»¯Ñ§¾ÍÔÚÎÒÃÇÉí±ß£¬ÓëÎÒÃǵÄÉú»îÓÐ×ÅÃÜÇеÄÁªÏµ£®
£¨1£©ÏÖÓеªÆø¡¢ÑõÆø¡¢Éúʯ»Ò¡¢Êìʯ»Ò¡¢Ì¼ËáÇâÄƵÈÎïÖÊ£¬ÇëÄãÑ¡ÔñºÏÊʵÄÎïÖÊ°´ÒªÇóÌî¿Õ£º
¢Ùº¸½Ó½ðÊôʱ³£ÓÃ×ö±£»¤ÆøµÄÊÇ______£»ÂÁ²­ËÜÁÏҩƬ£¨Ö÷Òª³É·Ö£ºÌ¼Ëá¸Æ¡¢µí·Û£©
¢Ú³£ÓÃ×öʳƷ¸ÉÔï¼ÁµÄÊÇ______£¬Æä¸ÉÔïÔ­ÀíÊÇ______£¨Óû¯Ñ§·½³Ìʽ±íʾ£©£»
¢Û¿ÉÓÃÀ´ÖÎÁÆθËá¹ý¶àµÄÊÇ______£®
£¨2£©ÈçͼΪij¿¹ËáÒ©µÄʵÎïͼ£¬»Ø´ðÏÂÁÐÎÊÌ⣮
¢ÙͼÖбêʾµÄÎïÖÊÖУ¬ÊôÓÚÓлúºÏ³É²ÄÁϵÄÊÇ______£»
¢Úµí·ÛÊôÓÚ______£¨Ìî×ÖĸÐòºÅ£©£®
A£®µ°°×ÖÊ¡¡ B£®ÌÇÀà¡¡ C£®ÓÍÖ¬¡¡ D£®Î¬ÉúËØ
¢ÛͼÖÐËÜÁÏ°ü×°·ÏÆúºó¿ÉÓÃÓÚ΢ÐÍʵÑ飮ÆäÓŵãÊÇ______£®

½â£º£¨1£©¢ÙµªÆø»¯Ñ§ÐÔÖʲ»»îÆ㬺¸½Ó½ðÊôʱ³£ÓÃÓÚ±£»¤Æø£¬¹Ê´ð°¸Îª£ºµªÆø£»
¢ÚÉúʯ»ÒÄÜÓëË®·´Ó¦Éú³ÉÇâÑõ»¯¸Æ£¬³£ÓÃ×öʳƷ¸ÉÔï¼Á£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇCaO+H2O=Ca£¨OH£©2£»
¹Ê´ð°¸Îª£ºÉúʯ»Ò£»CaO+H2O=Ca£¨OH£©2£»
¢Û̼ËáÇâÄÆÄܺÍθËáÀïµÄÑÎËá·´Ó¦£¬¿ÉÓÃÀ´ÖÎÁÆθËá¹ý¶à£¬¹Ê´ð°¸Îª£ºÌ¼ËáÇâÄÆ£»
£¨2£©¢Ù¸ù¾Ýͼʾ²»ÄÑ¿´³ö£¬ÂÁ²­ÊǽðÊô²ÄÁÏ£¬ËÜÁÏÊÇÓлú¸ß·Ö×ӺϳɲÄÁÏ£¬¹Ê´ð°¸Îª£ºËÜÁÏ£»
¢Úµí·ÛÊÇÌÇÀ࣬¹Ê´ð°¸Îª£ºB£»
¢ÛËÜÁÏ°ü×°·ÏÆúºó¿ÉÓÃÓÚ΢ÐÍʵÑ飬ÆäÓŵãÊǽÚÔ¼Ò©Æ·£»ÏÖÏóÃ÷ÏÔ£¬±ãÓڶԱȵȣ¬¹Ê´ð°¸Îª£º½ÚÔ¼Ò©Æ·£»ÏÖÏóÃ÷ÏÔ£¬±ãÓڶԱȵȣ®
·ÖÎö£º£¨1£©ÎïÖʵÄÐÔÖʾö¶¨ÎïÖʵÄÓÃ;£¬¸ù¾Ý³£¼ûÎïÖʵÄÐÔÖÊÓëÓÃ;½øÐзÖÎö½â´ð¼´¿É£»
£¨2£©¸ù¾Ý²ÄÁϵķÖÀà·ÖÎö£»¸ù¾ÝÈËÌåÓªÑøËØ·ÖÎö£®
µãÆÀ£º±¾ÌâÄѶȲ»ÊǺܴó£¬Ö÷Òª¿¼²éÁËÎïÖʵÄÐÔÖÊÓëÓÃ;¡¢²ÄÁϵķÖÀà¡¢ÓªÑøËصȻ¯Ñ§ÖªÊ¶£¬´Ó¶øÅàÑøѧÉú·ÖÎöÎÊÌâ¡¢½â¾öÎÊÌâµÄÄÜÁ¦£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

22¡¢»¯Ñ§¾ÍÔÚÎÒÃÇÉí±ß£¬ÓëÎÒÃǵÄÉú»îϢϢÏà¹Ø£®Çë»Ø´ðÒÔÏÂÉú»îÖеÄÎÊÌ⣺
£¨1£©¡°É³ÀïÌÔ½ð¡±ËµÃ÷»Æ½ðÔÚ×ÔÈ»½çÖÐÄܹ»ÒÔ
µ¥ÖÊ
£¨Ìî¡°µ¥ÖÊ¡±»ò¡°»¯ºÏÎ£©ÐÎʽ´æÔÚ£®
£¨2£©ÆÏÌѱíƤÉÏÒòÅçÈ÷¡°²¨¶û¶àÒº¡±¶ø³ÊÏÖµÄÀ¶É«Ä§µã£¬¿ÉÒÔÓóø·¿µ÷ÁÏ
ʳ´×
Ï´¾»£®
£¨3£©Áõ´óÒ¯ÖÖµÄСÂó³öÏÖµ¹·üÏÖÏó£¬Äã»á½¨ÒéËûʹÓû¯·ÊÖеÄ
¼Ø
·Ê£®
£¨4£©³¤ÆÚʹÓõÄůˮƿÄÚµ¨³£ÓÐÒ»²ãË®¹¸£¨Ö÷Òª³É·ÖΪ̼Ëá¸Æ£©£¬¿ÉÒÔÓÃÏ¡ÑÎËá³ýÈ¥£®Çëд³ö·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
CaCO3+2HCl¨TCaCl2+CO2¡ü+H2O
£®
£¨5£©Í¨³£Ëù˵µÄúÆøÖж¾ÊÇÖ¸ÓÉ
CO
£¨ÌîÎïÖÊ»¯Ñ§Ê½£©ÒýÆðµÄÖж¾£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

16¡¢»¯Ñ§¾ÍÔÚÎÒÃÇÉí±ß£¬ÓëÎÒÃǵÄÉú»îϢϢÏà¹Ø£®Çë»Ø´ðÒÔÏÂÉú»îÖеÄÎÊÌ⣺
£¨1£©¼ÒÓÃÌìÈ»Æø£¨»òÒº»¯Æø£©Öг£Ìí¼ÓÒ»ÖÖ¾ßÓÐÌØÊâÆøζµÄÎïÖÊÒÒÁò´¼£®µ±Îŵ½¸ÃÆøζʱ£¬ÌáÐÑÈË×¢ÒâÌìÈ»Æø£¨»òÒº»¯Æø£©ÒѾ­ÔÚй¶£®ÄÜÎŵ½ÒÒÁò´¼µÄÆø棬Õâ֤ʵÁË·Ö×Ó
£¨²»¶Ï£© Ô˶¯
µÄÐÔÖÊ£®
£¨2£©¡°É³ÀïÌÔ½ð¡±ËµÃ÷»Æ½ðÔÚ×ÔÈ»½çÖÐÄܹ»ÒÔ
µ¥ÖÊ
£¨Ìî¡°µ¥ÖÊ¡±»ò¡°»¯ºÏÎ£©ÐÎʽ´æÔÚ£®
£¨3£©Í¨³£Ëù˵µÄúÆøÖж¾ÊÇÖ¸ÓÉ
CO
£¨ÌîÎïÖÊ»¯Ñ§Ê½£©ÒýÆðµÄÖж¾£®
£¨4£©³¤ÆÚʹÓõÄůˮƿÄÚµ¨³£ÓÐÒ»²ãË®¹¸£¨Ö÷Òª³É·ÖΪCaCO3¡¢Mg£¨OH£©2£©£¬¿ÉÒÔÓóø·¿ÖеÄ
£¨Ê³£©´×
µ÷ÁϳýÈ¥£®
£¨5£©ÎóʳÖؽðÊôÑΣ¨È磺CuSO4¡¢BaCl2µÈ£©¿Éͨ¹ýºÈ´óÁ¿¶¹½¬µÄ·½·¨½â¶¾£®ÄãÈÏΪÒûÓÃÏÂÁÐ
A¡¢C
ÎïÖÊÒ²¿ÉÒԴﵽͬÑùµÄÄ¿µÄ£¨Ìî×Öĸ£©£®
A¡¢Å£ÄÌ      B¡¢¹ûÖ­       C¡¢µ°Çå       D¡¢¿óȪˮ      E¡¢Ê³ÑÎË®£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

»¯Ñ§¾ÍÔÚÎÒÃÇÉí±ß£¬ÓëÎÒÃǵÄÉú»îϢϢÏà¹Ø£®Çë»Ø´ðÏÂÁÐÉú»îÖеĻ¯Ñ§ÎÊÌ⣮
£¨1£©¡°µÍ̼Éú»î¡±·½Ê½Öеġ°Ì¼¡±Ö¸µÄÊÇ̼
ÔªËØ
ÔªËØ
 £¨Ìî¡°µ¥ÖÊ¡±¡°Ô­×Ó¡±¡°Àë×Ó¡±»ò¡°ÔªËØ¡±£©£»ÈÕ³£Éú»îÖÐÄãµÄ×ö·¨ÊÇ
ËæÊֹصƣ¬²½ÐÐÌæ´ú³Ë³µµÈ
ËæÊֹصƣ¬²½ÐÐÌæ´ú³Ë³µµÈ
£®£¨¾ÙÒ»Àý£©
£¨2£©ÔÚ¼ÒÖÐÅݲèʱ£¬ÒªÏëÖªµÀËùÓÃÊÇӲˮ»¹ÊÇÈíË®µÄ·½·¨ÊÇ
¼Ó·ÊÔíË®
¼Ó·ÊÔíË®
£¬²è±­ÄÚµÄÉ´Íø¿É½«²èÒ¶Óë²èË®·ÖÀë±ãÓÚÒûÓ㬸ÃÉè¼ÆÔ­ÀíÀàËÆ»¯Ñ§ÊµÑé²Ù×÷ÖеÄ
¹ýÂË
¹ýÂË
£®
£¨3£©ÎïÖʵĽṹ¾ö¶¨ÎïÖʵÄÐÔÖʺÍÓÃ;£®ÏÖÓÐÒÔϳ£¼ûµÄÎïÖÊ£º
A£®Ã÷·¯   B£®»îÐÔÌ¿   C£®ÌìÈ»Æø    D£®C60   E£®Ï¡ÓÐÆøÌå   F£®¸É±ù
Çë°´ÒªÇóÌî¿Õ£¨ÌîÉÏÊöÏàÓ¦ÎïÖʵÄ×Öĸ£¬²»Öظ´Ê¹Óã©£º
¢ÙÄ¿Ç°ÎÒÊÐÕý¹ã·ºÊ¹ÓõÄÇå½àȼÁÏÊÇ
C
C
£» ¢Ú³£ÓÃÓÚ±ùÏä³ý³ôµÄÊÇ
B
B
£»
¢Û¿É×÷µÆÅÝÌî³äÆøµÄÊÇ
E
E
£»          ¢ÜÓÃÓÚÀä²Ø±£´æʳƷµÄÊÇ
F
F
£»
¢Ý¿ÉÓÃÓÚÏç´å»òÒ°Í⾻ˮµÄÊÇ
A
A
£»     ¢ÞÓ¦ÓÃÓÚ³¬µ¼ÌåµÄÊÇ
D
D
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

»¯Ñ§¾ÍÔÚÎÒÃÇÉí±ß£¬ÓëÎÒÃǵÄÉú»îϢϢÏà¹Ø£®Çë»Ø´ðÏÂÁÐÉú»îÖеĻ¯Ñ§ÎÊÌ⣮
£¨1£©¡°µÍ̼Éú»î¡±·½Ê½Öеġ±¡°Ì¼¡±Ö¸µÄÊÇ̼
ÔªËØ
ÔªËØ
£¨Ìî¡°µ¥ÖÊ¡±¡°Ô­×Ó¡±¡°Àë×Ó¡±»ò¡°ÔªËØ¡±£©£»ÈÕ³£Éú»îÖÐÄãµÄ×ö·¨ÊÇ
ËæÊֹصƣ¬²½ÐÐÌæ´ú³Ë³µµÈ£¨ºÏÀí´ð°¸¾ù¿É£©
ËæÊֹصƣ¬²½ÐÐÌæ´ú³Ë³µµÈ£¨ºÏÀí´ð°¸¾ù¿É£©
£®£¨¾ÙÒ»Àý£©
£¨2£©ÔÚ¼ÒÖÐÅݲèʱ£¬ÒªÏëÖªµÀËùÓÃÊÇӲˮ»¹ÊÇÈíË®µÄ·½·¨ÊÇ
ÏòË®ÖмÓÈë·ÊÔíË®£¬ÅÝÄ­¶àµÄÊÇÈíË®£¬¸¡Ôü½Ï¶àµÄÊÇӲˮ
ÏòË®ÖмÓÈë·ÊÔíË®£¬ÅÝÄ­¶àµÄÊÇÈíË®£¬¸¡Ôü½Ï¶àµÄÊÇӲˮ
£¬²è±­ÄÚµÄÉ´Íø¿É½«²èÒ¶Óë²èË®·ÖÀë±ãÓÚÒûÓ㬸ÃÉè¼ÆÔ­ÀíÀàËÆ»¯Ñ§ÊµÑé²Ù×÷ÖеÄ
¹ýÂË
¹ýÂË
£»
£¨3£©ÎïÖʵĽṹ¾ö¶¨ÎïÖʵÄÐÔÖʺÍÓÃ;£®ÔÚ¼Òͥú¯ȼÉÕúʱ×ö°Ñ³É·äÎÑú£¬ÕâÑù×ö±ãÓÚȼÉÕµÄÔ­ÀíÊÇ
Ôö´óÁËúÓëÑõÆøµÄ½Ó´¥Ãæ»ý
Ôö´óÁËúÓëÑõÆøµÄ½Ó´¥Ãæ»ý
£¬ÓùýµÄ·ÏÆúÎïúÇò·Åµ½ÎÛË®³ØÖУ¬·¢ÏÖÓÐÈ¥³ýÒìζºÍÉ«ËصÄЧ¹û£®·ÏÆúÎïúÇòµÄ×÷ÓÃÓ뾻ˮÆ÷ÖеÄ
»îÐÔÌ¿µÄÎü¸½
»îÐÔÌ¿µÄÎü¸½
ÏàËÆ£¨ÌîÓйØÎïÖʼ°Ïà¹ØÐÔÖÊ£©£¬ÎªÁ˼õÇá¶Ô´óÆøµÄÎÛȾ£¬ºÃ¶à¼ÒÍ¥ÓÃÉÏÁ˸üΪÇå½àµÄÐÂÐÍȼÁÏ--ÌìÈ»Æø£¬ÊÔд³öÌìÈ»ÆøÖ÷Òª³É·ÖȼÉյĻ¯Ñ§·½³Ìʽ
CH4+2O2
 µãȼ 
.
 
2H2O+CO2
CH4+2O2
 µãȼ 
.
 
2H2O+CO2
£®
£¨4£©Ò»ÊÜÍâÉË»¼Õßµ½Ò½Ôº£¬Ò½Éú³£»á¸ø»¼ÕßÊ×ÏÈÓõâ¾ÆÈÜÒºÏû¶¾£¬ËùÓõâ¾ÆÈÜÒºµÄÈܼÁÊÇ
C2H5OH
C2H5OH
£¨Ìîд»¯Ñ§Ê½£©£¬2013ÄêÆðÈ«¹úʵʩеĽ»Í¨·¨¹æ£¬¼Ó´óÁ˾ƺó¼ÝÊ»µÄ¼ì²é¡¢´¦ÀíÁ¦¶È£¬½»¾¯¼ì²éÊÇ·ñ¾Æºó¼ÝÊ»³£Óõ½Ò»ÖÖ»¯Ñ§ÎïÖÊ--ÖظõËá¼Ø£¨K2Cr2O7£©£¬ÖظõËá¼ØÖиõÔªËصĻ¯ºÏ¼ÛΪ
+6
+6
¼Û£®
£¨5£©Ê³ÑÎÊdzø·¿Öв»¿ÉȱÉٵĵ÷ζƷ£¬ÆäÖ÷Òª³É·ÖÊÇÂÈ»¯ÄÆ£¬¹¹³ÉÂÈ»¯ÄƵÄÑôÀë×Ó·ûºÅÊÇ
Na+
Na+
£¬ÔÚÇåÏ´¼ÒÖÐÓÍÎÛʱ¿ÉÓÃÆûÓÍÏ´£¬Ò²¿ÉÓÃÏ´µÓ¼ÁÏ´£¬¶þÕߵIJ»Í¬Ö®´¦ÊÇ
Ç°ÕßΪÈܽ⣬ºóÕßΪÈ黯
Ç°ÕßΪÈܽ⣬ºóÕßΪÈ黯
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

»¯Ñ§¾ÍÔÚÎÒÃÇÉí±ß£¬ÓëÎÒÃǵÄÉú»îÓÐ×ÅÃÜÇÐÏà¹Ø£¬ÇëÄãÑ¡ÔñÊʵ±ÎïÖʵÄÐòºÅ°´ÒªÇóÌî¿Õ£º
£¨1£©º¸½Ó½ðÊôʱ³£ÓÃÓÚ±£»¤ÆøµÄÊÇ
µªÆø
µªÆø
£®
£¨2£©³£ÓÃÓÚ¸ÄÁ¼ËáÐÔÍÁÈÀµÄ¼îÊÇ
ÇâÑõ»¯¸Æ
ÇâÑõ»¯¸Æ
£®
£¨3£©ÎªÊ¹ÂøÍ·¡¢Ãæ°üµÈ·¢ÃæʳƷÊèËɶà¿×¡¢ËÉÈí¿É¿Ú£¬³£¼ÓÈëµÄÎïÖÊÊÇ
̼ËáÇâÄÆ
̼ËáÇâÄÆ
£®
£¨4£©ÖÎÁÆ»¼ÓÐθ´©¿×²¡È˵ÄθËá¹ý¶àµÄÊÇ£º
ÇâÑõ»¯ÂÁ
ÇâÑõ»¯ÂÁ
£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸