13£®¼¦µ°¿ÇµÄÖ÷Òª³É·ÖÊÇ̼Ëá¸Æ£¬ÎªÁ˲ⶨij¼¦µ°¿ÇÖÐ̼Ëá¸ÆµÄº¬Á¿£¬Ð¡ÈºÍ¬Ñ§½øÐÐÁËÈçÏÂʵÑ飻½«¼¦µ°¿ÇÏ´¾»£¬¸ÉÔï²¢µ·Ëéºó£¬³ÆÈ¡5g·ÅÔÚÉÕ±­ÖУ¬È»ºóÍùÉÕ±­ÖмÓÈë×ãÁ¿µÄÏ¡ÑÎËá45g£®³ä·Ö·´Ó¦ºó£¬³ÆµÃ±­ÖеÄÎïÖʵÄ×ÜÖÊÁ¿Îª48.9g£¨¼ÙÉ輦µ°¿ÇÖеÄÆäËûÎïÖʲ»ÓëÑÎËá·´Ó¦£©
£¨1£©²úÉú¶þÑõ»¯Ì¼ÆøÌå1.1g
£¨2£©¼ÆËã5g¸Ã¼¦µ°¿ÇÖÐ̼Ëá¸ÆµÄÖÊÁ¿£¬ÒªÇóд³öÍêÕûµÄ½âÌâ²½Ö裮

·ÖÎö ¼¦µ°¿ÇµÄÖ÷Òª³É·ÖÊÇ̼Ëá¸Æ£¬Ì¼Ëá¸ÆÓëÏ¡ÑÎËá·´Ó¦Éú³ÉÂÈ»¯¸Æ¡¢Ë®ºÍ¶þÑõ»¯Ì¼£¬ÓÉÖÊÁ¿Êغ㶨ÂÉ£¬ÉÕ±­ÖÐÎïÖʼõÉÙµÄÖÊÁ¿¼´ÎªÉú³É¶þÑõ»¯Ì¼µÄÖÊÁ¿£¬ÓÉ·´Ó¦µÄ»¯Ñ§·½³ÌʽÁÐʽ¼ÆËã³ö²Î¼Ó·´Ó¦µÄ̼Ëá¸ÆµÄÖÊÁ¿£¬½øÐзÖÎö½â´ð£®

½â´ð ½â£º£¨1£©ÓÉÖÊÁ¿Êغ㶨ÂÉ£¬Éú³É¶þÑõ»¯Ì¼µÄÖÊÁ¿Îª5g+45g-48.9g=1.1g£®
£¨2£©Éè²Î¼Ó·´Ó¦µÄ̼Ëá¸ÆµÄÖÊÁ¿Îªx
CaCO3+2HCl=CaCl2+H2O+CO2¡ü
100                  44
x                    1.1g
$\frac{100}{44}=\frac{x}{1.1g}$     x=2.5g
´ð£º£¨1£©1.1£»£¨2£©5g¸Ã¼¦µ°¿ÇÖÐ̼Ëá¸ÆµÄÖÊÁ¿Îª2.5g£®

µãÆÀ ±¾ÌâÄѶȲ»´ó£¬ÕÆÎÕ¸ù¾Ý»¯Ñ§·½³ÌʽµÄ¼ÆËã¼´¿ÉÕýÈ·½â´ð±¾Ì⣬½âÌâʱҪעÒâ½âÌâµÄ¹æ·¶ÐÔ£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

3£®ÊµÑéÊÒÀïÓÐÎåÖÖÎïÖÊ£º¢ÙO2  ¢ÚH2 ¢ÛCO  ¢ÜCO2  ¢ÝCH4  Çë°´ÏÂÁÐÒªÇó½«ÎåÖÖÎïÖʵÄÐòºÅÌîÈëÏàÓ¦µÄ¿Õ¸ñÖУ¬²¢Ð´³öÓйط´Ó¦µÄ»¯Ñ§·½³Ìʽ£®
£¨1£©ÄÜʹ´ø»ðÐǵÄľÌõ¸´È¼µÄÊÇ¢Ù£®
£¨2£©¡°Î÷Æø¶«Ê䡱ÆøÌåµÄÖ÷Òª³É·ÖÊǢݣ¬ÆäȼÉյĻ¯Ñ§·½³ÌʽΪCH4+2O2$\frac{\underline{\;µãȼ\;}}{\;}$CO2+2H2O
£¨3£©ÄÜʹ³ÎÇåµÄʯ»ÒË®±ä»ë×ǵÄÆøÌåÊǢܣ¬Æä·´Ó¦µÄ»¯Ñ§·½³ÌʽΪCO2+Ca£¨OH£©2¨TCaCO3¡ý+H2O£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

4£®ÕýÈ·µÄ»¯Ñ§·½³ÌʽÊÇ£¨¡¡¡¡£©
A£®2KClO3$\frac{\underline{\;MnO_2\;}}{\;}$2KCl+3O2¡üB£®3CO+Fe2O3$\frac{\underline{\;¸ßÎÂ\;}}{\;}$2Fe+3CO2
C£®4Fe+3O2$\frac{\underline{\;µãȼ\;}}{\;}$2Fe2O3D£®Mg+O2$\frac{\underline{\;µãȼ\;}}{\;}$MgO2

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

1£®Ä¿Ç°£¬ÈËÃÇʹÓõÄȼÁÏ£¬´ó¶àÊýÀ´×ÔʯȼÁÏ£®
£¨1£©»¯Ê¯È¼ÁÏÊDz»¿ÉÔÙÉúÄÜÔ´£¬ÆäÖгÊÕ³³í×´µÄÒºÌåÊÇʯÓÍ£®ÔÚ»¯Ê¯È¼ÁÏÖУ¬ÌìÈ»ÆøÊDZȽÏÇå½àµÄȼÁÏ£®º£µ×Âñ²Ø×Å´óÁ¿¿ÉȼÉÕµÄÎïÖʳÆΪ¿Éȼ±ù£®½«³ÉΪδÀ´ÐÂÄÜÔ´£®
£¨2£©»¯Ê¯È¼ÁÏȼÉÕ£¬ÓÐʱ»á²úÉúºÚÑÌ£¬Ô­ÒòÊÇÑõÆø²»×㣬ȼÁÏÖÐ̼²»ÄÜÍêȫȼÉÕ£®¼ÈÀË·ÑÔ­ÁÏÓÖÎÛȾ¿ÕÆø£¬´ËʱºÚÑÌÖаéÓÐÒ»ÖÖÄÜÓëÈËÌåѪҺÖÐѪºìµ°°×½áºÏµÄÓж¾ÆøÌåÊÇÒ»Ñõ»¯Ì¼£®
£¨3£©´Ó20161ÔÂ1ÈÕÆ𣬴óÁ¬Êи÷¼ÓÓÍÕ¾¿ªÊ¼³öÊÛ¡°¹úÎ塱±ê×¼³µÓÃÆû²ñÓÍ£®ÒÔÌæ´úÄ¿Ç°ÔÚÓõġ°¹úËÄ'±ê×¼µÄ³µÓÃÆû²ñÓÍ£¬ÆäÖ÷ҪĿµÄÊÇÌá¸ßÆûÓÍÆ·ÖÊ£¬¼õÉÙ»·¾³ÎÛȾ£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

8£®ÏÂÁÐA¡¢BÁ½Í¼·Ö±ð±íʾÁò·Û¡¢ÌúË¿ÔÚÑõÆøÖÐȼÉÕµÄʾÒâͼ£®
£¨1£©Ìú˿ȼÉÕµÄÏÖÏóÊÇ£¨Ð´³öÁ½µã¼´¿É£©»ðÐÇËÄÉä¡¢Éú³ÉºÚÉ«¹ÌÌ壮
£¨2£©ÊµÑéÇ°£¬A¡¢BÁ½Æ¿Öоù×°ÓÐÉÙÁ¿Ë®£¬AÖÐË®µÄ×÷ÓõÄÊÇÎüÊÕ¶þÑõ»¯Áò£¬·ÀÖ¹ÎÛȾ¿ÕÆø£¬BÖÐË®µÄ×÷ÓÃÊÇ·ÀÖ¹¸ßÎÂÉú³ÉÎヲÂäÆ¿µ×£¬Ê¹Æ¿µ×Õ¨ÁÑ£®
£¨3£©Áò¡¢ÌúȼÉյĻ¯Ñ§·½³Ìʽ·Ö±ðΪS+O2$\frac{\underline{\;µãȼ\;}}{\;}$SO2¡¢3Fe+2O2$\frac{\underline{\;µãȼ\;}}{\;}$Fe3O4£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

18£®¸ÆÔªËصÄÔ­×ӽṹʾÒâͼºÍÔÚÔªËØÖÜÆÚ±íÖÐÏÔʾµÄÐÅÏ¢ÈçͼËùʾ£¬Çë»Ø´ðÓйØÎÊÌ⣺
£¨1£©¸ÆÔªËصÄÔ­×ÓÐòÊýΪ20£¬¸ÆÔªËصÄÏà¶ÔÔ­×ÓÖÊÁ¿ÊÇ40.08£®
£¨2£©¸ÆÔ­×ÓºËÍâµÄµç×Ó²ãÊýΪ4£¬¸ÆÔ­×ÓÔÚ»¯Ñ§·´Ó¦ÖÐÒ×ʧȥ£¨Ìî¡°µÃµ½¡±»ò¡°Ê§È¥¡±£©µç×Ó£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

5£®¼ÓÈÈÂÈËá¼ØºÍ¶þÑõ»¯Ã̵ĻìºÏÎï14.8gÖÁÍêÈ«·´Ó¦ºó£¬ÀäÈ´£¬³ÆÖØ£¬Ê£Óà¹ÌÌåÎïÖʵÄÖÊÁ¿Îª10.0g£¬Çó£º
£¨1£©Éú³ÉÑõÆøµÄÖÊÁ¿Îª4.8g£»
£¨2£©»ìºÏÎïÖжþÑõ»¯Ã̵ÄÖÊÁ¿·ÖÊý£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

7£®Ë®ÊÇÉúÃüÖ®Ô´£¬°®»¤Ë®×ÊÔ´ÊÇÿ¸ö¹«ÃñÓ¦¾¡µÄÔðÈΣ®Çë»Ø´ðÏÂÁÐÓйØÎÊÌ⣺
£¨1£©ÏÂÁÐÊÇһЩ¾»Ë®µÄ·½·¨£¬ÆäÖпÉʹӲˮ±ä³ÉÈíË®µÄÊÇ£º¢Ú¢Ü£¨ÌîÐòºÅ£©£®
¢Ù¹ýÂË      ¢ÚÕôÁó      ¢Û³Áµí       ¢ÜÖó·Ð
£¨2£©°´Í¼ËùʾװÖýøÐÐʵÑ飬һ¶Îʱ¼äºó£¬ÀíÂÛÉÏAÊÔ¹ÜÖвúÉúµÄÆøÌåÓëBÊÔ¹ÜÖвúÉúµÄÆøÌåµÄÌå»ý±ÈΪ1£º2£¬¸ÃʵÑé¹ý³ÌÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ2H2O$\frac{\underline{\;ͨµç\;}}{\;}$2H2¡ü+O2¡ü£®
£¨3£©´Ó×é³É½Ç¶È¿´£ºË®ÊÇÓÉÇâÔªËغÍÑõÔªËØ×é³ÉµÄ£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

8£®ÏÈÁÐʽ¡¢ºó¼ÆËãÏÂÁÐÎïÖʵÄÏà¶Ô·Ö×ÓÖÊÁ¿£º
CO2                           CH4                                  H2O£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸