3; NH4OH and HCl; NH4OH and H2SO4. Describe the experiment
represented by each equation, and be sure you can perform it if
asked to do so. What is the usual action of a salt on litmus? How
is a salt made? What else is formed at the same time? Have all
salts a saline taste? Does every salt contain a positive element
or radical? A negative?

73. A Salt is the product of the union of a positive and a
negative element or radical; it may be made by mixing a base and
an acid.

The salt KI represents what acid? What base, or hydrate? Write
the equation for making KI from its acid and base. Describe the
experiment in full. Classify, as to acids, bases, or salts: KBr,
Fe(OH)2, HI, NaBr, HNO2, Al2(OH)6, KClO3, HClO3, H2S, K2S, H2S03,
K2SO4, Ca(OH)2, CaCO3, NaBr03, CaSO4, H2CO3, K2CO3, Cu(OH)2,
Cu(NO3)2, PbSO4, H3P04, Na2P04. In the SALTS above, draw a light
vertical line, separating the positive from the negative part of
the symbol. Now state what acid each represents. What base. Write
the reaction in the preparation of each salt above from its acid
and base; then state the experiment for producing it.

74. Naming Salts.--(NO3) is the nitrate radical; KNO3 is
potassium nitrate. From what acid? (NO2) is the nitrite radical;
KN02 is potassium nitrite. From what acid? Note that the endings
of the acids are OUS and IC; also that the names of their salts
end in ITE and ATE. From which acid--IC or OUS--is the salt
ending in ATE derived? That ending in ITE?

Name these salts, the acids from which they are derived, and the
endings of both acids and salts: NaNO3, NaNO2, K2SO4, K2SO3,
CaSO4, CaSO3, KClO3, KClO2, KClO, KClO4 (use prefixes HYPO and
PER, as with acids), Ca3(PO4)2, Ca3(P03)2, CuSO4, CuSO3, AgNO3,
Cu(NO3)2. FeS, FeS2, are respectively FERROUS SULPHIDE and FERRIC
SULPHIDE. Name: HgCl, HgCl2, FeCl2, Fe2Cl6, FeSO4, Fe2(SO4)3.75.
Acid Salts.--Write symbols for nitric, sulphuric, phosphoric
acids. How many H atoms in each? Replace all the H in the symbol
of each with Na, and name the products. Again, in sulphuric acid
replace one atom of H with Na; then in phosphoric replace first
one, then two, and finally three H atoms with Na. HNaSO4 is
hydrogen sodium sulphate; HNa2P04 is hydrogen di-sodium
phosphate. Name the other salts symbolized. Name HNaNH4P04.
Though these products are all salts, some contain replaceable H,
and are called acid salts. Those which have all the H replaced by
a metal are normal salts. Name and classify, as to normal or acid
salts: Na2CO3, HNaCO3, K2SO4, HKSO4, (NH4)2SO4, HNH4SO4, Na3P04,
HNa2P04, H2NaP04.

The BASICITY of an acid is determined by the number of
replaceable H atoms in its molecule. It is called MONOBASIC if it
has one; DIBASIC if two; TRI- if three, etc. Note the basicity of
each acid named above. How many possible salts of H2SO4 with Na?
Of H3P04 with Na? Which are normal and which acid? What is the
basicity of H4Si04?

Some normal, as well as acid, salts change litmus. Na2CO3,
representing a strong base and a weak acid, turns it blue. There
are other modes of obtaining salts, but this is the only one
which we sball consider.

76. Salts Occur Abundantly in Nature, such as NaCl, MgSO4, CaCO3.
Acids and bases are found in small quantities only. Why is this?
Why are there not springs of H2SO4 and NH4OH? We have seen that
acids and bases are extremely active, have opposite characters,
and combine to form relatively inactive salts. If they existed in
the free state, they would soon combine by reason of their strong
affinities. This is what in all ages of the world has taken
place, and this is why salts are common, acids and bases rare.
Active agents rarely exist in the free state in large quantities.
Oxygen seems to be an exception, but this is because there is a
superabundance of it. While vast quantities are locked up in
compounds in rocks, water, and salts of the earth, much remains
with which there is nothing to combine.

CHAPTER XVII.

CHLORHYDRIC ACID.

77. We have seen that salts are made by the union of acids and
bases. Can these last be obtained from salts?

78. Preparation of HCl.

Experiment 47.--Into a flask put 10 g. coarse NaCl, and add 20
cc. H2SO4. Connect with Woulff bottles [Woulff bottles may be
made by fitting to wide-mouthed bottles corks with three holes,
through which pass two delivery tubes, and a central safety tube
dipping into the liquid, as in Figures 22 and 23.] partly filled
with water, as in Figure 22. One bottle is enough to collect the
HCl; but in that case it is less pure, since some H2SO4 and other
impurities are carried over. Several may be connected, as in
Figure 23. The water in the first bottle must be nearly saturated
before much gas will pass into the second. Heat the mixture 15 or
20 minutes, not very strongly, to prevent too much foaming.
Notice any current in the first bottle. NaCl + H2SO4 = HNaSO4 +
HCl. Intense heat would have given: 2NaCl + H2SO4 = Na2SO4 +
2HCl. Compare these equations with those for HNO3. In which
equation above is H2SO4 used most economically? Both reactions
take place when HCl is made on the large scale.

(Fig. 22)

79. Tests. Experiment 48.--(1) Test with litmus the liquid in
each Woulffbottle. (2) Put a piece of Zn into a t.t. and cover it
with liquid from the first bottle. Write the reaction, and test
the gas. (3) To 2 cc.solution AgNO3 in a t.t. add 2 cc.of the
acid. Describe, and write the reaction. Is AgCl soluble in water?
(4) Into a t.t. pour 5 cc.Pb(NO3)2 solution, and add the same
amount of prepared acid. Give the description and the reaction.
(5) In the same way test the acid with Hg2(NO3)2 solution, giving
the reaction. (6) Drake a little HCl in a t.t., and bring the gas
escaping from the d.t. in contact with a burning stick. Does it
support the combustion of C? (7) Hold a piece of dry litmus paper
against it. [figure 23] (8) Hold it over 2 cc.of NH4OH in an
evaporating-dish. Describe, name the product, and write the
reaction. (3), (4), (5), (8), are characteristic tests for this
acid.

80. Chlorhydric, Hydrochloric or Muriatic, Acid is a Gas.--As
used, it is dissolved, in water, for which it has great affinity.
Water will hold, according to temperature, from 400 to 500 times
its volume of HCl. Hundreds of thousands of tons of the acid are
annually made, mostly in Europe, as a bye-product in Na2CO3
manufacture. The gas is passed into towers through which a spray
of water falls; this absorbs it. The yellow color in most
commercial HCl indicates impurities, some of which are Fe, S, As,
and organic matter. As, S, etc., come from the pyrites used in
making H2SO4. Chemically pure (C.P.) acid is freed from these,
and is without color. The gas may be dried by passing it through
a glass tube holding CaCl2 (Fig. 16) and collecting it over
mercury.

The muriatic acid of commerce consists of about two- thirds water
by weight. HCl can also be made by direct union of its
constituents.81. Uses.--HCl is used to make Cl, and also
bleaching- powder. Its use as a reagent in the laboratory is
illustrated by the following experiment:-- Experiment 49.--Put
into a t.t. 2 cc. AgNO3 solution, add 5 cc. H2O, then add slowly
HCl so long as a ppt. (precipitate) is formed. This ppt. is AgCl.
Now in another t.t. put 2 cc. Cu(NO3)2, solution, add 5 cc. H2O,
then a little HCl. No ppt. is formed. Now if a solution of AgNO3
and a solution of Cu(NO3)2 were mixed, and HCl added, it is
evident that the silver would be precipitated as chloride of
silver, while the copper would remain in solution. If now this be
filtered, the silver will remain on the filter paper, while in
the filtrate will be the copper. Thus we shall have performed an
analysis, or separated one metal from another. Perform it. Note,
however, that any soluble chloride, as NaCl, would produce the
same result as HCl.

BROMHYDRIC AND IODIHYDRIC ACIDS.

82. NaCl, being the most abundant compound of Cl, is the source
of commercial HCl. KCl treated in the same way would give a like
product. Theoretically HBr and HI might be made in the same way
from NaBr and NaI, but the affinity of H for Br and I is weak,
and the acids separate into their elements, when thus prepared.

83. To make HI.

Experiment 50.--Drop into a t.t. three or four crystals of I, and
add 10 cc. H2O. Hold in the water the end of a d.t. from which
H2S gas is escaping. Observe any deposit, and write the reaction.

FLUORHYDRIC ACID.

84. Preparation and Action.

Experiment 51.--Put 3 or 4 g. powdered CaF2, i.e. fluor spar or
fluorite, into a shallow lead tray, e.g. 4x5 cm, and pour over it
4 or 5 cc. H2SO4. A piece of glass large enough to cover this
should previously be warmed and covered on one side with a very
thin coat of beeswax. To distribute itevenly, warm the other side
of the glass over a flame. When cool, scratch a design (Fig. 24)
through the wax with a sharp metallic point. Lay the glass, film
side down, over the lead tray. Warm this five minutes or more by
placing it high over a small flame (Fig. 25) to avoid melting the
wax. Do not inhale the fumes. Take away the lamp, and leave the
tray and glass where it is not cold, for half an hour or more.
Then remove the wax and clean the glass with naphtha or benzine.
Look for the etching.

Two things should have occurred: (1) the generation of HF. Write
the equation for it. (2) Its etching action on glass. In this
last process HF acts on SiO2 of the glass, forming H2O and SiF4.
Why cannot HF be kept in glass bottles?

A dilute solution of HF, which is a gas, may be kept in gutta
percha bottles, the anhydrous acid in platinum only; but for the
most part, it is used as soon as made, its chief use being to
etch designs on glass-ware. Glass is also often etched by a blast
of sand (SiO2).

Notice the absence of O in the acids HF, HCI, HBr, HI, and that
each is a gas. HF is the only acid that will dissolve or act
appreciably on glass.

Chapter XVIII.

NITRIC ACID.

85. Preparation. Experiment 52.--To 10 g. KNO3 or NaNO3, in a
flask, add 15 cc. H2SO4. Securely fasten the cork of the d.t., as
HNO3 is likely to loosen it, and pass the other end to the bottom
of a t.t. held deep in a bottle of water (Fig. 26). Apply heat,
and collect 4 or 5 cc.of the liquid. The usual reaction is: KNO3
+ H2SO4 = HKSO4 + HNO3. With greater heat, 2 KNO3 + H2SO4 = K2SO4
+ 2HNO3. Which is most economical of KNO3? Of H2SO4? Instead of a
flask, a t.t. may be used if desired (Fig. 27).

86. Properties and Tests.

Experiment 53.--(1) Note the color of the prepared liquid. (2)
Put a drop on the finger; then wash it off at once. (3) Dip a
quill or piece of white silk into it; then wash off the acid.
What color is imparted to animal substances? (4) Add a little to
a few bits of Cu turnings, or to a Cu coin. Write the equation.
(5) To 2 cc.indigo solution, add 2 cc. HNO3. State the leading
properties of HNO3, from these tests.

87. Chemically Pure HNO3 is a Colorless Liquid.-- The yellow
color of that prepared in Experiment 52 is due to liquid NO2
dissolved in it. It is then called fuming HNO3, and is very
strong. NO2 is formed at a high temperature.

Commercial or ordinary HNO3, is made from NaNO3, this being
cheaper than KNO3; it is about half water.

88. Uses. HNO3 is the basis of many nitrates, as AgNO3, used for
photography, Ba(NO3)2 and Sr(NO3)2 for fire-works, and others for
dyeing and printing calico; it is employed in making aqua regia,
sulphuric acid, nitro-glycerine, gun-cotton, aniline colors,
zylonite, etc.

Enough experiments have been performed to answer the question
whether some acids can be prepared from their salts. H2SO4 is not
so made, because no acid is strong enough to act on its salts. In
making HCl, HNO3, etc., sulphuric acid was used, being the
strongest.

AQUA REGIA.

89. Preparation and Action. Experiment 54.--Into a t.t. put 2 cc.
HNO3, and 14 qcm. of either Au leaf or Pt. Warm in a flame. If
the metal is pure, no action takes place. Into another tube put 6
cc. HCl and add a similar leaf. Heat this also. There should be
no action. Pour the contents of one t.t. into the other. Note the
effect. Which is stronger, one of the acids, or the combination
of the two? Note the odor. It is that of Cl. 3HCl + HNO3 = NOCl +
2H2O + Cl2. This reaction is approximate only. The strength is
owing to nascent chlorine, which unites with Au. Au + 3Cl =
AuCl3. If Pt be used, PtCl4 is produced. No other acid except
nitro-hydrochloric will dissolve Au or Pt; hence the ancients
called it aqua regia, or king of liquids. It must be made as
wanted, since it cannot be kept and retain its strength.

CHAPTER XIX.

SULPHURIC ACID.

90. Preparation.

Experiment 55.--Having fitted a cork with four or five
perforations to a large t.t., pass a d.t. from three of these to
three smaller t.t., leaving the others open to the air, as in
Figure 28. Into one t.t. put 5 cc. H2O, into another 5 g. Cu
turnings and 10 cc. H2SO4, into the third 5 g. Cu turnings and 10
cc. dilute HNO3, half water. Hang on a ring stand, and slowly
heat the tubes containing H2O and H2SO4. Notice the fumes that
pass into the large t.t.

Trace out and apply to Figure 28 these reactions:--

(1) Cu + 2 H2SO4 = CuSO4 + 2 H2O + SO2.

(2) 3 Cu + 8 HNO3 = 3 Cu(NO3)2+ 4 H2O + 2 NO.

(3) NO + O = NO2.

(4) SO2 + H2O + NO2 =H2SO4 + NO.

(4) comes from combining the gaseous products in (1), (2), (3).
In (3), NO takes an atom of O from the air, becoming NO2, and at
once gives it up, to the H2SO3 (H2O + SO2), making H2SO4, and
again goes through the same operation of taking up O and passing
it along. NO is thus called a carrier of O. It is a reducing
agent, while NO2 is an oxidizing agent. This is a continuous
process, and very important, since it changes useless H2SO3 into
valuable H2SO4. If exposed to the air, H2SO3 would very slowly
take up O and become H2SO4.

Instead of the last experiment, this may be employed if
preferred: Burn a little S in a receiver. Put into an
evaporating-dish, 5 cc. HNO3, and dip a paper or piece of cloth
into it. Hang the paper in the receiver of SO2, letting no HNO3
drop from it. Continue this operation till a small quantity of
liquid is found in the bottle. The fumes show that HNO3 has lost
O. 2 HNO3 + SO2 = H2SO4 + 2 NO2.

91. Tests for H2SO4.

Experiment 56.--(1) Test the liquid with litmus. (2) Transfer it
to a t.t., and add an equal volume of BaCl2 solution. H2SO4 +
BaCl2 = ? Is BaSO4 soluble? (3) Put one drop H2SO4 from the
reagent bottle in 10 cc. H2O in a clean t.t., and add 1 cc. BaCl2
solution. Look for any cloudiness. This is the characteristic
test for H2SO4 and soluble sulphates, and so delicate that one
drop in a liter of H2O can be detected. (4) Instead of H2SO4, try
a little Na2SO4 solution. (5) Put two or three drops of strong
H2SO4 on writing-paper, and evaporate, high over a flame, so as
not to burn the paper. Examine it when dry. (6) Put a stick into
a t.t. containing 2 cc. H2SO4, and note the effect. (7) Review
Experiment 5. (8) Into an e.d. pour 5 cc. H2O, and then 15 cc.
H2SO4. Stir it meantime with a small t.t. containing 2 or 3 cc.
NH4OH, and notice what takes place in the latter; also note the
heat of the e.d.

The effects of (5), (6), (7), and (8) are due to the intense
affinity which H2SO4 has for H2O. So thirsty is it that it even
abstracts H and O from oxalic acid in the right proportion to
form H2O, combines them, and then absorbs the water.

92. Affinity for Water.--This acid is a desiccator or dryer, and
is used to take moisture from the air and prevent metallic
substances from rusting. In this way it dilutes itself, and may
increase its weight threefold. In diluting, the acid must always
be poured into the water slowly and with stirring, not water into
the acid, since, as H2O is lighter than H2SO4, heat enough may be
set free at the surface of contact to cause an explosion.
Contraction also takes place, as may be shown by accurately
measuring each liquid in a graduate, before mixing, and again
when cold. The mixture occupies less volume than the sum of the
two volumes. For the best results the volume of the acid should
be about three times that of the water.

93. Sulphuric Acid made on a Large Scale involves the same
principles as shown in Experiment 55, excepting that S02 is
obtained by burning S or roasting FeS2 (pyrite),

[Fig. 29.]

and HNO3 is made on the spot from NaNO3 and H2SO4. SO2 enters a
large leaden chamber, often 100 to 300 feet long, and jets of
steam and small portions of HNO3 are also forced in. The "chamber
acid" thus formed is very dilute, and must be evaporated first in
leaden pans, and finally in glass or platinum retorts, since
strong H2SO4, especially if hot, dissolves lead. See Experiment
124. Study Figure 29, and write the reactions. 2 HNO3 breaks up
into 2 NO2, H2O, and O. 94. Importance.--Sulphuric acid has been
called, next to human food, the most indispensable article known.
There is hardly a product of modern civilization in the
manufacture of which it is not directly or indirectly used.
Nearly a million tons are made yearly in Great Britain alone. It
is the basis of all acids, as Na2CO3 is of alkalies. It is the
life of chemical industry, and the quantity of it consumed is an
index of a people's civilization. Only a few of its uses can be
stated here. The two leading ones are the reduction of Ca3(PO4)2
for artificial manures and the sodium carbonate manufacture.
Foods depend on the productiveness of soils and on fertilizers,
and thus indirectly our daily bread is supplied by means of this
acid; and from sodium carbonate glass, soap, saleratus, baking-
powders, and most alkalies are made directly or indirectly. H2SO4
is employed in bleaching, dyeing, printing, telegraphy,
electroplating, galvanizing iron and wire, cleaning metals,
refining Au and Ag, making alum, blacking, vitriols, glucose,
mineral waters, ether, indigo, madder, nitroglycerine, gun-
cotton, parchment, celluloid, etc., etc.

FUMING SULPHURIC ACID.

95. Nordhausen or Fuming Sulphuric Acid, H2S207 used in
dissolving indigo and preparing coal-tar pigments, is made by
distilling FeSO4. 4FeSO4 + H2O = H2S207 + 2Fe203 + 2S02. This was
the original sulphuric acid. It is also formed when S03 is
dissolved in H2SO4. When exposed to the air, S03 escapes with
fuming.

CHAPTER XX.

AMMONIUM HYDRATE.

96. Preparation of Bases.--We have seen that many acids are made
by acting on a salt of the acid required, with a stronger acid.
This is the direct way. The following experiments will show that
bases may be prepared in a similar way by acting on salts of the
base required with other bases, which we may regard as stronger
than the ones to be obtained.

97. Preparation of NH4OH and NH3.

Experiment 57.--Powder 10 g. ammonium chloride, NH4Cl, in a mortar
and mix with 10 g. calcium hydrate, Ca(OH)2; recently slaked lime
is the best. Cover with water in a flask, and connect with Woulff
bottles, as for making HCl (Fig. 22); heat the flask for fifteen
minutes or more. The experiment may be tried on a smaller scale
with a t.t. if desired.

The reaction is: 2NH4Cl + Ca(OH)2 = CaCl2 + 2NH4OH. NH4OH is
broken up into NH3, ammonia gas, and water. NH4OH = NH3 + H2O.
These pass over into the first bottle, where the water takes up
the NH3, for which it has great affinity. One volume of water at
0 will absorb more than 1000 volumes of NH3. Thus NH4OH may be
called a solution of NH3, in H2O. Write the reaction.

Experiment 58.--Powder and mix 2 or 3 g. each of ammonium
nitrate, NH4NO3, and Ca(OH)2; put them into a t.t., and heat
slowly. Note the odor. 2NH4NO3 + Ca(OH)2 = ?

98. Tests.

Experiment 59.--(1) Generate a little of the gas in a t.t., and
note the odor. (2) Test the gas with wet red litmus paper. (3)
Put a little HCl into an e.d., and pass over it the fumes of NH3
from a d.t. Note the result, and write the equation. (4) Fill a
small t.t. with the gas by upward displacement; then, while still
inverted, put the mouth of the t.t. into water. Explain the rise
of the water. (5) How might NH4Cl be obtained from the NH4OH in
the Woulff bottles? (6) Test the liquid in each bottle with red
litmus paper. (7) Add some from the first bottle to 5 or 10 cc.
of a solution of FeSO4 or FeCl2, and look for a ppt. State the
reaction.

99. Formation.--Ammonia, hartshorn, exists in animal and
vegetable compounds, in salts, and, in small quantities, in the
atmosphere. Rain washes it from the atmosphere into the soil;
plants take it from the soil; animals extract it from plants.
Coal, bones, horns, etc., are the chief sources of it, and from
them it is obtained by distillation. It results also from
decomposing animal matter. NH3 can be produced by the direct
union of N and H, only by an electric discharge or by ozone. It
may be collected over Hg like other gases that are very soluble
in water.

100. Uses. --Ammonium hydrate, NH4OH, and ammonia, NH3, are used
in chemical operations, in making artificial ice, and to some
extent in medicine; from them also may be obtained ammonium
salts. State what you would put with NH4OH to obtain (NH4)2SO4.
To obtain NH4NO3. The use of NH4OH in the laboratory may be
illustrated by the following experiment:--

Experiment 60.--Into a t.t. put 10 cc. of a solution of ferrous
sulphate, FeSO4. Into another put 10 cc. of sodium sulphate
solution, Na2SO4. Add a little NH4OH to each. Notice a ppt. in
the one case but none in the other. If solutions of these two
compounds were mixed, the metals Fe and Na could be separated by
the addition of NH4OH, similar to the separation of Ag and Cu by
HCl. Try the experiment.

CHAPTER XXI.

SODIUM HYDRATE.

101. Preparation.

Experiment 61.--Dissolve 3 g. sodium carbonate, Na2CO3, in 10 or
15 cc. H2O in an e.d., and bring it to the boiling-point. Then
add to this a mixture of 1 or 2 g. calcium hydrate, Ca(OH)2, in 5
or 10cc. H2O. It will not dissolve. Boil the whole for five
minutes. Then pour off the liquid which holds NaOH in solution.
Evaporate if desired. This is the usual mode of preparing NaOH.

The reaction is Na2CO3 + Ca(OH)2 = 2NaOH + CaCO3. The residue is
Ca(OH)2 and CaCO3; the solution contains NaOH, which can be
solidified by evaporating the water. Sodium hydrate is an
ingredient in the manufacture of hard soap, and for this use
thousands of tons are made annually, mostly in Europe. It is an
important laboratory reagent, its use being similar to that of
ammonium hydrate. Exposed to the air, it takes up water and CO2,
forming a mixture of NaOH and Na2CO3. It is one of the strongest
alkalies, and corrodes the skin.

Experiment 62.--Put 20 cc. of H2O in a receiver. With the forceps
take a piece of Na, not larger than half a pea, from the naphtha
in which it is kept, drop it into the H2O, and at once cover the
receiver loosely with paper or cardboard. Watch the action, as
the Na decomposes H2O. HOH + Na = NaOH + H. If the water be hot
the action is so rapid that enough heat is produced to set the H
on fire. That the gas is H can be shown by putting the Na under
the mouth of a small inverted t.t., filled with cold water, in a
water-pan. Na rises to the top, and the t.t. fills with H, which
can be tested. NaOH dissolves in the water.102. Properties.

Experiment 63.--(1) Test with red litmus paper the solutions
obtained in the last two experiments. (2) To 5cc.of alum
solution, K2A12(SO4)4, add 2cc.of the liquid, and notice the
color and form of the ppt.

POTASSIUM HYDRATE.

103. KOH is made in the Same Way as NaOH.

Describe the process in full (Experiment 61), and give the
equation.

Experiment 64.--Drop a small piece of K into a receiver of H2O,
as in Experiment 62. The K must be very small, and the experiment
should not be watched at too close a range. The receiver should
not be covered with glass, but with paper. The H burns, uniting
with O of the air. The purple color is imparted by the burning,
or oxidation of small particles of K. Write the equation for the
combustion of each.

H2O might be considered the symbol of an acid, since it is the
union of H and a negative element; or, if written HOH, it might
be called a base, since it has a positive element and the (OH)
radical. It is neutral to litmus, and on this account might be
called a salt. It is better, however, to call it simply an oxide.

Potassium hydrate, caustic potash, is employed for the
manufacture of soft soap. As a chemical reagent its action is
almost precisely like that of caustic soda, though it is usually
considered a stronger base, as K is a more electro-positive
element than Na.

CALCIUM HYDRATE.

104. Calcium Hydrate, the Most Common of the Bases, is nearly as
important to them as H2SO4 is to acids. Since it is used to make
the other bases, it might be called the strongest base; as H2SO4
is often called the strongest acid. The strength of an acid or
base, however,depends on the substance to which it is applied, as
well as on itself, and for most purposes this one is classified
as a weaker base than the three previously described.

Sulphuric acid, the most useful of the acids, is not made
directly from its salts, but has to be synthesized. Calcium
hydrate is also made by an indirect process, as follows:

CaCO3, i.e. limestone, marble, etc., is burnt in kilns with C, a
process which separates the gas, CO2, according to the reaction:
CaCO3 = CaO + CO2. CaO is unslaked lime, or quick-lime. On
treating this with water, slaked lime, Ca(OH)2 is formed, with
generation of great heat. CaO + H2O = Ca(OH)2. Its affinity for
H2O is so great that it takes the latter from the air, if
exposed.

Experiment 65.--Saturate some unslaked lime with water, in an
e.d., and look for the results stated above, leaving it as long
as may be necessary.

105. Resume.--From the experiments in the last few chapters on
the three divisions of chemical compounds, acids, bases and
salts, we have seen (1) that acids and bases are the chemical
opposites of each other; (2) that salts are formed by the union
of acids and bases; (3) that some acids can be obtained from
their salts by the action of a stronger acid; (4) that some bases
can be got from salts by the similar action of other bases; (5)
that the strongest acids and bases, as well as others, may be
obtained in an indirect way by synthesis.

CHAPTER XXII.

OXIDES OF NITROGEN.

106. There are five oxides of N, only two of which are important.

NITROGEN MONOXIDE (N2O).

107. Preparation.

Experiment 66.--Put into a flask, holding 200cc, lOg of ammonium
nitrate, NH4NO3; heat it over wire gauze or asbestus in an iron
plate, having a d.t. connected with a large t.t., which is held
in a receiver of water, and from this t.t., another d.t. passing
into a pneumatic trough, so as to collect the gas over water
(Fig. 30). Have all the bearings tight. The reaction is NH4NO3 =
2H2O + N2O. The t.t. is for collecting the H2O.

[Fig. 30.]

Note the color of the liquid in the t.t.; taste a drop, and test
it with litmus. If the flask is heated too fast, some NO is
formed, and this taking O from the air makes NO2, which liquefies
and gives an acid reaction and a red color. Some NH4NO3 is also
liable to be carried over.

108. Properties.

Experiment 67.--Test the gas in the receiver with a burning stick
and a glowing one, and compare the combustion with that in O.
N20may also be tested with S and P, if desired. N is set free in
each case. Write the reactions.

Nitrogen monoxide or protoxide, the nitrous oxide of dentists,
when inhaled, produces insensibility to pain,-- anaesthesia,--
and, if continued, death from suffocation. Birds die in half a
minute from breathing it. Mixed with one-fourth O, and inhaled
for a minute or two, it produces intoxication and laughter, and
hence is called laughing gas. As made in Experiment 66, it
contains Cl and NO, as impurities, and should not be breathed.

NITROGEN DIOXIDE (NO, OR N2O2).

109. Preparation.

Experiment 68.--Into a t.t. or receiver put 5g Cu turnings, add 5
cc. H2O and 5 cc. HNO3. Collect the gas like H, over water. 3Cu +
8HNO3 = ? What two products will be left in the generator? Notice
the color of the liquid. This color is characteristic of Cu
salts. Notice also the red fumes of NO2.

110. Properties.

Experiment 69.--Test the gas with a burning stick, admitting as
little air as possible. Test it with burning S. NO is not a
supporter of C and S combustion. Put a small bit of P in a
deflagrating-spoon, and when it is vigorously burning, lower it
into the gas. It should continue to burn. State the reaction.
What combustion will NO support? Note that NO is half N, while
N2O is two-thirds N, and account for the difference in supporting
combustion.

NITROGEN TETROXIDE (NO2 or N2O4).

111. Preparation.

Experiment 70.--Lift from the water-pan a receiver of NO, and
note the colored fumes. They are NO2, or N2O4, nitrogen
tetroxide. NO + O = NO2. Is NO combustible? What is the source of
O in the experiment?OXIDES OF NITROGEN.

NITROGEN TRIOXIDE (N2O3).

112. Preparation.

Experiment 71.--Put into a t.t. 1 g. of starch and 1 cc. of HNO3.
Heat the mixture for a minute. The red fumes are N2O3 and NO2.

Nitrogen pentoxide, N2O5, is an unimportant solid. United with
water it forms HNO3. N2O5 + H2O = 2HNO3.

CHAPTER XXIII.

LAWS OF DEFINITE AND OF MULTIPLE PROPORTION.

113. Weight and Volume.--We have seen that water contains two
parts of H by volume to one part of O; or, by weight, two parts
of H to sixteen of O. These proportions are invariable, or no
symbol for water would be possible. Every compound in the same
way has an unvarying proportion of elements.

114. Law of Definite Proportion.--In a given compound the
proportion of any element by weight, or, if a gas, by volume is
always constant. Apply the law, by weight and by volume, to
these: HCl, NH3, H2S, N2O.

There is another law of equal importance in chemistry, which the
compounds of N and O well illustrate.

Weight. Volume.
N. O. N. O.
Nitrogen protoxide N2O 28 16 2 1
Nitrogen dioxide N2O2 28 32 2 2
Nitrogen trioxide. N2O3 28 48 2 3
Nitrogen tetroxide N2O4 28 64 2 4
Nitrogen pentoxide N2O5 28 80 2 5

Note that the proportion of O by weight is in each case a
multiple of the first, 16. Also that the proportion by volume of
O is a multiple of that in the first compound. In this example
the N remains the same. If that had varied in the different
compounds, it would also havevaried by a multiple of the smallest
proportion. This is true in all compounds.

115. Law of Multiple Proportion.--Whenever one element combines
with another in more than one proportion, it always combines in
some multiple, one or more, of its least combining weight, or, if
a gas, of its least combining volume.

The least combining weight of an element is its atomic weight;
and it is this fact of a least combining weight that leads us to
believe the atom to be indivisible.

Apply the law in the case of P2O, P2O3, P2O5; in HClO, HClO2,
HClO3, HClO4, arranging the symbols, weights, and volumes in a
table, as above.

The volumetric proportions of each element in the oxides of
nitrogen are exhibited below.

_ + _ + _ = __
N + N + O = N2O

_ + _ + _ + _ = __
N + N + O + O = N2O2

_ + _ + _ + _ + _ = __
N + N + O + O + O = N2O3

_ + _ + _ + _ + _ + _ = __
N + N + O + O + O + O = N2O4

_ + _ + _ + _ + _ + _ + _ = __
N + N + O + O + O + O + O = N2O5

CHAPTER XXIV.

CARBON PROTOXIDE.

116. Preparation.

Experiment 72.--Put into a flask, of 200 cc., 5 g. of oxalic acid
crystals, H2C2O4, and 25 cc. H2SO4. Have the d.t. pass into a
solution of NaOH in a Woulff bottle (Fig. 31), and collect
the gas over water. Heat the flask slowly, and avoid inhaling the
gas.

117. Tests.

Experiment 73.--Remove a receiver of the gas, and try to light
the latter with a splinter. Is it combustible, or a supporter of
(C) combustion? What is the color of the flame? When the
combustion ceases, shake up a little lime water with the gas left
in the receiver. What gas has been formed by the combustion, as
shown by the test? See page 80. Give the reaction for the
combustion.

We have seen that H2SO4 has great affinity for H2O. Oxalic acid
consists of H, C, O in the right proportion to form H2O, CO2, and
CO. H2SO4 withdraws H and O in the right proportion to form
water, unites them, and then absorbs the water, leaving the C and
O to combine and form CO2 and CO. NaOH solution removes CO2 from
the mixture, forming Na2CO3, and leaves CO. Write both reactions.

118. Carbon Protoxide, called also carbon monoxide, carbonic
oxide, etc., is a gas, having no color or taste, butpossessing a
faint odor. It is very poisonous. Being the lesser oxide of C, it
is formed when C is burned in a limited supply of O, whereas CO2
is always produced when O is abundant. The formation of each is
well shown by tracing the combustion in a coal fire. Air enters
at the bottom, and CO2 is first formed. C + 2O = CO2. As this gas
passes up, the white-hot coal removes one atom of O, leaving CO.
CO2 + C - 2CO. At the top, if the draft be open, a blue flame
shows the combustion of CO. CO + O = CO2. The same reduction of
CO2 takes place in the iron furnace, and whenever there is not
enough oxygen to form CO2, the product is CO.

Great care should be taken that this gas does not escape into the
room, as one per cent has proved fatal. Not all of it is burned
at the top of the coal; and when the stove door is open, the
upper drafts should be open also. It is the most poisonous of the
gases from coal; hence the danger from sleeping in a room having
a coal fire.

119. Water Gas.--CO is one of the constituents of "water gas,"
which, by reason of its cheapness, is supplanting gas made from
coal, as an illuminator, in some cities. It is made by passing
superheated steam over red-hot charcoal or coke. C unites with
the O of H2O, forming CO, and sets H free, thus producing two
inflammable gases. C + H2O --? As neither of these gives much
light, naphtha is distilled and mixed with them in small
quantities to furnish illuminating power See page 183.

CHAPTER XXV.

CARBON DIOXIDE.

120. Preparation.

Experiment 74.--Put into a t.t., or a bottle with a d.t. and a
thistle-tube, 10 or 20 g. CaCO3, marble in lumps; add as many
cubic centimeters of H2O, and half as much HCl, and collect the
gas by downward displacement (Fig. 39). Add more acid as needed.
CaCO3 + 2 HCl = CaCl2 + H2CO3. H2CO3 = H2O + CO2. H2CO3 is a very
weak compound, and at once breaks up. By some, its existence as a
compound is doubted.

121. Tests.

Experiment 75.--(1) Put a burning and a glowing stick into the
t.t. or bottle. (2) Hold the end of the d.t. directly against the
flame of a small burning stick. Does the gas support combustion?
(3) Pour a receiver of the gas over a candle flame. What does
this show of the weight of the gas? (4) Pass a little CO2 into
some H2O (Fig. 32), and test it with litmus. Give the reaction
for the solution of CO2 in H2O.

Experiment 76.--Put into a t.t. 51 cc. of clear Ca(OH)2 solution,
i.e. lime water; insert in this the end of a d.t. from a CO2
generator (Fig. 32). Notice any ppt. formed. It is CaCO3. Let the
action continue until the ppt. disappears and the liquid is
clear. Then remove the d.t., boil the clear liquid for a minute,
and notice whether the ppt. reappears.

122. Explanation.

Ca(OH)2 + CO2 = CaCO3 + H2O. The curious phenomena of this
experiment are explained by the solubility of CaCO3 in water
containing CO2, and its insolu-bility in water, having no CO2.
When all the Ca(OH)3 is combined, or changed to CaCO3, the excess
of CO2 unites with H2O, forming the weak acid H2CO3, which
dissolves the precipitate, CaCO3, and gives a clear liquid. On
heating this, H2CO3 gives up its CO2, and CaCO3 is
reprecipitated, not being soluble in pure water.

Lime water, Ca(OH)2 solution, is therefore a test for the
presence of CO2. To show that carbon dioxide is formed in
breathing, and in the combustion of C, and that it is present in
the air, perform the following experiment:

Experiment 77.--(1) Put a little lime water into a t.t., and blow
into it through a piece of glass tubing. Any turbidity shows
what? (2) Burn a candle for a few minutes in a receiver of air,
then take out the candle and shake up lime water with the gas.
(3) Expose some lime water in an e.d. to the air for some time.

133. Oxidation in the Human System.--Carbon dioxide, or carbonic
anhydride, carbonic acid, etc., CO2, is a heavy gas, without
color or odor. It has a sharp, prickly taste, and is commonly
reckoned as poisonous if inhaled in large quantities, though it
does not chemically combine with the blood as CO does. Ten per
cent in the air will sometimes produce death, and five per cent
produces drowsiness. It exists in minute portions in the
atmosphere, and often accumulates at the bottom of old wells and
caverns, owing to its slow diffusive power. Before going down
into one of these, the air should always be tested by lowering a
lighted candle. If this is extinguished, there is danger. CO2 is
the deadly "choke damp" after a mine explosion, CH4 being
converted into CO2 and H2O; a great deal is liberated during
volcanic eruptions, and it is formed in breathing by the union of
O in the air with C in the system. This union of C and O takes
place in the lungs and in all the tissues of the body, even on
the surface. Oxygen is taken into the lungs, passes through the
thin membrane into the blood, forms a weak chemical union with
the red corpuscles, and is conveyed by them to all parts of the
system. Throughout the body, wherever necessary, C and H are
supplied for the O, and unite with it to form CO2 and H2O. These
are taken up by the blood though they do not form a chemical
union with it, are carried to the lungs, and pass out, together
with the unused N and surplus O. The system is thus purified, and
the waste must be supplied by food. The process also keeps up the
heat of the body as really as the combustion of C or P in O
produces heat. The temperature of the body does not vary much
from 99 degrees F., any excess of heat passing off through
perspiration, and being changed into other forms of energy.

If, as in some fevers, the temperature rises above about 105
degrees F., the blood corpuscles are killed, and the person dies.
During violent exercise much material is consumed, circulation is
rapid, and quick breathing ensues. Oxygen is necessary for life.
A healthy person inhales plentifully; and this element is one of
nature's best remedies for disease. Deep and continued
inhalations in cold weather are better than furnace fires to heat
the system. All animals breathe O and exhale CO2. Fishes and
other aquatic animals obtain it, not by decomposing H2O, but from
air dissolved in water. Being cold-blooded, they need relatively
little; but if no fresh water is supplied to those in captivity,
they soon die of O starvation.

124. Oxidation in Water.--Swift-running streams are clear and
comparatively pure, because their organic impurities are
constantly brought to the surface and oxidized, whereas in
stagnant pools these impurities accumulate. Reservoirs of water
for city supply have sometimes been freed from impurities by
aeration, i.e. by forcing air into the water.

125. Deoxidation in Plants.--Since CO2 is so constantly poured
into the atmosphere, why does it not accumulate there in large
quantity? Why is there not less free O in the air to-day than
there was a thousand years ago? The answer to these questions is
found in the growth of vegetation. In the leaf of every plant are
thousands of little chemical laboratories; CO2 diffused in small
quantities in the air passes, together with a very little H2O,
into the leaf, usually from its under side, and is decomposed by
the radiant energy of the sun. The C is built into the woody
fiber of the tree, and the O is ready to be re-breathed or burned
again. CO2 contributes to the growth of plants, O to that of
animals; and the constituents of the atmosphere vary little from
one age to another. The compensation of nature is here well
shown. Plants feed upon what animals discard, transforming it
into material for the sustenance of the latter, while animals
prepare food for plants. All the C in plants is supposed to come
from the CO2 in the atmosphere. Animals obtain their supply from
plants. The utility of the small percentage of CO2 in the air is
thus seen.

126. Uses.--CO2 is used in making "soda-water," and in chemical
engines to put out fires in their early stages. In either case it
may be prepared by treating Na2CO3 or CaCO3 with H2SO4. Give the
reactions. On a small scale CO2 is made from HNaCO3. CO2 has a
very weak affinity for water, but probably forms with it H2CO3.
Much carbon dioxide can be forced into water under pressure. This
forms soda-water, which really contains no soda. The
justification for the name is the material from which it is
sometimes made. Salts from H2CO3, called carbonates, are
numerous, Na2CO3 and CaCO3 being the most important.

Chapter XXVI.

OZONE.

127. Preparation.

Experiment 78.--Scrape off the oxide from the surface of a piece
of phosphorus 2 cm long, put it into a wide-mouthed bottle, half
cover the P with water, cover the bottle with a glass, and leave
it for half an hour or more.

128. Tests.

Experiment 79.--Remove the glass cover, smell the gas, and hold
in it some wet iodo-starch paper. Look for any blue color. Iodine
has been set free, according to the reaction, 2 KI + 03= K20 + O2
+ I2, and has imparted a blue color to the starch, and ordinary
oxygen has been formed. Why will not oxygen set iodine free from
KI?. What besides ozone will liberate it?

129. Ozone, oxidized oxygen, active oxygen, etc., is an
allotropic form of O. Its molecule is 03, while that of ordinary
oxygen is 02.

Three atoms of oxygen are condensed into the space of two atoms
of ozone, or three molecules of O are condensed into two
molecules of ozone, or three liters of O are condensed into two
liters of ozone. Ozone is thus formed by oxidizing ordinary
oxygen. 02 + O = 03. This takes place during thunder storms and
in artificial electrical discharges. The quantity of ozone
produced is small, five per cent being the maximum, and the usual
quantity is far less than that.

Ozone is a powerful oxidizing agent, and will change S, P, and As
into their ic acids. Cotton cloth was formerly bleached, and
linen is now bleached, by spreading it on the grass and leaving
it for weeks to be acted on by ozone, which is usually present in
the air in small quantities, especially in the country. Ozone is
a disinfectant, like other bleaching agents, and serves to clear
the air of noxious gases and germs of infectious diseases. So
much ozone is reduced in this way that the air of cities contains
less of it than country air. A third is consumed in uniting with
the substance which it oxidizes, while two-thirds are changed
into oxygen, as in Experiment 79.

It is unhealthful to breathe much ozone, but a little in the air
is desirable for disinfection.

Ozone will cause the inert N of the air to unite with H, to form
ammonia. No other agent capable of doing this is known, so that
all the NH3 in the air, in fact all ammonium compounds taken up
by plants from soils and fertilizers, may have been made
originally through the agency of ozone. At a low temperature
ozone has been liquefied. It is then distinctly blue.

Electrolysis of water is the best mode of preparing this
substance in quantity. When prepared from P it is mixed with
P2O3.

Chapter XXVII.

CHEMISTRY OF THE ATMOSPHERE.

130. Constituents.--The four chief constituents of the atmosphere
are N, O, H2O, CO2, in the order of their abundance. What
experiments show the presence of N, O, and CO2 in the air? Set a
pitcher of ice water in a warm room, and the moisture that
collects on the outside is deposited from the air. This shows the
presence of H2O. Rain, clouds, fog, and dew prove the same. H2SO4
and CaCl2, on exposure to air, take up water. Experiment 18 shows
that there is not far from four times as much N as O by volume in
air. Hence if the atmosphere were a compound of N and O, and the
proportion of four to one were exact, its symbol would be N4O.

131. Air not a Compound.--The following facts show that air is
not a compound, but rather a mixture of these gases.

1. The proportion of N and O in the air, though it does not vary
much, is not always exactly the same. This could not be true if
it were a compound. Why?

2. If N4O were dissolved in water, the N would be four times the
O in volume; but when air is dissolved, less than twice as much N
as O is taken up.

3. No heat or condensation takes place when four measures of N
are brought in contact with one of O. It cannot then be N4O, for
the vapor density of N4O would be 36--i.e. (14 x 4 + 16) / 2; but
that of air is 14 1/2 nearly --i.e. (14 x 4 + 16) / 5. Analysis
shows about 79 parts of N to 21 parts of O by volume in air.

132. Water.--The volume of H2O, watery vapor, in the atmosphere
is very variable. Warm air will hold more than cold, and at any
temperature air may be near saturation, i.e. having all it will
hold at that temperature, or it may have little. But some is
always present; though the hot desert winds of North Africa are
not more than 1/15 saturated. A cubic meter of air at 25 degrees,
when saturated, contains more than 22 g. of water.

133. Carbon Dioxide.--Carbon dioxide does not make up more than
three or four parts in ten thousand of the air; but, in the whole
of the atmosphere, this gives a very large aggregate. Why does
not CO2 form a layer below the O and N?

134. Other Ingredients.--Other substances are found in the air in
minute portions, e.g. NH3 constitutes nearly one-millionth. Air
is also impregnated with living and dead germs, dust particles,
unburned carbon, etc., but these for the most part are confined
to the portion near the earth's surface. In pestilential regions
the germs of disease are said sometimes to contaminate the air
for miles around.

Chapter XXVIII.

THE CHEMISTRY OF WATER.

135. Pure Water.--Review the experiments for electrolysis, and
for burning H. Pure water is obtained by distillation.

Experiment 80.--Provide a glass tube 40 or 50 cm long and 3 or 4
cm in diameter. Fit to each end a cork with two perforations,
through one of which a long tube passes the entire length of the
larger tube (Fig. 32a). Connect one end of this with a flask of
water arranged for heating; pass the other end into an open
receptacle for collecting the distilled water. Into the other
perforations lead short tubes,-- the one for water to flow into
the large tube from a jet; the other, for the same to flow out.
This condenses the steam by circulating cold water around it. The
apparatus is called a Liebig's condenser. Put water into the
flask, boil it, and notice the condensed liquid. It is
comparatively pure water; for most of the substances in solution
have a higher boiling-point than water, and are left behind when
it is vaporized.

(Fig. 32a.)

136. Test.

Experiment 81.--Test the purity of distilled water by slowly
evaporating a few drops on Pt foil in a room free from dust.
There should be no spot or residue left on the foil. Test in the
same way undistilled water. 137. Water exists in Three States,--
solid, liquid, and vaporous. It freezes at 0 degrees, suddenly
expanding considerably as it passes into the solid state. It
boils, i.e. overcomes atmospheric pressure and is vaporized, at
100 degrees (760 mm pressure). If the pressure is greater, the
boiling-point is raised, i.e. it takes a higher temperature to
overcome a greater pressure. If there be less pressure, as on a
mountain, the boiling-point is lowered below 100 degrees. Salts
dissolved in water raise its boiling-point, and lower its
freezing-point to an extent depending on the kind and quantity of
the salt. Water, however, evaporates at all temperatures, even
from ice.

Pure water has no taste or smell, and, in small quantities, no
color. It is rarely if ever found on the earth. What is taken up
by the air in evaporation is nearly pure; but when it falls as
rain or snow, impurities are absorbed from the atmosphere. Water
falling after a long rain, especially in the country, is
tolerably free from impurities. Some springs have also nearly
pure water; but to separate all foreign matter from it, water
must be distilled. Even then it is liable to contain traces of
ammonia, or some other substance which vaporizes at a lower
temperature than water.

138. Sea-Water.--The ocean is the ultimate source of all water.
From it and from lakes, rivers, and soils, water is taken into
the atmosphere, falls as rain or snow, and sinks into the ground,
reappearing in springs, or flowing off in brooks and rivers to
the ocean or inland seas. Ocean water must naturally contain
soluble salts; and many salts which are not soluble in pure water
are dissolved in sea-water. In fact, there is a probability that
all elements exist to some extent in sea-water, but many of them
in extremely minute quantities. Sodium and magnesium salts are
the two most abundant, and the bitter taste is due to MgSO4 and
MgCl2. A liter of sea- water, nearly 1000 g., holds over 37 g. of
various salts, 29 of which are NaCl. See Hard Water.

139. River Water.--River water holds fewer salts, but has a great
deal of organic matter, living and dead, derived from the regions
through which it flows. To render this harmless for drinking,
such water should be boiled, or filtered through unglazed
porcelain. Carbon filters are now thought to possess but little
virtue for separating harmful germs.

140. Spring Water.--The water of springs varies as widely in
composition as do the rocks whence it bubbles forth. Sulphur
springs contain much H2S; many geysers hold SiO2 in solution;
chalybeate waters have compounds of Fe; others have Na2SO4, MgSO4
NaCl, etc.

CHAPTER XXIX.

THE CHEMISTRY OF FLAME.

141. Candle Flame.

Experiment 82.--Examine a candle flame, holding a dark object
behind it. Note three distinct portions: (1) a colorless interior
about the wick, (2) a yellow light-giving portion beyond that,
(3) a thin blue envelope outside of all, and scarcely
discernible. Hold a small stick across the flame so that it may
lie in all three parts, and observe that no combustion takes
place in the inner portion.

142. Explanation.--A candle of paraffine, or tallow, is chiefly
composed of compounds of C and H, in the solid state. The burning
wick melts the solid; the liquid is then drawn up by the wick
till the heat vaporizes and decomposes it, and O of the air comes
in contact with the outer heated portion of gas, and burns it
completely. Air tends to penetrate the whole body of the flame,
but only N can pass through uncombined, for the O that is left
after combustion in the outer portion seizes upon the compounds
of C and H in the next, or yellow, part. There is not enough O
here for complete combustion; at this temperature H burns before
C, and the latter is set free. In that state it is of course a
solid. Now an incandescent solid, or one glowing with heat, gives
light, while the combustion of a gas gives scarcely any light,
though it may produce great heat. While C in the middle flame is
glowing, during the moment of its dissociation from H, it gives
light. In the outer flame the temperature is high enough to burn
entirely the gaseous compounds of C and H together, so that no
solid C is set free, and hence no light is given except the faint
blue. No combustion takes place in the inner blue cone, because
no O reaches there.

By packing a wick into a cylindrical tin cup 5 or 10 cm high and
4 cm in diameter, containing alcohol, and lighting it, gunpowder
can be held in the middle of the flame in a def. spoon, without
burning. This shows the low temperature of that portion. Burning
P will also be extinguished, thus showing the exclusion of O.

143. Bunsen Flame.

Experiment 83.--Examine a Bunsen burner. Unscrew the top, and
note the orifices for the admission of gas and of air. Make a
drawing. Replace the parts; then light the gas at the top,
opening the air-holes at the base. Notice that the flame burns
with very little color. Try to distinguish the three parts, as in
the candle flame. These parts can best be seen by allowing direct
sunlight to fall on the flame and observing its shadow on a white
ground. Make a drawing of the flame. Hold across it a Pt wire and
note at what part the wire glows most. Also press down on the
flame for an instant with a cardboard or piece of paper; remove
before it takes fire, and notice the charred circle. Put the end
of a match into the blue cone, and note that it does not burn.
Put the end of a Pt wire into this blue cone, and observe that it
glows when near the top of the cone. What do these experiments
show? Ascertain whether this inner portion contains a combustible
material, by holding in it one end of a small d.t., and trying to
ignite any gas escaping at the other end. It should burn. This
shows that no combustion takes place in the interior of the
flame, because sufficient free O is not present.

Next, close the air-holes, and note that the flame is yellow and
gives much light. From this we infer the presence of solid
particles in an incandescent state. But these could not come from
the air. They must be C particles which have been set free from
the C and H compounds of the gas, just as in the candle flame.
The smoke that rises proves this. Hold an e.d. in the flame and
collect some C. Try the same with the air-holes open. 144. Light
and Heat of Flame.--Which of the two flames is hotter, the one
with the air-holes open, or that with them closed? Evidently the
former; for air is drawn in and mixes with the gas as it rises in
the tube, and, on reaching the flame at the top, the two are well
mingled, and the gaseous compounds of C and H burn at so high a
temperature that solid C is not freed; hence there is little
light. On closing the air-holes, no O can reach the flame except
from the outside, and the heat is much less intense.

(Fig 33.) (Fig 34.)

The H burns first, and sets the C free, which, while glowing,
gives the light. This again illustrates the facts (1) that flame
is caused by burning gas; (2) that light is produced by
incandescent solids. Charcoal, coke, and anthracite coal burn
without flame, or with very little, because of the absence of
gases.

145. Temperature of Combustion.

Experiment 84.--Light a Bunsen flame, with the basal orifices
open, and hold over it a fine wire gauze. Notice that the flame
does not rise above the gauze. Extinguish the light, and try to
ignite the gas above the gauze, holding the latter within 5 or 6
cm of the burner tube. Notice that it does not burn below the
gauze (Fig. 33).

Gas and O are both present. Evidently, then, the only condition
wanting for combustion is a sufficiently high temperature. The
gauze cools the gas below its kindling- point.

This principle is made use of in the miner's lamp of Davy (Fig.
34). In coal mines a very inflammable gas, CH4, called fire-damp,
issues from the coal. If this collects in large quantities and
mixes with O of the air, a kindling-point is all that is needed
to make a violent explosion. An ordinary lamp would produce this,
but the gauze lamp prevents it; for, though the inside may be
filled with burning gas, CH4, the flame cannot communicate with
the outside.

(Fig 35.) (Fig 36.)
a, reducing flame b, oxidizing flame

146. Oxidizing and Reducing Flames.--The hottest part of a Bunsen
flame is just above the inner blue cone (b, Fig. 36). Evidently
there is more O at that point. If a reducing agent, i.e. a
substance which takes up O, be put into this part of the flame,
the latter will remove the O and appropriate it, forming an
oxide. Cu heated there would become copper oxide. This part is
called the oxidizing flame. The inner blue part of the Bunsen
flame is devoid of O. It ought to remove O from an oxidizing
agent, i.e. a substance which supplies O. If copper oxide be
heated there (a, Fig. 36) by means of a mouth blow-pipe (Fig.
35), the flame will appropriate the O and leave the copper. This
is called the reducing flame. Only the upper part of this blue
central cone has heat enough to act in this way. By using a
prepared piece of metal, to make the flame thin and to shut off
the air, and then blowing the flame with a blow-pipe, greater
strength can be obtained in both oxidizing and reducing flames
(Fig. 36).

147. Combustible and Supporter Interchangeable.-- H was found to
burn in O. H was the combustible, O the supporter. Would O itself
burn in H?--i.e. would the combustible become the supporter, and
the supporter the combustible? As illuminating gas consists
largely of H, and as air is part O, we may try the experiment
with gas and air. Gas will burn in air. Will air burn in gas?

Experiment 85.--Fit a cork with two holes in it to the large end
of a lamp chimney. Through each hole pass a short piece of
tubing, and connect one of these with a rubber tube leading to a
gas-jet. Pass a metallic tube, long enough to reach the top of
the chimney, through the other, so that it will move easily up
and down. Turn on the gas, and light it at the top of the
chimney. Hold the end of the tube passing through the cork in the
flame for a minute, then draw it down to the middle of the
chimney (Fig. 37, a) and finally slowly remove it (b). Note that
O from the air is burning in the gas. Which is the supporter, and
which the combustible in this case? O will burn equally well in
an atmosphere of H, as can be shown by experiment.

148. Explosive Mixture of Gases.

Experiment 86.--Slowly turn down the burning gas of a Bunsen
lamp, having the orifices open, and notice that it suddenly
explodes and goes out at the top, but now burns at the base. As
the gas was gradually turned off, more air became mixed with it,
until there was the right proportion of each gas for an
explosion. Figure 38 shows the same thing. Light the gas at the
top a, when the tube c covers the jet b. Then gradually raise the
tube c. At a certain place there is the same explosion as with
the lamp.

149. Generalizations.--These experiments show (1) that three
conditions are necessary for combustion,--a combustible, a
supporter, and a burning temperature which varies for different
substances. Given these, "a fire" always results. The conditions
for "spontaneous combustion" do not differ from those of any
combustion. See Experiments 34, 112, 113, 114. (2) That
combustible and supporter are interchangeable. If H burns in O, O
will burn in H, the product, being the same in each case. (3) For
any combustion there must be a certain proportion of combustible
and of supporter. Twenty per cent of CO2 in the air dilutes the O
to such an extent that C will not burn. Hence the utility of the
chemical engine for putting out fires. (4) When two

gases, a combustible and a supporter, are mixed in the requisite
proportion, they form an explosive mixture, needing only the
kindling temperature to unite them.

Chemical combination is always accompanied by disengagement of
heat. Chemical dissociation is always accompanied by absorption
of heat. The disengagement, or the absorption, is not always
evident to the senses.

Combustion is the chemical combination of two or more substances
with the self-evident disengagement of great heat, and usually of
light.

The temperature of ignition varies greatly with different
substances. PH3 burns spontaneously at the usual temperatures of
the air. P takes fire at 60 degrees, but even at 10 degrees it
oxidizes with rapidity enough to produce phosphorescence. The
vapor of CS2 may be set on fire by a glass rod heated to 150
degrees, but a red-hot iron will not ignite illuminating gas.

Spontaneous combustion often takes place in woolen or cotton rags
which have been saturated with oil. The oil rapidly absorbs O,
and sets fire to the cloth. This is thought to be the origin of
some very destructive fires.

CHAPTER XXX.

CHLORINE.

150. Preparation.

Experiment 87.--Put into a t.t. 5 g. of fine granular MnO2 and 10
cc. HCl. Apply heat carefully, and collect the gas by downward
displacement in a receiver loosely covered with paper (Fig. 39).
Add more HCl if needed. Have a good draft of air, and do not
inhale the gas. If you have accidentally breathed it, inhale
alcohol vapor from a handkerchief; alcohol has great affinity for
Cl. Note the color of the gas, and compare its weight with that
of air.

MnO2 + 4 HCl = MnCl2 + 2 H2O + 2 Cl. How much Cl can be separated
with 5 g. MnO2?

If preferred, a flask may be used for a generator instead of a
t.t. Cl can be obtained directly from NaCl by adding H2SO4 (which
produces HCl) and MnO2. 2 NaCl + 2 H2SO4 + MnO2 = MnSO4 + Na2SO4
+ 2 H2O + 2 Cl. Try the experiment, using a t.t. and adding
water.

151. Cl from Bleaching-Powder.

Experiment 88.--Put a few grams of bleaching- powder into a small
beaker, and set this into a larger one. Cover the latter with
pasteboard or paper, through which passes a thistle-tube reaching
into the small beaker (Fig. 40). Pour through the tube a little
H2SO4 dilated with its volume of H2O.

152. Chlorine Water.--A solution of Cl in water is often useful,
and may be made as follows:-- Experiment 89.--To 3 or 4 crystals
of KClO3 add a few drops of HCl. Heat a minute, and when the gas
begins to disengage, pour in 10 cc. H2O, which dissolves the gas.
2 KClO3 + 4 HCl = 2 KCl + Cl2O4 + 2 H2O + 2 Cl.

153. Bleaching Properties.

Experiment 90.--Put into a receiver of Cl, preferably before
generating it, two pieces of Turkey red cloth, one wet, the other
dry; a small piece of printed paper and a written one; also a red
rose or a green leaf, each wet. Note from which the color is
discharged. If it is not discharged from all, put a little H2O
into the receiver, shake it well, and state what ones are
bleached.

Experiment 91.--(1) Add 5 cc. of Cl water to 5 cc. of indigo
solution. (2) Treat in the same way 5 cc. K2Cr2O7 (potassium
dichromate) solution, and record the results.

Indigo, writing-ink, and Turkey red or madder, are vegetable
pigments; printer's ink contains C, and K2Cr2O7 is a mineral
pigment. State what coloring matters Cl will bleach.

154. Disinfecting Power.

Experiment 92.--Pass a little H2S gas from a generator into a
t.t. containing Cl water. Look for a deposit of S. Notice that
the odor of H2S disappears. H2S + 2 Cl = 2 HCl + S.

155. A Supporter of Combustion.

Experiment 93.--Sprinkle into a receiver of Cl a very little fine
powder or filings of Cu, As, or Sb, and notice the combustion.
Observe that here is a case of combustion in which O does not
take part. Chlorides of the metals are of course formed. Write
the reactions. See whether Cl will support the combustion of
paper or of a stick of wood.

Experiment 94.--Warm 2 or 3 cc. of oil of turpentine (C1OH16) in
an evaporating-dish; dip a piece of tissue paper into it, and
very quickly thrust this into a receiver of Cl. It should take
fire and deposit carbon. C1OH16 + 16 Cl = ? Test the moisture on
the sides of the receiver with litmus. Clean the receiver with a
little petroleum.

Experiment 95.--Prepare a H generator with a lamp-tube bent as in
Figure 41. Light the H, observing the cautions in Experiment 23,
and when well burning, lower the flame into a receiver of Cl.
Observe the change of color which the flame undergoes as it comes
in contact with Cl. Give the reaction for the burning. Test with
litmus any moisture on the sides of the receiver. A mixture of Cl
and H, in direct sunlight combines with explosive violence;
whereas in diffused sunlight it combines slowly, and in darkness
it does not combine. From these experiments state the chief
properties of Cl, and what combustion it will support.

[Figure 41.]

156. Sources and Uses.--The great source of Cl is NaCl, though it
is often made from HCl. Its chief use is in making bleaching-
powder, one pound of which will bleach 300 to 500 pounds of
cloth. Cl is very easily liberated from this powder by a dilute
acid, or, slowly, by taking moisture from the air. Hence its use
as a disinfectant in destroying noxious gases and the germs of
infectious diseases. Cl attacks organic matter and germs as it
does the membrane of the throat or lungs, owing to its affinity
for H.

Cl is the best bleaching agent for cotton goods. It is not
suitable for animal materials, such as silk and wool, as it
attacks their fiber. It does not discharge either mineral or
carbon colors. The chemistry of bleaching is obscure.

As dry material will not bleach, Cl seems to unite with H in H2O
and to set O free. The O then unites with some portion of the
coloring matter, oxidizing it, and breaking up its molecule.
Colors bleached by Cl cannot be restored.

Chapter XXXI.

BROMINE.

Examine bromine, potassium bromide, sodium bromide, magnesium
bromide.

157. Preparation.

Experiment 96.--Pulverize 2 or 3 g. KBr, and mix it with about
the same bulk of MnO2. After putting this into a t.t, add as much
H2SO4, mix them together by shaking, attach a d.t., and conduct
the end of it into a t.t. that is immersed in a bottle of cold
water. Slowly heat the contents of the t.t., and notice the color
of the escaping vapor, and any liquid that condenses in the
receiver. Avoid inhaling the fumes, or getting them into the
eyes.

MnO2 + 2 KBr + 2 H2SO4 = ? Compare this with the equation for
making Cl from NaCl.

158. Tests.

Experiment 97.--Try the bleaching action of Br vapor as in the
case of Cl. Bleach a piece of litmus paper, and try to restore
the color with NH4OH. Explain its bleaching and disinfecting
action. Try the combustibility of As, Sb, and Cu.

159. Description.--Bromine at usual temperatures is a liquid
element; it is the only common one except Hg; it. quickly
evaporates on exposure to air. The chemistry of its manufacture
is like that of Cl; its bleaching and disinfecting powers are
similar to the latter, though they are not quite so strong as
those of Cl. Its affinity for H and for metals is also strongly
marked. A drop of Br on the skin produces a sore slow to heal.
Bromine salts are mainly KBr, NaBr, MgBr2. These in small
quantities accompany NaCl, and are most common in brine springs.
The world's supply of Br comes chiefly from West Virginia and
Ohio, over 300,000 pounds being produced from the salt (NaCl)
wells there in 1884. The water taken from these wells is nearly
evaporated, after which NaCl crystallizes out, leaving a thick
liquid--bittern, or mother liquor--which contains the salts of
Br. The bittern is treated with H2SO4 and Mn02, as above.

For transportation in large quantities, Br has to be made into
the salts NaBr and KBr, on account of the danger attending
leakage or breakage of the receptacles for Br.

160. Uses.--Its chief uses are in photography (page 167),
medicine, as KBr, and analytical chemistry.

Chapter XXXII.

IODINE.

Examine iodine, potassium iodide.

161. Preparation of I.

Experiment 98.--Put into a t.t. 2 or 3 g. of powdered KI mixed
with an equal bulk of MnO2, add H2SO4 enough to cover well, shake
together, complete the apparatus as for making Br, and heat.
Notice the color of the vapor, and any sublimate. The direct
product of the solidification of a vapor is called a sublimate.
The process is sublimation. Observe any crystals formed. Write
the reaction, and compare the process with that for making Br and
Cl. Compare the vapor density of I with that of Br and of Cl.
With that of air. What vapor is heavier than I? What acid and
what base are represented by KI?

162. Tests.

Experiment 99.--(1) Put a crystal of I in the palm of the hand
and watch it for a minute. (2) Put 2 or 3 crystals into a t.t.,
and warm it, meanwhile holding a stirring-rod half-way down the
tube. Notice the vapor, also a sublimate on the sides of the t.t.
and rod. (3) Add to 2 or 3 crystals in a t.t. 5 cc. of alcohol,
C2H5OH; warm it, and see whether a solution is formed. If so, add
5 cc. H2O and look for a ppt. of I. Does this show that I is not
at all soluble in H2O, or not so soluble as in alcohol?

163. Starch Solution and Iodine Test.

Experiment 100.--Pulverize a gram or two of starch, put it into
an evaporating-dish, add 4 or 5 drops of water, and mix; then
heat to the boiling-point 10 cc. H2O in a t.t., and pour it over
the starch, stirring it meanwhile.

(1) Dip into this starch paste a piece of paper, hold it in the
vapor of I, and look for a change of color. (2) Pour a drop of
the starch paste into a clean t.t., and add a drop or two of the
solution of I in alcohol. Add 5 cc. H2O, note the color, then
boil, and finally cool. (3) The presence of starch in a potato or
apple can be shown by putting a drop of I solution in alcohol on
a slice of either, and observing the color. (4) Try to dissolve a
few crystals of I in 5 cc. H2O by boiling. If it does not
disappear, see whether any has dissolved, by touching a drop of
the water to starch paste. This should show that I is slightly
soluble in water.

164. Iodo-Starch Paper.

Experiment 101.--Add to some starch paste that contains no I 5
cc. of a solution of KI, and stir the mixture. Why is it not
colored blue? Dip into this several strips of paper, dry them,
and save for use. This paper is called iodo-starch paper, and is
used as a test for ozone, chlorine, etc. Bring a piece of it in
contact with the vapor of chlorine, bromine, or ozone, and notice
the blue color.

Experiment 102.--Add a few drops of chlorine water to 2cc. of the
starch and KI solution in 10 cc. H2O. This should show the same
effect as the previous experiment.

165. Explanation.--Only free I, not compounds of it, will color
starch blue. It must first be set free from KI. Ozone, chlorine,
etc., have a strong affinity for K, and when brought in contact
with KI they unite with K and set free I, which then acts on the
starch present. Com- plete the equation: KI + Cl = ?

166. Occurrence.--The ultimate source of I is sea water, of which
it constitutes far too small a percentage to be separated
artificially. Sea-weeds, or algae, especially those growing in
the deep sea, absorb its salts--NaI, KI, etc.--from the water. It
thus forms a part of the plant, and from this much of the I of
commerce is obtained. Algae are collected in the spring, on the
coasts of Ireland, Scotland, and Normandy, where rough weather
throws them up. They are dried, and finally burned or distilled;
the ashes are leached to dissolve I salts; the water is nearly
evaporated, and the residue is treated with H2SO4, and MnO2, as
in the case of Br and Cl. I also occurs in Chili, as NaI and
NaIO3, mixed with NaNO3. This is an important source of the I
supply.

167. Uses.--I is much used in medicine, and was formerly employed
in taking daguerreotypes and photographs. Its solution in alcohol
or in ether is known as tincture of iodine.

168. Fluorine.--F, Cl, Br, I, are called halogens or haloids, and
exist in compounds--salts--in sea water. F is the most active of
all elements, combining with every element except O. Until
recently it has never been isolated, for as soon as set free from
one compound it attacks the nearest substance, and seems to be as
much averse to combining with itself, or to existing in the
elementary state, as to uniting with O. It is supposed to be a
gas, and, as is claimed, has lately been isolated by electrolysis
from HF in a Pt U-tube. Fluorite (CaF2) and cryolite (Al2F6 + 6
NaF) are its two principal mineral sources. The enamel of the
teeth contains F in composition.

CHAPTER XXXIII.

THE HALOGENS.

169. Halogens Compared.--The elements F, Cl, Br, I, form a
natural group. Their properties, as well as those of their
compounds, vary in a step-by-step way, as seen below. F is
sometimes an exception. They are best remembered by comparing
them with one another. Notice:

1. Similarity of name-ending. Each name ends in ine.

2. Similarity of origin. Salt water is the ultimate source of
all, except F.

3. Similarity of valence. Each is usually a monad.

4. Similarity of preparation. Cl, Br, I, are obtained from their
salts by means of MnO2 end H2SO4.

5. Variation in occurrence. Cl occurs in sea-salt, Br in sea-
water, I in sea-weed.

6. Variation in color; F being colorless, Cl green, Br red, I
violet.

7. Gradation in sp. gr.; F 19, Cl 35.5, Br 80, I 127.

8. Gradation in state, corresponding to sp. gr.; F being a light
gas, Cl a heavy gas, Br a liquid, I a solid.

9. Corresponding gradation in their usual chemical activity; F
being most active, then Cl, Br, and I.

10. Corresponding gradation in the strength of the H acids; the
strongest being HF, the next, HCl, etc.

11. Corresponding gradation in the explosibility of their N
compounds; the strongest NCl3, the next, NBr3, etc.

12. Corresponding gradation in the number of H and O acids; Cl 4,
Br 3, I 2.

170. Compounds.--The following are some of the oxides, acids, and
salts of the halogens. Name them.

CI2O (+H2O=) 2 HClO. The salts are hypochlorites, as Ca(ClO)2.
Cl2O3 (+H20=) 2 HClO2. The salts are chlorites, as KClO2.
Cl2O4
-- HClO3 The salts are chlorates, as KClO3.
-- HClO4 The salts are perchlorates, as KClO4,
-- HBrO The salts are ? KBrO,
-- -- The salts are wanting.
-- HBrO3. The salts are ? KBrO3,
-- HBrO4. The salts are ? KBrO4,
-- -- The salts are wanting.
-- -- The salts are wanting.
I2O5 (+H2O=) 2 HIO3. The salts are ? KIO3.
-- HIO4. The salts are ? KIO4.

F forms no oxides, and no acids except HF. HF, HCl, HBr, HI, are
striking illustrations of acids with no O. HClO4 is a very strong
oxidizing agent. A drop of it will set paper on fire, or with
powdered charcoal explode violently. This is owing to the ease
with which it gives up 0. Notice why its molecule is broken up
more readily than HC103. The higher the molecular tower, or the
more atoms it contains, the greater its liability to fall. Some
organic compounds contain hundreds of atoms, and hence are easily
broken down, or, as we say, are unstable. Inorganic compounds
are, as a rule, much more stable than organic ones. It is not
always true, however, that the compound with the least number of
atoms is the most stable. SO2 is more stable than SO3, but H2SO3
is less so than H2SO4.
Chapter XXXIV.

VAPOR DENSITY AND MOLECULAR WEIGHT.

Examine a liter measure, in the form of a cube,--cubic decimeter,
--and a cubic centimeter.

171. Gaseous Weights and Volumes.--A liter of H, at 0 degrees
and 760 mm., weighs nearly 0.09 g. This weight is called a crith.
Find the weight of H in the following, in criths and in grams: 15
1., 0.07 1., 50.3 1., 0.035 1., 0.6 1..

It has been estimated that there are (10) 24. molecules of H in a
liter. Does the number vary for different gases? The weight of a
molecule of H in parts of a crith is 1/(10) 24.; in parts of a
gram .09/(10) 24.. If the H molecule is composed of 2 atoms, what
is the weight of its atom in fractions of a crith? What in
fractions of a gram? The weight of the H atom is a microcrith.
What part of a crith is a microcrith?

172. Vapor Density.--Vapor density, or specific gravity referred
to H as the standard, (Physics) is the ratio of the weight of a
given volume of a gas or vapor to the weight of the same volume
of H. A liter of steam weighs nine times as much as a liter of H.
Its vapor density is therefore nine. For convenience, a definite
volume of H is usually taken as the standard, viz., the H atom.
The volume of the H atom and that of the half-molecule of H2O, or
of any gas are identical, each being represented by one square.
If, then, the standard of vapor density is the H atom, half the
molecular weight of a gas must be its vapor density, since it is
evident that we thus compare the weights of equal volumes. The
vapor density of H2O, steam, is found from the symbol as follows:
(2 + 16) / 2 = 9. To obtain the vapor density of any compound
from the formula, we have only to divide its molecular weight by
two. Find the vapor density of HCl, N2O, NO, C12H22O11, Cl, CO2,
HF, SO2. Explain each case.

The half-molecule, instead of the whole, is taken; because our
standard is the hydrogen atom, the smallest portion of matter, by
weight, known to science.

How many criths in a liter of HCl? How many grams? Compute the
number of criths and of grams in one liter of the compounds whose
symbols appear above.

PROBLEMS.

(1) A certain volume of H weighs 0.36 g. at standard temperature
and pressure. How many liters does it contain? If one liter
weighs 0.09 g., to weigh 0.36 g. it will take 0.36 / 0.09 = 4
liters.

(2) How many liters, or criths, of H in 63 g.? 2.7 g.? 1 g.? 5
g.? 250 g.? Explain each.

(3) Suppose the gas to be twice as heavy as H, how many liters in
0.36 g.? A liter of the gas will weigh 0.18 g. (0.09 X 2). In
0.36 g. there will be 0.36 / 0.18 = 2. Answer the question for 63
g., 2.7 g., etc.

(4) How many liters of Cl in each of the above numbers of grams?

(5) How many of HCl? H2O (steam)? CO2? Explain fully every case.

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