Óк˵çºÉÊýСÓÚ18µÄA B C DËÄÖÖÔªËØ£¬AÔªËصÄÔ­×Ó¹¹³ÉµÄµ¥ÖÊ¿ÉÒÔºÍð¤ÍÁÖÆǦ±Ê£¬BÔªËصĵ¥ÖÊÔÚ¿ÕÆøµ±ÖÐÌå»ýÕ¼21%£¬CÔ­×ÓÓÐ3¸öµç×Ӳ㣬×îÍâ²ãÓë×îÄÚ²ãµç×ÓÊýÏàͬ£¬BÔ­×Ó±ÈDÔ­×ÓÉÙÒ»¸öµç×Ӳ㣬Á½Õß×îÍâ²ãµç×ÓÊýÏàͬ£®
£¨1£©Ð´³öËÄÖÖÔªËصÄÔªËØ·ûºÅ  A£º
 
B£º
 
C£º
 
D£º
 

£¨2£©Ð´³öA¡¢BÐγɻ¯ºÏÎïµÄ»¯Ñ§Ê½
 

£¨3£©Dµ¥ÖÊÔÚBÖÐȼÉյķ½³Ìʽ
 
£¬ÆäÉú³ÉÎï¶Ô»·¾³µÄÓ°ÏìÊÇ
 

£¨4£©A¡¢C¡¢D·Ö±ðÓëBÐγɵĻ¯ºÏÎïͳ³Æ
 

£¨5£©B¡¢C ºÍ DÈýÖÖÔªËØÐγɵĻ¯ºÏÎïµÄ»¯Ñ§Ê½
 
·ÖÎö£º½âÌâʱÍùÍùÐèÒª´ÓÌâÄ¿ÖÐÍÚ³öһЩÃ÷ÏÔ»òÒþº¬µÄÌõ¼þ£¬×¥×¡Í»ÆÆ¿Ú£¨Í»ÆÆ¿ÚÍùÍùÊÇÏÖÏóÌØÕ÷¡¢·´Ó¦ÌØÕ÷¼°½á¹¹ÌØÕ÷£©£¬µ¼³ö½áÂÛ£¬×îºó±ðÍüÁ˰ѽáÂÛ´úÈëÔ­ÌâÖÐÑéÖ¤£¬Èô¡°Â·¡±×ßµÃͨÔòÒѾ­³É¹¦£®
£¨1£©¢ÙÓÉ¡°AÔªËصÄÔ­×Ó¹¹³ÉµÄµ¥ÖÊ¿ÉÒÔºÍð¤ÍÁÖÆǦ±Ê¡±¿ÉÍƲâ³öAÔªËØ£»¢Ú¡°BÔªËصĵ¥ÖÊÔÚ¿ÕÆøµ±ÖÐÌå»ýÕ¼21%£¬¡±¿ÉÍƲâ³öBÔªËØ£»¢Û¡°CÔ­×ÓÓÐ3¸öµç×Ӳ㣬×îÍâ²ãÓë×îÄÚ²ãµç×ÓÊýÏàͬ¡±¿ÉÍƲâ³öCÔªËØ£»¢Ü¡°BÔ­×Ó±ÈDÔ­×ÓÉÙÒ»¸öµç×Ӳ㣬Á½Õß×îÍâ²ãµç×ÓÊýÏàͬ¡±¿ÉÍƲâ³öDÔªËØ£®
£¨2£©ÍƲâ³öA¡¢BÔªËØ£¬¿Éд³öÐγɻ¯ºÏÎïµÄ»¯Ñ§Ê½£»
£¨3£©ÍƲâ³öBÓëDÔªËØ£¬¿Éд³öȼÉյķ½³Ìʽ£¬²¢ÇÒ¸ù¾ÝÉú³ÉÎïµÄÐÔÖÊ×÷´ð£»
£¨4£©ÏÈÍƲâ³öA¡¢C¡¢DÓëBÔªËØ£¬ÔÙÊéдA¡¢C¡¢D·Ö±ðÓëBÐγɵĻ¯ºÏÎÔÙ×ܽá¹æÂÉ£»
£¨5£©ÏÈÍƲâ³öC¡¢DÓëBÔªËØ£¬ÔÙÊéдC¡¢DÓëBÐγɵĻ¯ºÏÎïµÄ»¯Ñ§Ê½£®
½â´ð£º½â£º£¨1£©ÓÉÌâÒ⣬ÍƲâ³öA¡¢B¡¢C¡¢DËÄÖÖÔªËصÄÔªËØ·ûºÅ·Ö±ðΪC¡¢O¡¢Mg¡¢S£»
£¨2£©ÒòΪA¡¢BÔªËØΪC¡¢O£¬ËùÒÔÐγÉÒ»Ñõ»¯Ì¼ºÍ¶þÑõ»¯Ì¼£¬¹Ê»¯Ñ§Ê½ÎªCO¡¢CO2£»
£¨3£©ÒòΪBÓëDÔªËØΪO¡¢S£¬ËùÒÔÁòÔÚÑõÆøÖÐȼÉյĻ¯Ñ§·½³ÌʽΪS+O2
 µãȼ 
.
 
SO2£¬¶þÑõ»¯Áò»áÐγÉËáÓꣻ
£¨4£©ÒòΪA¡¢B¡¢C¡¢DËÄÖÖÔªËØ·Ö±ðΪC¡¢O¡¢Mg¡¢S£¬ÔòA¡¢C¡¢D·Ö±ðÓëBÐγɵĻ¯ºÏÎï·Ö±ðΪ¶þÑõ»¯Ì¼¡¢Ñõ»¯Ã¾¡¢¶þÑõ»¯Áò£¬¿É¼ûËüÃǶ¼ÓÉÁ½ÖÖÔªËØ×é³É£¬ÇÒÆäÖÐÒ»ÖÖΪÑõÔªËØ£¬¹ÊΪÑõ»¯Î
£¨5£©ÒòΪC¡¢DÓëB·Ö±ðΪþԪËØ¡¢ÁòÔªËØ¡¢ÑõÔªËØ£¬¹ÊC¡¢DÓëBÐγɵĻ¯ºÏÎïΪÁòËáþ£¬Æ仯ѧʽΪMgSO4
¹Ê´ðΪ£º£¨1£©C¡¢O¡¢Mg¡¢S£»£¨2£©CO¡¢CO2£»£¨3£©S+O2
 µãȼ 
.
 
SO2¡¢ËáÓꣻ£¨4£©Ñõ»¯Î£¨5£©MgSO4£¨ÌîдMgSO3Ò²¸ø·Ö£©
µãÆÀ£º±¾ÌâÖ÷Òª¿¼²éѧÉú¶Ô»¯Ñ§ÓÃÓïµÄÊéдºÍÀí½âÄÜÁ¦£¬ÌâÄ¿Éè¼Æ¼È°üº¬¶Ô»¯Ñ§·ûºÅÒâÒåµÄÁ˽⣬ÓÖ¿¼²éÁËѧÉú¶Ô»¯Ñ§·ûºÅµÄÊéд£¬¿¼²éÈ«Ã棻Á˽âÎïÖʵÄÍƶϺͼø±ð·½·¨£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

16¡¢Óк˵çºÉÊýСÓÚ18µÄA¡¢B¡¢C¡¢DËÄÖÖÔªËØ£¬AÔªËصÄÔ­×Óʧȥһ¸öµç×Óºó£¬BÔªËصÄÔ­×ӵõ½Á½¸öµç×Óºó£¬ËüÃÇÐγÉÀë×ӵĵç×ÓÅŲ¼·Ö±ðÓëÄÊ¡¢ë²Ô­×ÓÏàͬ£»CÔªËصÄÔ­×ÓµÚÈý²ãµÄµç×ӱȵڶþ²ãÉϵĵç×ÓÉÙÒ»¸ö£»DÔªËصÄÔ­×Ӻ˱ÈAÔªËØÔ­×Ӻ˶àÒ»¸öÖÊ×Ó£®ÊÔÍƶÏÕâËÄÖÖÔªËصÄÃû³Æ£º
A¡¢B·Ö±ðΪ
ÄƺÍÁò
£»C¡¢D·Ö±ðΪ
ÂȺÍþ
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

Óк˵çºÉÊýСÓÚ18µÄA¡¢B¡¢C¡¢DËÄÖÖÔªËØ£¬AÔªËØ×é³ÉµÄµ¥ÖÊÊÇÇå½àµÄÄÜÔ´£»Bµ¥ÖʵÄÌå»ýÔÚ¿ÕÆøÖÐÕ¼21%£»CÔ­×ÓÓÐ3¸öµç×Ӳ㣬×îÄÚ²ãµç×ÓÊýÊÇ×îÍâ²ãµÄ2±¶£»DÔªËØÊÇÅ©×÷ÎïÉú³¤ËùÐèÒªµÄÓªÑøÔªËØ£¬Æä×é³ÉµÄµ¥ÖÊ¿ÉÒÔ±»¶¹¿ÆÖ²ÎïµÄ¸ùÁö¾úÎüÊÕ¡¢ÀûÓã®
£¨1£©Ð´³öB¡¢DÔªËصÄÃû³Æ£ºB
Ñõ
Ñõ
£¬D
µª
µª
£®
£¨2£©Ð´³öA¡¢B¿ÉÄÜÐγɻ¯ºÏÎïµÄ»¯Ñ§Ê½
H2O
H2O
¡¢
H2O2
H2O2
£®
£¨3£©Ð´³öÓÉA¡¢B¡¢DÈýÖÖÔªËØ×é³ÉµÄÁ½ÖÖ»¯ºÏÎïÖ®¼ä·¢ÉúÖкͷ´Ó¦µÄ»¯Ñ§·½³Ìʽ
HNO3+NH3?H2O=NH4NO3+H2O
HNO3+NH3?H2O=NH4NO3+H2O
£®
£¨4£©»îÆÃÐÔÇ¿µÄ½ðÊôÇ⻯ÎïµÄ»¯Ñ§Ê½¿É±íʾΪMHn£¬ËüÓëË®¾çÁÒ·´Ó¦Éú³É¼îºÍAµÄµ¥ÖÊ£¬Çëд³öCµÄÇ⻯ÎïÓëË®·¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ
NaH+H2O=NaOH+H2¡ü
NaH+H2O=NaOH+H2¡ü
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

Óк˵çºÉÊýСÓÚ18µÄA¡¢B¡¢C¡¢DËÄÖÖÔªËØ£¬AÔªËص¥ÖÊÔÚ³£ÎÂÏÂÊÇ×îÇáµÄÆøÌ壻BÔªËصĵ¥ÖʺÍCÔªËصĵ¥ÖÊ·´Ó¦µÄÏÖÏóΪÃ÷ÁÁµÄÀ¶×ÏÉ«»ðÑ棻CÔªËص¥ÖʵÄÌå»ýÔÚ¿ÕÆøÖÐÕ¼21%£»DÔªËصÄÔ­×ÓÓÐ3¸öµç×Ӳ㣬µÚÈý²ãÉϵĵç×ÓÊýÓëµÚÒ»²ãÉϵĵç×ÓÊýÏàµÈ£®
£¨1£©Ð´³öÕâËÄÖÖÔªËصķûºÅ£ºA
H
H
£¬B
S
S
£¬C
O
O
£¬D
Mg
Mg
£®
£¨2£©Ð´³öC¡¢DÐγɻ¯ºÏÎïµÄ»¯Ñ§Ê½
MgO
MgO
£®
£¨3£©Bµ¥ÖÊÔÚCµ¥ÖÊÖÐȼÉյķûºÅ±í´ïʽ
S+O2
 µãȼ 
.
 
SO2
S+O2
 µãȼ 
.
 
SO2
£¬ÆäÉú³ÉÎï¶Ô»·¾³µÄÆÆ»µÐÎʽÊÇÐγÉ
ËáÓê
ËáÓê
£®
£¨4£©A¡¢B¡¢D·Ö±ðÓëCÐγɵĻ¯ºÏÎï¿ÉÒÔͳ³ÆΪ
Ñõ»¯Îï
Ñõ»¯Îï
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

Óк˵çºÉÊýСÓÚ18µÄA¡¢B¡¢C¡¢DËÄÖÖÔªËØ£¬AÔªËصÄÔ­×Ó¹¹³ÉÒ»ÖÖµ¥ÖÊ£¬¿ÉÒÔºÍð¤ÍÁ»ìºÏÖƳÉǦ±Êо£»BÔªËصÄijÑõ»¯Îï·Ö×Ó×é³ÉΪBO3£¬ÆäÖÐBÔªËØÓëÑõÔªËصÄÖÊÁ¿±ÈΪ2£º3£¬ÇÒBÔ­×ÓºËÄÚÖÊ×ÓÊýÓëÖÐ×ÓÊýÏàµÈ£»Cµ¥ÖʵÄÌå»ýÔÚ¿ÕÆøÖÐÕ¼21%£»DÔªËصÄÔ­×ÓºËÍâÓÐÈý¸öµç×Ӳ㣬µÚÈý²ãÉϵĵç×ÓÊýÓëµÚÒ»²ãÉϵĵç×ÓÊýÏàµÈ£®
£¨1£©Ð´³öÕâËÄÖÖÔªËصÄÃû³Æ£ºA
̼
̼
£¬B
Áò
Áò
£¬C
Ñõ
Ñõ
£¬D
þ
þ
£®
£¨2£©Ð´³öC¡¢DÐγɻ¯ºÏÎïµÄ»¯Ñ§Ê½
MgO
MgO
£®
£¨3£©Bµ¥ÖÊÔÚCµ¥ÖÊÖÐȼÉÕʱµÄ»¯Ñ§·½³Ìʽ
S+O2
 µãȼ 
.
 
SO2
S+O2
 µãȼ 
.
 
SO2
£¬ÆäÉú³ÉÎï¶Ô»·¾³µÄÆÆ»µÊÇÐγÉ
ËáÓê
ËáÓê
£®
£¨4£©A¡¢B¡¢D·Ö±ðÓëCÐγɵĻ¯ºÏÎï¿ÉÒÔͳ³ÆΪ
Ñõ»¯Îï
Ñõ»¯Îï
£®
£¨5£©A¡¢CºÍDÈýÖÖÔªËØÐγɵĻ¯ºÏÎïµÄ»¯Ñ§Ê½ÊÇ
MgCO3
MgCO3
£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸