This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1883 Excerpt: ...11-16 litres of chlorine can be obtained from 86-5 grams of hydrochloric acid, how many litres of chlorine can be obtained from 50 grams of that acid? grams grams litres 86-5: 50:: 11-16: x 50 x 11-16 = 15-287 litres of chlorine. To find the volume resulting from the combination of gaseous bodies. Example 1.--If a ...
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This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1883 Excerpt: ...11-16 litres of chlorine can be obtained from 86-5 grams of hydrochloric acid, how many litres of chlorine can be obtained from 50 grams of that acid? grams grams litres 86-5: 50:: 11-16: x 50 x 11-16 = 15-287 litres of chlorine. To find the volume resulting from the combination of gaseous bodies. Example 1.--If a mixture of 20 c.c. of hydrogen and 10 c.c. of oxygen at 140C. be exploded, what volume of steam will be formed? Answer. 20 c.c of steam. Example 2.--If 100 c.c. of hydrogen are mixed with 150 c.c of chlorine, and exploded, what volume of hydrochloric acid is produced? Which gas, and how much of it, remains uncombined? By the equation--H + CI = HC1 hydrogen chlorine hydrochloric acid. and 1 vol. + 1 vol. = 2 vols, therefore 100 c.c. + 100 c.c. = 200 c.c. From this it will be seen that 100 c.c. of the chlorine unite with all the hydrogen forming 200 c.c. of hydrochloric acid gas; the remaining 50 c.c. of chlorine are left uncombined. Answer. 200 o.o. of HC1; 50 c.c. CI. Example 8.--One volume of carbon monoxide (carbonic oxide) gas is mixed with two volumes of oxygen, and an electric spark passed through the mixture. What volume of carbon dioxide gas (carbonic anhydride) is formed, and what volume of oxygen left? (Science and Art Department Examination. 1877.) By the equation--CO + 0 = CO2 carbon monoxide oxygen carbon dioxide and 2 vols. + 1 vol. = 2 vols. therefore 1 vol. + % vol. = 1 vol. (dividing by 2) From this it will be seen that 1 volume of carbon monoxide requires only-J volume of oxygen to unite with it to form carbon dioxide; the additional l volumes of oxygen remain behind uncombined. Answer: 1 volume of carbon dioxide; 1J volumes of oxygen left. It should be remembered that a molecule is always written 2 volumes, whatever number ...
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