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CHEMICAL
PROBLEMS AND  REACTIONS,

             TO ACCOMPANY

   STOCKHAEDT'S ELEMENTS OF CHEMISTRY.



                 "    1/
       JOSIAH P. COOKE, Jr.
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        CAMBRIDGE:
PUBLISHED BY JOHN BARTLETT,
      Boofcsellev to tje Sftnrberstta?.
            1857. 
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                CAMBRIDGK :



BLECTEOTTPED AND PRINTED BY METCAI.F AND  COMPANY. 
PREFACE.
  This book has been prepared  solely for the  use  of the
undergraduates of Harvard College.   It contains a collection
of chemical problems and reactions, with references to the sec-
tions of Stockhardt's Elements of Chemistry, and  also a few
chapters on the chemical nomenclature and the use of chemical
symbols,  subjects which are not sufficiently developed in that
text-book  for the purposes of college instruction.  In writing
chemical symbols  the author has  adopted  a uniform  system
throughout the volume, which,  as  he hopes, will be  found to
be at once  expressive and clear.  The problems and  reactions
cover the  Inorganic portion of  Stockhardt's Elements ; the
problems have only been extended to the section on the  Heavy
petals.  Beyond this, the reactions alone have been given, as it
was  supposed  that,  before reaching  this section, the student
will  easily be able  to  propose   problems for himself.    In
solving many of the problems  it will be found convenient to
use logarithmic tables of four places, which, with several other
tables, will be  found at  the end of the volume.  The student
is advised to  remove the tables of logarithms, and paste them
for use on a card.   The difficulty of insuring complete accu-
racy in the printing  of chemical formulae can be known  only
to those who  have  had to see a book  of this kind through the
press.   Several  errors  have  been  already  discovered, and
corrected, but others unquestionably  exist.
Cambridge, May 15th, 1857. 
 
NOMENCLATURE  OF  CHEMISTRY.
   Origin of the Nomenclature. — Previous  to the year 1787
the names  given  by  chemists or alchemists to substances
were not conformed to any general rules.   Many of these old
names, such as Oil of Vitriol, Calomel, Corrosive Sublimate,
Red-Precipitate, Saltpetre, Liver of Sulphur, Cream of Tartar,
 Glauber's and Epsom Salts, are still retained in common use.
As chemical  science advanced, and  the  number  of known
substances increased, it became important to adopt a scientific
nomenclature.  The admirable system  now in use is due al-
most entirely to Lavoisier, who reported to  the French Acad-
emy on the subject, in the  name of  a committee,  in  1787.
This system, now known as the  Lavoisierian nomenclature,
was  generally  adopted  by scientific men soon after its pub-
lication, and has  not been materially  modified since.  In it
the name of a substance is made to indicate the composi-
tion.

  Names of the Elements. — The  names of the elements  are
the only ones which are now  independent of any rule.  Those
which were known before the adoption of  the nomenclature,
                  1 
2
NOMENCLATURE OF CHEMISTRY.
such as  Sulphur, Phosphorus,  Iron,  Lead,  retain their old
names.   Several  of the more  recently discovered elements
have been  named in allusion to some prominent property or
some  circumstance  connected with their history;  as,  Oxygen
from 6&s, yewaa (aeid-generator) ;  Hydrogen, from vScop, yewda
(water-generator) ;  Chlorine,  from  x^P0'* (g1"6611) »  Iodine,
from  Icobrjs (violet); Bromine, from fipuuos  (fetid odor),  &c.
The names of the  newly discovered metals  have a common
termination, um,  as Platinum,  Potassium,  Sodium; and  the
names of a class of the metalloids terminate  in  ine, as Chlo-
rine, Bromine, &c; but except in these respects the  names
of the elements are entirely arbitrary.

   Classification of Compounds. — There are three orders of
chemical compounds: — 1st, Binary Compounds, consisting of
two elements,  or  of the  representatives of two elements ;  2d,
 Ternary Compounds, consisting  of three elements, or of their
representative; and 3d,  Quaternary  Compounds,  consisting of
four elements, or  their representatives.   There are some chem-
ical compounds containing  more than four elements;  but in
most cases two or more of these elements are representatives,
i. e. occupy the place, of only one, as will be explained farther
on.  Binary compounds  are subdivided into two  classes, Elec-
tro-Positive Binaries, or Bases, and Electro-Negative Binaries,
or Acids.  Each  of these classes is distinguished by a peculiar
set of  properties, or at least  this is the case with the  promi-
nent  members of either class;  but the  two classes merge so
gradually into each  other,  that it is impossible to draw a line
of demarcation between them;  and there is a large  class of
intermediate compounds, which  either partake of the proper- 
NOMENCLATURE OF  CHEMISTRT.
3
ties of both,  or are entirely  indifferent.  Indeed, the binary
compounds may best be regarded as forming a continuous series
of substances, varying in their properties from those of strong
acids on  the  one hand to those of strong bases on  the other,
and with every possible grade of qualities between the two
extremes.  In  this series each binary may be  considered as
an electro-positive compound or base towards all those which
precede  it, and as an electro-negative compound  or acid to-
wards all those which follow it.  Ternary compounds are  gen-
erally, at  least in  Inorganic Chemistry,  composed  of  two
binaries,  i. e. of  an  acid  and a  base,  and are then called
Salts.  The quaternary compounds are generally composed of
two salts, and  are called Double  Salts.

  Names of Binaries. — The  most important  binaries, as well
as those which  have been the best studied, are the compounds
of oxygen with the  other  elements.   To  these the  generic
term Oxide has been applied.  The electro-positive binaries
are called  simply  Oxides of the elements  of which they con-
sist.   Thus we have
   Oxide of Hydrogen, consisting of oxygen and hydrogen.
   Oxide of Potassium,     "      "      "     potassium.
   Oxide of Sodium,       "      "            sodium.
When  the name of the  metal ends in um, the name of the
compound with oxygen is frequently formed by changing this
termination into a, with such other modifications of the termi-
nal letters as euphony may require.   Thus we  use, instead of
         Oxide of Sodium,                 Soda.
      ■   Oxide of Potassium,             Potassa. 
4             NOMENCLATURE OF  CHEMISTRY.
    Oxide of Calcium,               Calcia (or Lime).*
    Oxide of Barium,               Baryta.
    Oxide of Strontium,             Strontia.
    Oxide of Magnesium,            Magnesia.
    Oxide of Aluminum,             Alumina.
  The two  names  are  in  all cases synonymous.   Generally
oxygen  combines with an element  in more than one  propor-
tion ; then, in order to distinguish between the different oxides
of the same  element, we use various Latin and Greek prefixes,
such as sub, proto, sesqui, deuto, hyper.   This is well illustrated
by the  names  of the different oxides of  mercury and man-
ganese,  which are as follows.
                                          Composition.
            Names.                     Mercury.     Oxygen.
     ■SkSoxide of Mercury                 100          4
     Protoxide  of Mercury                100          8
                                      Manganese.    Oxygen.
     Protoxide  of Manganese             27.6          8
     Seso/moxide of Manganese           27.6         12
     ffyperoiddie of Manganese           27.6         16
   The electro-negative binary compounds of oxygen (the acids)
are named on a different principle.  These are called different
kinds of acids.   If the element forms but one acid with oxy-
gen, the name is formed by adding to the name of the  acid the
termination  ic, with such  changes  of the final  letters as eu-
phony may require.   Thus, carbon  and oxygen form Carbonic

  * The common  name Lime is much more frequently used than either of its
scientific synonymes, Oxide of Calcium, or Calcia.  Indeed, the last has never
been in general use. 
NOMENCLATURE OF CHEMISTRY.
5
Acid.  When  the element forms two acids, by combining with
different  amounts of oxygen, the termination ic is reserved for
that  containing the most oxygen, while the termination ous is
given to the other.  We have, for example,
                                      Arsenic.     Oxygen.
     ArsenioMs Acid                     75         24
     Arsem'c Acid                       75         40
                                     Phosphorus.    Oxygen.
     Phosphoroiw Acid                   32         24
     Phosphoric Acid                    32         40
   If oxygen combines with an element in more than two pro-
portions, to form acids, the names  are  formed with the  Greek
prefix hypo, indicating a less,  or the Latin prefix per, indicating
a greater, amount of oxygen  than  that contained in the acids
to whose names  they  are prefixed.  The  acid compounds of
sulphur and oxygen are
                                      Sulphur.      Oxygen.
     HyposulphuTous Acid                16           8
     Sulphurous Acid                     16          16
     iZ^posulphuric Acid                  16          20
     Sulphuric Acid                      16          24
 The acid compounds of chlorine and oxygen are
                                      Chlorine.     Oxygen.
     Hypochlorous Acid                35.5           8
     Chlorows Acid                    35.5         24
     Bypochloric Acid -                 35.5         32
     Chloric Acid                      35.5         40
     Perchloric Acid                   35.5         56
   Very frequently the higher degrees of oxidation of an ele-
                     1* 
6
NOMENCLATURE OF  CHEMISTRY.
ment are acids, when the lower degrees are bases, or indifferent
compounds.   This is the  case with the oxides of manganese.
Besides the three already mentioned, there are also
                                    Manganese.     Oxygen.
    Manganic Acid                    27.6         24
    Permanganic Acid                 27.6         28
  The different oxides of nitrogen are, in like manner,
	Nitrogen.	Oxyg
Protoxide of Nitrogen	14	8
Pewtoxide of Nitrogen	14 x	16
Nitrows Acid	14	24
Hyponitrous Acid	14	32
Nitric Acid	14	40
  It will be noticed that here, as in other  places,  the  term
oxides is used generically for all the compounds of an element
with oxygen, whether acids or bases, and the  terms protoxide,
&c. to designate the specific compounds.   It is not unfre-
quently the case that the same oxide acts as an acid under
some circumstances, and as  a  base under others, and accord-
ingly is  known  under two different names.  Water, when  a
base, is  called  Oxide of Hydrogen, but when an acid,  it  is
named  Hydric Acid.   The  oxide  of aluminum, when a  base,
is called Sesquioxide  of Aluminum,  but when an acid,  Alu-
minic Acid.
  Next  to the  oxides, the  compounds of sulphur  with the
elements are the most important binaries.  These are called
sulphides, and, as a  general rule,  for every oxide there is  a
corresponding sulphide, which is named after the analogy of
the name of the oxide.  Thus  we have  two sulphides of  mer-
cury:— 
NOMENCLATURE OF CHEMISTRY.
7
Mercury. 100	Sulphur. 8
100	16
Iron. 28	Oxygen. 8
26	12
28	16
     £w5sulphide of Mercury
     Protosulphide of Mercury.
There are three sulphides of iron: —

     Protosulphide of Iron
     /Sesgwisulphide of Iron
     Persulphide of Iron

The last  corresponds to no oxide.  The sulphides, like the
oxides, may be divided into  basic or indifferent sulphides, and
into acid sulphides.  The  sulphur bases are named as above.
The sulphur acids have been named by prefixing to the name
of the  oxygen acid the letters sulpho.  Thus we have, cor-
responding to
     Arsenious Acid,               Sulphoaxsemous Acid.
     Arsenic Acid,                 Sulphoarsemc Acid.
     Carbonic  Acid,                /Sw^oAocarbonic Acid.
     Hydric Acid,                  Sulphohydric Acid.

  The names  of the binary compounds of the other  elements
are named after  the same analogy as the oxides and sulphides.
Thus with the other elements
           Oxygen        forms      Oxides.
           Fluorine         "        Fluorides.
           Chlorine         "        Chlorides.
           Bromine         "        Bromides.
           Iodine           "        Iodides.
           Sulphur          "        Sulphides.
           Selenium        "        Selenides. 
8             NOMENCLATURE OF  CHEMISTRY.
            Tellurium      forms      Tellurides.
            Nitrogen         "        Nitrides.
            Phosphorus      "        Phosphides.
            Arsenic          "        Arsenides, &c.
   We  distinguish the  different fluorides,  bromides, &c. by
 prefixes, as  we do the oxides.   Each of these classes of com-
 pounds may be subdivided, like the  oxides or sulphides;  but
 only in a few cases  has the nomenclature of the oxygen and
 sulphur  acids been  extended to them.   Almost  the only
 instances are  the  names of the compounds of the  first few
 elements of the last  list  with hydrogen, which  are some-
 times  called Fluohydric,  Chlorohydric, Bromohydric  Acids,
 &c.;  but even  here the  synonymous names Hydrofluoric,
 Hydrochloric, Hydrobromic Acids,  &c. are  more  generally
 used.

   Names of Ternaries. — The name of a ternary compound,
 or salt, is very simply formed from the  names of the acid and
 base of which it consists.  The termination of the  name of
 the acid is changed,  if ic, into ate; if  ous, into ite ; and the
 name of the base added.  Thus, sulphuric acid and oxide of
 sodium form Sulphate of the Oxide of Sodium (or of Soda);
 sulphurous  acid, Sulphite of the  Oxide of Sodium.  In like
 manner,  hyposulphuric acid forms Hyposulphate, and  hypo-
 sulphurows  acid,  Hyposulphite  of the Oxide  of Sodium.  So,
 also, Chlorite of Oxide of Potassium, and Chlorate of Oxide
of Potassium; Selenite of Oxide of Barium, and Selenate of
Oxide of Barium; Nitrite of Oxide of Calcium, and Nitrate of
Oxide of Calcium; Arsenite of Oxide of Lead, and  Arsenate
of Oxide of  Lead.  When the  base is  an oxide of one of the 
NOMENCLATURE OF  CHEMISTRY.
9
heavy metals, the name of the  salt is frequently  abbreviated
by leaving out the word oxide ; as, Sulphate of Lead, for  Sul-
phate of  the Oxide  of Lead; and Nitrate of  Copper, for
Nitrate of the Oxide of Copper.  When the base is  an oxide
of one of the light metals, by  substituting for  Oxide  of So-
dium, Oxide of Potassium, &c, the single  words Soda, Po-
tassa, &c, as already explained on  page 3.  Instead of Sul-
phate of  Oxide of Barium, chemists more frequently  write
the shorter Sulphate of Baryta; instead of Nitrate  of Oxide
of Calcium, Nitrate of Lime.  When the same  acid  combines
with two or more different oxides of the same element to form
salts, these  are distinguished by introducing  the specific name
of the oxide into the  name of the salt; as, Sulphate of Prot-
oxide of  Iron, and Sulphate  of Sesquioxide of Iron; Nitrate
of Suboxide of Mercury,  and  Nitrate of Protoxide of  Mer-
cury.  It is frequently the case that an  acid  combines with
the same base in two or  more different proportions, forming
two or more different salts.   In  order to distinguish  these
salts, the one is called the neutral salt; that containing  more
base than the neutral salt, a basic ;  and that containing less, an
acid.  Thus we have
                 Basic Phosphate of Lime.
                 Neutral Phosphate of Lime.
                 Acid Phosphate of Lime.
 So, also,
                 Basic Chromate of Lead.
                 Neutral Chromate of Lead.
   When  there are several basic salts, they are  distinguished
 by Latin prefixes.  Thus there  are five acetates of lead: — 
10
NOMENCLATURE OF CHEMISTRY.
Oxide of Lead.	Acetic Acid
112	51
224	51
280	51
336	51
672	51
31	35
31	70
Potassa.	Chromic Aci
47	51
47	102
47	153
     Neutral Acetate of Lead
     Bibasic Acetate of Lead
     Sesquibasic Acetate of Lead
     Tribasic Acetate of Lead
     Sexbasic Acetate of Lead
The acid salts  are distinguished by placing the Latin prefixes
directly before the name of the neutral salts ; as,
                                     Soda.       Boracic Acid.
     Borate of Soda
     Biborate of Soda

     Neutral Chromate of Potassa
     Bichromate of Potassa
     Trichromate of Potassa
   Most of the oxygen  acids are commonly used only when in
combination with  water.  These  compounds  are true  salts, in
which  the water plays the part  of a .base.   The  common
Sulphuric Acid is  a Sulphate  of Water, and the  common
Nitric  Acid, Nitrate of Water.   Properly speaking, the terms
Sulphuric and  Nitric Acid ought only to be applied to the
anhydrous  compounds; but  custom  authorizes us  to extend
these names to the  hydrates.
   The names  of the ternary sulphur compounds, the sulphur
salts, are formed in the same way as those of the oxygen salts;
thus,  the compound of sulphocarbom'c  acid and sulphide of
sodium  is called  the Sulphocarbonate of the Sulphide of So-
dium ;  the compound of sulphoarsemc  acid and  sulphide of
potassium, the Sulphoarsenate of the Sulphide of Potassium;
and the compound of sulphoarseniow* acid  and  the  same base^
Sulphoarsem'te of the Sulphide of Potassium. 
NOMENCLATURE OF  CHEMISTRY.            11
  The names of the ternary fluorine,  chlorine,  bromine, and
iodine compounds are formed after a different analogy.  These
are generally regarded  as double  salts, and  are called  the
double fluorides, chlorides, bromides,  or iodides of the two
metals which they contain.   Hence the names
         Double Chloride of Aluminum and Sodium.
         Double Chloride of Platinam and Ammonium.

  Names  of Quaternaries. — The double  salts  formed from
two  salts  each  containing the same acid but  different bases,
are called double salts of the two bases.  Thus,  the compound
of sulphate  of alumina  and sulphate  of potassa is  called the
Double Sulphate of Alumina and Potassa; the compound of
sulphate of soda and sulphate of zinc, the Double Sulphate
of Soda  and Zinc.  So, also, the Double Sulphate of Potassa
and Magnesia, and the Double Sulphate of Potassa and Water.
The last is  more frequently called Bisulphate of Potassa, as if
it were composed of one  equivalent of base and two equiva-
lents of acid, instead of being, as  it  probably is, a compound
of two salts. Similar incongruities hi the nomenclature, aris-
ing frequently from  different theories  in regard to the consti-
tution of substance, are not uncommon among the higher orders
of compounds.
   The  Lavoisierian  nomenclature  has not been found  suffi-
ciently  expansive to meet  the  requirements  of modern sci-
ence, not only on account of the complex character of many
of  the newly  discovered  compounds,  but more especially
because  it  involves the  ideas of  a  peculiar  theory, which,
although once almost universally received, is  not now so gen-
erally admitted.   The  inadequacy of  the old nomenclature to 
12           NOMENCLATURE  OF CHEMISTRY.

afford names for  the great variety of chemical compounds, or
to describe the  infinite  changes to which they are  liable, has
given rise to a peculiar chemical language, analogous to the
language of mathematics,  called Chemical Symbols.  To ex-
plain the  use of this language  will be the object of the next
chapter. 
CHEMICAL  SYMBOLS.
   Since all matter is composed  of one or more of sixty-two
 different substances, never as yet decomposed, and hence called
 Elements, it is evident that, if we adopt an arbitrary symbol for
 each of these elements, we shall be able, by combining them to-
 gether, to express all possible varieties of combination.   More-
 over, since  the elements  always combine  in certain fixed and
 definite proportions by weight, it is equally evident that, if we
 assign to each of these symbols  a certain weight, we shall be
 able to indicate the relative quantities of the different elements
 which enter into any compound.

   Symbols of Elements. — It has  been  agreed by  chemists of
 different nations  to use,  as  symbols of the elements, the  first
 letters of their Latin  names.   When two  or more  names com-
 mence with  the same letter, a second letter is added for dis-
 tinction.   The  first  letter is printed or written in capitals, and
 the second, when used, in small letters, immediately following
 the first.   A list of the Elements, with  their Symbols, is given
on the following page.
                    2 
14             CHEMICAL SYMBOLS.
CHEMICAL SYMBOLS AND EQUIVALENTS.
Aluminum	Al =	13.7	Nickel	Ni =	29.G
Antimony (Stibium)	Sb =	129	Niobium	Nb	
Arsenic	As =	75	Nitrogen	N =	14
Barium	Ba =	68.5	Norium	No	
Bismuth	Bi =	213	Osmium	Os =	99.6
Boron	B =	10.9	Oxygen	O =	8
Bromine	Br =	80	Palladium	Pd =	53.3
Cadmium	Cd =	56	Pelopium	Pe	
Calcium	Ca =	20	Phosphorus	P =	32
Carbon	C =	6	Platinum	Pt =	98.7
Cerium	Ce =	47	Potassium (Kalium)	K =	39.2
Chlorine	CI =	35.5	Bhodium	K =	52.2
Chromium	Cr =	26.7	Ruthenium	PvU =	52.2
Cobalt	Co =	29.5	Selenium	Se =	S9.5
Copper (Cuprum)	Cu =	31.7	Silicon	Si =	21.3
Didymium	D		Silver (Argentum)	Ag =	108.1
Erbium	E		Sodium (Natrium)	Na =	23
Fluorine	Fl =	18.9	Strontium	Sr =	43.8
Glucinum	G =	4.7	Sulphur	S =	16
Gold (Aurum)	Au =	197	Tantalum	Ta =	184
Hydrogen	H =	1	Tellurium	Te =	64.2
Iodine	I =	127.1	Terbium	Tb	
Iridium	Ir =	99	Thorium	Th =	59.6
Iron (Ferrum)	Fe =	28	Tin (Stannum)	Sn =	59
Lanthanium	La		Titanium	Ti =	25
Lead (Plumbum)	Pb =	103.7	Tungsten (Wolfram)	W =	95
Lithium	Li =	6.5	Uranium	U =	60
Magnesium	Mg =	12.2	Vanadium	V =	68.6
Manganese	Mn =	27.6	Yttrium	Y	
Mercury (Hydrargyrum) Hg =		100	Zinc	Zn =	32.6
Molybdenum	Mo =	46	Zirconium	Zr =	22.4
 
CHEMICAL  SYMBOLS.
15
   The  student will  do well to notice in the foregoing list the
 symbols of those elements whose Latin names commence with
 letters differing from the initial letters of the  English names,
 since they are not so easily remembered as the others.

   Chemical  Equivalents. — The  chemical  symbols not  only
 stand for the names of the  elements, but also for a fixed pro-
 portional weight of each.  These  weights  are given in  the
 above table,  opposite to the symbols.   They have only rela-
 tive values;  if one  is in pounds,  all the rest  are in pounds;
 and if one is in ounces, all the rest are  in  ounces.  We may
 leave the standard indefinite, and express the weight in parts ;
 then Al  stands for 13.7 parts of aluminum; Sb stands for 129
 parts of antimony, &c.  The weight of an element indicated
 by  its symbol is called  one equivalent, and  it  is  a  law  of
 chemistry that elements always combine by  equivalents; that
 is, one equivalent of one  combines with  one equivalent of an-
 other, or else several equivalents of one combine with one or
 with several equivalents of another.
  As stands  for 75 parts  of  Arsenic,  or  one equivalent.
  Bu     "      68.5     "    Barium,      "      "
  H      "       1       "    Hydrogen,    "      "
  0      "       8       "    Oxygen,      "      «
  In order to express two, three,  or more  equivalents of an
element,  we place a  figure just below the symbol at its  right
hand; thus,
     02 stands for 16 parts of Oxygen, or two equivalents.
     08     "     24     "     "      three equivalents.
  These figures merely multiply  the symbols beneath which 
16
CHEMICAL  SYMBOLS.
they stand, and must not be confounded with algebraic powers,
which are sometimes written in a similar way.  We sometimes
place the figure, though larger, before  the symbol; 2 O means
exactly the same thing as 02.

  Symbols of Compounds. — In order to form the symbol of a
compound, we write the  symbols of the elements of which it
consists  one  after the other, indicating by means of figures the
number of equivalents of  each  which have entered into com-
bination.   Thus, H 0 is the symbol of water, a compound con-
sisting of one equivalent  or one  part of hydrogen, and of one
equivalent or  eight parts of oxygen; S 03  is the symbol  of
sulphuric acid, a compound  consisting of  one equivalent or six-
teen parts of sulphur, and of three equivalents or twenty-four
parts of oxygen ; Qn Hu  Ou is the symbol of common sugar,
a compound  consisting of  twelve  equivalents of carbon, eleven
equivalents of hydrogen, and eleven equivalents of oxygen.

  Binary Compounds. — The  symbols of  binary compounds
are  formed by  writing  the symbols of the two  elements to-
gether, taking care to  place  the symbol of the metal, or  of
the most electro-positive element, first.  The binary symbol thus
obtained  represents always one  equivalent  of the compound.
The  weight  of this equivalent is evidently the sum of the
weights  of the equivalents  of the elements entering into the
compound.
  14 + 40 =  54.
  N 05  stands for one equiv. or 54 parts of Nitric Acid.
  16 + 24 =  40.
  s°3        "        "        40    «    Sulphuric Acid. 
CHEMICAL  SYMBOLS.
17
6 + 16 = 22.				
C 02 stands for one	equiv.	or 22	parts	of Carbonic Acid.
1+8 = 9. HO "	>.	9	a	Water.
28 + 8 = 36. FeO "	a	36	«	Oxide of Iron.
20 + 8 = 28. CaO	«	28 •	ii	Lime.
23 + 8 = 31. Na 0 "	..	31	a	Oxide of Sodium.
6 + 32 = 38. cs2	u	38	a	Sulphocarbonic Acid.
75 + 48 = 123. AsS3	a	123	a	Sulphoarsenious Acid.
28 + 16 = 44. FeS	a	44	a	Sulphide of Iron.
39 + 16 = 55. KS	a	55	a	Sulphide of Potassium.
  In order to express two or more equivalents of a binary, we
place a figure immediately before the symbol, like an algebraic
coefficient.   A figure  so  placed always multiplies the whole
binary.
  3 S 03 stands for 3 equiv. or 120 parts of Sulphuric Acid.
  5 Pb O    "     5    "    5(50    "    Oxide of Lead.

   Ternary Compounds. — The symbol of a ternary compound
is formed  by writing together the  symbols of the two binaries
of which  it  consists, separated by a. comma,  taking  care to
place the most electro-positive binary, the base, first.   If the
salt is composed of more than one equivalent of either base or
acid, then the number of equivalents  must be indicated  by
                    2* 
18                  CHEMICAL SYMBOLS.

coefficients.  The ternary symbol thus obtained always stands
for  one equivalent  of the  compound,  and the weight  of this
equivalent is evidently the sum of the  weights of the  equiva-
lents of the  elements entering into it.

l + 8 + 16 + 24 = 49.
H O, S 03 stands for 1 equiv. or 49 parts of Sulphate of Water
     (common  Sulphuric Acid).
1 + 8 + 14 + 40 = 63.
H 0,N05 stands for 1 equiv. or 63 parts of Nitrate of Water
     (common  Nitric Acid).
39 + 8 -f 6 + 16 = 69.
K O, C 02 stands for 1  equiv. or  69 parts of Carbonate  of
     Potassa.
23 + 8 + 16 -f-  24 = 71.
Na 0, S 03 stands for 1 equiv. or 71 parts of Sulphate of Soda.
39 + 8 + 27 +  24 = 98.
K O, Cr 03 stands for 1 equiv. or 98 parts of Neutral Chromate
    of Potassa.
39 -4- 8 + 2 (27  + 24) = 149.
K 0,2 Cr 03 stands for 1  equiv. or 149 parts of Bichromate of
    Potassa.
39 + 8 -f- 3 (27  + 24) = 200.
K 0,3 Cr 03 stands for 1  equiv. or 200 parts of Trichromate of
    Potassa.
104 + 8 + 27 + 24 = 163.
PbO,Cr03  stands for 1  equiv. or 163 parts of Neutral Chro-
    mate of Lead.  -
2 (104 + 8) + 27 + 24 = 275.
2 PbO, Cr03 stands for 1 equiv. or 275 parts of Basic Chromate
    of Lead.

  In order to  express two or  more equivalents of a ternary,
we  enclose  the  symbol in parentheses, and place before the
whole the required figure.   Thus, 
CHEMICAL  SYMBOLS.                 19
 3 (NaO, S03) stands for three equivalents of Sulphate of Soda.
 5 (2 Pb O, Cr 03) stands for five equivalents of Basic Chromate
     of Lead.

   Neutral Salts. — The larger number of inorganic acids com-
 bine most readily with  one equivalent of base, and the salts so
 formed will be called neutral salts.   If the salts contain more
 equivalents of acid or base than one, they are  called acid  or
 basic salts respectively.  There are, however, some acids which
 combine most readily with two or three equivalents of base, in
 the same way that the others combine with one.  Such acids
 are called bibasic or tribasic  acids, in order to distinguish them
 from the rest, which are frequently called  monobasic.   Of
 bibasic and  tribasic  acids, the most important in inorganic
 chemistry is Phosphoric Acid.  This  is known in three differ-
 ent  conditions.  In the first  of these it is monobasic, in the
 second bibasic, and in  the third tribasic, the last being the
 ordinary condition.  The three conditions are  designated by
 the symbol aP 05; bP 05; CP 05.  The acid CP 05 forms neutral
 salts when combined with three equivalents  of  base, the  acid
 bP Os when combined with two, and the acid aP 05 when com-
 bined with only one.  There are three compounds of the acid
 and water corresponding to the three conditions, which are rep-
 resented in symbols by H O, aP 05; 2 H O, bP 05; 3 H O, CP 05.
 In these compounds we can substitute for the  equivalents  of
 water  equivalents  of other bases, either in whole or in  part,
 forming  such compounds  as Na O, aP Os;  2  Na 0, bP 05;
 3NaO,cPOs ;  [HO, 2NaO] CP05; [2HO,Na0] CP05.
The  equivalents of water may even  be replaced by different
bases, as in the compounds [Na 0, Pb 0] bP 05 ; [H 0, Na 0, 
20
CHEMICAL  SYMBOLS.
K 0], CP 05 ;  [K O, 2 Mg 0] CP 05.  All the above are sym-
bols of neutral salts.
  As protoxide bases combine most readily with one equivalent
of acid, so sesquioxide bases  combine most readily with  three
equivalents.   A neutral salt of a sesquioxide base is therefore
one which contains for every equivalent of base three equiva-
lents of a monobasic acid, or one equivalent of a tribasic acid,
and for every two equivalents of base three  equivalents of a
bibasic  acid.   Hence, Fe2 03, 3 S 03 ;  2 Fe2 03 , 3 bP 05 ;
Fe2 03, CP Os are all symbols of  neutral salts.  On the  other
hand, Fe2 03, S 03; 4 Al2 03, 3 bP 05; 2 Al2 03, CP 05 are sym-
bols of basic salts.  It will be noticed, on examining the above
symbols, that neutral salts of monobasic acids contain  as many
equivalents of acid as there  are equivalents of oxygen in the
base, and neutral salts of bibasic and tribasic  acids  one half
and one third as many, respectively.  This rule must  be kept
in mind when writing the symbols of salts.

   Compound Radicals. — There is a large class of substances
which, although compound, nevertheless act in chemical changes
exactly as if they were  simple, frequently replacing  the ele-
ments  themselves.   Such substances  are termed compound
radicals.  Many  of  these radicals have, like the elements,
received  arbitrary  names,  such  as Cyanogen,  Ammonium,
Ethyle, Acetyle, &c.; and, moreover,  the  first letter or letters
of these names  are  frequently used as their symbols.  It is
best, however, to write out the symbols of the elements  form-
ing these compounds, and  enclose the whole in brackets ; thus,
[N H4] stands  for Ammonium, [C2 N]  for Cyanogen, [C4 Hs]
for Ethyle, [C4 H3] for Acetyle.   The oxides of the compound 
CHEMICAL  SYMBOLS.
21
radicals, like those of the elements, may be divided into acids
and bases, and their symbols are written exactly like those of
other binaries.  Thus,
        [C4 H3] 03 is the symbol of Acetic Acid.
        [N H4] O     "     "      Oxide of Ammonium.
        [C4H5]0     «     "      Oxide of Ethyle.
So, also, with the salts.
[N H4] 0, [C4 H3] 03 is the symbol of Acetate of Ammonia.
[C4 H5] 0, [C4 H3] 03     "     «   Acetate of Oxide of Ethyle.

   Water of Crystallization. — Besides the water of constitution,
which frequently forms  a part  or  the whole  of the base of a
salt, most salts combine with water as  a whole.   This water is
held  in  combination  by a  comparatively feeble affinity, and
may be generally driven off by  exposing the  salt to the tem-
perature of 100° C, and  sometimes escapes at  the ordinary
temperature  of the air,  the crystals of the salt in  all cases
falling into powder.   Its presence is essential to the crystalUne
condition  of many salts,  and hence the  name Water of Crys-
tallization.  The  presence of water of crystallization in a salt
is expressed in symbols, by writing after the symbol of the
salt, and separated  from it by a period, the number of equiva-
lents of water.  Thus,
Fe O, S 03 . 7 H O is the symbol of CrystaUized Sulphate of
     the Oxide of Iron (Green Vitriol).
H O, 2 Na O, CP Os . 24 H O  is  the  symbol of CrystaUized
     Phosphate of Soda.
   The same salt, when CrystaUized at different temperatures, not
unfrequently combines with different amounts of water of crys-
tallization, the less  amounts corresponding to  the higher tern- 
22
CHEMICAL  SYMBOLS.
 peratures.   Thus, the Sulphate of Manganese may be crystal-
 lized with three different amounts of water of crystallization.
   Mn O, S 03. 7 H O when crystallized below 6° Centigrade.
   Mn 0, S 03. 5 H 0     "     "     between 7° and 20°.
   Mn O, S 03. 4 H 0     "     "     between 20° and 30°.
 The crystalUne forms  of these three compounds are entirely
 different from  each other, proving that the form depends,  in
 part at least, on the amount of water which the salt contains.
   The symbols of other ternary  compounds are  written  like
 those of the  oxygen salts, and therefore require no  further
 explanation.   Below are a few of these symbols, together with
 those of the corresponding oxygen  salts, which may serve as
 examples.
 K 0, C 02        = Carbonate of Oxide of Potassium.
 K S, C S2        = Sulphocarbonate of Sulphide of Potassium.
 K O, As 03       = Arsenite of Oxide of Potassium.
 K S, As S3       = Sulphoarsenite of Sulphide of Potassium.
 3 Na CI, Sb Cl3   = Double Chloride of Antimony and  Sodium.
 [NH4] Cl,PtCl2 = Double Chloride of Platinum and Potassium.
 K I, Pt I2       = Double Iodide of Platinum and Potassium.

   Quaternaries. — The  symbols of the double salts are formed
 by writing together the  symbols of the two salts of which they
 consist, separated by a period.  Thus,
 K0, S 03. MgO, S03  .  6HO= Double Sulphate of Mag-
     nesia and Potassa.
 K 0, S 03. Al2 03,3 S 03. 24 H O = Double  Sulphate  of  Alu-
     mina and Potassa (Alum).
When the salts  contain water of crystallization, the amount of
this water expressed in equivalents is written after the  symbol
of the salt, as already explained. 
CHEMICAL  REACTIONS.
   The various chemical changes to which  all matter is more
 or  less Uable are termed, in the language of chemistry, reac-
 tions, and the agents which cause these changes, reagents.   In
 every chemical reaction we must distinguish  between the sub-
 stances which are involved in the change and those which are
 produced by it.   The first wiU  be termed the factors, and the
 last the products, of the reaction.  As matter is indestructible,
 it follows, that  The sum of the weights of the products of any
 reaction must always  be equal to the sum  of the weights of the
factors.  This statement seems at first sight to be contradicted by
 experience, since wood and many other combustible substances
 are apparently consumed by burning.   In all such cases, how-
 ever, the apparent  annihUation  of the substance arises from
 the fact that the  products of the change are invisible gases;
 and when these are  collected, their weight is found to be equal,
 not only to  that of the substance, but also, in addition, to the
 weight of the oxygen  from  the air consumed in the process.
 As the products and factors of every chemical change must be
 equal, it follows that A chemical  reaction may always be repre- 
24
CHEMICAL REACTIONS.
sented in an  equation  by writing the  symbols of the factors in
the first member, and those of the products in the second.  The
reaction of sulphuric acid on common salt may be represented
by the following equation:
         23 + 35 +  1+8 + 16+24  =  23+8+16+24  +  1+35 = 107.
         NaCI + i70,  S 03 = Na 0, S 03 + H CI.
The correctness of this may be proved by adding together the
equivalents of both sides, when  the  sums will be found to be
equal.  In like manner, the reaction  of  a solution of common
phosphate of soda on  a solution of  chloride of calcium  may be
represented by the equation
        1+8 + 2 (23 + 8)+ 32+40  +  3 (20+35) =
       H0,2Na 0, CP 05-j-3 Ca Cl+ Aq* =
           3(20 + 8) + 32+40  + 2(23 + 35)  + 1 + 35 = 308.
           3 CaO, 6P06 + 2iVa CH-\-HCl-\-Aq.
So, also, the reaction of hydrochloric acid on chalk, which  may
be proved Uke the other two:
  CaO,C02-f HCl-\-Aq=*  Ca CI + HO + Aq-\-C02.
   Although the equation is the most concise, and therefore
in most cases  the best  form of representing chemical reac-
tions, it is nevertheless  frequently • advantageous,  in  studying
complicated changes,  to adopt a  more  graphic  method,  by
which the various steps of the process may be indicated.   The
reactions represented by the preceding equations may be writ-
ten thus:—
  * The symbol Aq, for Aqua, merely indicates the condition of solution, and
is not to be regarded in adding up the equivalents in order to prove the equa-
tion. 
CHEMICAL REACTIONS.
25
ao
(2.)
(3.)
  Chemical reactions may be classed under three divisions.
  First, those reactions in which a compound is decomposed,
and  diviHes into simpler compounds or into  elements.  E. g.
when oxide of mercury is heated, it is decomposed into oxygen
gas and metallic mercury.   Thus,

                    Hg0 = ^ + O.

Again, when  Chlorate of Potassa is heated, it is resolved into
oxygen gas and chloride of potassium.   Thus,

                K0,C105=KCl+6O.
                   3 
26
CHEMICAL  REACTIONS.
So, also, when sulphate of lead is heated, it is resolved  into
anhydrous sulphuric acid and oxide of lead.  Thus,
                Pb0,S03=Pb0 + SO3.
Such reactions  as  these  will  be  called  analytical,  and the
process analysis.
   Second, those reactions in which the elements are united to
form compounds, or compounds of a lower order to form those
of a higher.  E. g. when hydrogen and carbon burn in the air,
they combine with oxygen to form  water or  carbonic  acid.
Thus,
       H + 0 = HO;        C + 02=C02.
Again, when anhydrous sulphuric  acid combines with lime to
form sulphate of lime.  Thus,
                CaO + S03=CaO,S03.
Reactions like these will be called synthetical, and the process
synthesis.
   Third, those reactions in which one element displaces an-
other.  E. G. when  sodium takes  the place of hydrogen in
water, or zinc  the place of hydrogen in dilute sulphuric  acid.
Thus,
                 HO + ~8&=NaO + Vi.
       Zn + HO, S 03 + Aq = Zn 0, S 03 + Aq + H.
This division includes  also those reactions in which there is
a,  mutual interchange of elements between two compounds.
E. g.  when  a solution of chloride  of barium  is added  to  a
solution of sulphate of  soda, the sodium and barium change
places, and we have formed an insoluble precipitate of sulphate 
CHEMICAL REACTIONS.
27
of baryta and chloride of sodium (common salt), which remains
in solution.   Thus,
  Ba CI-f Na 0, S03-\- Aq = BaO, S03 + Na CI-f Aq.
Reactions Uke these will be called metathetical, and the process
metathesis*
  Of the three  classes of chemical  reactions, the last is by far
the most important;  indeed, the larger number of reactions
described in  an elementary treatise  on chemistry are examples
of metathesis.   All metathetical reactions can be  illustrated
very  elegantly with  the  aid  of mechanical diagrams, as fol-
lows : —
     A        (*•)                         (B-)
   They are easily made by printing with stench's on the larger
piece of pasteboard, A B, Fig. 4, the symbols of the  elements
not disturbed in the reaction, and on the smaller piece, a b, the
symbols of the elements which exchange places.  The smaller
piece having been  fastened to the larger by means of an eye-
let at 0, the reaction  is represented  by merely turning it
half round.   (See Fig. 5.)   If the symbols of the interchang-
ing elements are  not symmetrical on aU sides, it is of course
necessary to  make  them reversible, by printing each on a sep-
arate small square  of pasteboard, fastened by an eyelet to the

       * From the Greek /ueraTi&j/u, to displace or to transpose. 
28
CHEMICAL  REACTIONS.
top or bottom  of the  revolving piece a b, since otherwise the
letters would be inverted when the diagram  is turned.  This
method of illustration  may, with a little ingenuity, be extended
to some of the  most comphcated cases of chemical change.
   The most important condition of chemical action is, that the
particles of the substances  involved in the change  should be
indued  with  freedom  of motion.   This condition  is generally
fulfilled, both in nature and in our  laboratories, by bringing the
substances together in  solution, either in water or in some other
fluid.  When substances are brought together  in solution, there
are two circumstances which, more than any others, determine
the nature and  extent of the resulting change.
   First, If by an interchange of analogous elements an insoluble
compound may  be formed, this  compound always separates from
the fluid as a precipitate.   As this circumstance is by far the
most  important of all in  determining chemical  reactions,  it
requires full iUustration.





        HO,S03  |+^-BaO,S03}^

   In these examples, and in general throughout the volume, the
symbols of the substances, when  in  solution, are printed in
italic letters, and the solid  precipitate in Roman letters.  The
symbol Aq, as  already stated, stands for an indefinite amount
of water, in which  the substances are supposed to be dissolved.
It is obvious, from the above examples, that, in order to ascer- 
CHEMICAL  REACTIONS.
29
tain whether two salts will react on each other, when brought
together in solution, so as to form a precipitate, it is only neces-
sary to write the symbol  of one under  that of the other, and
interchange the symbols of  the  metallic elements.  If either
compound whose symbols are thus formed  is insoluble in the
menstruum present, a reaction wiU take place, and the insoluble
compound will be precipitated.  At the end of the  volume will
be found a table, reprinted  from the English edition of Fre-
senius's Qualitative Analysis, by means  of which  the student
can easily ascertain  from inspection  the  solubility of any of
the more frequently  occurring binary compounds or salts, and
thus wiU be able to solve the following problems.
   Problem 1.  If chloride of barium and sulphate  of soda are
mixed together in solution, wiU there be a reaction  ; and if so,
what wiU be formed ?
   Problem 2. If chloride of sodium  and nitrate of silver are
mixed together in solution, will there be a reaction,  &c. ?
   Problem 3. If sulphide of hydrogen and nitrate of lead are
mixed together in solution, will there be a reaction ?
   Problem 4. If sulphide of hydrogen and sulphate of  zinc
are mixed  together in solution, will there be a reaction ?
   In  solving the last problem, it must be noticed that the fluid
which would result from a  reaction would be a weak acid, in
which many substances are soluble which would be insoluble
in pure water, as may be seen from the table.
   Problem 5. If chloride  of sodium  and sulphate of copper
are mixed  together in solution, will there be a reaction ?
   Problem 6. If sulphuric acid and borate of soda are mixed
together in solution, will there be a reaction ?
   It will be  found that, by an interchange of metallic elements
                    3* 
30
CHEMICAL REACTIONS.
in the last two examples, no insoluble compound will be formed,
and hence the conclusion follows from our data, that there wiU
be no precipitate.  We must not, however, conclude from this
that there will be  no reaction,  since, as can easily be seen, it
does not necessarily follow, because the possible formation of an
insoluble compound  always determines a reaction,  that the re-
verse is equally true, and that no reaction can take place unless
an insoluble  compound is formed.  Indeed, in the  last two ex-
amples, we are able, from incidental phenomena, to determine
satisfactorily  that  a  change does  result; thus,  in  Problem  5,
when the solutions are  mixed, the blue color of sulphate  of
copper  changes into the green  color of chloride  of copper;
and in Problem 6, if sulphuric  acid is not added in excess, the
claret  color  to which  blue litmus-paper turns in  the  mixed
solution proves that  it  is boracic acid,  and not sulphuric acid,
which is in a free  state ; nevertheless, in  most similar cases it
is impossible to determine,  unless an insoluble compound is
formed, whether any reaction  has taken place.
   Second.  The circumstance  which,  next  to insolubility,  is
most  important in determining metathetical reactions, is vola-
tility, and it may be  laid down as a general principle, that, If
by an interchange of analogous  elements between two substances
in solution, a substance can be formed, which  is volatile at the
temperature at which the experiment is conducted, such an inter-
change always takes  place, and the volatile product is set free.
In order to  illustrate  this principle,  a few examples may be
adduced.
   1. If diluted sulphuric acid is poured upon  granulated zinc,
a brisk evolution of hydrogen gas ensues, and sulphate of oxide
of zinc is retained in solution.  Thus, 
CHEMICAL  REACTIONS.
31
         Zn              )      j H
         HO,S03 + Aq] ~  \ZnO,S03 + Aq.
In this example, and those that  follow, the volatile or gaseous
products are always printed with a full-face type.
   2. If diluted sulphuric  acid is poured upon protosulphide of
iron, sulphide of hydrogen gas escapes, and sulphate of protox-
ide of iron remains in solution.   Thus,
         FeS           )     |HS
         HO,S03-\-Aq\  ~  \FeO,S03-{-Aq.
   It wiU be noticed  from the last two  reactions, that it is not
essential that more than  one of the factors of the reaction
should be fluid, or in solution.
   3. If strong sulphuric acid is poured upon common salt, and
the mixture sUghtly heated, chlorohydric  acid  gas  is evolved,
and bisulphate of soda  remains  dissolved in the excess  of sul-
phuric acid.  Thus,
     Na CI               \     f H CI
     HO,S03.HO,S03\  = \NaO,S03.HO,S03.
   4.  If strong  sulphuric acid is poured  upon nitre, and the
temperature of the mixture sUghtly elevated, the  vapor  of
nitric acid is given off, and  bisulphate of potassa is formed.
Thus,
K
H
0,N05            }      (HO,JV05
0,S03.HO,S03{ = \K0,S03.H0,S03
  5. If diluted nitric acid is poured upon chalk, or any other
analogous carbonate, carbonic acid gas is set free, and a salt
of nitric acid formed.  Thus,
         CaO,C02      I  _  t#0 + COf
         HO,NOs + Aq\ ~ \  CaO,N05-\-Aq. 
32
CHEMICAL REACTIONS.
  It is  very frequently the  case that two, or even  all of the
three, classes of chemical reactions are combined, and going on
simultaneously, in a single experiment.  In the last example,
for instance, the metathesis is succeeded by an analysis of one
of the products, owing to the want of affinity between C 02 and
H O.   Again, the reaction of nitric acid on copper is an ex-
ample where  all three varieties  of reactions are  combined.
Three equivalents of copper  react on four equivalents of nitric
acid, and the reaction may be conveniently studied in two parts.
In the first part, the copper is oxidized by one of the equiva-
lents of the acid; here we have analysis accompanied by syn-
thesis ; and in the second part the copper changes place with
the hydrogen of the acid, a case of metathesis.
  1st. 3 Cu + H0,N05  = 3 (CuO, HO) + N©2.
       HO,N05l_,A     <CuO,NOs  },A
  2d'  CuO, HO \-rA1- \HO + HO\^A?'
The whole reaction combined may be expressed thus:
3Cu + 4(ffO,NOs)+Aq = 8(CuO,N05)+Aq + NOa,
Such mixed processes are very common in all complex cases
of chemical change.

   Stochiometrical Problems. — The chemical  symbols enable
us not only to represent chemical changes, but also to calculate
exactly the amounts of the  substances required  in  any given
process, as well  as  the amounts of the  products which it will
yield. The  method of making such calculation  can best be
illustrated  by examples.  In these examples the weights and
measures will be given according to the French decimal system,
which is now very generally used in chemical laboratories, and 
CHEMICAL  REACTIONS.
33
which, on account of the very simple relation between the units
of measure and of  weight, greatly facilitates stochiometrical
calculations.  The French measures and weights  can,  when
required, be very easily reduced to the English standards, by
means of a table at the end of the volume.
  Problem 1.   We  have given  10 kilogrammes of common
salt, and it is required to  calculate  how much chlorohydric acid
gas can be obtained from it by treating  with sulphuric acid.
The reaction is
   23+35.5 = 58.5.                                  1+35.5 = 36.5.
  NaCl-f HO,S03 + Aq = NaO, S03 + Aq+H.Cl
Hence one equivalent, or 58.5 parts, of common salt, will yield
one equivalent, or 36.5 parts, of chlorohydric acid gas.  There-
fore  the amount which  10 kilogrammes will yield can be cal-
culated  from the proportion
       53.5    36.5
     Na CI: H CI  = 10 : x = 6.239 kilogrammes, Ans.
  Problem 2.   It is  required to calculate how much chlorine
gas can be obtained  from chlorohydric acid with 6 grammes of
hyperoxide  of manganese.   The  equation representing the
reaction is
  28.6+16      44.6.                                      35.5.
  Mn02 + 2 HCl+Aq = Mn CI + 2  HO + Aq + CI.
Hence the amount of hyperoxide of manganese represented by
Mn 02, or 44.6 parts, yields an amount of chlorine  gas repre-
sented by CI, or 35.5 parts, and we have the proportion,
            44.6   35.5
           Mn 02: CI =  6 : x = 4.775 grammes.
  Problem 3.  It is  required to calculate  how much sulphuric 
34
CHEMICAL REACTIONS.
acid and nitre must be  used to  make 250  grammes of the
strongest nitric acid.
  101.2           98                                63
KO,K05+2(HO,S03)=KO,S03.HO,S03+HO,1X05.

Hence,
               63            98
            HO,NOs:  2{H0,S 03) == 250 : x.
                 63         101.2
             HO,N05: KO, NOs = 250 : x.

  From the above examples we can deduce the following gen-
eral rule for calculating  from the amount of any given factor
of  a  chemical  reaction the amount of the  products,  or the
reverse.  Express the reaction in an  equation • make then the
proportion,  As the  symbol of the substance given is to the sym-
bol  of the substance required, so is the amount of the  substance
given to x, the amount of the substance required; reduce the
symbols to numbers, and calculate the value of x.
  On account of the very great lightness, the amount of a gas
is very much more frequently estimated by measure than by
weight.  At the end of the volume a table will be found giving
the  weight in  grammes of one thousand cubic centimetres, or
one litre, of  each of the most important gases.  With the aid
of this table such  problems as the foUowing may be solved.
  Problem 4,  How much chlorate of potassa must be used to
obtain  one  Utre  of oxygen  gas?   One Utre of oxygen  gas
weighs 1.43 grammes.

               K O, CI  05 = K CI + 6 O.
               48       122.7
              6 0  :  K O, CI 05 = 1.43 : x. 
CHEMICAL  REACTIONS.
35
   Problem 5. How much zinc and sulphuric acid must be used
 to obtain 4 litres of hydrogen gas ?   Four litres of hydrogen
 weigh 0.357 grammes.
     Zn + H 0, S 03 + Aq =  Zn 0, S 03 + Aq + H.
        1  32.6
       H : Zn = 0.357 : x = amount of zinc required.
  1      49
 H : HO, S 03 = 0.357 : x = amount of sulphuric acid required.
   Problem 6. If ten  grammes  of water  are  decomposed  by
 galvanism, how large a volume of mixed gases will they give ?
                      HO = H+0.
          9   1
         HO : M=10 : x — ty grammes of hydrogen.
           9   8
         HO : © = 10 : x = ££■ grammes of oxygen.
   ■V1. grammes of hydrogen occupy 12.429 cubic centimetres.
   ■^-grammes of oxygen occupy    6.214      "     "
   The mixed gases occupy        18.644      "     "
 And hence * water, when decomposed into its elements, expands
 1864 times.

   The use of chemical symbols,  both in expressing chemical
 reactions and  in  stochiometrical calculations, having been ex-
 plained,  they will be used in the foUowing pages to iUustrate
 the text of Stockhardt's Elements of Chemistry.  The reac-
 tions  described in that work  are represented  in the  form of
 equations, the first member containing always the factors, and

  * It must be remembered that one cubic centimetre of water weighs one
gramme. 
36
CHEMICAL  REACTIONS.
the second,  the products of the process.   It remains for  the
student to work out the reaction, and represent it in the manner
explained on page 25.  The equivalents of water which are set
free or formed during a  reaction,  are not generally indicated,
but are merged in the general symbol Aq; and the student
will frequently be obUged to supply these equivalents in  order
to work out the reaction.  The  symbols of  solids are printed
in Roman letters, those of fluids in italics, and those of gases
in full-faced type.  When, however, the solid or  gas is dis-
solved in water,  the symbol  is  printed in itaUcs, foUowed by
the general symbol Aq, or aq.   Aq always stands  for an indefi-
nite and large  amount of  water; aq, for an indefinite but small
amount of water.  The color of a  substance, especiaUy of pre-
cipitates, is  frequently printed above its symbol.   The  head-
ings and  figures, or letters at the side of the page, refer to the
sections of the above-mentioned book.
   In  order still further to illustrate the subject, a large number
of problems have been  added to  the reactions, which the stu-
dent is expected to solve.  The  method of solving these  prob-
lems  can, in  most  cases, be  deduced  from the  explanations
already given,  and  from the sections of the text-book; in all
other cases the method is  explained under   the  problem.
Throughout the foUowing pages, aU gases and vapors are sup-
posed to be measured  at the temperature of 0° and when  the
barometer stands at 76 centimetres, unless some other tempera-
ture or barometric pressure  is  expressly stated.   All specific
gravities are referred to  water at its maximum density (at 4°).
The  temperatures are all given on the Centigrade scale, and
the weights and measures  according  to  the French  decimal
system. 
REACTIONS AND PROBLEMS.
4 
 
WEIGHING  AND  MEASURING.
10.  Problems on French System of Weights and Measures.

   1. Reduce by means of the table at the end of the book,—
   a. 30 inches to fractions of a metre.
   J. 76 centimetres to EngUsh inches.
   c. 36 feet to metres.
   d. 10 metres to feet and inches.
   2. Reduce by means of the table at the end of the book, —
   a. 8 lbs. 6 oz. to grammes.
   b. 7640 grammes to English apothecaries' weight.
   c. 45 grains to grammes.
   3. What is  the diameter, and what  is the circumference, of
the globe in French measure ?
   4. What is  the distance  from Dunkirk in France to Barce-
lona in Spain ?  The latitude of Dunkirk = 51° 3', that of
Barcelona = 41° 22',  and the two places are on  the  same
meridian.
   5. Were our globe composed entirely of water at its great-
est density, what would be its weight in kuogrammes?
   6. What is the weight of one cubic decimetre of water ?
   7. Reduce by means of the table at the end of the book, —
   a. 4 pints to litres and cubic centimetres.
   b. 5 gallons to litres and cubic centimetres. 
40
WEIGHING AND MEASURING.
  c. 5 litres to English measure.
  d. 4 cubic centimetres to EngUsh measure.

11-17.           Problems on Specific Gravity.

  1. The specific gravity of iron = 7.84.  What is the weight
of 1 cubic centimetre, 4 cubic  centimetres, &c. of the metal in
grammes ?
  2. The specific gravity  of  alcohol = 0.81.   What is  the
weight of one  litre in grammes ? of 45 cubic centimetres, &c. ?
  3. The specific gravity of  sulphuric acid = 1.85.   If you
wish to use in  a chemical experiment 250 grammes, how much
must you measure out ?
  4. Knowing the specific gravity of any given substance, how
can you calculate  the weight corresponding to any given meas-
ure, or the measure corresponding to any given weight ?  Give
a general algebraic formula  for the purpose, representing spe-
cific gravity, weight, and volume, by Sp. Gr., W., and V.
  5.  Determine  the  Sp. Gr. of absolute alcohol from  the
following data: —
                                                  Grammes.
  Weight of bottle empty,                           4.326
     "       "    fiUed with  water at 4°,     *     19.654
     "       "        "      alcohol at 0°,         16.741
  6. Determine the  Sp. Gr. of lead  shot  from the foUowing
data: —
                                                  Grammes.
  Weight of bottle empty,                           4.326
     "   "   "   filled with water at 4°,           19.654
     "   " shot,                                 15.456
     "   " bottle, shot, and water,                 33.766
  7. Determine the Sp. Gr. of iron from the following data: —
                                                    Grammes.
  Weight of iron in air,                               3.92
     "     "    "  water,                            3.42 
WEIGHING  AND MEASURING.
41
  8. Determine the Sp. Gr. of copper from
                                                     Grammes.
     Weight of copper in air,                        10.000
         "     "    "     water,                       8.864
  9. Determine the Sp. Gr. of ash wood from
                                                     Grammes.
     Weight of wood in air,                          25.350
          "     a copper sinker,                      11.000
          "     wood and sinker under water,         5.100
  10.  How  much buUc  must a hollow vessel of  copper fill,
weighing one kilogramme, which will just float  in water ?
  11.  How  much bulk must a hollow vessel  of iron occupy,
weighing ten kilogrammes, which sinks one hah" in water ?
  12.  An aUoy of gold  and silver  weighs ten kilogrammes
in the air,  and 9.735  kilogrammes in water.  What are the
proportions of gold and  silver ?  Sp. Gr. of gold = 19.2,  of
sUver = 10.5.
  Solution. — In the French system the volume of a solid in cubic centimetres
                                                 w
equals its weight in grammes divided by its Sp. Gr., or V = g—gp  Since one
cubic centimetre of water weighs one gramme, the volume of a solid in cubic
centimetres is equal to the weight of water it displaces in grammes.  Hence the
weight of water displaced = g—jr-. Put, then, x = weight of gold in alloy,
10 — x will equal weight of silver.  -^ = weight of water displaced by gold,
and 1Q~* = weight of water displaced by silver.  Hence —  +   ~* = 0.265.

  13.  An aUoy of copper and  silver weighs 37 kilogrammes
in the air, and  loses 3.666 kilogrammes when weighed in water.
What  are the proportions of  silver and copper ?

22.       Problems on Expansion of Liquids and Gases.

  1. If  34.562 cubic centimetres of mercury at  0°  are heated
to 100°, what increase of volume do  they undergo, and what
is the increased volume ?
  Solution. — The  small fraction of its  volume by which one c. c of  a liquid
                     4* 
42
WEIGHING AND MEASURING.
or gas increases when heated from 0° to 1°, is called the coefficient of expansion
of that liquid or gas.  The coefficient of expansion of mercury, for example,
= 0.00018, that is, one c. c. of mercury at 0° becomes 1.00018 c. c. at 1°.  If
we assume that the expansion  is proportional to  the temperature, then one
c. c. at 0° becomes 1.00036 at 2°, 1.0009 at 5°, and 1.018 at 100°.   Hence,
34.562 c. c. of mercury would become, at 100°, (34.562 X 1.018) c. c. To make
this solution general, let k = coefficient of expansion; then (1 + k) = increased
volume of one c. c. when heated  from 0° to 1°,  and (1 + t k) = increased
volume of one  c. c. when heated from 0° to t°, and V (1 + t k) = increased
volume of V c. c. when heated from 0° to t°; representing by V the increased
volume, we have
                          V'=V(l + t°k),
from which the increased volume of any liquid or gas may be calculated when
the volume at 0°, the coefficient of expansion, and the temperature are known.
  It is not true, as we have assumed, that liquids expand twice as much for two
degrees, three times as much for three degrees, &c, as they do for one; but,
on the contrary, the rate of expansion slowly increases with the temperature.
For example, one c. c. of mercury  at 0° becomes 1.000179 at 1°, but one c. c.
at 300Q becomes 1.000194 at 331°.  This  difference of rate, however, is  so
small, that we  can neglect it, except in the most  refined experiments, more
especially  if we use not the coefficient observed at any particular temperature,
but a mean coefficient obtained  by observing  the total amount of expansion
between 0° and 100°, and then dividing the result by 100, by which we averrge
the error.  Again, experiments on the expansion of fluids are commonly con-
ducted in glass vessels, which expand themselves by heat, and therefore cause
the expansion  to appear less than it is.  If they expanded as much  as the
fluid, it is  evident that the fluid would not appear to expand at all.  They in
fact expand much less than the  fluids, but, nevertheless, sufficiently to make
a material difference between the absolute expansion of a fluid, and its apparent
expansion in glass vessels.   The mean absolute coefficient of mercury between
0° and 100° is  1.000181.  The apparent expansion  in glass  between 0° and
100° is 1.000156.  In Table IV., at the end of the volume, will be found the mean
coefficients of expansion  in glass of some of the more important fluids.  The
rate of expansion of water varies so rapidly and so anomalously, that  no use
should  be made of its coefficient except in experiments extending over 100°.
When it is required to determine the amount of expansion between narrower
limits, use should be made of Table V., which gives the volume to which one
cubic centimetre of water at 0° or at 4° increases,  when heated to the  tem-
peratures at the side of the table. It also gives the Sp. Gr. of water at different
temperatures, when either water at 0° or at 4° is taken as the unit.

   2. What  wUl be  the volume of 5.346 c. c. of water at 0°,
when heated to 100° ?

   3. What will be the volume of 250 c. c. of oil  of turpentine,
at 0°, when  heated to 50° ? 
WEIGHING AND MEASURING.
43
  4. What wiU be the volume of 35 c. c. of water at 0°, when
heated to 4°, to 25°, to 40°, and to 84° ?

17.  Problems on  reducing Specific Gravities  to the Standard
                        Temperature.
  The specific gravities of both Uquids and solids are supposed
to be referred to water at 4°, its greatest density; but in prac-
tice we always use water at a much higher temperature, and it
becomes  therefore necessary to reduce the results to 4°, or in
other words, to calculate what would  have been  the  result had
the  temperature  been 4° during the experiment.   Hence an
important class of problems like the foUowing : —
  1. The Sp. Gr. of  zinc was found to be 7.1582  when the
temperature of the water was 15°.   What would  have  been
the  Sp. Gr. at 4° ?
  Solution. — Sp. Gr. of water at 4° : Sp. Gr. at 15° = Sp. Gr. of zinc at
15° : Sp. Gr. at 4°;  or   1 : 0.9992647 = 7.1582 : x.
  2. The Sp. Gr. of antimony was found to be  6.681 when
the  temperature  of  the water was  15°.  What  would  have
been the Sp. Gr. at 4° ?                       Ans. 6.677.
  3. The Sp. Gr. of  an aUoy of zinc and antimony  was found
from the following data : —
                                                   Grammes.
  Weight of the  alloy,                             4.4106
             Sp. Gr. bottle,                        9.0560
       «       "      «  full of water  at 4°,       19.0910
       "     bottle, alloy, and water at 14°.6,        22.8035
                                              Ans. 6.375.
  4.  Find  the  Sp.   Gr.  of  metaUic zinc  from the following
data: —
	Grammes.
Weight of the zinc,	12.4145
" bottle,	9.0560
« " full of water at 18°,	19.0790
" " zinc and water at 12°.4,	29.7663
	Ans. 7.153.
 
44
WEIGHING AND  MEASURING.
24. Problems  on reducing Centigrade Degrees to  Fahrenheit,
                        and the reverse.

   1. What do —40°, —20°, —15°, —9°, 0°, 4°,  10°, 13°, 15°,
80°, 100°, 150° Cent,  correspond to on the Fahrenheit scale ?

   Rule. — Double the  number of degrees,  subtract one tenth
of the whole, and add 32 if the degrees are  above 0°, or sub-
tract 32 if  they are below.

   2. What do—40°,—32°,—18°, —7°, 0°,  10°,  32°, 42°, 50°,
70°, 90°, 100°, 150°,  212°, 300°,  450° Fahr. correspond to on
the Centigrade scale ?

   Ride. — Subtract 32  if above  0°, or add 32  if below, add
one ninth to the result, and divide by two.


27.               Problems on Expansion of Solids.

   1. What wiU be the length of a rod of iron 424.56 metres
long at 0°,  when  heated to 20° ?

  Solution. — The small fraction of its length by which a rod  of iron, or of any
other solid, one metre long, expands when heated from 0° to 1°, is called the
coefficient  of linear  expansion  of the solid.  A bar of iron one metre long
at 0° becomes 1.0000122 at  1°, and the small  fraction  0.0000122 is  the
coefficient  of linear  expansion  of iron.  Representing this  coefficient by k,
we have for  the new length of the rod, at 1°, Z' = £ (1  + k), and at i°,
l' = l(l + tk).  We may assume, in  the case  of solids, that the expansion is
proportional to the temperature, especially if we deduce our coefficient from
experiments made between 0° and 100°, as described  under examples on the
expansion of fluids.  Substituting for I, k, and t the values given in the problem,
we have I' = 424.56  (1 + 20 x 0.0000122).  The coefficients  of linear expan-
sion for a number of solids are given in Table IV. at the end of the volume.

   2. What will be the length of a rod of copper 2.365 metres
long, at 0°, when heated to 100° ?

   3. What will  be  the length of a  rod of silver 0.760 metres
long, at 12°, when heated to 20° ?

  Solution. — Denoting by I the unknown length of the rod at 0°, by I' the
known length at t°, and I" the required length at tf°, we have as above,
                l' = l(l + tk),  and  l" = l(l + fk). 
WEIGHING  AND  MEASURING.
45
By combining these, we obtain

               l" = v (f+Ti) = l'D- + k («• -t)+ &c].
All the terms of the quotient may be neglected after the first, because they
contain powers of the already very small fraction k.  Substituting the values
in the above equation, we get
                  I" = 0.760  [1 + 0.000019 (20 — 12)].

  4.  One of the large iron tubes of the Britannia Bridge over
the Menai  Straits is  143.253 metres  long.   What  increase of
length does it undergo between 8° and 20° Centigrade ?

  5.  What is  the increased capacity of a globe of glass which
holds exactly one Utre, at 0°, when  heated to 250° ?

  Solution. — If we consider for a moment the glass globe as forming the out-
side shell of a solid globe of glass, it is evident that the increased capacity of
the globe will be equal to the increased volume of this solid globe, which we
have  supposed for  a moment  to fill the  interior.  The  problem, therefore,
resolves itself into calculating the amount of cubic expansion of a solid glass
globe having a volume equal to one litre, or  1000 c. c.  The coefficient of linear
expansion of glass is  given in  the tabte as 0.00001.  The coefficient of cubic
expansion  is always three  times* as  great as the linear expansion ; in the
case of glass, therefore, it is 0.00003.  By this  is meant  that, if a rod of glass
one metre long at 0° becomes  1.00001  long at  lc, then a cube of glass of one
cubic centimetre at 0° becomes  1.00003 c. c. at 1°.  Using, then, the equa-
tion
                          V' = V(l-Hfc),
we obtain by substitution
              V = 1000 (1 + 250 X 0.00003) = 1007.5 c. c.

   6. What  is the increased capacity  of a globe of glass which
holds exactly 495 c. c. at 15°, when heated to 300° ?
  * Each edge of a cube of glass one centimetre long at 0°, would become
(1 + k) c. m. long at 1°.   The increased volume of the cube would be equal to
(1 + k)3 = 1 + 3 k + 3 k2 + P;  but  as  k-\s an exceedingly small fraction,
k2 and ks may be neglected in comparison, so that a cube of glass of one c. c,
at 0° becomes (1 + 3 k) at 1°, which proves that the amount of cubic expan-
sion is three times as great as the linear. 
NON-METALLIC  ELEMENTS,  or
               METALLOIDS.
              FIRST GROUP :  ORGANOGENS.

                      Oxygen (0).
     Red.
56.  Hg 0 = Hg + O.
      1. How much oxygen can be obtained by heating 108
    grammes,  250  grammes, 25 grammes, or 5 grammes, of
    red oxide of mercury ?
      2. How many cubic centimetres of oxygen  can be
    obtained from  the  amounts of oxide of mercury  given
    in the last example ?
      3. How much oxide of  mercury would be required to
    yield one litre of oxygen by the process of 1 ?
      4. How much mercury would remain after the experi-
    ment of last example ?

59.  K O, C105 = K CI  + 6 O.
      1. How much oxygen  can be  obtained from 1  kilo-
    gramme,  50 grammes, or  5 grammes, of chlorate of po-
    tassa ?
      2. How much chloride of potassium would remain after
    the oxygen in the last examples had been driven off? 
        NON-METALLIC ELEMENTS,  OR METALLOIDS.     47

      3. How much chlorate of potassa would be required to
    make one Utre of oxygen ?

63. C + 2 O = C 02.

      1. How much carbonic acid gas will be formed by burn-
    ing 5 grammes of carbon ?
      2. How many cubic centimetres of carbonic acid will
    be formed by burning 5 grammes of carbon ?
      3. How much oxygen wiU be consumed in the last two
    examples ?
      4. How many  cubic centimetres of carbonic acid wiU
    be  formed from one Utre of oxygen, by burning in it
    carbon ?

64. S + 20 = SOa.

      1. How much sulphurous  acid gas wiU be formed by
    burning 10 grammes of sulphur ?
      2. How much  oxygen  wiU  be consumed in  the last
    example ?
      3. How many cubic centimetres of sulphurous acid wiU
    be formed by burning sulphur in one litre of oxygen ?
      4. Assuming that one Utre of oxygen yields exactly one
    Utre of sulphurous acid, what is the Sp. Gr. of S 02 gas ?

65. P-f5O=P05.

      1. How much phosphoric acid can be formed from 48
    grammes of phosphorus ?
      2. How much  phosphorus wiU exactly consume one
    Utre of oxygen ?
      3. How much phosphorus is required in order to make
    250 grammes of phosphoric acid ? 
48      NON-METALLIC  ELEMENTS, OR METALLOIDS.

67.  Na + O = Na O.
      1. How much oxide  of  sodium can be  made  from
    28.75 grammes of sodium ?
      2. How much  oxide of sodium can be made with one
    litre of oxygen ?  and how much sodium will be consumed
    in the experiment ?
68.  3Fe + 40  = FeO, Fe203.
71.  2 Na 0 + P 05 + Aq = HO, 2 Na 0, P 05 + Aq.
      The crystallized salt = [HO,  2NaO] P05 . 24HO.
72.  Ca  0 + G02 + Aq = CaO, C02 + Aq.
73.  Ca  0 + S 02 + Aq = Ca 0, S 02 + Aq
       1. How much  Ume would be required to neutraUze 20
     grammes of sulphurous acid ?
       2. How much lime would  be required to neutralize
     the sulphurous acid obtained  by burning  5 grammes  of
     sulphur ?
       3. How much Ume would be required to neutraUze one
     Utre of sulphurous acid gas ?

79. 3 Mn 02 = Mn 0, Mn2 03 + 2 O.
       1. How much oxygen can be obtained  from one kilo-
     gramme of hyperoxide of manganese by heating ?
       2. How much of the  oxide must be used in order to
     obtain 30 grammes of oxygen ?
      Mn02 + HO,  S 03 = MnO, S 03 + HO + O.
        1.  How much oxygen can be  obtained from one kilogr.
     of hyperoxide of manganese by the last process ? and how
     much sulphuric acid will be required to decompose it ?
        2.  What  per  cent of the whole amount, of oxygen in
     the mineral is obtained by the two processes ?  and by how
     much does the second exceed the first ? 
        NON-METALLIC ELEMENTS, OR METALLOIDS.    .49

                     Hydrogen (H).

81. Na + Aq = Na 0, HO -j- Aq -f- H.

       1.  How much hydrogen  will  be  set  free  by  5.25
     grammes  of sodium?   How many cubic centimetres will
     be set free?

82.  3 Fe -f 4 H O = FeO, Fe2 03 + 4 H.
       1.  How much hydrogen would be obtained in the last
     experiment  by the  decomposition of 39  grammes  of
     water ?
       2.  By how much would the weight of the tube increase
     in last example ?

83. Zn + HO, S03-\-Aq=ZnO,S03-\-Aq+U.
       1.  How much sulphuric acid and how much zinc must
     be used in order to make 2 grammes of hydrogen ?  How
     much in order to make  one litre ?
       2.  How much hydrogen can be obtained with one kilo-
     grammesof zinc ?
       3.  How much with one kilogramme of sulphuric acid ?
     Decomposed by Galvanism.
55. #0 = 11 + 0.
       1.  How many cubic  centimetres of hydrogen would be
     obtained by  the decomposition of one cubic  centimetre
     (one  gramme) of water?  How  many  cubic centimetres
     of oxygen ?   How many of mixed gases ?
85.    A sohd immersed in a liquid or a  gas  is buoyed  up
     by a force equal to the  weight of the liquid or gas which
     it  displaces.  The  excess of the buoyancy over its own
     weight is called its ascensional force.
       1. What would  be   the ascensional force of  a smaU
     baUoon fiUed with one Utre  of hydrogen gas, when the
     balloon itself weighs five centigrammes ?
                  5 
50      NON-METALLIC ELEMENTS, OR METALLOIDS.

       2.  What would be the ascensional force of a spherical
     balloon seven  metres in  diameter, two thirds  fiUed with
     hydrogen, when the baUoon and attachments weigh twenty
     kilogrammes ?

87. H + 0 = HO.
       1.  How much water would be  formed  by burning one
     thousand  Utres  of hydrogen? and how  much oxygen
     would be consumed in the process ?
       2. How much vapor of water would be formed in Ex-
     ample 1 ?

93.             Problems on the Barometer.

       1. When the surface of the column  of mercury  in  a
     barometer  stands at 76 centimetres  above  the mercury in
     the basin, with what weight is  the atmosphere pressing on
     every square centimetre of surface ?  Sp.  Gr. of mercury
     = 13.596.
       2. To what difference of pressure does  a difference of
     one centimetre in the  barometric column correspond ?
       3. When the water  barometer  stands  at  ten metres,
     what is the pressure of the air if the temperature is 4° ?
       4. How high would  an alcohol barometer,  and  how
     high  a  sulphuric-acid barometer,  stand under  the same
     circumstances,  disregarding  in each case  the  tension of
     the vapor ?  Sp.  Gr. of alcohol = 0.8095 ; Sp.  Gr. of
     sulphuric acid = 1.85.

94.  Problems on the  Compression and Expansion of Gases.

       Mariotte's Law. — It is an established principle of sci-
     ence that  The volume  of a given weight of gas is inversely
     as the pressure to which it is exposed; that is, the greater
     the pressure, the smaller  is the volume; and the less the 
    NON-METALLIC ELEMENTS,  OR METALLOIDS.      51

pressure, the larger is the volume. This may be illustrated
by an India-rubber bag holding one  litre of air, or of any
other gas.   This is exposed to a pressure, under the ordi-
nary conditions  of the atmosphere,  of a little over one
kilogramme on  every square centimetre of surface.  If
this  pressure is doubled, the volume of the  bag will be
reduced  to  one half; if trebled, to one third, &c.  On
the other hand, if the pressure is  reduced  to one half,
the volume will double; if to one third, the volume will
treble, &c*  The principle is expressed in mathematical
language by the proportion

                H' :  H  =  V : V;               (1.)
where H  and H' are the heights of the barometer which
measure the pressure to which  the  gas is exposed  under
the two conditions of volume V and V.
   Since  the density of a given weight of gas is inversely
as the volume, or D' : D = V : V, it follows that
                 H': H = D': D,                (2.)
or the density of a gas is  proportional to the pressure to
which  it  is exposed.  Moreover, since the weight  of a
given volume of gas  is proportional  to its density, and its
density, as  just proved, proportional to the  pressure, it
follows that The weight of a given volume of gas is directly
as the pressure to which it is exposed, or
                H' : H = W : W.                (3.)

These  three proportions are  very important, and will be
constantly referred to in the following pages.  The student
must be  careful to notice that in (1) the weight of gas is
supposed  to be  constant  and the volume to vary, and in
(2) the volume is supposed  to be constant and the weight
to vary.
* We suppose the bag to have no elasticity. 
52      NON-METALLIC  ELEMENTS, OR METALLOIDS.

       The variations in the pressure of the atmosphere, amount-
     ing at times, to one  tenth of the whole, necessarily cause
     equally  great changes in the volume of gases which are
     the objects of chemical experiment.   Hence,  in  order to
     compare together different volumes of gas, it  is essential
     that they should  have been measured when  exposed to
     the  same pressure.   A  standard pressure has therefore
     been  agreed  upon, that measured by 76 centimetres of
     mercury, to which  the  volume  of gases  measured under
     any other pressure  should  be reduced.  Hence a number
     of problems like  the following: —
       1.  A  volume of hydrogen  gas was measured and found
     to be equal to 250  c. c.   The height of the barometer,
     observed  at  the  same  time, was 74.2  centim.   What
     would have been the volume if observed when the barom-
     eter stood at 76 centim. ?
       Solution. — Proportion (1) gives, by substituting the data of the prob-
     lem, 74.2  : 76 =250 : V'=.Ans.
       2.  A  volume of nitrogen gas measured 756 c. c. when
     the barometer stood at 77.4 centim.   What would it have
     measured if the barometer had stood at 76 centim. ?
       3.  A  volume of air standing in a bell-glass over a mer-
     cury pneumatic trough  measured 568 c. c.  The barome-
     ter at the time stood at 75.4 centim., and the surface of the
     mercury in the bell was  found, by measurement, to be 6.5
     centim.  above the surface  of the mercury in  the trough.
     What would  have been  the volume had the air been ex-
     posed to the pressure of  76 centim. ?
       Solution. — It  can easily be seen, that the pressure of the air on the
     surface of the mercury in the  pneumatic trough, measured by the
     height of  the barometer at the time (75.4 centim.), was balanced first by
     the column of mercury in the bell, and secondly by the tension * of the
     confined air. Hence, the pressure * to which the air was exposed was
  * The tension of a gas is the force with which it tends to expand, and, when
the gas is at rest, must evidently be exactly equal to the pressure to which it
is exposed. 
    NON-METALLIC ELEMENTS,  OR METALLOIDS.       53


equal to the height of the barometer  less the height of the mercury
in the bell, or 75.4 — 6.5 = 68.9 centim.  We have then the proportion
68.9 : 76 = 568 :  V = Ans.

   4. A volume of air standing in a tall  beU-glass over a
mercury pneumatic trough measured 78 c. c.   The barom-
eter at the time stood at 74.6 centim., and the  mercury in
the bell at  57.4 centim. above the mercury in the trough.
What would have been the volume had^the pressure been
76 centim. ?

   5.  What would be the answers to the last two  problems,
had the  pneumatic trough  been fiUed with water instead
of mercury ?

      Problems on Expansion of Gases by Heat.

   1.  What will  be the volume of 250 cubic  centimetres
of air at 0° when heated to  300° ?

  Solution. — It has been found by Regnault and others, that the per-
manent gases expand so nearly equally for the same increase of tempera-
ture, that the differences may be entirely disregarded except in the most
refined investigations; and it has also been found, that their rate of
expansion does not materially vary from the lowest to the highest tem-
peratures at which experiments have been made.  The coefficient of
expansion for air, as  determined  by Regnault, is equal  to 0.00366, and
we can therefore calculate the volume, V, of a gas at any temperature,
from its volume, V, at 0°, by means of the equation, already explained,
                y = V(l + *X 0.00366);               (1.)
or if we know the volume V for a temperature t, we  can calculate the
volume V" foranother temperature tf, by means of  the equation
              V"=V'(1 + (t'—t) 0.00366).              (2.)
By transposing, we can obtain from equation (1)
                              v
                    V =_____-______                 (3.)
                        1 + (X 0.00366                  K '
As the volume of a gas varies very considerably with the temperature,
it is important, in comparing together different  measurements, that we
should adopt a standard temperature, as we have adopted a standard
pressure.  The temperature which has been agreed upon is 0°; but as it
would be inconvenient, and often impossible, to make our measurements
at this temperature, it becomes necessary to calculate, by means of equa- 
54       NON-METALLIC ELEMENTS, OR METALLOIDS.
tion (3), from a volume V measured at t°, what would be the volume at
0°.  This is called technically reducing the volume to 0°.  There can be
obtained also from equations (1) and (2) the equations
      V — V                             V" — V
(=VX 0.00366'    (4'>     and   </=< + V'X 0.00366'     (5l)
by means of  which  we can calculate the change of temperature when
we know the  change of volume.  Representing the coefficient of expan-
sion in the above formulae by k, we can  obtain, by transposing and re-
ducing the equation,
          V'-V                     V — V
      Jc==-JV~,    (60    and  t = V'(C-0'          (7°
From these we can calculate the coefficient of expansion when we know
the volume of a gas at two different temperatures.

   2.  A volume of gas  measured  560 c. c. at 15°.   What
would it measure at 95° ?
   3.  A glass  globe  holding  450 c. c.  of air  at  0° was
heated to 300°.   At  this temperature  the neck  was her-
metically sealed, and the globe cooled  again to  0°.   The
neck was then opened under  mercury, and the  air remain-
ing in the globe passed up into a  graduated jar, and meas-
ured.   How much was it found to measure ?
  Solution. — By substituting the values for V and /  in the equation
V' = V (1 +1 x 0.00003), we obtain the increased capacity of the globe,
and of course the number of cubic centimetres of expanded  air which it
contains at 300°.  It is then only necessary to substitute this value for
V and 300 for t, in equation (3), in order to find what will  be the volume
of this expanded air when cooled again to 0°.
   4.  What is the weight of air contained in an open  glass
globe of 250 c. c. capacity, at the temperature  of 20°, and
when the barometer stands at 74  centimetres ?
  Solution. — In order to make the solution general, we will represent the
capacity of the globe, the temperature, and  the height of the barometer,
by V, t,  and H, respectively.  One cubic  centimetre of air at 0°, and
when the barometer  stands at 76 centimetres, weighs 0.00129 grammes.
To find what one cubic  centimetre would weigh when the barometer
stands at H centimetres, we make use of  proportion (3), page 51:
       H : B7 = W : W;   or  76 : H = 0.00129 : W;
whence W = 0.00129 . =, the weight of one cubic centimetre at 0°, and
under a pressure of  H centimetres.   To find what one cubic centimetre 
         NON-METALLIC ELEMENTS, OR  METALLOIDS.       55


     would weigh at t°, it must be remembered that one cubic centimetre at 0°
     becomes (1 + t 0.00366) c. c. at t°; therefore, at t° and at H centimetres
     of the barometer, (1 + t 0.00366) c. c. weigh 0.00129 . ^ grammes.  By
     equating these two terms we obtain (1 +1 0.00366) = 0.00129. ^, whence
     1 = 0.00129 • i . , 0 00366 " ^i the weight of one cubic centimetre at P and
     under a pressure of  H centimetres.  The weight of V cubic centimetres,
     ic, is evidently
                    w = 0.W129V.TT7i355a-?.               (8.)

        5. What is the weight of air contained in an open glass
     globe  of 560 cubic  centimetres' capacity  at  0°,  at  the
     temperature  of  300°,  and  under a pressure of 77 centi-
     metres ?
       Solution. — In the solution of the last example we neglected the change
     of capacity of the glass globe due to the change of temperature.  This
     causes  no sensible  error when the change of temperature is small, but
     when,  as  in the present problem, the change of temperature is quite
     large, the change of capacity of the globe must  be considered.  If the
     capacity is V  c. c. at 0°, it becomes at P V (1 + t 0.00003). Introducing
     this value for  V into the equations of the last section, we obtain

             v = 0.00129 V (1 + t 0.00003) . lTT^Me • %.        (9.)


99.           Problems on Specific Gravity of Vapors.

        General Solution. — The specific gravity  of a vapor is its weight com-
     pared with the weight of the same volume of air under the  same con-
     ditions  of temperature and pres-
     sure.   To find, then, the specific
     gravity of a vapor, we must as-
     certain the weight  of a known
     volume, V, at a known tempera-
     ture, t, and under a known pres-
     sure, H,  and  divide this by the
     weight of the  same volume of air
     at  the same  temperature, and
     under  the same pressure.  The
     method may best be explained by
     an example.  Suppose, then, that
     we wish to ascertain the specific
     gravity of alcohol vapor. We take
     a  light glass  globe having a ca-
     pacity of from 400 to 500 c. c, and
 
56       NON-METALLIC  ELEMENTS,  OR METALLOIDS.


     draw the neck out in the flame of a blast lamp, so as to leave only a fine
     opening, as shown in the figure at a.  The first step is now to ascertain
     the  weight  of the glass globe when completely exhausted of air.   As
     this cannot  readily be done directly, we weigh the globe full of air, and
     then subtract the weight of the air, ascertained by calculation from the
     capacity of the  globe, and from  the  temperature and pressure of the air,
     by means of equation (8).  Call the weight of the globe and air W, and
     the weight of the air w, then W — w is the weight of the globe exhausted
     of  air.  The second step  is to  ascertain the  weight of the globe
     filled with alcohol vapor at a known temperature, and under a known
     pressure.  For  this purpose we introduce into the globe a few grammes
     of pure alcohol, and mount it on the support represented in Fig. 6.  By
     loosening the screw, r, we next sink the balloon beneath the oil contained
     in the iron vessel, V, and secure it in this position. We now slowly raise
     the temperature of the oil to between 300° and 400°, which we observe by
     means of the thermometer, B.   The alcohol changes to vapor and drives
     out  the air, which, with the excess of vapor, escapes  at a.  When the
     bath has attained the requisite temperature, we close the opening a, by
     suddenly melting the end of the tube at a with a mouth  blow-pipe, and
     as nearly as possible at the same moment observe the temperature of the
     bath and the height of the barometer.  We have now the globe filled with
     alcohol vapor at  a known temperature, and  under  a  known pressure.
     Since it is  hermetically sealed, its  weight  cannot  change,  and we
     can therefore allow  it to cool, clean it, and weigh it  at  our leisure.
     This will  give  us the weight of the globe filled with alcohol  vapor
     at a known  temperature, i', and under a known pressure, H'.  Call
     this weight W.   The  weight  of  the vapor is W — W + w.  The
     third step is to ascertain the weight of the same volume of air  at the
     same temperature and  under the same pressure. This can easily be
     found by calculation from equation (9).   The last step is to  find the
     capacity of the globe, which, although we have supposed  it known, is
     not  actually ascertained experimentally until the end of the process.
     For this purpose we break  off the tip of the tube (a), under mercury,
     which, if the experiment has been carefully conducted, rushes in and
     fills  the globe completely.  We then empty this mercury into a carefully
     graduated glass cylinder, and read off the volume.  We find then the
     specific gravity by dividing the weight of the vapor  by the weight  of
     the air.  The formulae for the calculation are then

     Weight of the globe and air,          W.
         "   "   air,                    w = 0.00129 V . r-r—^—w-.
                     '                                   1 + J0.OU366  76
         "   "   globe exhausted of air,   W — to.
        "   "      "  filled with vapor "J
         at a temperature t' and under a > W.
        pressure H',                     ) 
NON-METALLIC ELEMENTS, OR METALLOIDS.      57
Weight of the vapor,
   "    "    air at t'
and under a pressure H1

          Sp. Gr. =
;,}
H
 W — W + w.

■ 0.00129 V(l +00.00003) • r+Tr^osss " 76

      W' — W + u
0.00129 V (I + V 0.00003) ,
                                  l + t 0.00366  76
  1. Ascertain the Sp. Gr. of alcohol vapor from the fol-
lowing data: —
Weight of glass globe,      W
Height of barometer,       H
Temperature,               t
Weight of globe and vapor, W
Height of barometer,       H'
Temperature,               t'
Volume,                  V

   2. Ascertain the Sp.  Gr. of
foUowing data: —
Weight of glass globe,     W
Height of barometer,       H
Temperature,               t
Weight of globe and vapor, W
Height of barometer,       H'
Temperature,                t
Volume,                   V
  50.8039 grammes.
  74.754 centim.
   18°
   50.8245 grammes.
   74.764 centim.
   167°
   351.5  cubic centim.
      Ans. 1.5795.
camphor vapor from the

   50.1342 grammes.
   74.2   centim.
   13°.5
   50.8422 grammes.
   74.2   centim.
   244°
   295    cubic centim.
       Ans.  5.298.
                       Carbon (C).

109. C + 20 = C02.
     2Hg0-fC = 2J^ + CO2.
       1. How many grammes  and  how many  cubic  centi-
    metres of carbonic acid gas are formed by burning 10
    grammes of charcoal ? 
   NON-METALLIC  ELEMENTS, OR METALLOIDS.

  2.  How many grammes  and  how many cubic  centi-
metres of oxygen are consumed in the process ?
  3.  Assuming that the volume of carbonic acid gas gen-
erated during combustion is exactly equal to the volume of
oxygen gas consumed, what is the  Sp. Gr. of carbonic
acid gas ?
  4.  How much oxide of mercury is required to burn up
5.672 grammes of charcoal ?

 C + © = C O.      C + COa=2CO.
  1.  How many grammes and how many  cubic centime-
tres  of oxide of carbon gas are  formed  by burning 10
grammes of charcoal ?
  2.  How many grammes and how many  cubic centime-
tres of oxygen are consumed in the process ?
  3.  Ten  cubic centimetres of oxygen yield how  many
cubic centimetres of carbonic acid gas, and how many of
oxide of carbon  gas ?  What expansion 'does oxygen un-
dergo in combining  with carbon to form oxide of carbon ?
 Spermaceti.
 C^ H64 04 + 188 © = 64 C 02 + 64 HO.

  1.  How many grammes of carbonic acid, and how many
grammes of water, are formed by burning  10 grammes of
spermaceti ?
  2.  The carbonic  acid and water given off by a burning
spermaceti candle were carefully collected and weighed.
The  water weighed  0.564  grammes, the carbonic acid
weighed 1.3786.  How much of the candle was burned? 
NON-METALLIC ELEMENTS, OR METALLOIDS.     59
             SECOND GROUP OF METALLOIDS.

                       Sulphur (S).

132. Fe S -f- HO, S03-\-Aq = FeO,S03 + Aq + H S.
       1. How much  sulphide of hydrogen  can be made from
     15 grammes of sulphide of iron ?  How many cubic centi-
     metres ?
       2. How much sulphide of iron, and how much sulphuric
     acid, is  required to generate sufficient gas to  saturate one
     Utre of  water ?
     HS-\-Aq + 0 = S-\-Aq.
     HS + 3©= M© + S©2.
                            Black.
133. a. Pb -f- HS-\- Aq  = Pb S + Aq + H.
        Yellow.                  Black.
     b. Pb 0 + HS+ Aq = Pb S + Aq.
                                                Black.
     c.PbO,  [04^] 03 +  HS +  Aq   =  Pb S  +
        HOilCiH^Ot + Aq.
                                                 Black.
     d. Fe 0, S 03 -f- Ca  0, HO + HS + Aq  = Fe S -f-
         Ca 0,S03 + Aq.
                                                Black.
        Cu 0,  [<74 H3~] 03  +  HS  -f Aq  =  Cu  S  +
         HCIC.H,-] 03 + Aq.
                               Orange.
       SbCk + 3 HS-\-Aq = Sb S3 + 3 HCl + Aq.
                               Yellow.
       As Cl3-\-3HS-\-Aq = AsS3 + 3HCl-\-Aq.
                              White.
       Zn Cl+ Ca SA Aq =  Zn S + Ca Cl+ Aq. 
60      NON-METALLIC ELEMENTS, OR METALLOIDS.

                     Phosphorus (P).

140.   P -f- 3 © = P 03 (by slow combustion).
       p _|_ 5 4) = p Os (by rapid combustion).
144.   Burnt bones consist chiefly of 3 Ca O, CP 05.
       3 CaO, c-P0s-\-2 (HO, S03) -\-aq =  2 (CaO,S03)
         -\-[2HO, Ca C] CP 05 + aq.
        Heated to a red heat.
       [2HO, Ca 01 CP05 + Aq + xC =  CaO, P05
         -j-a;C + Aq + 2C© + 2H.
         Heated intensely.
       3 (CaO,aP05) + xC = 3 CaO,aP05 + a:C + 2P
         + 10C©.
       1. How much  phosphorus can be  manufactured from
    20 kilogrammes of burnt bones, of which four  fifths are
    phosphate of Ume ?

145.   4P + 3(CaO, HO) = 3 (CaO, PO) + PH3-
       Besides the above reaction, there take  place simulta-
    neously the two  following  reactions,  in the experiment
    described in the text-book.

       3 P + 2 (Ca 0, H 0) = 2 (Ca 0, P 0)  + P H2.
       P + CaO, H 0 == Ca 0, P 0 + H.
             THIRD GROUP OF METALLOIDS.

                     Chlorine (CI).

150. Mn 02 -f 2 HCl -f aq = Mn CI -f aq + CI.

       1. How much chlorine gas can be obtained from 2.467 
        NON-METALLIC ELEMENTS, OR METALLOIDS.      61

     grammes  of chlorohydric acid gas?  How  many  cubic
     centimetres ?
       2.  How much chlorine  can be obtained from an unde-
     termined  amount  of muriatic acid  by means of 4.567
     grammes of hyperoxide of manganese ?  How many cubic
     centimetres ?
       3.  The hyperoxide of manganese  of commerce is more
     or less adulterated.  What per cent of Mn 02 does an
     article contain, of which  10 grammes, when heated with
     strong muriatic acid, evolve 4.0135 grammes of chlorine ?
       4.  How much chlorine  can be obtained from 25  cubic
     centimetres of muriatic acid*  of Sp. Gr. = 1.16 ?  How
     many cubic centimetres ?
       5.  In order to prepare one litre of chlorine gas how much
     hyperoxide of manganese, and how much muriatic acid,
     must  be used ?  Calculate  the amounts for pure Mn 02 and
     H CI  gas, and also when the oxide used contains only
     70  per cent of pure Mn02, and when the liquid acid
     used  has a Sp. Gr. = 1.15.

151.  Mn02 + 2Na CI -f 2 (HO, S 03) -\-aq = MnCl +
        2(NaO,S03)+aq + €\.

      We might use one half  as much common  salt, but then
     we should find  sulphate of manganese instead of chloride
     of manganese in solution.  Thus,

      Mn 02 + Na CI + 2 (HO, S 03)  -\-aq=--Mn 0, S03
        + NaO, S03 + aq + CL

      1. How much chlorine gas can  be  obtained by the last
    process from 34 kilogrammes of salt ?
  * See Table VT., which gives the per cent of H CI in the fluid acid of dif-
ferent specific gravities.
                   6 
62      NON-METALLIC ELEMENTS, OR  METALLOIDS.
      2. How many cubic centimetres of chlorine can  be ob-
    tained from one cubic centimetre of rock salt ?  Sp. Gr.
    of salt = 2.15.

152./. 2(FeO,S03) -\-HO,S03 + CI + Aq = Fe2 03,
         3S03 + HCl+Aq.
       6 (Fe 0,S03)+3Cl+Aq=2 (Fe.2 03, 3 S 03)    *
         + Fe3 Oh + Aq.
    g. Au + 3 CI + Aq = Au Cl3 + Aq. 
ACIDS.
              FIRST GROUP : OXYGEN ACIDS.

                  Nitrogen and Oxygen.

                 Nitric Acid (HO,N05).
159. KO, N05 + 2 (HO,S03) =  KO, S03. HO, S03
       + H©,]¥©5.

     NaO, N05 + 2 (HO, S03) = NaO, S03. HO, S 03
       + H©,JV©5.
      1. How mueh nitric acid can be made from 250 kilo-
    grammes of potash  nitre, and  how much sulphuric acid
    must be used in the process ?
      2. How much more nitric acid  wiU the same weight of
    soda nitre yield ?
      3. How much nitric acid, containing 40  per cent, of
    N05, can be  made from  1700 kUogrammes of potash
    nitre ?
      4. How much soda nitre, and how much sulphuric acid,
    and how much water, must be  used  to make 450 kUo-
    grammes of nitric acid, which shall contain  60 per cent.
    of pure acid ? 
ACIDS.
. c. [NH,-] 0,HO + HO,NOs+Aq= [ JV^] 0, N 05

  d. Pb 0 + HO, N05 + Aq = Pb 0, NO, + Aq.
   1. How much  nitric acid,  of Sp. Gr.  1.14,  is required
 to dissolve 20 kilogrammes of oxide of lead ?
   2. How much nitric acid,  of Sp. Gr. 1.14, and  how
 much oxide of lead, must be used to make 10 kUogrammes
 of nitrate of lead ?
 e. 3 Pb + 4 (HO, N05)-\-aq = 3 (Pb 0, N05) + aq
     + N©2.
   3 Cu + 4 (HO, N05) + aq = 3 (Cu 0, N05) + aq
     + N©2.
   1. How much nitric acid, of Sp. Gr.  1.22,  is required
 to dissolve 450 grammes of lead ?   How much nitrate of
 lead would be formed ?
   2. How much nitric acid, of  Sp.  Gr. 1.362, is required
 to dissolve 450 grammes of copper ?

/.3P + 5 (HO, N05) + Aq = 3 (3H0, J> 05) + Aq
     + 5N©2.
   S + H0,N0S + Aq = HO,S03 -f- Aq + J¥©2.

               Nitric Oxide (]¥ ©).

  &FeCl+KO,N05 + 4HCI -f Aq = 3 Fe2 Cl3 -f
     KCl+aq + NOz.
     ColorlesB.        Red.
  nr©2 + ©2 = jjr©4.
   1. How much  ]V ©2  can  be  obtained by dissolving 10
 grammes of  copper in  nitric  acid? How many cubic
 centimetres ? 
ACIDS.
65
      2. How much If ©2 can be obtained from 10 grammes
    of iron by the last reaction but one ?
      3. What volume of oxygen must be mixed with one
    litre of If ©2 in order to change it into W ©t ?


                 Nitrous Oxide (IV ©).
        When healed.
163.  [NH4]0,-N05= 4H© + 2If©.

     4 Pb + 5 (HO, N05)+°Aq = 4 (Pb 0, NOs) + Aq
      -f-Jf ©.
     Na 0, S 02 + Aq + If ©2 = Na 0, S 03  + Aq +
       N ©•
      1. Ten grammes of nitrate of ammonia yield how many
    grammes and how many cubic centimetres of protoxide of
    nitrogen ?
      2. How much nitrate of ammonia must be used in order
    to make one Utre of the gas ?
      3. One litre of ]f ©2 yields how many cubic centime-
    tres of If © by the third reaction of this section ?
      4. One Utre  of  If © gives, when  decomposed,  what
    volume of nitrogen ?

                   Carbon and Oxygen.

                 Carbonic Acid (C ©2).

164.  CaO, C02 + HO, NO, -f  Aq = Ca 0, NOs -f Aq
       + C©2.
     CaO, CO2 + iT0, S03-\- Aq= CaO, S03 + Aq
       + c©2.
                 6* 
66
ACIDS.
     Ca 0, NO,  + HO, S03  -f aq =  CaO,  S 03+
       HO,N05 + aq.
       1. How much sulphuric acid and how  much nitric
    acid* must be used to drive out all the carbonic acid from
    25.462 grammes of chalk ?  How many grammes and how
    many cubic centimetres of gas would be obtained ?
       2. The specific gravity of Carrara  marble is  2.716.
    How  many cubic centimetres of carbonic acid  gas does
    one cubic centimetre of the marble contain in a condensed
    state ?
167.   1. Animals remove oxygen from the air, and return the
    whole  as carbonic  acid.   Plants  remove carbonic acid,
    and, having decomposed it, return the oxygen it contained.
    How does the volume of the  oxygen in either case com-
    pare with that of the carbonic acid ?

                   Sulphur and Oxygen.

                  Sulphuric Acid (S 03).

169. S©2 + ©t= S03.
       1.  How much anhydrous  sulphuric acid is formed by
    the oxidation of 10  grammes of sulphurous acid? and
    how much oxygen is required in the  process ?
       2. How much anhydrous sulphuric acid  is formed by
    the  oxidation of one  litre of sulphurous acid  gas ? and
    how many cubic centimetres of oxygen must be mixed
    with it in the experiment ?

    a. 2 S 03 + H © = H 0, 2 S 03 (the Nordhausen Acid).
  * When the strength of the nitric acid is not stated, monohydrated acid
(HO, N05) is always intended.
  f The two gases are mixed together and led over heated platinum sponge in
a glass tube. 
                           ACIDS.                        67

        S 03 -f- H O = H 0, S 03 (the common Acid).

 170.  Fe O, S 03. 6 H 0 is the  symbol of CrystaUized  green
          vitriol.
          When heated.
        2(FeO,  S03  .  6HO) = Fe203, S 03 -j- 12H©
          + S©2-
        By further heating.
        Fe203, S03=Fe203+ S©3.

        By conducting the anhydrous acid fumes into HO, S 03
          we get HO,2S03.
        1. How much  anhydrous acid, and how much Nord-
     hausen, can be made by the  above process from  20 kilo-
     grammes of green vitriol ?  If the Nordhausen acid has
     the specific gravity of 1.9, how many litres can be obtained
     from 20 kilogrammes of green vitriol ?

 171.   S©2 + H©,If©5 + Aq = ^0,AS'03-f^+If©4.

       1. How much  HO,  S 03  will be formed  from one
     gramme of S 02 ?  How much from one Utre ?
                                    White.
       Ba Cl-\-HO, S03-\-Aq =  BaO, S03-\-ffCl + Aq.

       BaO,NO, +  HO,  S03 + Aq = BaO, S03 +
          HCl-\-Aq.

       1. Why  must sulphuric acid or a soluble sulphate pro-
     duce a precipitate  when added, in solution, to the solution
     of any salt of baryta ?
       2. The  precipitate produced by adding  an  excess  of
     Ba CI to  a solution of  H 0,  S 03  was  collected, and
     weighed 4.567  grammes.  How  much  sulphuric  acid*
     was present in solution ?

  * When the name sulphuric acid is used, H 0, S O3 is always meant, unless
otherwise specified. 
68
ACIDS.
      3. The precipitate  produced  by adding an excess  of
    H 0, S 03 to a solution of Ba O, N 05 weighed 5.942
    grammes.  How much Ba 0, N 05 did the solution  con-
    tain?
172.  1st stage. S + ©2=S©2,and K0,N0S + 2(^T0, S03)
                = K 0, S 03. H 0, S 03 4- H ©, W ©5 •

     2d stage. S©2 4" HO, NO, = HO, S 03 + If ©4.

     3d stage. 3if ©44-zI^© = 2(II©,]V©5)4-N©8•
               f if ©24-©2 = wo4.
     4th stage.  1 2 S©24- 2 (H ©, If ©5) = 2 (HO, S 03)
               (   4-2 If ©4.
      The last  two stages are now repeated indefinitely,  so
    long as there is a supply of sulphurous acid, oxygen, and
    steam, with  the same amount of N 04.
      1. How much sulphuric acid may be made by the above
    process from  100 kUogrammes  of sulphur ?  How many
    litres of acid having the Sp. Gr. 1.842 ?   How many of
    acid of Sp.  Gr. 1.734?  How many  cubic metres of oxy-
    gen  must be used in the process ?  How many cubic
    metres of air must pass through the lead chamber, sup-
    posing all its oxygen to be removed ?

173.  a. HO, S03 4- Aq = HO, S 0 -f Aq.
      1. How much water must one kilogramme of the mono-
    hydrated acid withdraw from the air in order to reduce its
    Sp. Gr. to 1.398 ?
    /. Na 0, C02 + H 0, S 03 + Aq = Na 0,  S 03 + Aq
        4-C©2.
      1. How much  HO, S 03 is  required  to exactly neu-
    tralize 5.645 grammes of anhydrous carbonate of soda?
     How much  acid of the Sp. Gr. 1.306 ? 
ACIDS.
69
       2. How  much  NaO, C02 must  be dissolved in one
    Utre of water so as to make a solution such that one cubic
    centimetre  wiU exactly  neutralize 0.01  of a gramme of
    H 0, S 03 ?
       Yellow.                         White.
    g. Pb 0 4- HO, S 03 4- Aq = PbO, S 03 -f Aq.
       Black.                         Blue Solution.
    h. Cu 0 4- HO, S 03 4- Aq = Cu 0,S03 + Aq,

    which, when evaporated, gives  crystals of the composition
         Blue Vitriol.
    CuO, S03.5HO.
       1. How much sulphate of lead can be made from twenty
    grammes of litharge ?   How much sulphuric acid must be
    used in the process ?
       2. How  much  crystallized Blue Vitriol can  be  made
    from one  kUogramme of oxide of copper ?  How  much
    sulphuric acid of Sp. Gr. 1.615 must be measured out for
    the process ?
174. Cu 4- 2 (HO, S03) = CuO, S03 4- 2 H© -f S©2.

175. C4-2(#0,  £03) = 2H©4-2S©24-C©2.

     NaO, C02A-S02 + Aq = Na0,S03 + Aq + CO2.

       1. How much sulphurous acid can be made from 4.562
    grammes of sulphuric acid  by means  of copper ?   How
    much by means of charcoal ?   How much anhydrous sul-
    phate of copper would be formed in the first case ?   How
    much carbonic acid in the second ?  How much carbonate
    of soda will the sulphurous  acid in the two examples neu-
    tralize ?  What is the  volume of S 02, and what the
    volume of C 02 evolved in the second case ?
       2. How  much copper  and how much sulphuric acid
    must be  used to make one Utre of sulphurous acid gas ? 
70
ACIDS.
    How much to make 500 grammes of anhydrous sulphite
    of soda?
      3. By burning sulphur in one litre of oxygen, how much
    S 02 gas is obtained ?  What is the Sp. Gr. of S 02 ?

                Phosphorus and Oxygen.

                Phosphoric Acid (P05).

176. P 4- 5 © = P 05 (white powder).
                                      Colorless Fluid.
    3P 4_ 5 (HO, NO,)  4-  Aq = 3 HO, cPO,+ Aq
      4- 5 rf ©2.
    For preparation of phosphoric acid from bones, see § 144.
          At the ordinary temperature.
    3CaO,cP05 4-  2(HO,S03)-\-aq= 2(CaO, S 03)
      4- [2 H 0, Ca 0] CP 0, 4- aq.
                 Intensely heated.
     2(CaO, S03)+ [2HO,  CaO]cP05 =  3CaOcP05
       4-2(H©,S©3).
      Before ignition.
     3HO,P 0, + 3 (INH,-] 0,HO) 4- 3 (Ag 0,NO,)+Aq
             Yellow.
       = 3 AgO, cP054-3 ([NHt-] 0, NO,) + Aq.
     After ijnition.
     H0, PO,A- \.NH,-\  0, HO -\-AgO,NO,-\-Aq =
           White.
        Ag 0, ttP 05 4- [_NHA-\ 0,NO,-\- Aq.
     3HO, cPO, 4- 2(MgO, S 03) + 8 [JVH]  O, HO
        4- Aq =  [N H4]0, 2 Mg O, CP 05 . 12 H O 4-
        2([NHi']0, S03)+Aq. 
ACIDS.
71
    [N H4] 0, 2 Mg 0, CP 05. 12 H 0 when ignited resolves
      into 2 Mg O, bP 05 4- If H3 4- 13 H ©.
      1. How much P 05 and how much 3 H O, CP Os can be
    obtained from  16 grammes of phosphorus ?
      2. By boiling one gramme of phosphorus in nitric acid
    untU it dissolves, diluting, neutralizing with aqua ammonia,
    and precipitating with a solution of sulphate of magnesia,
    collecting and  igniting the precipitate, how much wiU it be
    found to weigh ?
      3. How much nitrate of silver is required to precipitate
    the phosphoric  acid  made  from  one gramme of  phos-
    phorus before  ignition ?   How much after ignition ?


                  Cyanogen and Oxygen.

179. Cy O = Cyanic Acid, which is monobasic.
     Cy2 02 = Fulminic Acid, which is bibasic.
     Cy3 03 = Cyanuric Acid, which is tribasic.


                    Boron and Oxygen.

                   Boracic Acid (B 03).

180. NaO, 2B03  +  H CI +  Aq  =  2(H0, B 03)
       -\-NaCl-\- Aq.
     The crystaUized boracic acid is H O, B 03. 2 H O.
     Dried at 100° it becomes H O, 2 B 03. 2 H O.
     At a red heat it loses its water and melts, and on cooling
   it hardens to a vitreous mass. 
72
ACIDS.
                SiUcon and Oxygen.

                Silicic Acid (Si 03).

  Na 0, Si 03 +HCI +Aq = HO, Si 03 -f Na CI
    + Aq.
  If the quantity of water is large,  the hydrated silicic
acid remains in solution.  If the amount of water is small,
it separates as gelatinous precipitate.
  SECOND  GROUP : HYDROGEN ACIDS, OR COMPOUNDS  OF
             THE HALOGENS  WITH HYDROGEN.

                  Chlorine and Hydrogen.

                 Chlorohydric Acid (H CI).

185. Na CI 4- HO, S 03 = Na 0, S 03 -\- IS CI; or
     NaCl  4- 2(H0,  S03) =  NaO,  S 03  . HO, S 03
       4-HC1.
       Only one equivalent of HO, S 03 is necessary to de-
    compose one equivalent of salt; but then  the last half of
    the H. CI can be driven off only at a temperature suffi-
    ciently high to melt glass, so that with these proportions
    the process cannot be conducted in glass vessels.   If  two
    equivalents of HO, S03 are used, the whole of the HC1
    is expelled at a moderate  temperature, and the process can
    then be conducted to  its end  in a glass  flask or retort.
    Hence, in the  manufactories, where the  acid is generally
    generated in iron retorts,  only one equivalent of HO, S03
    is used, while  in the laboratory,  where  glass vessels are
    employed in the  process, two equivalents  are  taken to 
ACIDS.
73
    each equivalent of salt.  The last we  wiU assume to be
    the case in the following problems.
      1. How much chlorohydric acid  gas  can be made from
    4.562 grammes;  from  25 kUogrammes; from  34.567
    grammes of common salt ?
      2. How much  sulphuric acid is  required to decompose
    the above amounts of common salt ? and how much bisul-
    phate of soda is in each case formed ?
      3. How many cubic centimetres  of H CI can be made
    from 1 gramme ;  from 5.643 grammes of Na CI ?
      4. How much  salt and  how much  sulphuric acid are
    required  in order to make  one  kilogramme,  to make
    5.463 grammes, and  to make one litre, of HO?
      5. How much H CI  is contained in one litre of the
    liquid acid  of Sp. Gr. 1.16,  of Sp. Gr. 1.17, of Sp. Gr.
    1.14 ?
      6. How many  cubic centimetres  of gas are dissolved in
    one litre of the Uquid acid of the above strengths ?
      7. How much Na CI, and how much HO, S 03, and how
    much water in the receiver, are required to make, —
      a. 20 kilogrammes of Uquid acid of Sp.  Gr. 1.13 ?
      b. 560.4 grammes of liquid acid  of Sp. Gr. 1.18 ?
      c. 4 Utres of Uquid acid of Sp. Gr. 1.16  ?

    H 4- CI = H CI.
      1. One litre of hydrogen gas  combines  with  what vol-
    ume of chlorine gas ?  and  what is the volume  of hydro-
    chloric acid gas formed ?

186. a. Fe 4- H CI + Aq = Fe CI + Aq + H.
      1. How much  liquid acid, by weight and by measure,
    of Sp. Gr. 1.16 is required to dissolve 250 grammes of
                  7 
74                       ACIDS.
iron ?  How much chloride of iron would be obtained,
and how much hydrogen gas, by measure, evolved ?

Sn 4- HCl + Aq = SnCl-{-Aq-\- H.

  1. Solve the last problem, substituting tin for iron.

b. ¥%03, 3KO+ 3HCI-\-Aq = Fe2 Cl3 + Aq.

c. 2 Fe CI -f CI 4- Aq = Fe2 Cl3 4- Aq.

d.NaO,  C 02 + H CI 4- Aq = Na CI 4- Aq + C©3.

e. Ag 0, NO, + H CI 4- Aq = Ag CI 4- HO, NO, + Aq.

  1. How  much liquid acid of  Sp. Gr.  1.16 is required
to dissolve 4 grammes of iron-rust ?
  2. How many cubic centimetres of chlorine are required
to convert one gramme of protochloride of iron into ses-
quichloride ?
  3. How much Uquid acid of Sp. Gr. 1.13 is required
to neutralize one gramme  of carbonate of soda ?
  4. How  much H  CI  do 50  c.  c.  of a liquid  contain
which is exactly neutraUzed by one  gramme of anhydrous
carbonate of soda ?
  5. How  much HCl  do 50  c.  c.  of a. Uquid contain
which gives, with an  excess of nitrate of silver, a precipi-
tate weighing 5.643 grammes ?


                  Aqua Regia.

HO, NO, -\-3HCl-\-aq = N02 Cl2 -f CI4~ aq.

Au 4- (N02 Cl2 4-  C7 4- aq) = Au Ck -f aq 4- N02. 
ACIDS.
75
     Bromohydric Acid (HBr).   Iodohydric Acid (HI).

     PBr34-3Jff-(9 = P034-3H Br.
     PI34-3iT0 = PO34-3HI.


                 Hydrofluoric Acid (HF1).

     CslT1 + H0, SO-\-aq = Qa,0,&03-\-HFl-\-aq.
     Si 03 4- 3 H Fl = 3 H O -f Si FI3.


              Tartaric Acid (2 H 0, C8 H4 O10).

194. 2 (IN H^  0,  HO)  + 2 H 0, C8 H, Ow  + Aq  =
        2[NIQ 0, C3H,0,n^Aq.
     2  (K 0,  C  02)  -f 2  H 0,   Cs H,  O10  -f  Aq  =
        2K0, C3H, Ow 4- Aq + 2 C ©2.
     2K0,  C3Hi Ol0 4- HCl-^Aq = HO, KO, C8H4O10*
        -\-KCl-\-Aq.
     2 (CaO,  C02) 4- 2 (HO, KO, Cs H,  O10) + Aq  =
        2 CaO, C4H8O10 + 2KO,C8Hi O10-f ^4- 2 C©2,
     2X0, <78#4 O104- 2 (7a C?4-J^= 2 Ca 0, C8H4O10
        4-2X07 4-^.
     2 CaO, C4 H8O10 4- 2 (#<9, £03) 4- Aq = 2(CaO, S03)
        4-2^0, 0,^0,0 4-^.

  * This salt is not absolutely insoluble, but only difficultly soluble in water,
and hence is not completely deposited in this reaction. 
76
ACIDS.
      1. How much KO, C02 is required to  exactly neu-
    tralize 5.462 grammes of tartaric acid ?
      2. How much H CI is required to convert 4.678 grammes
    of 2 K 0, C8 H4 0IO into H O, K O, C8 H4 O10 ?
      3. From ten kilogrammes of cream of tartar how much
    tartaric acid can be made ?

                Oxalic Acid (HO, C203).

      The above is  the symbol of the acid dried  at 100°.
    When crystallized,  it contains two  more equivalents of
    water, and corresponds to the formula H O, C2 03 . 2 H 0.

196.  H 0, C2 03 . 2 HO 4- x (HO,  S 03) = x (H0,S03)
        4-3#<9 4-C©24-C©.

197. b. KO, C02 4- HO, C2 03 + Aq = KO, C2 03 4- Aq
        4-C©2.
       KO, C2 03-\-HO, C2 03-\-Aq = KO,2C203 + Aq.

     d.  Ca 0,  S03 4-  HO, C2 03 -\-  Aq = Ca O,  C203
       + H0tS0a + Aq.

     CaO,S03 -f [NH^  0,  C2 03  + Aq = CaO,  C2 03
       4- (NH,) 0,S03 + Aq.

       1.  How much C ©2 and how much C © wiU be ob-
     tained by decomposing five grammes of crystallized oxalic
     acid by sulphuric acid ?  How many cubic centimetres of
     each gas ?
       2.  How much crystallized oxalic  acid must be used to
     yield one litre of C ©2 and one litre of C © ?
       3.  How much crystallized oxalic acid will exactly neu-
     tralize 1.456 grammes of carbonate of potassa? 
ACIDS.
77
              Acetic Acid (HO, [<74 iT3] 03).

198.  PbO 4-  HO, [CtH3-] 03 + Aq= Pb 0,  [C4 J73] 03
       -{-Aq.
     CrystalUzed acetate of lead = Pb 0, [C4 H3]  03. 3 H 0.
    Pb 0,  (C4H3) 03 4- HO, S03  4- Aq =  PbO, S03
       + HO,iCi H3-]03-\-Aq.
7* 
LIGHT  METALS.
             FIRST GROUP : ALKALI METALS.

                     Potassium (K).

202. KO, C02-\-HO,[CiH3'] 03+Aq=KO,[CiH3] 03
       4- Aq 4- C ©2.

     K0,C02+TI0,S03 + Aq=K0,S03ArAq+CO2.
       1. How  much  HO, S03 must be dissolved in water
    in order to make a litre of test acid such that one cubic
    centimetre  wiU  exactly neutralize  one decigramme of
    KO, C02?
       2. How much crystaUized oxaUc acid  must be dissolved
    in water in order to make a litre of test acid such that
    one hundred cubic centimetres wUl exactly neutraUze 6.92
    grammes of K 0,  C 02 ?

203. K 0,  C 02  -f  Ca 0, H 0  + Aq =  Ca O,  C Oa
       + K 0, HO  4- Aq.
         Melted together.
204. d. Si03 -f KO, HO = KO, Si 03 + HO.
                                               Light Blue.
     e. CuO, S03 4-  KO,  HO +  Aq = Cu O,  H 0
       + K0,S03 + Aq. 
LIGHT METALS.
79
        Intensely heated*
205. KO, C02 -j- 2 C = K 4- 3 C ©.
206. KO, C02 + 2(HO, S03) + aq = K0, S03.HO,S03
       + aq + C02.

207. KO, C02  4- HO, NO,  -+-  aq = KO,  NO, 4~ aq
       4-c©2.
       When heated.
     a. KO, N05 = KO, N03 4- ©2.

     b. KO, N05 4- 3 C 4- S = K S 4- 3 C ©2 4- N.
       1. How many cubic centimetres of mixed gases  are
     formed  by the burning of one kUogramme of gunpowder,
     when measured at the standard temperature and pressure ?
     Assuming that the temperature at the time is 1000°, what
     would be the  volume of the gases the moment after the
     explosion.
       2. Assuming that gunpowder occupies the  same bulk as
     an equal weight of water, into how many times its own
     volume does it expand on  burning ?   Calculate both for
     0° and for 1000°.
       3.  Assuming that the temperature, the moment after
     explosion, is 1000°, what would be the pressure on the in-
     terior surface of a bomb of 20 centimetres internal diame-
     ter when exploded filled with gunpowder ?
208. a. KO, C105  = KC14-6©.
       1. How much does the oxygen contained in chlorate of
     potassa expand when the salt  is decomposed ?  Sp.  Gr.
     of chlorate of potassa is 2 nearly.
       2. How much mechanical force would be required to
     reduce  oxygen gas to the same degree of condensation in
     which it exists in the salt ?
     c.3(KO,C105)4-4(J70,£<93) = 2(KO,S03.HO,S03)
       4-KO, C1074-2C1©4. 
80
LIGHT METALS.
    /. KO, C105 4- GHCl+Aq = KCl-\-Aq 4- 2 CI.

      6 (KO, HO) 4- 6 CI 4- Aq =  KO, C105 4- 5 KCI
         + Aq.
210. KI 4- Mn02 4-  2 (HO, S 03)  + aq = K 0, S 03
       -\-MnO,S03-\-aq-{-\.

211. H 0, K O, C8 H4 O10     = Cream of Tartar.
     Na O, K O, C8 H4 O10     = Rochelle Salts.
     [N H4] 0, K O, C8 H4 O10 = Ammoniated Tartar.
     Fe2 03, K 0, C8 H4 O10   = Tartarized Iron
     Sb2 03, K O, C8 H4 OI0   = Tartar Emetic.
     B 03 K 0, C8 H4 O10     = Soluble Cream of Tartar.

213. 4(KO,C02)-fl6S = KO,S034-3KS54-4C©2.

     3 (KO, C 02) 4-8 S = KO, S202 4-2 K S3 4-3 C ©2.
      The  first or the  last of  these  reactions  takes  place,
    according to the proportion of  sulphur and the degree
    of temperature to which the mixture  is exposed.  If  the
    sulphur is in excess,  and the temperature of the  mass
    raised to a red  heat, pentasulphuret of potassium and sul-
    phate of potassa are formed.  If, however, the sulphur is
    present in smaUer quantity, and the temperature of  the
    mass  is not raised above  its melting  point,  a mixture of
    tersulphide of potassium and hyposulphite of potassa  re-
    sults.   At the higher temperature, hyposulphite of potassa,
    which is always first formed,  splits up into sulphate of
    potassa and pentasulphide of potassium.  Thus,
             4 (KO, S2 02) = 3 KO, S03 -f KS5.
      Either of these  sulphides is  decomposed by dilute acids,
    forming sulphide of hydrogen, and setting free  as  many
    equivalents of sulphur, less  one,  as are contained in  the
    compound. 
LIGHT  METALS.
81
      KSn + HO,  S03 -{-Aq  =  sn_, 4- KO, S034-


      KO, S 03 4- 4 C = K S 4- 4C©•


                      Sodium (Na).

215.              Problems on Common Salt.
      1. One gramme of salt contains how much chlorine and
    how much sodium ?
      2. One  cubic  centimetre  of common salt, Sp. Gr.  =
    2.13, contains how many cubic centimetres of sodium, and
    how many of chlorine gas ?

218. Na CI 4- HO, S 03 = Na 0, S03 -f H CI.
219. NaO, S034-4C = NaS4-4C©.

220. Na S -f CaO, C 02 = Ca S  4- NaO, C02.

     3 (NaO, S03) 4-13 C -f- 4 (CaO, C02) = 3 NaO, C02
       4- 3 Ca S, Ca 0 4- 14 C ©•
      1. How much carbonate of soda can be made from 500
    kilogrammes of common salt ?  How much sulphuric acid ?
    How much charcoal and how  much chalk are required in
    the process, according to the theory ?
      2. If in a chemical process  carbonate of potassa or car-
    bonate of soda may be used indifferently, which would be
    employed most profitably if the price were the same ?
      3. What relation  ought the price of CrystalUzed  car-
    bonate of soda to bear to that of the dry salt, if the intrin-
    sic  value is alone considered ?

221. Na 0, C02 A- Ca  °> H °  +  Aa =  Ca °'  C °*
       A-Na 0,H02A~Aq. 
82                   LIGHT METALS.
222. Na 0, C 02 4- 2 C = Na -f 3 C ©.
223. 2 (Na 0, C 02) + 3 HO, P0, + Aq =
       HO, 2NaO,PO,-\-Aq + 2C ©2.
       The symbol  of crystallized  phosphate  of  soda =
         HO, 2NaO, P05.24HO.
          When heated.
     HO, 2NaO, P05.24HO=2NaO, POs4-25H©.
     3(Ag 0, NO,)  4-  HO, 2 Na 0, P 0, -\- Aq =
           Yellow.
       3 AgO, P0S 4- 2 (Na 0, N0,) + HO, NO, -{- Aq.
                                                White.
     2 (Ag 0, NO,) -f 2Na 0, P 0, + Aq = 2 AgO, P 05
       + 2 (Na 0, N 0,)  + Aq.
       1. How much sodium can be made from one kilogramme
    of anhydrous carbonate of soda ?
       2. How much  AgO, N05 is  required to precipitate
    completely the phosphoric  acid from 1.345  grammes  of
    crystallized  phosphate of  soda?   How  much, from the
    same amount of pyrophosphate of soda ?
224. NaO,  N05 4- 2(HO,S03) = Na 0, S03, H0,S03
       4-H©,]\©5.
    Na 0, N 0,4- K CI + aq = Na Cl 4- K 0, N 0, -f- aq.
225. Na O, 2 B 03. 10 H 0 =  Symbol of common borax.
    NaO, 2B03. 5HO =   "    "  octohedral borax.
       1. When soda nitre and potash nitre bear both the same
    price, from which can  nitric acid be most profitably ex-
    tracted ?
       2. When potash nitre is worth 25 cents the kUogramme,
    and chloride of potassium 8 cents the kilogramme, at what 
LIGHT  METALS.
83
     price of soda nitre will it  be profitable to convert it into
     potash nitre, assuming that the cost of the process amounts
     to two  cents on each kilogramme of potash  nitre manu-
     factured ?
       3. What  is  the percentage composition  of common
     borax ?  What is that of octohedral borax ?
       4. How much  borax  glass  can  be made  from  100
     grammes  of common borax ?


                  Ammonia ([N H4] 0).

       The  above is the symbol of the base of the ammonia
     salts, and corresponds  to K O  and Na O.  The  student
     must be careful not to confound it  with ammonia  gas,
     which has the symbol N H3.

227. KO, H0 4-Fe = K0 4-Fe0 4-H.
     K O, N 05 4- 5 Fe = K O -f 5 Fe O 4- J\.

     3 (K 0, H 0) 4-K 0, N 05 4-8 Fe = 4 K O 4-8 Fe 0
       4-l*H3.
229. JV H3 4- H CI = [N H4] CI.

     [N H4] CI -f CaO, HO = Ca CI 4- 2 H 0 + IV H3.

230. i? +  ^H3= [NHt] 0,HO-{- Aq.

231. [NH,] 0,HO-\-Aq + 2n&=iNHi-]S,HS-\-Aq.

      The  compound  which remains in  solution after pass-
    ing H S gas through aqua ammonia until saturation, is
    [N H4]  S, H S, the sulphohydrate of sulphide  of ammo-
    nium, which is the reagent so much used in the laboratory,
    and  incorrectly called  sulphide of ammonium.  On  ex-
    posure to the air,  the solution becomes yellow, owing to 
LIGHT METALS.
 the formation of bisulphide of ammonium, as is  shown in
 the reaction below.

 [NHQ S, HS 4-  [NHJ 0, HO-At M =
   2[NH4-]S+Aq.
     Colorless Solution.
 2([NH4-] S,  HS) +  Af  +  BO -
    Yellow Solution.
   \_NH,-\ S2 4- [NHA 0, S2 02 4- Aq.

. 2 [N H4] CI 4-  3  (Ca 0,  C 02) = 3 Ca CI 4~ J¥ H3
   4- 2 [N H4] O, 3 C 02 4- H ©.
   1. How many cubic centimetres do 17 grammes of
 ammonia  gas occupy ?  How many do 36.5  grammes of
 chlorohydric acid gas occupy ?  When the two gases com-
 bine, in what proportions, by volume, do they unite ?  How
 great  is  the condensation which results?   Sp.  Gr. of
 [N H4] CI = 1.5.
   2. How much ammonia gas can be obtained from 5
 grammes  of chloride of ammonium.  How much chloride
 of ammonium and how  much Ca O must be used in  order
 to prepare one  litre of ammonia gas ?
   3. How much chloride of ammonium and how  much
 lime must be used  in  order to prepare one litre of aqua
 ammonia of Sp. Gr. = 0.9 ?  How much water must be
 placed in the receiver ?
   4. How much ammonia gas is held in solution by one
 Utre of aqua ammonia of Sp. Gr.  0.9 ?  To how  much
 [N H4] 0, H O does this amount of gas correspond ?
   5. How much IIO, S 03 is required to neutralize four
 cubic centimetres of aqua ammonia of Sp. Gr. 0.9 ?
   6. How much H S  gas  will be  absorbed  by one litre
 of aqua  ammonia  of Sp. Gr. 0.9, assuming  that only so
 much is  taken up  as is necessary to form the compound 
LIGHT METALS.
85
[N H4] S, H S ?  How much H O, S 03 and Fe S will be
required  to  produce this  amount of gas ?  How much
additional aqua  ammonia  must be  added to  the  above
solution in  order to  change it to  a  solution of  proto-
sulphide of ammonium ?
         SECOND GROUP'. THE ALKALINE EARTHS.

                     Calcium (Ca).

241. Ca 0,S03 + Ba CI A~ M= Ba °> s °3+ Ca 01A~A9-

     CaO,  S 03 A-  H °,  C*  °s  + M =  Ca °» C* °3
       A-h o, s o3 A-Aq-
242. HO, 2Na 0,P0, + 2 CaCl A~ Aq~ HO,2Ca0, P05
       4- 2 Na CI 4-  Aq.
244. 2 (CaO, HO) 4-2 CI = Ca O, CI O 4-Ca CI.
246. CaO, C02A-HCIA-Aq=  Ca CIA-  AcL + C02.
       1. How much CaO  can  be ohtained  from  100 kilo-
     grammes  of carbonate of lime ?  How much Ca 0, H 0
     can be obtained from the same amount ?
       2. How much carbonate of lime must be burnt in order
     to yield 140 kilogrammes of quicklime?   How much to
     yield 185  kilogrammes of  Ca O, H O ?  How many cubic
     metres of C 02 would be set free during the process ?
       3.  How many cubic metres of C 02 can be absorbed by
     a quantity of milk  of lime  containing 5  kilogrammes of
     CaO?
       4. What is the percentage composition of unburnt and
     of burnt gypsum ?
               8 
86
LIGHT METALS.
      5. What is the percentage composition of phosphate of
    lime (3 CaO, P05)?
      6. How much chlorine and how much lime are re-
    quired to make 100 kUogrammes of chloride of lime,
    assuming that its composition is expressed by the symbol
    Ca 0, CI 0 -|- Ca CI ?   How much Mn 02 and how much
    muriatic  acid of Sp. Gr. 1.15 will yield the requisite
    amount of chlorine ?
      7. To how much Ca 0, how much Ca CI, and how much
    Ca O, S 03 do 2.5 grammes of carbonate of Ume correspond?

            Barium and  Strontium (Ba and Sr).

248. BaO, S 03 4- 4 C = Ba S 4- 4 C ©.
     Ba S 4- H CI 4- Aq = Ba CI -\- Aq -f H S.
     Cu O 4- Ba S4- Aq = Cu S 4- Ba 0, HO -f Aq.
     Ba CI + NaO, S03 + Aq =  BaO, S 03 4- Na CI
       -[-Aq.

     Sr O, S 03 4- 4 C = Sr S 4- 4 C ©•
     Sr s 4- HO, NO, -f Aq = Sr 0, N 0, A-  M + H S.
     Sr 0, NO, 4-  Na  0, S 03 + Aq  =  Sr O,  S 03
       Na 0,N0,-\-Aq.
       1.  What is the percentage composition of sulphate of
    baryta ?
       2.  How much chloride of barium can  be  made from 5
    kilogrammes  of sulphate of baryta ?  How  much nitrate
    of baryta, and  how much Ba 0, can be made from the
    same amoun^of sulphate of  baryta ?
       3.  To how much  sulphuric acid do 4.567  grammes of
    sulphate of baryta correspond ?   To how much chloride
     of barium do they correspond ? 
LIGHT  METALS.
87
       4. How much nitrate of strontia can be made from one
    kilogramme of sulphate of strontia ?  How much carbon
    and how  much  nitric acid of Sp. Gr. 1.4 is required in
    the process ?

                    Magnesium (Mg).

249.   The symbol of serpentine mineral is

               9 [Mg, Fe] O, 4 Si 03, 6 H 0.

       The symbol [Mg, Fe] O indicates  that a portion of the
    magnesium  in  the  base is replaced by oxide of iron, and
    the whole stands for  but  one equivalent  of base.   We
    frequently write, instead of the portion enclosed in brack-
    ets,  the general symbol R,  when  the symbol of  the
    mixed base becomes R O, and that of serpentine mineral
    9 R O, 4 Si 03.  It must be noticed, that Mg and Fe when
    enclosed in brackets, with a comma between, as above, no
    longer stand for an  equivalent  of each metal.  On  the
    other hand, the two together make but one equivalent of
    metal,  represented  by R in  the other  mode of writing.
    Moreover, nothing is intended to be indicated in regard to
    the relative proportions of the two metals mixed together
    to form one equivalent, as they vary in  different  speci-
    mens of the same mineral.  This method of writing the
    symbols of  compounds containing isomorphous constitu-
    ents will be  constantly used hereafter.
    9 [Mg, Fe] O, 4Si03. 6H0 4- 9 (H 0, S 03)-\-Aq =
       4 Si 03  4- 9 [Mg, Fe-] 0, S 03 -{-.Aq.
       Since sulphate of magnesia and sulphate of protoxide
    of iron have the same  crystalline form, we shaU obtain, on
    evaporating  the solution, crystals containing both salts.
    We can prevent the  sulphate of protoxide of  iron from 
88
LIGHT METALS.
     crystallizing, by converting it into sulphate of sesquioxide
     of iron by means of nitric acid.   Thus,

     6 (Fe 0, S03) + 3 (HO, S 03)  + HO, NO, + Aq =
      3 (Fe2 03, 3 S03) + Aq + N02.

     Talc        = 6MgO, 5 Si03,  2 H O.

     Meerschaum = Mg 0, Si 03, HO.

     Hornblende  = 4 [Mg, Ca, Fe] 0, 3 Si 03.

     Augite      = 3 [Ca, Fe, Mg] O, 2 Si 03.

250. 5(MgO,S03)+o(KO, C02) +Aq = 3(MgO, C02.aq)
       4- MgO, KO + MgO, 2 C02 + 5(KO,S03)+Aq.

      The relative proportions of carbonate of magnesia and
    of hydrate of magnesia vary with the  temperature and
    other  circumstances attending the precipitation.

251. MgO,co2 + HCi + Aq = MgCi + Aq + co2.

     2 (Mg 0, S03)-\-HO,2 Na 0,P0,-\- [NH^ 0,H0
      + Aq= [NH4] 0, 2MgO, P05 4-  2 (Na 0, S03)
       + Aq.
          When heated.
     (NH4)0,2MgO,P05 = 2MgO,P654-NH34-H©.

      1. What is the percentage composition of talc ?  What
     that of hornblende and augite, assuming that the  whole of
     the base in either case is Mg 0 ?
      2. From a solution of sulphate of magnesia the whole
     of the magnesia was  precipitated  by phosphate  of soda
     and ammonia.   This  precipitate,  after ignition, was found
     to weigh 2.456 grammes.  How much sulphate of mag-
     nesia was contained in the solution ? 
LIGHT METALS.
89
                    Aluminum (Al).

258. Al2 03, Si 03. 2 H O -f- 3 (HO, S 03) -f aq = Si 03
       4-AlaOg, 3S034-ay.
    AU 03, 3 S 03 4- 3 (Na 0, C 02) + Aq = A\03, 3 HO
       4- 3 (Na 0, S 03) 4- Aq 4- C ©2.
    A1203, 3 HO 4- A'O, HO-\- Aq = KO, Al2 03 A~ Aq.
    KO,S03. Al2  03, 3 S 03 4- 3 (Na 0, C 02) -\-Aq =
       A1203, 3H0 4- 3(i^aO>S03)4-A"0, >Sf034-^
       + C©2.
     Al2 03, 3 5 03  4- 3 (P6 0,  [<74 H3-\ 03) A~  Aq =
       3(PbO, S03) 4- Al2 03, 3 [04 H3] 03 4- Aq.


             Symbols of Isomorphous Alums.

                Potassa, Alumina, Alum.
            K 0, S 03. Al2 03, 3  S 03. 24 H 0.
                  Soda, Alumina, Alum.
           NaO, S03. A1203, 3 S03. 24 HO.
                Ammonia, Alumina, Alum.
          [N H4] 0,  S03. Al2 03, 3 S 03 . 24HO.
                 Potassa, Chrome, Alum.
            KO, S 03. Cr2 03, 3 S 03. 24 HO.
                  Soda, Chrome, Alum.
           NaO, S 03. Cr203, 3 S03 . 24HO,
               Ammonia, Chrome, Alum.
         [N H4] 0, S 03. Cr2 03, 3 S 03. 24 H 0.
                 8* 
90
LIGHT METALS.
                Potassa, Iron, Alum.
         K 0, S 03. Fe2 03, 3 S 03. 24 HO.
                 Soda, Iron, Alum.
        NaO, S03.Fe203, 3S03.24HO.
               Ammonia, Iron, Alum.
      [NH4] 0, S 03. Fe203, 3 S 03 .  24HO.

  Symbols of the most important  SiUcates of Alumina.
 2 Al2 03, Si 03,                  Staurotide.
 3 Ala 03, 2 Si 03,                 Andalusite.
 3A1203,2Si03,                 Kyanite.
 3 Al2 03, 2 Si [0, Fl]3,            Topaz.
 K 0, Si 03. Ala 03, 3 Si 03,       Common Felspar.
 Na O, Si 03. Alg 03, 3 Si 03,      Albite.
 [Ca, Na] 0, Si 03. Al2 03, Si 03,   Labradorite.
 K 0, Si 03. 4 (Alj 03, Si 03),      Common Mica.
 3 R O * Si 03. R2 03,t Si 03,     Garnet.
    1. What  is the percentage  composition of staurotide ?
 What is that of kyanite ?
    2. An analysis of one of the  above siUcates would give
 the following percentage composition.
             Silica,                64.76
             Potassa,              16.87
             Alumina,             18.37
                                 100.00
    What is the symbol of  the mineral ?
* R 0 = Fe 0, Mn 0, Mg 0, or Ca 0.
t R2 O3 = Ak 03, Fe2, 63, or Cr2 O3. 
LIGHT METALS.
91
  Solution. — This problem is evidently the reverse of deducing the per-
centage composition from the symbol; but it does not admit, like that, of
a definite solution, for while there is but one percentage composition
corresponding to a given symbol, there  may be  an infinite number of
symbols  corresponding to a given percentage composition.  This can
easily be made  clear by  an  example.  The commonly received symbol
of alcohol is [C4 H5] 0, H 0  = Ci H6 0*.  The percentage composition
is easily ascertained.  Thus,
                      C<  He   02
                      24 + 6 + 16 = 46.
         46 : 24 = 100 :  x = 52.18 per cent of carbon.
         46 :  6 = 100 :  x = 13.04 per cent of hydrogen.
         46 : 16 = 100 :  x = 34.78 per cent of oxygen.
                Percent.
    Carbon,     52.18 = C2 = 12  or  C4 = 24   or Cs = 36
    Hydrogen,   13.04 = H3 =   3  "  H6 =  6   "  H9 =  9
    Oxygen,     34.78 = O  =_8   "  O2 = 16   "  O3 = 24
               100.00        23            46            69
 This  percentage composition evidently corresponds not only to C4 Hs O2,
 but also to C2H3 O, to CsHs O3, and to  any other symbol which is a
 multiple of the first;  for, taking the per cent of carbon as an example,
 we have
         100 : 52.18 = 23 : 12 = 46 : 24 = 69 : 36 = 92 : 48, &c.
 If, then, we  had given the  percentage composition of alcohol, it would
 be impossible to determine, without other data, whether the symbol was
 C2 H3 O, or some multiple of it.  If, however, we had also given that the
 sum of the equivalents of the elements  of alcohol  equalled 46, then we
 could easily reverse the above process.  Thus,
   100 :  52.18 = 46 :  x = 24, the sum of the  equivalents of carbon.
   100 :  13.04 = 46 :  x =  6,    "      "      "     "    hydrogen.
   100 :  34.78 = 46 :  x = 16,    "      "      "      "   oxygen.

 And         — = 4, number of equivalents of carbon.
              6    '
             — = 6,    "    "     "     "   hydrogen.

             — = -2,    "    "     "     "   oxygen.
              8
   Assuming, however, that we had no means  of ascertaining the sum of
 the equivalents in alcohol, then, although we  could not definitely fix its
 symbol, yet  nevertheless we could easily find which of all the possible
 symbols expressed its composition in the simplest terms; in other words, 
92
LIGHT METALS.
 with the fewest number of whole equivalents.  For this purpose, assume
 for a moment that the sum of the equivalents is equal to 100, then

                  the sum of the equivalents of carbon.
                    "      "      "      "   hydrogen.
                    "      "      "      "   oxygen.

                  8.697, number of equivalents of carbon.

                  13.04,     "     "     "     "  hydrogen.

                  4.348,     "     "     "     " oxygen.

   These are the number of equivalents of each element on the supposition
 that the sum of the equivalents in alcohol is equal to 100.   Any other
 possible number of equivalents must be either a multiple or  a submul-
 tiple  of these, and we can easily find the fewest number of whole
 equivalents possible,  by seeking for  the three smallest whole numbers
 which stand to each other in the relation of      8.697 :13.08 : 4.348,
 which will be found to be                        2:3:   1.
 Hence the simplest possible symbol  is C2 H3 O, but, from anything we are
 assumed to know, the symbol may  be any multiple of this; and for con-
 siderations which cannot be discussed in this connection, chemists usually
 assign to alcohol the symbol C4 Hg O2, which is double the above.
   The symbol thus obtained expresses  merely the relative number of
 equivalents of each element present in the compound,  and gives no in-
 formation in regard to the grouping of the elements.  Such symbols are
 called empirical symbols, to distinguish them from the rational symbols,
 which indicate the manner in  which the elements are supposed to be
 arranged.  The rational symbol of  alcohol is [C4 Hs] O, H O. This in-
 dicates  not only that  alcohol consists of four equivalents of carbon, six
 of hydrogen, and two of oxygen, but also that it is the hydrated oxide
 of a compound radical called ethyle.  It must be carefully noticed, how-
 ever, that the empirical symbols fully represent all our positive  knowl-
 edge.    They  alone are not liable to be changed.  The  grouping of
 elements in a compound is a matter of theory, and the rational symbols
 are liable to constant changes, as the opinions of chemists on this sub-
ject vary.
  From the example just  discussed we can easily deduce the following
 rule for finding the empirical symbol of a compound from its percentage
 composition.   Divide the per cent of each element entering into the com-
pound by its chemical equivalent, and find the simplest series of whole num-
 bers to tohich these results correspond.  To apply this rule to the problem
 under consideration.
52.18
13.04
34.78

52.18
  6
13.04
  1
34.78
  8 
LIGHT METALS.
93
         64.76
         ■77-jr- = 1.43, number of equivalents of silica.
         16.87
         ■^- = 0.3575,  «    «    »    "   potassa.
         18 37
         -f-= 0.3575,  "    "    "    "   alumina.
          01.4
               143 : 0.3575 : 0.3575 = 4:1:1.
  Empirical symbol, AI2 O3, K 0, 4  Si O3.
  Rational symbol, K 0,  Si O3, Al2 O3, 3 Si O3.

  3.  An  analysis of one of the siUcates of alumina would
give the following percentage composition.
Silica,	Per cent. 53.29
Lime,	16.47
Alumina,	30.24
                                   100.00

   What is the symbol of the mineral ?
                          Ans.  Ca O, Si 03. Ala 03 Si 03.

  Second Method of Solution. — By inspecting the formula obtained by
solving the problem  according to the method just described, the student
will see that the amount of oxygen in the acid stands in a very simple
relation to that in the bases.  This  relation is 1: 3 : 6, corresponding to
Ca O, AI2 O3, and 2 Si O3.   It has been shown  in the text-book, § 200,
that a similar simple ratio exists between the amount of oxygen in the
acid  and that in the bases of all oxygen salts.  The ratio can easily be
found from the  percentage composition.  For  this purpose we  have
merely to calculate the amount of oxygen in the per cent of the acid and
bases indicated by analysis, and find the simplest ratio in which- these
amounts  stand to each other.  In our example,
     53.29 per cent of silica-contains 28.24 parts of oxygen.
     16.47  "    "  lime    "      4.71   "   "   "
     30.24  "    "   alumina "     14.12   "   "   "
According to the  principle just stated, these numbers ought to' stand to
each other in some simple ratio, and it can easily be seen that
                4.71 : 14.12 : 28.24 = 1:3:6.
From this ratio we can easily deduce the symbol, for one equivalent of
oxygen corresponds to one equivalent of  Ca O, three equivalents of oxy-
gen correspond to one equivalent of AI2 O3,  and six  of oxygen to two 
94
LIGHT METALS.
equivalents of Si O3. Hence the symbol is Ca 0, AI2 O3, 2 Si O3, which
we may write' as above, Ca 0, Si O3 . AI2 O3, Si O3.  For convenience
in calculating the amount of oxygen from the per cent of acids or bases
indicated by analysis, Table VII. has been added at the end of the book,
which gives the per cent of oxygen, together  with its logarithm con-
tained in the bases and acids of most common occurrence.
  In deducing empirical symbols from the results of actual analysis, it
must be remembered that our processes are not absolutely accurate, and
that therefore we must not expect to find  more  than a close approxima-
tion to a simple ratio between the oxygen in the base  and that in the
acid.  Again, in mineral compounds, it is very frequently the case that
isomorphous bases replace each other to a greater or less extent. This is
the case in common garnet, the symbol of which may be written thus:
        3  [Fe, Mn, Mg, Ca] 0, Si 03 . [Al2Fe2] 03, Si O3.
We generally, however, write the symbol as on page 90:
                  3R0, Si03.R2 03,Si03.
Here R 0  stands for the sum of all the protoxide bases, which make
together but one equivalent of base, and  R2 O3 for the sum of all the
sesquioxide bases,  which also make together but one equivalent of  base.
Such general  symbols  as these give all the information in regard to the
constitution of the mineral which is required.  In deducing such  sym-
bols, it is  evident that the oxygen of all the protoxide bases must be
added together to obtain the amount of oxygen in the assumed base
R 0, and all the oxygen of the sesquioxide bases must be added together
in the same way in order to obtain the amount of oxygen in the assumed
base R2 O3 •  From these sums we can easily obtain the required oxygen
ratio.

   4.  An analysis of andesine (a mineral allied to felspar)
yielded the following result.
Silicic Acid,	59.60		Proportion of Oxygen. 30.90 in Si O
Alumina, Sesquioxide of Iron, Lime,	24.28 1.58 5.77	11.221 0.48. 1.61'	-= 11.70 in RaO
Magnesia, Soda,	1.08 6.53	0.37 1.65	► = 3.79 in RO.
Potassa,	1.08	0.16.	
99.92
What is the symbol of  the mineral ? 
LIGHT  METALS.
95
  Solution. — The ratio of the oxygen in R 0, R2 O3, and Si O3 is
             3.79 :  11.70 : 30.90 = 1 : 3.08 :  8.1,
for which we may substitute, for reasons stated above,
                       1:3:8.
  One equivalent of oxygen corresponds to R O.
  Three equivalents of oxygen correspond to R2 O3.
  Eight equivalents of oxygen correspond to ■§• Si O3.
  But as we do not admit fractional equivalents, we may multiply the
whole by three, when we obtain the empirical symbol
                 3RO, 3R203,8Si03;
from which we may deduce the rational symbol
             3RO,2Si03.3(R203,2Si03).

   5. Deduce the symbols of the  siUcious minerals of which
the following are analyses.
                        i.
Silicic Acid,          65.72
Sesquioxide of Iron,
Alumina,             18.57
Lime,                0.34
Magnesia,            0.10
Potassa,              14.02
Soda,                 1.25
                    100.00    100.0    100.04    100.16
2. 68.4	3. 44.12	4. 63.70
0.1	0.70	0.50
20.8	35.12	23.95
0.2	19.02	2.05
	0.56	0.65
	0.25	1.20
10.5	0.27	8.11
 
HEAVY  METALS.
        FIRST  GROUP OF THE HEAVY METALS.
                    Iron (Fe).
286. a.  4 F^ + O = Fe4 O.
     b.  3 Fe4  O +13 O = 4 (Fe O, Fe2 03).
     c.  2 (Fe  O, Fe2 03) + O = 3 Fe2 03.
     d.  2 (Fe  O, S 03.  6 H O) when  heated resolves into
          F2 03, S 03 -f- S 02 +12IIO; by further heat-
          ing, Fe2 03, S 03 = Fe2 03+SO,
                          Red.
     e.  3 Fe + 4 O = Fe 0, Fe2 03.
                                             Red.
    /.  2 (Fe 0, Fe2 03) + 9 HO+O = 3 (Fe2 03,3 H 0).
     g.  Fe  O,  Fe2  03 + 2  C 02  + Aq = Fe2  03
          + Fe 0, 2 0 02 A- Aq.
        2 (Fe 0, 2 C 02) + Aq +  O = Fe*. 03,% H 0
          A-±C02 + Aq.
285.    FeA-ff0,S03A-Aq = Fe 0, S03-\-Aq + H. 
HEAVY METALS.
97
285.    Fe + Cu  0,  S 03 + Aq = Cu +  Fe 0, S 03

          + Aq.
                                            Red.
     a. 6 (Fe  0, S 03) -f Aq + 3 O = Fe2 03,  3 H 0
                  Yellow-Brown.
          -\-2(Fe203, 3 S 03)-\-Aq.

     b. 6 (Fe 0, S 03) -f- 3  (HO, S 03) -f- H0, N 05
                       Yellow-Brown.
          -\-Aq = 3  (Fe2 03, 3 S 03)-\-Aq+ N  02.

     c. Fe 0,S03-{- [NHt] 0,HOA~Aq = Fe'^HO

          A-INH^O, S03-\-Aq.

       Fe2 03,  3 S 03 A- 3  ([JVUS] 0, H 0) + Aq =

          Fe, <C 3 H O +  3 ([JVJ74] 0, S 03) + Aq.

     d,  2Fe-\-±HO,NO,A-Aq = Fe203,3N05 + Aq



288.    [HO, 2NaO-]P05-\-3 (Fe 0, S 03) + Aq =
                White.              _  „ A     __  _  ~ --.
           3 Fe O, P 05  +  2 iVa 0, £ 03 + #0, £ 03

           A-Aq.

        [HO, 2Na0-]P0,-{-  Fe2 03, 3 S 03 + Aq =

           Fe203h3P05.4H0+2 (Na 0, S03)-\-H0, S03+Aq.

290.    3 Fe O  +  2 Fe2 03  +  9  H Cy +  Aq  =
                   Blue.
           3 Fe Cy, 2 Fe2 Cy3 + %HO + Aq.

291.    3 Fe CyT'2 Fe* Cy3 + 6 (K 0, H 0) + Aq =

           2 (Fe2 03,% H 0) + 3 (2 K Cy, Fe Cy) + Aq.

292. a.  3(2K Cy, Fe Cy) + 2 (Fe2  03, 3 S 03) + ^ =

           3 Fe Cy^'Feo Cy3 + 6 (K 0, S  03) -f Aq.

                9 
HEAVY METALS.
. b.  2K Cy,  Fe Cy +  2 (Fe 0,  S 03)  +  Aq  =
          White.
       3 Fe Cy + 2 K 0, S 03 + Aq.
       White.                            Blue.
     9 Fe Cy + 3 O = Fe2 03 + 3 Fe Cy, 2 Fea Cy3.

 c.  2 K Cy, Fe  Cy + 2 (Cu 0, S 03)  +  Aq  =
              Purple.
       2 Cu Cy, Fe  Cy.  Aq + 2 (K 0, S 03) -f Aq.

     2 KCy, Fe  Cy +  2 (Pi 0,  JV 05)  +  -4y  =
                White.
       2 Pb Cy, Fe  Cy.  Aq + 2 (K0,  NO,) -f Aq.

     A large number of simUar  compounds  having the

       general  symbol  2 R Cy, Fe Cy. Aq, may  be

       formed thus: —
            White.                   Pale Yellow.
     2 H Cy, Fe Cy. Aq.     2 [N H4] Cy, Fe Cy. Aq.
            Yellow.                   Yellow.
     2 Na Cy,  Fe Cy. Aq.   2 Ba Cy, Fe Cy. Aq.
         Pale Yellow.                  White.
     2 Mg Cy,  Fe Cy. Aq.   2 Zn Cy, Fe Cy. Aq.

     There are also compounds in which the two  equiva-

       lents of R in the general symbol are replaced  by

       different metals  thus: —
           White.                   Yellow.


           Red.


         Yellow Salt.                       Red Salt.
.     2(2KCy, FeCy)+ClA-Aq=3KCy,  Fea Cy3

       -\-KClA- Aq.

     3 K Cy, Fe2  Cy3  + 3 (Fe 0,  S 03)  + Aq  =
               Blue.
       3 Fe Cy,  Fe2 Cy3 + 3 (K 0,  S 03) + Aq. 
HEAVY  METALS.                    99
There also may  be  formed  a large number  of

  simUar compounds  having  the  general symbol

  3 R Cy, Fea Cy3, such as: —
        Brown.                    n  „  Light-Red.
  3 H Cy, Fe2 Cy3.             3  Ca Cy, Fe^ Cy3.
                      Red.





   + [JVi5] 0, SO, + Aq.
 Black                     Light-Green.
 Fe S + aq + 4 O = Fe  0, S 03 + aq.
              Manganese (Mn).

   Mn o2 + H 0, S 03 -\- aq = Mn 0, S03Ar-aq

     + ©
    Black                        Light-Pink Solution.
   Mn'02 + 2 H CI + Aq = Mn CI -\- Aq + CI.
    Light-Pink Solution.                         j^r0Wn,o TT 1^
a.  6 (Mn 0, S 03) + Aq + 3 O = Mtl, 03,  3 H 0
     + 2 (ilfw, 03, 3 S 03) + ^?.
   Light-Pink Solution.           „ „ _ ,   .     >rWAlWTTA
b.  MnO,S03-\-lNHi-]0,HOA-Aq=MnO,KO

     A-  [NHA-] 0, S03 + Aq.
           White                            Brown.
   2 (Mn 0,  H 0) + aq + O = Mn2 03,  3 H 0

      A-aq.
    Light-Pink Solution.       „,,.„,«,    ..     Pleih-eolored.
   M 0,  ^3+  [^ -HJ] # +  ^ =  Mn  S

      + {NH^ 0,S03A-Aq.
     Melted together.                          _  .  __ ^^
   Mn02 + ^0,iT0 + O = ^:0,if«03 + HO. 
HEAVY  METALS.
         Green Solution.
    3 (K 0,  Mn 03)  + Aq +  2 C 02 =  Mn 02
          Crimson Solution.
      -f K 0, Mn2 0i + 2(K0,C 02) + Aq.
        Green Solution.
    3 (K 0, Mn  03)  -f  2 (H 0,  S  03) +  Aq =
                Crimson Solution.
      Mn 02 + KO, Mn2 07 + 2  (K0, S 03)-\-Aq.





         Cobalt (Co) + Nickel (Ni).

       Fink Solution.
    Co  0,  S  03  +  [N #4] S -\-  Aq

      [NH4-] 0,S03-\-Aq.
         Green Solution.
    Ni  0,  S  03 + [N Ht-] S A-  Aq

      A-WH,-] 0,S03A-Aq.





                 Zinc (Zn).



    Zn + H 0,  S 03 A- Aq = Zn 0,  S 03  + Aq

      + H.
                                           White.
a.  Zn 0, S 03 + K 0,  H 0 + Aq  = Zn O, H 0

      + K  0, S 03 + Aq.

    This precipitate dissolves in an excess of K 0, HO

      A-Aq.
    „   „                                      White.
b.  Zn 0,   S 03  -\-   [N H,-\ S +  Aq =  Zn S

      + [^J74]  0,S03ArAq.
     Black.
=  Co  S
    Black.
=  Ni  S 
HEAVY METALS.
101
312.  c.  5  (Zn 0,  S 03)  + 5  (Na 0, C 02) + Aq  =
                  White.                      White.___
          2  (Zn  0,   C 02)  +  3  (Zn  O,  H  0)

          -j- 5 (Na 0,  S 03) +  Aq + 3 C 02.

        This precipitate is a mixture of Zn O, C 02, and

          Zn 0, H 0, but in variable proportions.



                   Cadmium (Cd).
                                    Yellow.
315.    Cd 0, S 03 A- HS-\- Aq = Cd S + H 0, S 03

          + Aq.



                      Tin  (Sn).
                           Dark-Brown. Gray-White.      White.
317.    The oxides of tin are Sn O ; Sn2 03 ;  and Sn 02.

319.    Sn + HCl+aq= Sn  Cl + aqA-VL.
                                              White.
320.    Sn CI -\- [NHA 0, H 0 + Aq = Sn O, H 0

           + [tf\SQ ClArAq.

321.    Sn CIA- Cl-\-Aq= Sn  Cl2 -\- Aq.

         Sn + 2 HCl A-  HO,  NO,  + Aq  —  Sn Ck

           + Aq + H ©3-
                                              White.
         Sn 04 + 2 ([NHJ 0, HO) + Jf-HO, Sn02

           -\-2\NHA  Cl + Aq.

         ^nOA-KO,HO-\-Aq = KO,Sn 0 + Aq..

         HO, Sn O2+K0, H0+Aq*=KQ, Sn 02

           A-Aq.

                   .9* 
102
HEAVY METALS.
        By evaporation of the last solution we  can obtain

          crystals of         K O, Sn 02.  4 H 0.

        We may also prepare Na 0, Sn 02. 4 H O.
                                           White.
324.    5 Sn + 10 (HO, NO,) +Aq = Sn5 O10.  10 H O

          -f-^+lOWO,.

        Sn5 O10.  10 H O  +  K O, H O  +  Aq =

          K 0, Sn,  Ol0 A- Aq.

        By evaporation  we  can  obtain  crystals of
                  White.
          KO, Sn5O10.  4 HO.
                               Brown.
325.    Sn  CI A- HS+ Aq = Sn S  + HCl -f Aq.
                                  Yellow.
        Sn  Cl2A-2-&S + Aq = SnS2-\-2HCl-\- Aq.
       SECOND GROUP  OP THE HEAVY  METALS.


                     Lead (Pb).

                                 Black.   Red or Yellow. Red-Yellow.
331.    The oxides  of lead are Pba O;  Pb O;  Pb2 03;
              Dark-Brown.
          and Pb 02.

334.    3 Pb + 4 (HO, NO,)-\-aq = 3 (Pb O,  N O,)

          A-aq + 'N o2.

        3 Pb O + 3 (H O, N O,) -f aq = 3 (Pb O, N O,)

          -\-aq. 
HEAVY METALS.                  103
335.     PbO, NO, 4- HO,  S 03A-Aq = Pb O, S 03

          A-HO,NO,A-Aq.

        Ak 03,3S03-\-3 (Pb 0, [C4 H3-\ 03) A~Aq =
               White.
          3 (Pb O, S 0,) + Ak 03, 3 [<74 H31 03 + Aq.
                                  White.
336.     Pb O + H CI + aq = Pb CI -j- aq.

        Pb O + H CI A- Aq = Pb CI -\- Aq.

337.     Acetate of oxide of lead = Pb O, [C4 H3] 03. 3 H O.

        Basic acetate of oxide of lead =

          3 PbO,  [C4H3]03. HO.

338.     2(PbO,  [C7425] 03)-\-2HO,  C3H, O10 + Aq =
                White.
          2 PbO, C8H4Ol0 + 2 (HO, [CtH3-]  03)A~Aq.
                                              White.
        Pb 0, N0,+ [NHt] 0, H0A-Aq = YbO,lI0

          -{-INH,-]  0, NO,  + Aq.

339.    Pb  0,  [C4  H,]  03  +  Aq  A-  2  Pb O =

          3Pb 0, [<74 H3~\  03-\-Aq.

        3  Pb 0,  [C4 H3~\  03  A- Aq A- 2  C Oa =
                 White.                     „
          2  (Pb O, C 02) + Pb 0, [C4 H3-\ 03 + Aq.

340.    Zn  +   Pb  0,  [C4 #3]  03  +  Aq =  Pb

          + ^0, [04#3] 03 + ^.
                                                Black.
341.    Pi 0, [<74 iT3] 03 + H S  A-  Aq = Pb  S

          A-HO, \_CtH3-\ 03A-Aq.

342.    Pb S + 3 O = Pb O  +  S 02,  also  Pb  S
                        White.
          + 4O = Pb0, S03. 
104
HEAVY METALS.
342.    By roasting galena we obtain a mixture of Pb 0
          with a smaU amount of Pb O, S 03.  By then
          melting together Pb O or Pb O,  S 03 and  an
          excess of Pb S, we obtain metaUic lead and sul-
          phurous acid, thus: —
        2 Pb O + Pb S = 3 Pb + S 02, also Pb 0, S 03
          + Pb S = 2 Pb -f 2 S 08.

                    Bismuth (Bi).
347.    2 Bi + 4 (H 0,  N 0,) +aq = Bi2 03, 3 N 0,
          A- aq + W 02.
                                            White.
        4 (Bi2 03,  3 N 0,) -f- Aq = 3 (Bi8 03, N 05)
          + Bi2 03, 9 N 06 -f- Aq.
                                              Brownish-Black.
        Bi2 0,,  9NOsA-3HS+Aq  =  Bi2  S3
          -f 9 (H 0, N 05) + Aq.

                    Copper  (Cu).
                                        Green.
349.    Malachite =           Cu O, HO.  CuO, C08.
        Blue  Carbonate  of  Copper  =
          Cu O, H O. B2 Cu O, C 02.
350.    The oxides of copper are Cu2 O and Cu 6.
           Blue Solution.
          <u 0, S  03 -f K 0, R
          A-KO, S03Ar Aq.
           Blue Solution.                            Blue.
352.    CuO, S  03-\-K0, H 0 + Aq = Cu O, H O 
HEAVY  METALS.
105
352.    By boiling, Cu 0, H O + Aq = Cu 6 -f Aq.
         Light Blue Solution.
353.     Cu 0,  S 03 + 2 ([NH,-] 0,  HO)  + Aq =
                     Very deep Blue Solution.
           Cu 0,  S  03.  2  N H3.   H  0  +  Aq  =


           {&%} 0, 5 03. [##,] 0, #0 + ^, a


           solution which, when treated  with alcohol, yields
crystals having the composition  ^
            Deep Blue.

  N^j- 0,  S03.  [NH4]0,  HO.


These,  when heated,  are resolved into
        Green Powder.

 JN*H O,  S  03 + [NT H4] O, H O.
{
                Cu|

354.    2 (CuO, S 03)A-2(KO, H0)-\-Aq — O* =
          Yellow-Red.
           Cu2 O  + 2 (K 0,  S 03) A- Aq.
                     Heated together.
355.    2 (Cu O, S 03) -j- 2 (Na 0,  C 02) + C = 2 Cu

           + 2 (Na O,  S 03) + 3 C  ©2.

356.    Zn -{-  Cu 0, S 03 -f Aq = Cu  + Zn 0, S 03

           A-Aq.

357.    Cu O, H O + II  = Cu + 2 H O.
         Black.                  Green Solution.
359.    Cu O +  H CI -\- Aq = Cu CI + Aq.
                                              Blue Solution.
360.    3 Cu + 4 (H 0, N 0,) + aq = 3 (Cu 0, N 0,)

           A- aq -\- I¥ 02.
* The oxygen is removed by adding grape sugar to the solution. 
           HEAVY METALS.

         Blue Salt.                  Black.      White.
2 (CuO, N 05. 4H 0) -f- Sn =2 Cu 0 + Sn 02

  + 8HO + 2NO,.

2 (Cu 0,  S 03) +  2 (Na  0, C 02) + Aq =
            Blue or Green.               __         _^ .
  Cu O, HO.  Cu O,  C 02  + 2 (ifa 0,  £ 03)

  + ^ + co2.

3(OwO, £03) + [#0, 2iVaO],P05 + ^ =
      Greenish-Blue.                „,_.  .  -™- ^-»  <-« y»
  3 Cu O, P 05 + 2 (Na 0, 5 03) + H 0, £ 03

  + Aq.
                   Emerald-Green.
Dioptase = 3 Cu O,  2 Si 03. 3 H O.

Basic  Acetate  of  Copper =
                   Green.
   Cu O, [C4 H3] 03. Cu O,  H 0.  5 H O.

Neutral Acetate of Copper =
              Green.
   CuO[C4H3]03. 5 HO.
  Blue Solution.               ,      J5lackA  ,  „ /,  „ s\
Cu 0, SO.ArHS+Aq^CnSAr-HO,  S 03

   + Aq.
  Black                  Green Solution.       -
Cu S'+ H CI + aq  =  Cu CI + aq + H S.
            Mercury (Hg).


6 Hg+±(HO, NO,)+aq = Z(Hg%0, NO,)

  A-aqA-N 02.
                                           Black.
Hg2  0,  NO, + K 0,  H 0 + Aq  = Hg2 0

  + K 0,  N 0, + Aq. 
HEAVY METALS.
107
368.    Hahnemann's  Suboxide  of  Mercury has  not  a

          constant composition.   Some chemists, however,

          assign to it the symbol  2 Hg2 O, N 05. N H3.
                                                 White.
370.    Hg2  0,  N 0,  +  Na  CI + Aq  = Hg2  CI

          A-NaO, NO,-\- Aq.

371.    3 Hg A- 4 (H 0, N 05) + aq = 3 (Hg 0, N 0,)

          + aq A- N 02.
                                                  Yellow.
        HgO,  NO,  A- KO,  H0 + Aq  =  Hg 0

          + K O, N 05 + Aq.
          When heated.       Red.
372.    Hg O, N 05 =  Hg O + 1¥ 04 + O.

373.    .% 0,  iV 05 +  iVa CI + -4?, gives no precipitate,

           because Hg CI is soluble in water.
                                White.
         Hg O  + H CI + aq = Hg CI 4- H 0 + aq.
                Heated together.       _ Residue       Sublimate
         Hg O, S 03 4- Na CI = Na O,  S 03 + Hg CI.
                                         Black.
         Hg2 CI 4- K 0, H 0 4- Aq = Hg2 O 4- K CI

           A-Aq.
                                        Yellow.
         Hg  CI 4- KO, HO 4- Aq = Hg O + K CI

           A-Aq.
            •    ■*                              White.

 374.    2 Hg CIA- [##*] 0,H0 + Aq = {n^J CI

           A-HCl A-Aq.
                                Black Powder.
 375.    Hg  CI A-Sn  Cl +Aq  =  Kg  + Sn  Ck + Aq.
                                 Black.
 376.    J5r>  CI + HS-{-Aq = Hg S + H CI -\- Aq. 
108
HEAVY METALS.
                    SUver (Ag).

380.    3 Ag + 4 (HO, NO,)-\-Aq = 3 (Ag 0, NO,)

          A-Aq + N02.
               Heated together.
381. a.  Ag O, N 05 4- 2 C = Ag 4- 2 C 08 4- ]¥ Oa.
                                              Brown.
     b.  Ag 0, N 0, + K 0, H 0  4- Aq = Ag O, H O

          4- K 0, N  0, 4- Aq.
           1                                   White.
        AgO, HO+[MJ 0, HO + Aq = |nH»]-

          + Aq.
                                                White.
     c.  Ag  0,  N 0, A-  Na  Gl +  Aq  =  Ag CI

          -f- Na  0, N 0, A- Aq.
                                    Black.
     e.  AgO, N05 + HS -\-Aq = Ag S 4- HO, NO,

          + Aq.


                    Gold (Au).

385.    Au 4- 3 H CI + H 0, N 0, -f Aq = Au Ck

          A- Aq a- w o2.
                                       Dark-Brown.
387.    Au Cl3 -\- 6 (Fe 0, S 03) -f- Aq = Au + Fe2 Cl3

          4-2 (Fe2 03, 3 S 03)+Aq.

388.    Aurate of Potassa (K O, Au 03 4- 4 H O) is  a

          compound  of oxide  of potassium and teroxide

          of gold, in which the last plays the part of an

          acid. 
HEAVY  METALS.
109
                    Platinum (Pt).

391.    3 Pt -\-  6 H CI +  2 (H 0,  N 0,) 4- Aq =

          3 Pt Ck 4- Aq -f 2 W 02.
                                            Yellow.
392.    [NH,-] CI A-Pt CkA~Aq= [N H4] CI,  Pt Cl2

          A-Aq.
                                      Yellow.
394.    K  Cl + Pt Cl2 + Aq = K  CI, Pt Cl2 + ^4?.
                                                 Black.
        The Oxides of Platinum are Pt 0* and Pt 02.
                                     Greenish-Brown. Reddish-Brown.
        The Chlorides of Platinum are Pt CI and Pt Cl2.
                                      Black.       Black.
        The Sulphides of Platinum are Pt S  and Pt S2.
                                Black.
        Pt  CI A- HS A- Aq = Pt S -f H CI A- Aq.
                                   Black.
        Pt  Ck -\-  2 HSA- Aq = Pt S2 + 2 HCl + Aq.
        THIRD  GROUP  OF THE HEAVY METALS.

                    Chromium (Cr).

397.    The symbol of chrome iron  ore is  Fe 0, Cr2 03;

           but  almost invariably a portion of the  Fe O is

           replaced  by Mg O, and a portion of the  Cr2 03
        Heated together in contact with air.
2 (Fe O, Cr2 03) + 2 (K O, C 02)  4-  7 O
by Al2 03.
      Heated
(Fe O, Cr

Fe2 03 + 2 (K O, 2 Cr 03). -f 2 C Oa.

* Only known in combination with water.
      10 
110
HEAVY METALS.
 397.     The above process is hastened by mixing with the

           pulverized mineral a portion of nitre, which yields

           when heated a large supply of oxygen.
               Red.                              Yellow.
 398.     KO, 2 Cr  03 + K0, C02 + Aq=2 (KO, Cr 03)

           + ^4-co2.
             Yellow.                             Red.
         2 (KO, Cr 03)+HO, NO, + Aq=KO,2Cr03

           A-KO, NO, + Aq.

 399.     K 0,  Cr  03 + Pb 0, [04  H3-]  03 A~ Aq =
             Yellow.
           Pb 0, Cr 03 + K 0,  [04 H3] 03 + Aq.
             Yellow.                              Red.
         2 (PbO, Cr03) -\-K0,HO-\-Aq = 2 PbO, Cr03

           A-KO,  Cr 03 4- Aq.
              Yellow.                              White.
400.     2 (Pb  O, Cr 03)  -f 8 H CI -f  Aq  = 2 Pb CI
               Green.
           + Cr2 ci3 4-^4.3 ci.

         Cr2 Cl3 -f  3  (IN H,-]  0,  H 0)  A~  Aq  =

           Cr2 03, 10 H O -f 3 [N774]  CI -f Aq.

         KO,  2  Cr 03A-HO,  S 03-{-3 S 02A-Aq =

          KO,  S 03.  Cr2 03, 3 S  03 4- Aq.

         The symbol  of  crystaUized chrome alum is

          K O, S  03.  Cr2 03, 3 S 03. 24  H O.

401.     K O, 2 Cr 03 4- x* (H O,  S03)+aq=2 Cr 03

          A- KO, S 03. HO, S03-\- x(HO, S 03)

          A-aq.
* x is here used to express an indefinite amount. 
HEAVY  METALS.
Ill
           Red.              Green.
401. a. 2 Cr 03 — 3 O = Cr2 03.

        The oxygen in the last reaction may be removed by

          alcohol or any other reducing agent.



                   Antimony (Sb).


        The oxides of antimony are as foUows : —
                             White.
403.    Oxide of Antimony, Sb 03.
                           White.
        Antimonious Acid,  Sb 04 = £ (Sb 03, Sb 05).
                        Pale-Yellow.
        Antimonic Acid,  Sb 05.

404.    3Sb4-4(ZT0,^05)=3Sb044-4HO+4I¥O2.

        When antimony is treated with an excess of con-

          centrated nitric acid,  only  Sb 04 appears to be

          formed.  If, however, the acid is dilute, the anti-

          monious acid is mixed with more or less of basic

          nitrate of antimony (2 Sb 03, N 05) according to

          the degree of dilution.

        By heating together one part of metallic antimony

          and four parts  of nitre in a  crucible,  there is

          formed a white mass, which is a mixture of anti-

          moniate  of  potassa  (K O,  Sb 05) ;  nitrite of

          potassa  (K  O,  N 03) ; and undecomposed nitre

          (K O, N 05).   Warm  water  will dissolve  the

          two last, but not the  antimoniate of potassa.   If, 
112
HEAVY METALS.
          however, this anhydrous  salt is boUed with water
          for one or two hours, it combines with five equiv-
          alents  of water, forming  a soluble compound
          (K O, Sb 05.  5 H 0).  The white  mass, which
          seemed at first insoluble, dissolves in great meas-
          ure, leaving in suspension only a small amount of
          binantimoniate of potassa.
405.    SbS3A-3HCl + aq=SbCl3A-aqA-3HS.
        SbA-3HCl+(HO,  N 0,) +aq=Sb Cl3
          A-aqA-N 02.
        Sb 4- x ci = Sb ci, 4- x ci.
        Sb  Ck -\- Aq  = Sb 03 -|-  3 H CI -f Aq.
        The precipitate which is first formed on dUuting  a
          concentrated solution of Sb Cl3 with water, always
          contains some chloride with the oxide; but by con-
          tinued washing with water, or  stiU better, with a
          weak solution of Na O, C 02, the whole wUl be
          converted into oxide.
        Sb  CI, -\- Aq  = Sb 05 +  5 H CI 4- Aq.
406.    Sb  03 4-  [H 0, K 0]  08 H,  O10  +  Aq =
          [K 0, Sb 03]  08 H4 OI0 4- Aq.
        The symbol of crystallized  tartar emetic is
          [K O, Sb 03] C3 H4 O10.  2 H O. 
HEAVY METALS.                  113
Red.
407.     [K 0, Sb 03]  08 H, O10 + 3 HS+ Aq = Sb S3
          [H 0, K 0] 08 Ht O10 4" Aq.
        Kermes mineral is an  amorphous  modification of
          SbS3.
                                Bright-Yellow.
        SbCl5 + 5HS+Aq = Sb$, + 5HCl + Aq.
        Golden Sulphuret is a mixture of Sb S3 and Sb S5.
          Melted together.
 408.    Sb S3 + 3 Fe = Sb + 3 Fe S.

                    Arsenic (As).
 412.    As + 3 O = As 03 (arsenious acid).
 413.    2 As 03 -4- 3 C = 2 As + 3 C 02.
 414.    As03-\-KO, HO + Aq  = KO, As03 + Aq.
      a. K O,  As 03  + 2  (Cu  O,  S  03)  + Aq =
           2Cu6TAs034-^> #4.. #0,  so34-^.
      b.  The  symbol of  Schweinfurth  green is
           Cu O, [C4 H3] 03.  2 Cu O, As 03.
 415.    As 03 + 2 (H O,  N O,)  + aq  = As  O,-\- aq
           + 2W04.
         The  symbol of crystallized binarseniate of potassa
           is  [2 H O, K O] As 05.
                                   Yellow.
 416.    As 03-\-3HS + Aq = AsS3+Aq.
                                   Yellow.
         As o, 4- 5 HS 4- Aq = As S5 4- Aq.
           Yellow.           .Red>v
         2 As S3 + S = 2 As S2.
                   10* 
114
HEAVY METALS.
416.    The symbol  of Mispickel (arsenical  pyrites)  is
          Fe[As,  SJ.
        2 (Fe [As, SJ) + 13 O =  Fe2 03  + 2 As 03
          + 2SO*.
417.    As Zn3 + 3 (HO, S03)-\-Aq = 3  (Zn 0, S 0S)
          + Aq+A*mls.
418.    SbZn3 + 3 (HO, S 03) +Aq= 3 (Zn 0, S 03)
          A- Aq 4- Sb H3.
        The compounds of arsenic and  antimony with hy-
          drogen are always mixed with more or less free
          hydrogen.   When  prepared  as  described  in
          sections 417, 418 of Stockhardt's Elements, the
          gas  consists almost entirely of  hydrogen, con-
          taining only a very minute amount of either me-
          tallic compound. 
TABLES. 
 
EXPLANATION   OF   TABLES.
  Table I. — This table,  -which has  been reprinted from the " Elementary
Instructions  in  Chemical Analysis" by Fresenius,  indicates by  means  of
figures the solubility or insolubility in water and acids of some of the more fre-
quently  occurring compounds; thus, 1 means  a substance soluble  in water;
2, a substance insoluble in water, but soluble in chlorohydric or nitric acid;
3, a substance insoluble either in water or acids.  For those substances stand-
ing on the limits between these three classes, the figures are jointly expressed;
thus 1-2 signifies a  substance difficultly  soluble  in water,  but soluble  in
chlorohydric or nitric acid;  1-3, a body difficultly soluble in water, and the
solubility of which is  not increased on the addition  of acids; and 2 - 3, a sub-
stance insoluble in water, and difficultly soluble in the acids.  When the rela-
tion of a substance to  hydrochloric acid is  different from that to nitric acid,
this is stated in the notes.  The figure indicating the solubility of a  given salt
will be found opposite to the  symbol of its acid, in the column headed by the
symbol of its base; that of a  given binary, under the symbol of the correspond-
ing oxide, and opposite to the symbol of its electro-negative element.  *

  Table II. — The values of the French measures and weights, in terms of the
corresponding English units, given in this table, were taken from the  second
volume  of the  Cavendish Edition  of " Gmelin's Hand-Book  of  Chemistry."
The logarithms of these values and their arithmetical complements  have been
added to facilitate the reduction from one system to the other.   The use of the
table can be illustrated best by a few examples.

  1. It is required to reduce  560.367 metres to English feet.
    Solution. — No. of feet = No. of metres  X No. of feet in one metre.
           log.  No. of feet = log. No. of metres + log. No. of feet in one metre.
           log.  560.367                                          2.7484726
           log.  3.2809 (value in feet of one metre from Table II.)    0.5159930
                                                               3.2644656
                           Ans. = 1838.51 feet. 
118
EXPLANATION OP  TABLES.
  2. It is required to reduce 30.964 inches to centimetres.
    Solution. —No. of centimetres = No. of inches -*■ No. of inches in one cen-
                  timetre.
          log. No. of centimetres = log. No. of inches + log. (ar. co.) No. of
                  inches in one centimetre.
          log. 30.964                                           1.4908571
          log. (ar. co.) 0.3937 (value of one centimetre in inches)  0.4048258
                                                               1.8956829
                        Ans. 78.6471 centimetres.

  3. It is required to reduce 23.576 kilometres to feet.
    Solution.—23.576 kilometres = 23576 metres.
          No. of feet = No. of metres X No. of feet in one metre.
          log. 2.3576                                           4.3724701
          log. 3.2809                                           0.5159930
                                                               4.8884631
                            Ans. 77350.5 feet.

  In the above examples logarithms of seven places have been used; but where
great  accuracy is not required, logarithms of four places are sufficient.  In
such cases the last three figures of the  logarithm given in the table may be
neglected, and the problems solved with great expedition by means of the table
of four-place logarithms which accompanies this book.

  Table III. — This table has been taken, with some few alterations, from
Weber's " Atomgewichts-Tabellen."  The atomic volumes assigned to the ele-
ments are the same  as those generally  given in English and American text-
books on Chemistry, with the exception of those of Carbon, Boron, and Silicon,
which are assumed to yield a one-volume gas like oxygen for convenience in
calculation.   The calculated  specific gravities are deduced from the observed
specific gravity of oxygen and the chemical equivalent of the  given substance
by means of the proportion,

Equiv. of Oxygen : Equiv. of given substance = Sp. Gr. of Oxygen :  Sp. Gr.
                         of given substance.
  This proportion yields the specific gravity  directly when one  equivalent of
the substance occupies the same volume as one equivalent of oxygen.  If it
occupies twice, three times,  or four times this volume,  the  results must be
divided by two, three, or four, as the case may be.  The method of calculating
may best be illustrated by a few examples.
   1. It is required to calculate the Specific Gravity of Nitrogen.
               Equiv. of O. Equiv. ofN.     Sp. Gr. of O.
    Solution.        8   :    14   =    1.10563  :    1.93485.

   This would be the specific gravity if 14 parts of nitrogen occupied the same
volume as 8 parts of oxygen; or, in other words, if the equivalent volume of 
EXPLANATION  OF TABLES.                119
nitrogen was 1, the same as that of oxygen.  The fact is that it is 2, so that
the true specific gravity of nitrogen = £ (1.93485) = 0.967428.

  2. It is required to calculate the specific gravity of ammonia gas.
              Equiv. of O.  Equiv.ofNH,.     Sp. Gr. of 0.
    Solution.      8     :     17    =     1.10563  :   2.34946.

  Hence the specific gravity of ammonia gas would be  2.34946 if one equiva-
lent (or 17 parts) occupied the same volume as one equivalent of oxygen; but
on referring to the table, it will be found that the equivalent volume of this gas
is 4, or, in other words, that  one equivalent occupies a volume four times as
large as that occupied by one equivalent  of oxygen, so that to find the true
specific gravity of ammonia gas we  must  divide 2.34946 by 4, which will give
a quotient 0.58736.   The slight difference between this result and that given
in the table arises from the fact that in Weber's table the equivalent of nitro-
gen used is 14.005, and not 14, as in the solutions of the above examples.
  Were the law of equivalent  volumes absolutely rigorous, (that is, did one
equivalent of every gas precisely occupy either the same volume, or else a vol-
ume two, three, or four times as great as  the volume of oxygen,) then the cal-
culated specific gravities ought to agree exactly with those obtained by experi-
ment.   On comparing  together the two  columns of observed and calculated
specific gravities in the table, it will be found that  the numbers,  although
approximatively equal, do not absolutely coincide.  Part of these differences
are unquestionably owing to errors of observation; but  after making the great-
est possible allowance for all errors  of that sort, there still remains (especially
in the case of the easily condensed gases, such as alcohol  vapor, sulphurous
acid, and carbonic acid) large differences to be accounted for.  The most prob-
able explanation of these differences seems to be found in the assumption that
the law of equivalent volumes holds rigorously only when the gases are in the
state of extreme expansion.   As we experiment upon them, they are more or
less condensed by the pressure of the atmosphere, and it is supposed that they
are not all condensed equally, or, in  other words, that even under this pressure
they do not obey absolutely the law of Marriotte.  The more easily a gas may
be reduced to a fluid, the greater is  it condensed by the atmospheric pressure,
and hence the greater  is its  specific gravity.   This view is confirmed by the
fact that the observed specific gravity of carbonic acid gas at  0°, and under a
more feeble pressure than that of the atmosphere, approaches more nearly to
that obtained by experiment.
  The  Specific gravity  of Carbonic Acid Gas, at 0° (air = 1), was,
      Under  the pressure of 76.000 centimetres,              1.52910
          "     "     "     37.413     "                    1.52366
          "     "     "     22.417*   "                     1.52145
      Theoretical specific gravity,                          1.52024
  * It must be remembered that the specific gravity of a gas is equal to its weight, di-
vided by the weight of an equal volume of air under the same conditions of temperature
and pressure. 
120
TABLES.
o	CO (M CM CO GO |<NCO | CO rH CO CO CO CO i-H CO | CO rH rH rH
o to	CO NHNMH |h(N«NN N H (N rH
o s	-r-1MH (NH(NHN(NNN(NlN rH j-h
O o	CO CO (Mr-1 1—1 CO 1—1 H(N CO | CO 1—1 1-1 rH
o c	GO rH rH CO rH H (N CO CO CO GO rH CO
o	GO GOrHrHGOrHrHrHCOCO G<l GO GO rH rH
o .£5 PH	CO CO CO | GO GO CO | rH CO CO CO CO GO CO i-l CO rH CO
0	HHHHWNHINNIMINNN rH CO
O u XII	HHHHCOINHNNHININCT rH CO
0 eS	CM CO CO | rH tH | |rHrHC0C0COGOG0CO rH CO rH rH rH
O 6 M	GO # CO | CO CO CO | (NH(N<M(M<M(M(M rH CO rH rH
	rHrHi—IrHrHrHrHrHi—li—IHHH rH r-i
0	
	
0 1—1	
	
 
TABLE8.
121
6	CO rH	CO rH GO rH	CO	CO CO	GO	rH	GO
M							
o	-t-r	++			CO		
X	CO rH	GO CO CO	CO CO		1	rH	i—1
O	I—1						
rH	GO rH	W/3 CO rH r^					
	-						
o a	CO CO	GO r-				rH	
c		+-r					GO
a	GO rH	GO rH GO		GO	CO	r-\	1
x							i—{
O	CO H 2 rH rH	rH rH r-+ GO	GO	GO	GO 1	r-{	i—1
rH
o
o
o
C
o
 S3
3
co
rH CO rH CO rH      CO r- rH rH
   rt
I  co
■<   I   S
GOrHCOGO      GOCOrHrH
GOrH    CO rH r-. rH GO GO CO   GO CO rH-
COrHGOrHGOCOCOGOGOCO  rH  rH
                       GO     GO

GOrH    rHGOGOGO-GOCO  |   rH
rHrH GOrHrHrHGOCO   COGOGO
O
     'Od(50 0
co co O £ Ph <j<j O PQ O l<J  iH
om^ oouu do o
3.9
03 rt _ fc'
11 
122
TABLES.
TABLE  II.
FRENCH MEASURES AND WEIGHTS.
1 Kilometre
1 Hectometre
1 Decimetre
1 Metre
1 Kilometre
1 Metre
1 Centimetre
Measures of Length.
1000 Metres.	1	Metre =	1.000 Metre.
100 "	1	Decimetre =	0.100 "
10 "	1	Centimetre =	0.010 "
1 "	1	Millimetre =	0.001 "
		Logarithms.	Ar. Co. Log.
0.6214 Miles.		9.7933712	0.2066188
3.2809 Feet.		0.5159930	9.4840070
0.3937 Inches.		9.5951742	0.4048258
       Measures of Volume.

1 Cubic Metre      =   1000.000 Litres.
1 Cubic Decimetre  =     1.000   "
1 Cubic Centimetre  =     0.001   "
1 Cubic Metre
= 35.31660 Cubic Feet.
1 Cubic Decimetre  = 61.02709 Cubic Inches.
1 Cubic Centimetre  =  0.06103    "     "
1 Litre
Litre
Litre
0.22017 Gallons.
0.88066 Quarts.
1.76133 Pints.
Logarithms.
1.5479790
1.7855226
8.7855226
9.3427581
9.9448083
0.2458407
Ar. Co. Log.
 8.4520210
 8.2144774
 1.2144774
 0.6572419
 0.0551917
 9.7541593
Weights.
1 Kilogramme   = 1000 Grammes.
1 Hectogramme =100     "
1 Decagramme  =   10     "
1 Gramme      =    1     "
1 Gramme     = 1.000 Gramme.
1 Decigramme = 0.100    "
1 Centigramme = 0.010    "
1 Milligramme = 0.001    "
                                           Logarithms.    Ar. Co. Log.
1 Kilogramme  —   2.67951 Pounds (Troy).    0.4280554    9.5719446
1 Gramme     =  15.44242 Grains.            1.1887154    8.8112846 
TABLES.
123
TABLE  III.
SPECD7IC  GRAVITY  AXD ABSOLUTE  WEIGHT OF ONE
  LITRE OF  SOME OF  THE  MOST  IMPORTANT  GASES
  AXD VAPORS.
Names of Gases.	> I'f	Sp. Gr. Observed.	Sp. Gr. Calculated.	Weight of 1 Litre = 1000 c. c.	Loga-rithms.	Ar. Co. Log-arithms.
Air,		1.	1.00000	1.29363	0.1118101	9.8881899
Alcohol,	4	1.613	1.58934	2.05602	0.3130273	9.6869727
Ammonia Gas,	4	0.5967	0.58753	0.76005	9.8808422	0.1191578
Antimony,	1		17.83274	23.06897	1.3630282	8.6369718
Antimonide of Hydr.	4		4.56239	5.90204	0.7710022	9.2289978
Arsenic,	1	10.65	10.36528	13.40884	1.1273912	8.8726088
Arsenide of Hydr.	4	2.695	2.69553	3.48702	0.5424519	9.4575481
Boron,	1		1.50591	1.94809	0.2896090	9.7103910
Bromine,	2	5.54	5.52605	7.14866	0.8542247	9.1457753
Bromohydric Acid,	4		2.79758	3.61903	0.5585922	9.4414078
Carbon,	1	0.8469*	0.82922	1.07270	0.0304783	9.9695217
Carbonic Acid,	2	1.52908	1.52024	1.96663	0.2937226	9.7062774
Carbonic Oxide,	2	0.96779 0.96743		1.25150	0.0974309	9.9025691
Chlorine,	2	2.47	2.45052	3.17007	0.5010689 9.4989311	
Chloride of Boron,	4	3.942	4.05226	5.24213	0.7195078 9.2804922	
Chloride of Silicon,	3	5.939	5.92477	7.66446	0.8844816 9.1155184	
Chlorohydric Acid,	4	1.2474	1.25981	1.62973	0.2121157^.7878843	
Cyanogen,	2	1.8064	1.79698	2.32463	0.366353819.6336462	
Cyanhydric Acid,	4	0.9476	0.93304	1.20701	0.0817109 9.9182891	
Ether,	2	2.586	2.55677	3.30751	0.5195012	9.4804988
Fluorine,	2		1.30151	1.68367	0.2262570	9.7737430
Fluoride of Boron,	4	2.3124	2.32875	3.01254	0.4789329	9.5210671
Fluoride of Silicon,	3	3.600	3.62677	4.69170	0.6713302	9.3286698
Fluohydric Acid,	4		0.68531	0.88654	9.9476983	0.0523017
Hydrogen,	2	0.06927	0.06910	0.08939	8.9512889	1.0487111
Iodine,	2	8.716	8.76760	11.34203	1.0547011	8.9452989
Iodohydric Acid,	4	4.443	4.41835	5.71571	0.7570702	9.2429298
Marsh Gas,	4	0.5576	0.55282	0.71514	9.8543911	0.1456089
Mercury,	2	6.976	6.91732	8.94845	0.9517478	9.0482522
Nitrogen,	2	0.97136	0.96776	1.25192	0.0975765	9.9024235
Xitrous Oxide,	2	1.5269	1.58951	2.05624	0.3130738	9.6869262
Nitric Oxide,	4	1.0388	1.03669	1.34109	0.1274580 9.8725420	
Olefiant Gas,	4	0.9852	0.96743	1.25150 0.0974309!9.9025691		
Oxygen,	1	1.10563		1.43028 0.15542109.8445790		
Phosphorus,	1	4.42	4.33452	5.60727 0.7487515 9.2512485		
Phosphide of Hydr.	4	1.178	1.18728	1.53590 0.1863629:9.8136371		
Selenium,	1		2.73801	3.54197 0.5492448 9.4507552		
Silicon,	1		3.07120	3.97300	0.5991186 9.4008814	
Sulphur,	i	6.5635	6.65866	8.61384	0.93519689.0648032	
Sulphide of Hydr.	2	1.1912	1.17888	1.52503	0.1832784 9.8167216	
Sulphurous Acid,	2|	2.247 1 2.21541		2.86592 0.457264019.5427360		
* Calculated from the Sp. Gr. of Carbonic Acid Ga=, observed by Regnault. 
124
TABLES.
                   TABLE  IV.

MEAN COEFFICIENTS OF LINEAR EXPANSION OF SOLIDS
      FOR ONE DEGREE BETWEEN 0° AND 100°.
Name of Substance.	Coefficients.	Name of Observer.
Glass (flint of Choisy le Roi),	0.00000760	Regnault.
Platinum,	0.00000884	Dulong and Petit.
Glass (common of Paris),	0.00000920	Regnault.
Palladium,	O.00001000	Wollaston.
Antimony,	0.00001083	Smeaton.
Iron (soft forged),	0.00001220	Lavoisier and Laplace.
Bismuth,	0.00001392	Smeaton.
Gold,	0.00001466	Lavoisier and Laplace.
Brass (English, in rods),	0.00001893	S0y' T.
Copper,	0.00001919	Troughton.
Silver,	0.00002083	
Tin (fine),	0.00002283	u
Lead,	0.00002866	It
Zinc,	0.00002942	it
APPARENT CUBIC EXPANSION OF LIQUIDS IN GLASS
              BETWEEN 0° AND  100°.
Name of Substance.	Expansion from 0© to 100©.	
Water,	1	= 0.0466
Chlorohydric Acid, Sp. Gr. 1.137,	*	= 0.0600
Nitric Acid, Sp. Gr. 1.40,	1 9	= 0.0100
Sulphuric Acid, Sp. Gr. 1.85,	1	= 0.0600
Common Ether,	1	= 0.0700
Olive Oil,	A	= 0.0800
Oil of Turpentine,	i	= 0.0700
Water saturated with salt,	1 To"	= 0.0500
Alcohol,	1 9	= 0.1100
Mercury,	1 6?	= 0.0156
COEFFICIENTS OF CUBIC EXPANSION OF GASES.
             Observed by V. REGNAULT.
Name of Substance.	Under constant Volume.	Under constant Pressure.
Air,	0.003665	0.003670
Nitrogen,	0.003668	0.003670
Hydrogen,	0.003667	0.003661
Oxide of Carbon,	0.003667	0.003669
Carbonic Acid,	0.003688	0.003710
Cyanogen,	0.003829	0.003877
Sulphurous Acid,	0.003845	0.003903
 
TABLES.
125
           TABLE  V.





DENSITIES AXD VOLUMES OF WATER.




           By M. DESPRETZ.
	Volumes.	1 , Densities. =T ~		Volumes.	i . 1 a -1 Densities. ! p* = ■ Volumes.			Densities.
0	1.0001269	0.999873 '34		1.00555 0.994480! 68| 1.02144				0.979010
1	1.0000730	0.999927	35	1.00593 1	0.994104! 69 j 1.02200			0.978473
2	1.0000331	0.999966!	36	1.00624	0.993799 70 1.02255			0.977947
3	1.0000083	0.999999-	37	1.00661	0.993433 !	71 1.02315		0.977373
4	1.0000000	1.000000!	38	1.00699	0.993058	72 1.02375		0.976800
5	1.0000082	0.999999	39	1.00734	0.992713	73 1.02440		0.976181
6	1.0000309	0.999969;	40	1.00773	0.992329 || 74 1.02499			0.975619
I 7 1.0000708		0.999929 i J	41	1.00812	0.991945 || 75j		1.02562	0.975018
8	1.0001216	0.9998781 42		1.00853	0.991542 76		1.02631	0.974364
9	1.0001879	0.999812J	43	1.00894	0.991139:1 77		1.02694	0.973766
10	1.0002684	0.999731	44	1.00938	0.990707! 78		1.02761	0.973132
11	1.0003598	0.999640 45		1.00985	0.990246'I 79		1.02823	0.972545
12	1.0004724	0.999527]: 46		1.01020	0.989903 |	80	1.02885	0.971959
13	1.0005862	0.999414i' 47		1.01067	0.989442 ;	81	1.02954	0.971307
14	1.0007146	0.999285 j	48	1.01109	0.989032 i	82	1.03022	0.970666
15	1.0008751	0.999125!	49	1.01157	0.988562 j 83		1.03090	0970027
16	1.0010215	0.998979	50	1.01205	0.988093	84	1.03156	0.969405
17	1.0012067	0.998794	51	1.01248	0.987674	85	1.03225	0.968757
18	1.00139	0.998612 52		1.01297	0.987196	86	1.03293	0.968120
19	1.00158	0.998422	J53	1.01345	0.986728	87	1.03351	0.967482
20	1.00179	0.998213	54	1.01395	0.986243	! 88	1.03430	0.966837
21	1.00200	0.998004	1 55	1.01445	0.985756	89	1.03500	0.966183
22	1.00222	0.997784	'56	1.01495	0.985270	90	1.03566	0.965567
23	1.00244	0.997566	!57	1.01547	0.984766	91	1.03639	0.964887
24	1.00271	0.997297	; 58	1.01597	0.984281	92	1.03710	0.964227
25	1.00293	0.997078	59	1.01647	0.983798	93	103782	0.963558
26	1.00321	0.996800	60	1.01698	0.983303	94	1.03852	0.962908
27	1.00345	0.9965621 61		1.01752	0.982782	95	1.03925	0.962232
28	1.00374	0.996274	62	1.01809	0.982231	96	1.03999	0.961547
29	1.00403	0.995986	63	1.01862	0.981720	97	1.04077	0.960827
30	1.00433	0.995688	64	1.01913	0.981229	98	1.04153	0.960125
31	1.00463	0.995391	65	1.01967	0.980709 ! 99		1.04228	0.959434
32	1.00494	0.995084	66	1.02025	0.980152 100		1.04315	0.958634
33	1.00525	0.994777	67	1.02085	0.979576			1
 
126
TABLES.
TABLE  VI.
PER  CENT OF NOs IN AQUEOUS  SOLUTIONS OF DIF-
   FERENT  SPECIFIC GRAVITIES.  By URE.  15° C.
Specific Gravity.	Per Cent	Specific	Per Cent	Specific	Per Cent	Specific	Per Cent
	of N 05.	Gravity.	ofN05.	Gravity.	of N 05.	Gravity.	of N 05.
1.500	79.7	1.419	59.8	1.295	39.8	1.140	19.9
1.498	78.9	1.415	59.0	1.289	39.0	1.134	19.1
1.496	78.1	1.411	58.2	1.283	38.3	1.129	18.3
1.494	77.3	1.406	57.4	1.276	37.5	1.123	17.5
1.491	76.5	1.402	56.6	1.270	36.7	1.117	16.7
1.488	75.7	1.398	55.8	1.264	35.9	1.111	15.9
1.485	74.9	1.394	55.0	1.258	35.1	1.105	15.1
1.482	74.1	1.388	54.2	1.252	34.3	1.099	14.3
1.479	73.3	1.383	53.4	1.246	33.5	1.093	13.5
1.476	72.5	1.378	52.6	1.240	32.7	1.088	12.7
1.473	71.7	1.373	51.8	1.234	31.9	1.082	11.9
1.470	70.9	1.368	51.1	1.228	31.1	1.076	11.2
1.467	70.1	1.363	50.2	1.221	30.3	1.071	10.4
1.464	69.3	1.358	49.4	1.215	29.5	1.065	9.6
1.460	68.5	1.353	48.6	1.208	28.7	1.059	8.8
1.457	67.7	1.348	47.9	1.202	27.9	1.054	8.0
1.453	66.9	1.343	47.0	1.196	27.1	1.048	7.2
1.450	66.1	1.338	46.2	1.189	26.3	1.043	6.4
1.446	65.3	1.332	45.4	1.183	25.5	1.037	5.6
1.442	64.5	1.327	44.6	1.177	24.7	1.032	4.8
1.439	63.8	1.322	43.8	1.171	23.9	1.027	4.0
1.435	63.0	1.316	43.0	1.165	23.1	1.021	3.2
1.431	62.2	1.311	42.2	1.159	22.3	1.016	2.4
1.427	61.4	1.306	41.4	1.153	21.5	1.011	1.6
1.423	60.6	1.300	40.4	1.146	20.7	1.005	0.8
PER CENT OF H CI IN AQUEOUS  SOLUTIONS  OF DIF-
  FERENT SPECIFIC GRAVITIES. By E DAVY  15° C.
Specific Gravity.	Per Cent of H CI.	Specific Gravity.	Per Cent of H CI.
1.21	42.43	110	20.20
1.20	40.80	1.09	18.18
1.19	38.38	1.08	16.16
1.18	36.36	1.07	14.14
1.17	34.34	1.06	12.12
1.16	32.32	1.05	10.10
1.15	30.30	1.04	8.08
1.14	28.28	1.03	6.06
1.13	26.26	1.02	4.04
1.12	24.24	1.01	2.02
1.11	22.22		
 
TABLES.
127
PER CENT OF H O, S 03 AND  S 03 IN  AQUEOUS SO-
  LUTIONS OF DIFFERENT  SPECIFIC  GRAVITIES.  By
  BLNEAU. 15° C.
Per Cent of HO, S03. 100	Specific Gravity.	Per Cent of S03.	Per Cent of H O, S 03.	Specific Gravity.	Per Cent of S03.
	1.8426	81.63	56	1.4586	45.71
99	1.842	80.81	55	1.448	44.89
98	1.8406	80.00	54	1.438	44.07
97	1.840	79.18	53	1.428	43.26
96	1.8384	78.36	52	1.418	42.45
95	1.8376	77.55	51	1.408.	41.63
94	1.8356	76.73	50	1.398	40.81
93	1.834	75.91	49	1.3886	40.00
92	1.831	75.10	48	1.379	39.18
91	1.827	74.28	47	1.370	38.36
90	1.822	73.47	46	1.361	37.55
89	1.816	72.65	45	1.351	36.73
88	1.809	71.83	44	1.342	35.82
87	1.802	71.02	43	1.333	35.10
86	1.794	70.10	42	1.324	34.28
85	1.786	69.38	41	1.315	33.47
84	1.777	68.57	40	1.306	32.65
83	1.767	67.75	39	1.2976	31.83
82	1.756	66.94	38	1.289	31.02
81	1.745	66.12	37	1.281	30.20
80	1.734	65.30	36,	1.272	29.38
79	1.722	64.48	35	1.264	28.57
78	1.710	63.67	34	1.256	27.75
77	1.698	62.85	33	1.2476	26.94
76	1.686	62.04	32	1.239	26.12
75	1.675	61.22	31	1.231	25.30
74	1.663	60.40	30	1.223	25.49
73	1.651	59.59	29	1.215	23.67
72	1.639	58.77	28	1.2066	22.85
71	1.637	57.95	27	1.198	22.03
70	1.615	57.14	26	1.190	21.22
69	1.604	56.32	25	1.182	20.40
68	1.592	55.59	24	1.174	19.58
67	1.580	54.69	22	1.159	17.95
66	1.578	53.87	20	1.144	16.32
65	1.557	53.05	18	1.129	14.69
64	1.545	52.24	16	1.1136	13.06
63	1.534	51.42	14	1.098	11.42
62	1.523	50.61	12	1.083	9.79
61	1.512	49.79	10	1.068	8.16
60	1.501	48.98	8	1.0536	6.53
59	1.490	48.16	6	1.039	4.89
58	1.480	47.34	4	1.0256	3.26
57	1469	46.53	2	1.013	1.63
 
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ANTILOGARITHMS.
Proportional Parts.
a?	\j			1007							1	2	3	4 _	5 _	6	7	8 —	9
.00	1000	1002	1005		1009	1012	1014	1016	1019	1021	0	0			l	1	2	2	2
.01	1023	1026	1028	1030	1033	1035	1038	1040	1042	1045	0	0			l	1	2	2	2
.02	1047	1050	1052	1054	1057	1059	1062	1064	1067	1069	0	0			l	1	2	2	2
.03	1072	1074	1076	1079	1081	1084	1086	1089	1091	1094	0	0			l	1	2	2	2
.04	1096	1099	1102	1104	1107	1109	1112	1114	1117	1119	0	1			l	2	2	2	2
.05	1122	1125	1127	1130	1132	1135	1138	1140	1143	1146	0				l	2	2	2	2
.06	1148	1151	1153	1156	1159	1161	1164	1167	1169'1172		0				l	2	2	2	2
.07	1175	1178	1180	1183	1186	1189	1191	1194	1197|1199		0				l	2	2	2	2
.08	1202	1205	1208	1211	1213	1216	1219	1222	122511227		0				i	2	2	2	3
.09	1230	1233	1236	1239	1242	1245	1247	1250	1253	1256	0				l	2	2	2	3
.10	1259	1262	1265	1268	1271	1274	1276	1279	1282	1285	0				l	2	2	2	3
.11	1288	1291	1294	1297	1300	1303	1306	1309	1312 1315		0				2	2	2	2	3
.12	1318	1321	1324	1327	1330	1334	1337	1340	1343 1346		0				2	2	2	2	3
.13	1349	1352	1355	1358	1361	1365	1368	1371	137411377		0				2	2	2	3	3
.14	1380	1384	1387	1390	1393	1396	1400	1403	1406, 1409		0				2	2	2	3	3
.15	1413	1416	1419	1422	1426	1429	1432	1435	1439 1442		0				2	2	2	3	3
.16	1445	1449	1452	1455	1459	1462	1466	1469	1472	1476	0				2	2	2	3	3
.17	1479	1483	1486	1489	1493	1496	1500	1503	1507	1510	0				2	2	2	3	3
.18	1514	1517	1521	1524	1528	1531	1535	1538	1542	1545	0				2	2	2	3	3
.19	1549	1552	1556	1560	1563	1567	1570	1574	1578	1581	0				2	2	3	3	3
.20	1585	1589	1592	1596	1600	1603	1607	1611	1614	1618	0			1	2	2	3	3	3
.21	1622	1626	1629	1633	1637	1641	1644	1648	1652	1656	0			2	2	2	3	3	3
.22	1660	1663	1667	1671	1675	1679	1683	1687	1690	1694	0			2	2	2	3	3	3
.23	1698	1702	1706	1710	1714	1718	1722	1726	1730	1734	0			2	2	2	3	3	4
.24	1738	1742	1746	1750	1754	1758	1762	1766	1770 1774		0			2	2	2	3	3	4
.25	1778	1782	1786	1791	1795	1799	1803	1807	1811;1816		0			2	2	2	3	3	4
.26	1820	1824	1828	1832	1837	1841	1845	1849	1854!1858		0			2	2	3	3	3	4
.27	1862	1866	1871	1875	1879	1884	1888	1892	1897| 1901		0			2	2	3	3	3	4
.28	1905	1910	1914	1919	1923	1928	1932	1936	1941!1945		0			2	2	3	3	4	4
.29	1950	1954	1959	1963	1968	1972	1977	1982	1986 1991		0			2	2	3	3	4	4
.30	1995	2000	2004	2009	2014	2018	2023	2028	2032	2037	0			2	2	3	3	4	4
.31	2042	2046	2051	2056	2061	2065	2070	2075	2080	2084	0			2	2	3	3	4	4
■32	2089	2094	2099	2104	2109	2113	2118	2123	2128	2133	0			2	2	3	3	4	4
.33	2138	2143	2148	2153	2158	2163	2168	2173	2178	2183	0			2	2	3	3	4	4
.34	2188	2193	2198	2203	2208	2213	2218 2223		2228	2234	1		2	2	3	3	4	4	5
.35	2239	2244	2249	2254	2259	2265	2270 2275		2280	2286			2	2	3	3	4	4	5
.36	2291	2296	2301	2307	2312	2317	2323	2328	2333	2339			2	2	3	3	4	4	5
.37	2344	2350	2355	2360 2366		2371	2377	2382	2388	2393			2	2	3	3	4	4	5
.38	2399	2404	2410	2415 2421		2427	2432	2438	2443	2449			2	2	3	3	4	4	5
.39	2455	2460	2466	2472 2477		2483	2489	2495	2500	2506			2	2	3	3	4	5	5
.40	2512	2518	2523	2529 2535		2541	2547	2553	2559	2564			2	2	3	4	4	5	5
.41	2570	2576	2582	2588 2594		2600	2606	2612	2618	2624		ll	2	2	3	4	4	5	5
.42	2630	2636	2642	2649 2655		2661	2667	2673	2679	2685			2	2	3	4	4	5 6	
.43	2692	2698	2704	2710 2716		2723	2729	2735	2742	2748			2	3	3	4	4	5	6
.44	2754	2761	2767	2773 2780		2786	2793	2799	2805	2812			2	3	3	4	4	5 6	
.45	2818	2825	2831	28381 2844		2851	2858	2864	2871	2877			2	3	3	4	5	i 5{ 6	
.46	2884	2891	2897	2904 2911		2917	2924	2931	2938	2944			2	3	3	4	5	5 6	
.47	2951	2958	2965	2972 2979		2985	2992	2999	3006	3013			2	3	3	4	5	5	6
.48	3020	3027	3034	3041 3048		3055	3062	3069	3076	3083			2 3		4	4	5	6	6
.49	3090	3097	3105	3112	3119	3126	3133	3141	3148	3155	1 1		2	3	4	4	5	6	6
ANTILOGARITHMS.
.50 9 3162
.51 | 3236
.52 | 3311
.53 1 3388
.54  3467
1  ,  2  I  3
3548
3631
3715
3802
3890

3981
4074
4169
4266
4365

4467
4571
4677
4786
4898

5012
5129
5248
5370
5495

5623
5754
5888
6026
6166

6310
6457
6607
6761
6918
3170  3177
3243  3251
3319|  3327
3396  3404
3475!  3483
3184:3192
3556	3565
3639	3648
3724	3733
3811	3819
3899	3908
3990	3999
4083	4093
4178	4188
4276	4285
4375	4385
4477	4487
4581	4592
4688	4699
4797	4808
4909	4920
5023	5035
5140	5152
52CO	5272
5383	5395
5508	5521
 3258
 3334
 3412
 3491

 3573
 3656
 3741
 3828
3917

4009'
4102
4198
4295
4395
 3266
 3342
 3420
 3499

 3581
 3664
 3750
3837
3926

4018
4111
4207
4305
4406
5636) 5649
5768! 5781
5902| 5916
6039! 6053
61801 6194
6324
6471
6622
6776
6934'
4498 4508
4603, 4613
4710 4721
4819, 4831
4932 4943

5047, 5058
5164 5176
5284: 5297
54081 5420
5534! 5546

5662! 5675
5794 5808
5929! 5943
6067 6081
6209' 6223
6339 6353 6368
6486: 65011 6516
6637'6653 6668
6792 6808
6950,6966
    7079  7096
    7244  7261
    7413j  74301
.88| 7586|  7603
    ?762  7780
6823
6982
.9017943  7962
.911 8128,  8147
.92? 8318  8337
.931 8511  8531
.94F 8710!  8730!
7112 7129 7145
7278 729517311
7447 7464 7482
7621 76381 7656
7798J 7816 7834

798o' 7998! 8017
8166, 8185 8204
8356 8375 8395
8551
8750
.95 [i 8913,  89331 8954 8974:8995
.96? 9120  9141  9162 9183 9204
.97|9333:  9354  9376 9397 9419
.9819550  9572  9594 9616 9638
.99 U 9772  9795  981719840 9863
8570,8590
8770:, 8790
3199
3273
3350
3428
3508
S    9
Proportional Parts.
3206
3281
3357
3436
3516
3589  3597
3673  3681
3758  3767
3846J  3855
3936: 3945

4027,4036
4121!4130
4217; 4227
4315 4325
4416 4426

45191 4529
4624: 4634
47321 4742
4842 4853
4955 4966

5070 5082
5188
5309
5433
5559

5689
5821
5957
5200
5321
5445
5572

5702
5834
5970
6095 6109
6237 6252
    1
6383 6397
6531 6546
6683 6699
6839 6855
6998, 7015
 3214  3221  3228
 3289  3296  3304
 3365  3373: 3381
 3443  345l!3459
 3524 ■  3532 J 3540

 3606!3614  3622
 3690  369813707
 3776  3784  3793
 3864  387313882
3954  3963!3972

4046  4055  4064
4140
4236
4335
4436

4539
4645
4753
4864
4977

5093
5212
5333
5458
5585

5715
5848
5984
6124
6266

6412
6561
6714
6871
7031
7161i7178 7194
7328  7345 7362
7499  7516 7534
7674  7691: 7709
7852  7870: 7889

8035' 8054 8072
8222;8241 8260
8414  8433 8453
8610  8630: 8650
8810  8831 8851
9016; 9036 9057 90781 9099
9226 9247 9268 9290  9311
9441'9462' 9484 9506: 9528
9661 9683 9705 9727  9750
9886 9908 9931 9954  9977
4150 4159
4246 4256
4345 4355
4446 4457

4550 4560
4656 4667
4764 4775
4875 4887
4989 5000
5105
5224
5346
5117
5236
5358
5470 5483
5598 5610

5728^ 5741
5861 5875
5998 6012
6138 6152
6281 6295

6427 6442
6577 6592
6730 6745
68871 6902
70471 7063

7211: 7228
7379 7396
7551: 7568
7727 7745
7907! 7925

8091!8110
8279 8299
8472 8492
8670! 8690
8872 8892
1   2' 3  4  5  6
1  2i 3
2  2  3
2  2  3
2  2  3
2|  2j 3

2  2! 3
2:  3  3
2|  3  3
2  3  4
2  3  4
S  9

 6  7
 6  7
 6  7
 6  7
 6  7
8 il
8
6|  7  8
7j  8l  9
71  8  9
7  8  9
7  8  9

7  8!  9
7|  9 10
8  9 10
8  9 10
8  9 10

b|  9| 11
8 10 11
9 10: 11
9 10|ll
9 10 12
 1
9 lo| 12
9j 11 12
     12
     13
     13
71  8 10] 11
7  8 10 11
7  9J 10 11

7  9 10 12
8!  9 11 12
8  9 11 12
8  9' 11 13
8 10 11 13
81 10
8; 10
9] 10
9 11
9 11
12 13 15
13 15
14 16
14 16
14 16

  .1
15 17
15' 17
 9 11 13
 9 111 13
10 12 14 15 17
10'12 14 16 18
10 12 14 16 18
   12 15 17 19
   13 15 17 19
   13 15 17 20
U  13 16 18 20
1114 16 18 20 
 
								---—== ^=			-—										--																				|
[ LOGARITHMS OF NUMBERS.																					LOGARITHMS OF NUMBERS.																				
ill 0123456								7	8 0334	:. 9	Proportional Parts,										li 152 : 55	0	1	2 7419	3 7427	4	5	6	7	8	9 7474	Proportional						Parts.			
											1 4	11	3 12	1 4 17	21	6 25	7 29	8 32	9 37																						
																																1	2 j 3 2			4.1		71 8 1 9 5, 5 6 7			
lOfJOOOO 0043 0086 0128 017010212 025S								0294		0374												7404	7412			7435	7443	7451	7459	' 7466											
1110414 0453 0492 0531 056910607 0645								0682	0719	0755	4	8	11	15	19	23	26	30	34		56	7482	7490	7497	7505	7573	7520	7528	1 7536 7543		7551	1	•	2 3		4!		1 ■ 5 5' 6 7			
12|079 13 f 113! 1411461		2 082J > 1173 1492	086J 1206 1523	089S 1239 1553	0934 1271 1584	10969 1004 1303 1335 1614 1644		1038 1367 1673	1072 1399 1703	1106 1430 1732	3 3 3	7 6 6	10 10 9	14 13 12	17 16 15	21 19 18	24 23 21	28 26 24	31 29 27		57 58 59	7559 7634 7709	7566 7642 7716	7574 7649 7723	7582 7657 7731	7589 7664 7738	7597 7672 7745	7604 7679 7752	7612 7686 7760	7619 7694 7767	7627 7701 7774	i i 1	2 1 1	2 3 2 3 1 2| 3		1 1 ■ 4 5 5,[ 6 7 4 4 5 6.' 7 4| 4| 5 6| 7					
15 16 17 18	11761 2041 2304 2553	1790 2068 2330 2577	1818 2095 2355 2601	1847 2122 2380 2625	1875 2148 2405 2648	1903 2175 2430 2672	1931 2201 2455 2695	1959 2227 2480 2718	1987 2253 2504 2742	2014 2279 2529 2765	3 3 2 2	6 : 5	8 ; 7	11 11 10 9	14 13 12 12	17 16 15 14	20 18 17 16	22 21 20 19	25 24 22 21		60 61 62 63	7782 7853 7924 7993	7789 7860 7931 8000	7796 7868 7938 8007	7803 7875 7945 8014	7810 7882 7952 8021	7818 7889 7959 8028	7825 7896 7966 8035	7832 7903 7973 8041	7839 7910 7980 8048	7846 7917 7987 8055	1 1 1 1	i 1 1 1	2 ?	3 3 3 3	4 4 3 <?	4 4 4 4	5 1 6 1 5 ft	€ 6 6 r>	6 6 I 6	
19	2788	2810	2833	2856	2878	2900	2923	2945	2967	2989	2	4	7	9	11	13	16	18	20		64	8062	8069	8075	8082	8089	8096	8102	8109	8116	8122	1	1	2	3	3	4	6	5	6	
20	3010	3032	3054 3075		3096	3118	3139	3160	3181	3201	2	4	6	8	11	13	15	17	19		65	8129	8136	8142	8149	8156	8162	8169	8176	8182	8189	1	1	?	3	3	4	5	6	6	
21	3222	3243	3263 3284		3304	3324	3345	3365	3385	3404	2	4	6	8	10	12	14	16	18		66	8195	8202	8209	8215	8222	8228	8235	8241	8248	8254	1	1	?	3	3	4	5	6	6	
22	3424	3444	3464	3483	3502	3522	3541	3560	3579	3598	2	4	6	8	10	12	14	15	17		67	8261	8267	8274	8280	8287	8293	8299	8306	8312	8319	1	1	?,	3	»	4	5	5	6	
23	3617	3636	3655	3674	3692	3711	3729	3747	3766	3784	2	4	6	7	9	11	13	15	17		68	8325	8331	833S	8344	8351	8357	8363	8370	8376	8382	1	1	?,	3	3	4	4	5	6	
24	3802	3820	3838	3856	3874	3892	3909	3927	3945	3962	2	4	5	7	9	11	12	14	16		69	8388	8395	8401	8407	8414	8420	8426	8432	8439	8445	1	1	2	2	3	4	4	5	6	
25	3979	3997	4014	4031	4048	4065	4082	4099	4116	4133	2	3	5	7	9	10	12	14	15		70	8451	8457	8463	8470	8476	848?,	8488	8494	8500	8506	1	1	?,	?.	3	4	4	5	a	
26	4150	4166	4183	4200	4216	4232	4249	4265	4281	4298	2	3	5	7	8	10	11	13	15		71	8513	8519	8525	8531	8537	8543	8549	8555	8561	8567	1	1	?,	2	3	4	4	R	R	
27	4314	4330	4346	4362	4378	4393	4409	4425	4440	4456	2	3	5	6	8	9	11	13	14		72	8573	8579	8585	8591 8597		8603	8609	8615	8621	8627	1	1	t.	2	3	4	4	R	5	
28	4472	4487	4502	4518	4533	4548	4564	4579	4594	4609	2	3	5	6	8	9	11	12	14		73	8633	8639	8645	8651 8657		8663	8669	8675	8681	8686	1	1	?.	9	3 4		4	5	R	
29	4624	4639	4654	4669	4683	4698	4713	4728	4742	4757	1	3	4	6	7	9	10	12	13		74	8692	8698	8704	87101 8716		8722	8727	8733	8739	8745	1	1	2	2	3	4	4	R	5	
30	4771	4786	4800	4814	4829	4843	4857	4871	4886	4900	1	3	4	6	7	9	10	11	13		75	8751	8756	8762	8768	8774	8779	8785	8791	8797	8802	1	1	2	2	3	3	4	R	R	
31	4914	4928	4942	4955	4969	4983	4997	5011	5024	5038	1	3	4	6	7	8	10	11	12		76	8808	8814	8820	8825	8831	8837	8842	8848	8854	8859	1	1	2	2	3	3	4	R	R	
32	5051	5065	5079	5092	5105	5119	5132	5145	5159	5172	1	3	4	5	7	8	9	11	12		77	8865	8871	8876	8882	8887	8893	8899	8904	8910	8915	1	1	2	2	3	3	4	4	R	
33	5185	5198	5211	5224	5237	5250	5263	5276	5289	5302	1	3	4	5	6	8	9	10	12		78	8921	8927	8932	8938	8943	8949	8954	8960	8965	8971	1	1	2	2	3	3	4	4	5	
34	5315	5328	5340	5353	5366	5378	5391	5403	5416	5428	1	3	4	5	6	8	9	10	11		79	8976	8982	8987	8993	8998	9004	9009	9015	9020	9025	1	1	2	2	3	3	4	4	R	
35	5441	5453	5465	5478	5490	5502	5514	5527	5539	5551	1	2	4	5	6	7	9	10	11		80	9031	9036'	9042	9047	9053	9058	9063	9069	9074	9079	1	1	2	2	3	3	4	4	R	
36	5563	5575	5587	5599	5611	5623	5635	5647	5658	5670	1	2	4	5	6	7	8	10	11		81	9085	9090	9096	9101	9106	9112	9117	9122	9128	9133	1	1	2	2	3	3	4	4	R	
37	5682	5694	5705	5717	5729	5740	5752	5763	5775	5786	1	2	3	5	6	7	8	9	10		82	9138	9143	9149	9154	9159	9165	9170	9175	9180	9186	1	1	2	2	3	3	4	4	R	
38	5798	5809	5821	5832	5843	5855	5866	5877	5888	5899	1	2	3	5	6	7	8	9	10		83	9191	9196	9201	9206	9212	9217	9222	9227	9232	9238	1	1	2	2	3	3	4	4	R	
39	5911	5922	5933	5944	5955	5966	5977	5988	5999	6010	1	2	3	4	5	7	8	9	10		84	9243	9248	9253	9258	9263	9269	9274	9279	9284	9289	1	1	2	2	3	3	4	4	R	
40	6021	6031	6042	6053	6064	6075	6085	6096	6107	6117	1	2	3	4	5	6	8	9	10		85	9294	9299	9304	9309	9315	9320	9325	9330	9335	9340	1	1	2	2	3	3	4	4	fi	
41	6128	6138	6149	6160	6170	6180	6191	6201	6212	6222	1	2	3	4	5	6	7	8	9		86	9345	9350	9355	9360	9365	9370	9375	9380	9385	9390	1	1	2	2	3	3	4	4	R	
42	6232	6243	6253	6263	6274	6284	6294	6304	6314	6325	1	2	3	4	5	6	7	8	9		87	9395	9400	9405	9410	9415	9420	9425	9430	9435	9440	0	1	1	2	2	3	3	4	4	
43	6335	6345	6355	6365	6375	6385	6395	6405	6415	6425	1	2	3	4	5	6	7	8	9		88	9445	9450	9455	9460	9465	9469	9474	9479	9484	9489	0	1	1	2	2	3	3	4	4	
44	6435	6444	6454	6464	6474	6484	6493	6503	6513	6522	1	2	3	4	5	6	7	8	9		89	9494	9499	9504	9509	9513	9518	9523	9528	9533	9538	0	1	1	2	2	3	3	4	4	
45	6532	6542	6551	6561	6571	6580	6590	6599	6609	6618	1	2	3	4	5	6	7	8	9		90	9542	9547	9552	9557	9562	9566	9571	9576	9581	9586	0	1! 1		2	2	3	3	4	4	
46	6628	6637	6646	6656	6665	6675	6684	6693	6702	6712	1	2	3	4	5	6	7	7	8		91	9590	9595	9600	9605	9609	9614	9619	9624	9628	9633	0 1 1			2	2	3	3	4	4	
47	6721	6730	6739	6749	6758	6767	6776	6785	6794	6803	1	2	3	4	5	5	6	7	8		92	9638	9643	9647	9652	9657	9661	9666	9671	9675	9680	0] 1! 1			2	2	3	3	4	4	
48	6812	6821	6830	6839	6848	6857	6866	6875	6884	6893	1	2	3	4	4	5	6	7	8		93	9685	9689	9694	9699	9703	9708	9713	9717	9722	9727	0 11 1			2	2	3	3	4	4	
49	6902	6911	6920	6928	6937	6946	6955	6964	6972	6981	1	2	3	4	4	5	6	7	8		94	9731	9736	9741	9745	9750	9754	9759	9763	9768	9773	0	1 1		2	2	3	3	4	4	
50	6990	6998	7007	7016	7024	7033	7042	7050	7059	7067	1	2	3	3	4	5	6	7	8		95	9777	9782	9786	9791	9795	9800	9805	9809	9814	9818	0	1 1		2	2	3	3	4	4	
51	7076	7084	7093	7101	7110	7118	7126	7135	7143	7152	1	2	3	3	4	5	6	7	8		96	9823	9827	9832	9836	9841	9845	9850	9854	9859	9863	0 1 1			2	2	3	3	4	4	
52	7160	7168	7177	7185	7193	7202	7210	7218	7226	7235	1	2	2	3	4	5	6	7	7		97	9868	9872	9877	9881	9886	9890	9894	9899	9903 9908		0 1 1			2	2	8	3	4 4		
53	7243	7251	7259	7267	7275	7284	7292	7300	7308	7316	1	2	2	3	4	5	6	6	7		98	9912	9917	9921	9926	9930	9934	9939	9943 9918 9952			0 1 1			2	2	3	3	4	4	
54	7324	7332	7340	7348	7356	7364	7372	7380	73881	7S96	1	2	2	3	4	5	«	6	7		99	9956	9961	9965	9969	9974 9978		9983	9987 9991 9996			0! 1 1			2	2	S	3 3		4 --'