Elementary Chemical Arithmetic: With 1100 ProblemsMacmillan and Company, 1882 - 286 páginas |
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Página 6
... oxide , how much oxide will be formed by 80 lbs . of mercury ? Since 100 lbs . of mercury form 108 lbs . of mercuric oxide , 108 1 lb. forms " " " " 100 " 9 " 9 108 and 80 lbs . form × 80 = 86.4 lbs . of 99 99 100 mercuric oxide . * In ...
... oxide , how much oxide will be formed by 80 lbs . of mercury ? Since 100 lbs . of mercury form 108 lbs . of mercuric oxide , 108 1 lb. forms " " " " 100 " 9 " 9 108 and 80 lbs . form × 80 = 86.4 lbs . of 99 99 100 mercuric oxide . * In ...
Página 7
... oxide , how much oxide will be formed by 80 lbs . of mercury ? X = 100 80 108 : ? Ans . ( or x ) . 108 x 80 = 86.4 lbs . mercuric oxide . or Since 100 a α с Ъ = d 2 + 1 = 21 , ± 1 , a ± b c + d d Thus , if abb :: cd : d . a : 5 : 12 : 4 ...
... oxide , how much oxide will be formed by 80 lbs . of mercury ? X = 100 80 108 : ? Ans . ( or x ) . 108 x 80 = 86.4 lbs . mercuric oxide . or Since 100 a α с Ъ = d 2 + 1 = 21 , ± 1 , a ± b c + d d Thus , if abb :: cd : d . a : 5 : 12 : 4 ...
Página 27
... oxide CO , 100 Carbon C = 12 × = 27.27 . 44 100 Oxygen 0,32 × = 72.73 44 44 100.00 For since 44 parts of carbon ... oxide . And if it be required to find how many grams of magnesium oxide are formed on burning 10 grams of magnesium , the ...
... oxide CO , 100 Carbon C = 12 × = 27.27 . 44 100 Oxygen 0,32 × = 72.73 44 44 100.00 For since 44 parts of carbon ... oxide . And if it be required to find how many grams of magnesium oxide are formed on burning 10 grams of magnesium , the ...
Página 28
... oxide , 40 1 gram forms " 9 29 24 99 99 " " 40 × 10 & 10 grams ,, form = 16.6 " 9 24 99 " " The same result may be reached with still greater rapidity by the use of simple proportion . Find the multiples of the atomic or molecular ...
... oxide , 40 1 gram forms " 9 29 24 99 99 " " 40 × 10 & 10 grams ,, form = 16.6 " 9 24 99 " " The same result may be reached with still greater rapidity by the use of simple proportion . Find the multiples of the atomic or molecular ...
Página 38
... density referred to air from that referred to hydrogen , multiply the density referred to hydrogen by 06926 , the density of hydrogen referred to air . Thus nitric oxide is 15 times as heavy as hydrogen 38 INTRODUCTION . Oxygen VI Hydrogen.
... density referred to air from that referred to hydrogen , multiply the density referred to hydrogen by 06926 , the density of hydrogen referred to air . Thus nitric oxide is 15 times as heavy as hydrogen 38 INTRODUCTION . Oxygen VI Hydrogen.
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Términos y frases comunes
20 grams 2H₂O 2HCl acid alloy aluminium ammonia antimony arsenic atomic weight barium bromine burnt c.c. of gas c.c. of hydrogen carbon dioxide cent CHAPTER CO₂ College copper Crown 8vo cubic cupric oxide Edition evolved fcap ferrous Find its formula Find the atomic Find the density Find the formula Find the mass Find the percentage formed grams grams of hydrogen grams of lead grams of phosphorus grams of potassium grams of silver grams of water H₂ H₂O H₂SO H₂SO₁ heavy as air Hence hydrate hydrogen chloride hydrogen sulphate iron kilograms kilos litres litres of hydrogen logarithm m.m. Find m.m. is given magnesium manganese dioxide mercury metres molecular weight multiplied nitric oxide obtained oxygen percentage composition potassium chlorate potassium dichromate pressure salt silicon sodium solution specific heat substance contains sulphide sulphur dioxide temperature vapour volume of carbon volume of chlorine volume of hydrogen volume of oxygen zinc
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