ÂÛ̳
ѧϰÖܱ¨
Ê×Ò³ ¸ß¿¼ÆµµÀ Öп¼ÆµµÀ ½ÌʦƵµÀ ¼Ò½ÌƵµÀ ×ÊÔ´¼ÓÓÍÕ¾ È«¹úѧϰ¿ÆÑ§Ñо¿»á
ÄúÏÖÔÚµÄλÖ㺠ÌìÀû½ÌÓýÍø >> ÊÔÌâÖÐÐÄ >> Öп¼Ä£ÄâÌâ >> »¯Ñ§ >> ÎÄÕÂÕýÎÄ
[×éͼ]06Æ½Ô­ÏØ³õÈýÁ·±ø¿¼ÊÔ»¯Ñ§ÊÔÌâ¼°´ð°¸    ÈÈ     ¡ï¡ï¡ï
06Æ½Ô­ÏØ³õÈýÁ·±ø¿¼ÊÔ»¯Ñ§ÊÔÌâ¼°´ð°¸
×÷ÕߣºØýÃû    ÎÄÕÂÀ´Ô´£ºÌìÀû¿¼ÊÔÐÅÏ¢Íø    µã»÷Êý£º    ¸üÐÂʱ¼ä£º2006-9-6

06ÄêÆ½Ô­ÏØ³õÈýÁ·±ø¿¼ÊÔ»¯Ñ§ÊÔÌâ

ÓÑÇéÌáʾ£º£±¡¢»¯Ñ§¾ÍÔÚÎÒÃÇÉí±ß£¬Éú»îÖд¦´¦³äÂú»¯Ñ§£»»¯Ñ§ÊÇ21ÊÀ¼Í×îÓÐÓã¬×îÓд´ÔìÁ¦µÄÖÐÐÄѧ¿Æ£»ÊµÑé̽¾¿ÊÇѧϰ»¯Ñ§µÄÖØÒªÊֶΣ¬ÏàО­¹ýһѧÆÚµÄѧϰ£¬ÄãµÄ×ÔÖ÷ѧϰÄÜÁ¦ºÍ̽¾¿Ñ§Ï°ÄÜÁ¦ÓÐÁ˺ܴóÌá¸ß£¬Çë³ä·ÖÀûÓÃÄãµÄ˼ά£¬»ý¼«´ðÌâ¡£ÏàÐÅÖ»ÒªÄãÈÏÕæ»Ø´ðÁËÎÊÌ⣬һ¶¨»áµÃµ½ÂúÒâ³É¼¨¡£

£²¡¢¿ÉÄÜÓõ½µÄÏà¶ÔÔ­×ÓÖÊÁ¿£ºCl¡ª35.5    O¡ª16    K¡ª39 

µÃ·Ö

ÆÀ¾íÈË

 

 

 

Ò»¡¢Ñ¡ÔñÌ⣨±¾Ìâ¹²11СÌ⣬ÿСÌâÖ»ÓÐÒ»¸öÕýÈ·Ñ¡Ï1-4СÌâÿСÌâ1·Ö£¬5-11СÌâÿСÌâ2·Ö£¬¹²18·Ö£©

 

1¡¢ÏÂÁС°¹ØÓÚ¼ÒÍ¥ÓÃË®µÄÎʾíµ÷²é¡±µÄÎÊÌ⣬ÐèÒª½øÐÐʵÑéºó²ÅÄܻشðµÄÊÇ£º

   A. Äú¼ÒʹÓõÄÊÇ×ÔÀ´Ë®³§µÄ×ÔÀ´Ë®Âð£¿

   B. Äú¼ÒʹÓõÄÊÇӲˮ£¬»¹ÊÇÈíË®£¿

   C. Äú¼ÒÓÃË®ÊÇ·ñÓжþ´ÎʹÓõÄÇé¿ö£¨ÈçÏ´Íê²Ëºó£¬³å²ÞËù£©?

   D. Çë̸̸Äú¼ÒµÄ½ÚË®´ëÊ©£¿

2¡¢¹ú¼Ò¾ö¶¨ÍƹãÒÒ´¼ÆûÓ͵ÄÓ¦Óã¬ËùνÒÒ´¼ÆûÓ;ÍÊÇÔÚÆûÓÍÖмÓÈëÊÊÁ¿ÒÒ´¼»ìºÏ¶ø³ÉµÄÒ»ÖÖȼÁÏ£¬ÏÂÁÐÐðÊö´íÎóµÄÊÇ£º                                  

  A. ÒÒ´¼ÆûÓÍÊÇÒ»ÖÖÐÂÐÍ»¯ºÏÎï

  B. Æû³µÊ¹ÓÃÒÒ´¼ÆûÓÍÄܼõÉÙÓк¦ÆøÌåµÄÅÅ·Å

  C. ÓÃʯÓÍ¿ÉÒÔÖÆµÃÆûÓÍ

  D. ÓÃÓñÃס¢¸ßÁ»·¢½Í¿ÉÒÔÖÆµÃÒÒ´¼

3¡¢ÏÂÁб仯һ¶¨ÊôÓÚ»¯Ñ§±ä»¯µÄÊÇ£º

¢ÙÁ¸Ê³Äð³É¾Æ  ¢Ú±¬Õ¨  ¢ÛÓûîÐÔÌ¿³ýÈ¥Óж¾ÆøÌå  ¢ÜÖ²ÎïµÄ¹âºÏ×÷Óà ¢Ý×ÔÐгµÆïÒ»¶Îʱ¼äºó£¬³µÈ¦ÉúÐâÁË  ¢ÞÏ´ÍêµÄÒ·þÔÚÑô¹âÏÂÁÀɹ£¬ºÜ¿ì»á¸É

A£®¢Ù¢Ú¢Û     B£®¢Ú¢Û¢Ý   C£®¢Ù¢Ü¢Ý   D£®¢Û¢Ü¢Ý

4¡¢¡°ÂÌÉ«»¯Ñ§¡±µÄºËÐÄÊÇÔÚ»¯Ñ§·´Ó¦¹ý³Ì»ò»¯¹¤Éú²úÖУ¬¾¡Á¿¼õÉÙʹÓûò³¹µ×Ïû³ýÓк¦ÎïÖÊ¡£ÏÂÁÐ×ö·¨ÖУ¬·ûºÏÂÌÉ«»¯Ñ§µÄÊÇ£º

A Éú²úºÍʹÓþ綾ũҩ                B ÔìÖ½³§ÓöþÑõ»¯Áò½øÐÐÖ½½¬Æ¯°× 

C »¯¹¤³§²úÉúµÄ·ÏÆøÏò¸ß¿ÕÅÅ·Å        D ÀûÓÃË«ÑõË®ÖÆÑõÆø

5£®Íõ°²Ê¯µÄ¡¶Ã·»¨¡·Ê«£º¡°Ç½½ÇÊý֦÷£¬Á躮¶À×Ô¿ª¡£Ò£Öª²»ÊÇÑ©£¬ÎªÓаµÏãÀ´¡£¡±Ê«ÈËÔÚÔ¶´¦¾ÍÄÜÎŵ½µ­µ­µÄ÷»¨ÏãζµÄÔ­ÒòÊÇ£º

  A£®·Ö×ÓºÜС                   B£®·Ö×ÓÊÇ¿ÉÒÔ·ÖµÄ

C£®·Ö×ÓÖ®¼äÓмä϶             D£®·Ö×ÓÔÚ²»¶ÏµØÔ˶¯

6£®ÉÆÓÚÊáÀí»¯Ñ§ÖªÊ¶£¬ÄÜʹÄãÍ·ÄÔ¸ü´ÏÃ÷¡£ÒÔÏÂÍêÈ«ÕýÈ·µÄÒ»×éÊÇ£º

 

 

A

ÎïÖʵÄÐÔÖÊÓëÓÃ;

 

 

B

°²È«³£Ê¶

N2ÐÔÖÊÎȶ¨¡ª¡ª¿É×÷µÆÅÝÌî³äÆø

ÒÒ´¼¾ßÓпÉȼÐÔ¡ª¡ª¿É×÷ȼÁÏ

ʯīºÜÈí¡ª¡ª¿É×÷µç¼«

¼Ù¾ÆÖж¾¡ª¡ªÓɼ×È©ÒýÆð

ú¿ó±¬Õ¨¡ª¡ªÓÉÍß˹ÒýÆð

¼ÙÑÎÖж¾¡ª¡ªÓÉNaNO2ÒýÆð

 

 

C

ÔªËØÓëÈËÌ彡¿µ

 

 

D

ÈÕ³£Éú»î¾­Ñé

ȱάÉúËØC¡ª¡ªÒ×ÒýÆð»µÑª²¡

ȱ¸Æ¡ª¡ªÒ×¹ÇÖÊÊèËÉ»òµÃØþÙͲ¡

ȱµâ¡ª¡ªÒ×¼××´ÏÙÖ×´ó

ʳƷ¸ÉÔï¼Á¡ª¡ª³£ÓÃCuO

Çø±ðӲˮÓëÈíË®¡ª¡ª³£Ó÷ÊÔíË®¼ìÑé

ʹúȼÉÕ¸üÍú¡ª¡ª°Ñú×÷³É·äÎÑ×´

7£®½üÀ´ÓÐÑо¿±¨¸æ³Æ£º³ýÈ¥¡°ÆÕͨˮ¡±ÀﺬÓеĵªÆøºÍÑõÆøºó£¬Ë®µÄÈ¥ÎÛÄÜÁ¦½«´ó´ó¼ÓÇ¿¡£¶Ô´ËµÄÏÂÁÐÀí½â²»ÕýÈ·µÄÊÇ£º 

A£®¡°ÆÕͨˮ¡±º¬Óеª·Ö×Ó                 B£®¡°ÆÕͨˮ¡±º¬ÓÐÑõ·Ö×Ó

 C£®³ýÈ¥ÑõÆøºóµÄË®¾Í²»ÔÙº¬ÓÐÑõÔªËØÁË   D£®µªÆøºÍÑõÆøÔÚË®ÖÐÓÐÒ»¶¨µÄÈܽâÐÔ

8£®Ä³ÊµÑéÖУ¬ÀÏʦÇëͬѧÃÇÓÃÊÔ¹ÜȡϡÑÎËá×öÐÔÖÊʵÑé(ûÓиæËßÓÃÁ¿)£¬ÏÂÃæËÄλͬѧµÄ²Ù×÷ÖÐ×î·ûºÏÒªÇóµÄÊÇ£º

¡¡¡¡A£®ÏòÊÔ¹ÜÖеÎÈë2µÎÏ¡ÑÎËá
¡¡¡¡B£®ÏòÊÔ¹ÜÖÐ×¢ÈëÔ¼2mLÏ¡ÑÎËá
¡¡¡¡C£®ÏòÊÔ¹ÜÖÐ×¢Èë10mLÏ¡ÑÎËá
¡¡¡¡D£®×¢ÈëµÄÏ¡ÑÎËáÊÇÊÔ¹ÜÈÝ»ýµÄ1/2

9£®Ä³Í¬Ñ§ÔÚʵÑéÊÒ½«Ë®ÕôÆøÍ¨¹ýÊ¢ÓÐÌú·ÛµÄ²£Á§¹Ü£¬Í¬Ê±¶ÔÌú·Û³ÖÐø¸ßμÓÈÈ£¬Ò»¶Îʱ¼äºó£¬·¢ÏÖ¹ÜÄÚÓкÚÉ«¹ÌÌ壬¸Ãͬѧ¶ÔºÚÉ«¹ÌÌåµÄ×é³É×÷ÁËÈçϼ¸Öֲ²⣺

¢ÙFe  ¢ÚFe2O3   ¢ÛFe3O4  ¢ÜFe(OH)3 ¡£ÄãÈÏΪ¸ÃͬѧµÄÉÏÊö²Â²â¿ÉÄܺÏÀíµÄÊÇ£º

A£®¢Ú¢Ü           B£®¢Ù¢Û           C£®¢Ù¢Ú          D£®¢Û¢Ü

10£®Ëæ×Å¿ÆÑ§µÄ½ø²½£¬»¯Ñ§ÓëÈËÀཡ¿µµÄ¹ØÏµÔ½À´Ô½ÃÜÇÐÁË¡£ÏÂÁÐ˵·¨´íÎóµÄÊÇ£º

A¡¢ÌÇ¡¢µ°°×ÖÊ¡¢ÎÞ»úÑεȶ¼ÊÇÈËÀàά³ÖÉúÃüºÍ½¡¿µËù±ØÐëµÄÓªÑø¡£

B¡¢ÈËÌåÄÚÈç¹ûȱÉÙÌúÔªËØÒ×»¼È±ÌúÐÔÆ¶Ñª¡£

C¡¢¿ÉÒÔÓþÛÂÈÒÒÏ©ËÜÁϰüװʳƷ¡£

D¡¢´¦ÓÚÉú³¤·¢ÓýÆÚµÄÇàÉÙÄêÓ¦¶à³Ô¸»º¬Î¬ÉúËØµÄË®¹û¡¢Ê߲ˡ£

11£®Ä³Í¬Ñ§ÔÚ»¯Ñ§Íí»áÉÏΪ´ó¼ÒÏÖ³¡×÷ÁËÒ»·ù¡¶¾µºþµÆÓ°¡·µÄ¾°É«»­£ºËûÔÚ°×Ö½ÉÏÓøɾ»µÄë±Êպȡ¼×ÈÜÒº¡°»­ÉÏ¡±Ò»´®µÆÓ°£¬ÔÙÓÃÁíһ֧ë±ÊպȡÒÒÈÜÒºÔÚµÆÓ°ÖÜΧ¡°»­ÉÏ¡±Ò»Æ¬ºþË®£¬½«°×Ö½¹ÒÔÚǽÉÏ£¬´ýÁÀ¸Éºó£¬ÓÃ×°ÓбûÈÜÒºµÄÅçÎíÆ÷Ïò°×Ö½ÉÏÅçÈ÷¡£½á¹û³öÏÖÁË¡°À¶É«µÄºþÃæÉϵ¹Ó³×źìÉ«µÄµÆÓ°¡±µÄ»­Ãæ¡£¸ÃͬѧËùÓõļס¢ÒÒ¡¢±ûÈýÖÖÈÜÒº¿ÉÄÜÊÇÏÂÁÐËÄÏîÖеģº                                            

         ¼×                 ÒÒ             ±û

    A£®Ï¡ÑÎËá           ÇâÑõ»¯ÄÆÈÜÒº    ʯÈïÈÜÒº

    B£®ÇâÑõ»¯¼ØÈÜÒº     Ï¡ÑÎËá          Ê¯ÈïÈÜÒº

    C£®ÂÈ»¯ÄÆÈÜÒº       ÁòËá¼ØÈÜÒº      ÁòËáÄÆÈÜÒº

    D£®Ê¯»ÒË®           Ï¡ÁòËá          ÂÈ»¯ÄÆÈÜÒº   

µÃ·Ö

ÆÀ¾íÈË

 

 

 

 

¶þ¡¢Ìî¿Õ¼°¼ò´ðÌ⣨¹²15·Ö£©

 

12¡¢£¨6·Ö£©¢ÅÔÚC¡¢H¡¢O¡¢S¡¢NaÎåÖÖÔªËØÖУ¬Ñ¡ÔñÊʵ±ÔªËØ£¬×é³É·ûºÏÏÂÁÐÒªÇóµÄÎïÖÊ£¬Ç뽫Æä»¯Ñ§Ê½ÌîÈë¿Õ¸ñÖС£

  ¢ÙÌìÈ»ÆøµÄÖ÷Òª³É·Ö             £»¢Ú×îÀíÏëµÄ¿É×÷¸ßÄÜȼÁϵĵ¥ÖÊ            £»

  ¢ÛÒ×ÓëѪºìµ°°×½áºÏʹÈËÖж¾µÄÑõ»¯Îï            £»¢ÜÒ»ÖÖ¼î               £»

  ¢ÝʵÑéÊÒ³£ÓõÄËá            £»¢ÞÒ»ÖÖ³£¼ûµÄÑΠ                 ¡£

13£®£¨3·Ö£©¶Ô֪ʶµÄ¹éÄÉÓëÕûÀíÊÇѧϰ»¯Ñ§µÄÒ»ÖÖÖØÒª·½·¨¡£ÏÖÓÐÈý¸ö»¯Ñ§·´Ó¦ÈçÏ£º



¡÷

 
(1)ͨ¹ý±È½ÏÎÒ·¢ÏÖ£ºËüÃÇÓÐÐí¶àÏàËÆÖ®´¦£¬ÆäÒ»¶¼ÓëÑõÆø·´Ó¦£¬Æä¶þÉú³ÉÎï¶¼ÊÇÒ»ÖÖ¡­¡­ ÎÒÄÜÁíдһ¸ö·ûºÏÕâÁ½µãµÄ»¯Ñ§·½³Ìʽ                                        £»ËüÃÇÖ®¼äÒ²´æÔÚÏàÒìÖ®´¦£¬ÆäÖÐÒ»¸ö·´Ó¦ÓëÖÚ²»Í¬£¬Õâ¸ö·´Ó¦ºÍËüµÄ²»Í¬Ö®´¦ÊÇ                                           ¡£

(2)¿ÎÍâѧϰCu2(OH)2CO3 ==== 2CuO£«H2O£«CO2¡üºó£¬ÎÒ·¢ÏÖËüÒ²ºÍÉÏÊöÈý¸ö·´Ó¦ÓÐÏàËÆÖ®´¦£¬ÆäÏàËÆÖ®´¦ÊÇ                                          ¡£

14£®(3·Ö)ÈËÃÇΪ½Òʾԭ×ӽṹµÄ°ÂÃØ£¬¾­ÀúÁËÂþ³¤µÄ̽¾¿¹ý³Ì¡£×Ô1 897ÄêÌÀÄ·Éú·¢ÏÖµç×Ó²¢Ìá³öÀàËÆ¡°Î÷¹Ï¡±µÄÔ­×ÓÄ£ÐÍ£¬1911ÄêÖøÃûÎïÀíѧ¼Ò¬ɪ¸£µÈÈËΪ̽Ë÷Ô­×ÓµÄÄÚ²¿½á¹¹ÓÖ½øÐÐÁËÏÂÃæµÄʵÑé¡£ËûÃÇÔÚÓÃÒ»Êø´øÕýµçµÄ¡¢ÖÊÁ¿±Èµç×Ó´óµÃ¶àµÄ¸ßËÙÔ˶¯µÄ¦ÁÁ£×Óºä»÷½ð²­Ê±£¬·¢ÏÖ£º

 

 

 

 

 

 

 ¢Ù´ó¶àÊý¦ÁÁ£×ÓÄÜ´©Í¸½ð²­¶ø²»¸Ä±äÔ­À´µÄÔ˶¯·½Ïò

  ¢ÚһС²¿·Ö¦ÁÁ£×ӸıäÁËÔ­À´µÄÔ˶¯·½Ïò

  ¢ÛÓм«ÉÙÊý¦ÁÁ£×Ó±»µ¯ÁË»ØÀ´

ÇëÄã¸ù¾Ý¶ÔÔ­×ӽṹµÄÈÏʶ£¬·ÖÎö³öÏÖÉÏÊöÏÖÏóµÄÔ­Òò£º

  (1)ÏÖÏóÒ»£º                                                       £º

  (2)ÏÖÏó¶þ£º                                                         £»

  (3)ÏÖÏóÈý£º                                                         ¡£

15£®£¨3·Ö£©Ð¡¸Õͬѧ»æÖÆÁËÈçÓÒͼËùʾA¡¢BÁ½ÖÖ¹ÌÌåÎïÖʵÄÈܽâ¶ÈÇúÏß¡£

£¨1£©Î¶ÈΪt1¡æÊ±£¬AÎïÖÊÓëBÎïÖʵÄÈܽâ¶È¡¡¡¡¡¡¡¡¡¡£¨Ìî¡°ÏàµÈ¡±¡¢¡°²»ÏàµÈ¡±»ò¡°²»ÄܱȽϡ±£©¡£

£¨2£©½«AÎïÖÊt2¡æÊ±µÄ±¥ºÍÈÜÒº½µµÍζÈÖÁt1¡æÊ±£¬ÆäÈÜÖÊÖÊÁ¿·Ö

Êý                £¨Ì¡°±ä´ó¡±¡¢¡°±äС¡±¡¢¡°²»±ä¡±£©¡£

£¨3£©ÎÒ¹úÓÐÐí¶àÑμîºþ£¬ºþÖÐÈÜÓдóÁ¿µÄNaClºÍNa2CO3,ÄÇÀïµÄÈËÃǶ¬ÌìÀ̼ÏÄÌìɹÑΡ£¾Ý´ËÎÒÈÏΪÓÒͼÖУ¨Ìî×Öĸ£©¡¡¡¡¡¡¡¡¡¡ÇúÏßÓë´¿¼îµÄÈܽâ¶ÈÇúÏßÏàËÆ¡£

 

µÃ·Ö

ÆÀ¾íÈË

 

 

Èý¡¢ÊµÑéÓë̽¾¿Ì⣨¹²10·Ö£©

 

 

16£®£¨5·Ö£©Ð¡º£Ì½¾¿ÇâÑõ»¯ÄƹÌÌåÈÜÓÚˮʱÆäζȵı仯ºó£¬ÓÉÓÚÊèºö£¬Î´½«Ê¢ÇâÑõ»¯ÄƹÌÌåÊÔ¼ÁµÄÆ¿ÈûÈû½ô£¬Ò»¶Îʱ¼äºó£¬Ëü¿ÉÄܱäÖÊÁË¡£¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º                                                       ¡£

Сº£¶Ô¸ÃÑùÆ·½øÐÐÁËϵÁÐ̽¾¿£º

[Ìá³öÎÊÌâ1]£º¸ÃÑùÆ·ÕæµÄ±äÖÊÁËÂð£¿

[ʵÑé·½°¸1]£ºÈ¡¸ÃÑùÆ·ÉÙÁ¿ÓÚÊÔ¹ÜÀ¼ÓÈëÊÊÁ¿µÄË®£¬Õñµ´£¬ÑùÆ·È«²¿ÈÜÓÚË®£¬ÔÙÏòÆäÖеμÓÇâÑõ»¯¸ÆÈÜÒº£¬¹Û²ìµ½                   ˵Ã÷ÑùÆ·ÕæµÄ±äÖÊÁË¡£

[Ìá³öÎÊÌâ2]£º¸ÃÑùÆ·ÊDz¿·Ö±äÖÊ»¹ÊÇÈ«²¿±äÖÊ£¿

[ʵÑé·½°¸2]£ºÈ¡¸ÃÑùÆ·ÉÙÁ¿ÓÚÉÕ±­À¼ÓÊÊÁ¿µÄË®£¬½Á°è£¬È«²¿Èܽâºó£¬ÔÙÏòÆäÖмÓÈë¹ýÁ¿µÄ              ÈÜÒº£¬È»ºó¹ýÂË£¬ÔÙÏòÂËÒºÀïµÎ¼ÓÎÞÉ«·Ó̪ÊÔÒº£¬¹Û²ìµ½±ä³ÉºìÉ«£¬ËµÃ÷¸ÃÑùÆ·ÊDz¿·Ö±äÖÊ¡£

[Ìá³öÎÊÌâ3]£ºÔõÑù³ýÈ¥ÑùÆ·ÖеÄÔÓÖÊ£¿

[ʵÑé·½°¸3]£º½«ÑùÆ·È«²¿ÈÜÓÚË®£¬ÏòÆäÖмÓÈëÊÊÁ¿µÄ           ÈÜÒº£¬·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º                                          £»È»ºó¹ýÂË£¬ÔÙ½«ÂËÒºÕô¸É£¬¼´µÃµ½´¿¾»µÄÇâÑõ»¯ÄƹÌÌå¡£

17£®£¨5·Ö£©ÄãÏëºÈ×ÔÖÆµÄÆûË®Âð?СÓêͬѧΪÄãÌṩÁË¡°Æßϲ¡±ÆûË®³É·Ö±í(¼ûÏÂͼ)¡£

 

 

 

 

 

 

 

(1)¸ù¾Ý¡°Æßϲ¡±ÆûË®µÄ³É·Ö±í£¬·ÖÎö×ÔÖÆÆûË®ËùÐèµÄÔ­ÁÏÊÇ£º´¿¾»Ë®¡¢°×ɰÌÇ¡¢¹ûÖ­¡¢Ê³ÓÃÏãÁÏ¡¢ ̼ËáÇâÄÆºÍ                ¡£

(2)´ò¿ªÆûˮƿ¸Ç£¬ÓдóÁ¿ÆøÌåÒݳö¡£ÇëÄã¶Ô¸ÃÆøÌåµÄ³É·Ö½øÐÐ̽¾¿£¬Íê³ÉÏÂ±í£º

²ÂÏëÓë¼ÙÉè

ÑéÖ¤·½·¨¼°²Ù×÷

  ¹Û²ìµ½µÄÏÖÏó

½áÂÛ(Óû¯Ñ§·½³Ìʽ±íʾ)

 

 

 

 

 

µÃ·Ö

ÆÀ¾íÈË

 

 

 

ËÄ¡¢¼ÆËãÓë·ÖÎöÌ⣨¹²7·Ö£©

 

 

18£®Ð¡ÒâºÍС˼ͬѧ¶Ô»¯Ñ§¼ÆËãºÜÓÐÐĵá£ÒÔÏÂÊÇËûÃǽâ´ðÒ»µÀ¼ÆËãÌâµÄʵ¼¡£

ÇëÄãÒ»Æð²ÎÓëÑо¿²¢Íê³ÉÏà¹ØÎÊÌâ¡££¨¼ÆËã½á¹û¾«È·µ½0.01£©

[ÌâÄ¿] ÒÑÖª:  2KClO3 ===  2KCl  +  3O2¡ü,½«10gÂÈËá¼ØºÍ2g¶þÑõ»¯ÃÌ»ìºÏºó·ÅÈëÊÔ¹ÜÖмÓÈÈ£¬ÊÕ¼¯ËùÐèÁ¿µÄÑõÆøºó£¬Í£Ö¹¼ÓÈÈÈÃÊÔ¹ÜÀäÈ´£¬³ÆµÃÊÔ¹ÜÄÚÊ£Óà¹ÌÌåµÄÖÊÁ¿Îª7.2g¡£ÇóÉú³ÉÂÈËá¼ØµÄÖÊÁ¿¡£

£¨1£©Ð¡ÒâºÜ¿ìµÃµ½£¨10g + 2g ¨C7.2g£©ÊÇ                  £¨Ìѧʽ£©µÄÖÊÁ¿£¬½ø¶øÇó³öKCl µÄÖÊÁ¿ÊÇ              g.

 ÇëÄã¼òµ¥Ð´³öСÒâ¼ÆËãKCl µÄÖÊÁ¿µÄ¹ý³Ì¡£

 

 

 

 

 

(2)С˼ÒÀ¾ÝÉÏÊö¼ÆËãµÄ½á¹û·¢ÏÖÌâÄ¿Êý¾ÝÓÐÎÊÌâ¡£ÇëÄãͨ¹ý¼òµ¥µÄ¼ÆË㣬ÂÛÖ¤ËûµÄ·¢ÏÖ¡£

 

 

 

£¨3£©ÈçºÎ²ÅÄܸüÕý¸ÃÌâÄØ£¿Ð¡ÒâºÍС˼ÈÏΪÓÐÐí¶à·½·¨£¬ÀýÈ罫ÌâÄ¿ÖС°10gÂÈËá¼Ø¡±¸ÄΪ¡°agÂÈËá¼Ø¡±£¬ÆäËüÎïÖʵÄÖÊÁ¿²»±ä£¬ÔòaµÄȡֵ·¶Î§ÊÇ                           ¡£


     »¯Ñ§ÊÔÌâ²Î¿¼´ð°¸¼°ÆÀ·Ö±ê×¼

Ò»¡¢Ñ¡ÔñÌ⣨±¾Ìâ¹²11СÌ⣬ÿСÌâÖ»ÓÐÒ»¸öÕýÈ·Ñ¡Ï1-4СÌâÿСÌâ1·Ö£¬5-11СÌâÿСÌâ2·Ö£¬¹²18·Ö£©

ÌâºÅ

1

2

3

4

5

6

7

8

9

10

11

´ð°¸

B

A

C

D

D

A

C

B

B

C

A

¶þ¡¢Ìî¿Õ¼°¼ò´ðÌ⣨¹²15·Ö£©

12¡¢£¨6·Ö£©CH4    H2    CO     H2SO4     NaOH   Na2CO3 £¨»ò NaHCO»òNa2SO4£©

µãȼ

 
 


13¡¢£¨3·Ö£©£¨1£©C+O2 ====CO2     ÆäÖÐÒ»ÖÖ·´Ó¦ÎïΪ»¯ºÏÎï  £¨2£©Éú³ÉÎï¶¼ÊÇÑõ»¯Îï

£¨Ã¿¿Õ1·Ö£¬ÆäËü´ð°¸ºÏÀíͬÑù¸ø·Ö£©

14¡¢£¨3·Ö£©(1)Ô­×Ӻ˺ÜС£¬Ô­×ÓÄÚ²¿Óкܴó¿Õ¼ä  (2)Ô­×Ӻ˴øÕýµç£¬¦ÁÁ£×Ó;¾­½ðÔ­×Ӻ˸½½üʱ£¬Êܵ½³âÁ¦¶ø¸Ä±äÁËÔ˶¯·½Ïò  (3)½ðÔ­×ÓºËÖÊÁ¿±È¦ÁÁ£×Ó´óºÜ¶à£¬µ±¦ÁÁ£×ÓÕýÅöµ½½ðÔ­×Ӻ˱»µ¯ÁË»ØÀ´¡££¨ËµÃ÷£ºÑ§Éú´ð³ö¢ÙÔ­×ÓÄÚ²¿Óкܴó¿Õ¼ä»òÔ­×Ӻ˺ÜС£¬¢ÚÔ­×Ӻ˴øÕýµç£¬¢Û½ðÔ­×ÓºËÖÊÁ¿´ó£¬¼´¿ÉµÃÂú·Ö¡££©£¨Ã¿¿Õ1·Ö£©

15¡¢£¨3·Ö£©£¨1£©ÏàµÈ¡£¡¡£¨2£©±äС¡£¡¡£¨3£©£¨Ã¿¿Õ1·Ö£©

Èý¡¢ÊµÑéÓë̽¾¿Ì⣨¹²10·Ö£©

16¡¢£¨5·Ö£©2NaOH+ CO2 =Na2CO3+ H2O£»£¬Éú³É°×É«³Áµí£»ÂÈ»¯¸Æ£»ÇâÑõ»¯¸ÆÈÜÒº£¬

 Na2CO3+ Ca(OH)2 =CaCO3 ¡ý+ 2NaOH £¨Ã¿¿Õ1·Ö£©

 

17¡¢£¨5·Ö£©(1) ÄûÃÊËá 

(2)CO2  Í¨Èë³ÎÇåʯ»ÒË®  ʯ»ÒË®±ä»ë×Ç  CO2+Ca(OH)2=CaCO3¡ý+H2O£¨Ã¿¿Õ1·Ö£©£©

ËÄ¡¢¼ÆËãÓë·ÖÎöÌ⣨¹²7·Ö£©

18¡¢£¨7·Ö£©£¨1£©£¨3·Ö£¬Ã¿¿Õ1·Ö£¬¼ÆËã¹ý³Ì1·Ö£©·Ö    O2    7.45g

       ÉèÉú³ÉÂÈ»¯¼ØµÄÖÊÁ¿Îªx

 2KClO3 ===  2KCl  +  3O2¡ü

                     2¡Á74.5    3¡Á32

                       X     10g + 2g ¨C7.2g

             2¡Á74.5 £º3¡Á32  =  X  £º10g + 2g ¨C7.2g

      ½âµÃ£ºX  =  7.45g

 (2)(2·Ö)

   ËùÊ£¹ÌÌåµÄÖÊÁ¿=7.45g + 2g =9.45g >7.2g

   ËùÒÔÌâÄ¿ÓдíÎó¡£

£¨ÂÈËá¼ØÖÊÁ¿= 7.45g + 4.8 g = 12.25g > 10g    ËùÒÔÌâÄ¿ÓдíÎó¡£ £©

(3) (2 ·Ö)  5.2g  < a ¡Ü8.55g

 

 

 

 

 

 

 

 

 

 

18£®£¨5·Ö£©ÓÃÓÒͼËùʾµÄ×°ÖýøÐÐʵÑé¿ÉÒÔÊ¹ÆøÇòÅòÕÍ»òËõС¡£

×°ÖÃ˵Ã÷£º

Ò»¸öÈÝ»ý½Ï´óµÄ¹ã¿ÚÆ¿£¬Æ¿ÈûÉÏ×°ÓÐA¡¢BÁ½¸ö·ÖҺ©¶·ºÍ²£Á§µ¼¹Ü£¬¿ØÖÆÂ©¶·ÉϵĻîÈû¿ª¹Ø¿ÉÒÔËæÊ±µÎ¼ÓÈÜÒº£¬µ¼¹ÜÉìÈëÆ¿ÄڵIJ¿·ÖÓëÆøÇò½ôÃÜÏàÁ¬£¬µ¼¹ÜÁíÒ»¶ËÓë¿ÕÆøÏàͨ¡£

¢Å×°ÖÃµÄÆøÃÜÐÔÊDZ¾ÊµÑéÄÜ·ñ³É¹¦µÄ¹Ø¼ü¡£¹Ø±ÕA¡¢BµÄ»îÈû£¬½«¹ã¿ÚÆ¿½þûÔÚ±ùË®ÖУ¨ÊÒÄÚζÈΪ25¡æ£©£¬Èç¹û×°ÖÃÆøÃÜÐÔÁ¼ºÃ£¬»á³öÏÖʲôÏÖÏó¡£                             

                                               ¡£

¢ÆÈÃÆ¿ÄÚ³äÂú¶þÑõ»¯Ì¼ÆøÌ壬ÏÈÏò¹ã¿ÚÆ¿ÖеμÓAÖеÄÈÜÒº£¬Õñµ´¹ã¿ÚÆ¿£¬ÆøÇòÅòÕÍ£»ÔÙÏò¹ã¿ÚÆ¿ÖеμÓBÖеÄÈÜÒº£¬Õñµ´¹ã¿ÚÆ¿£¬ÆøÇòÓÖËõС£¬Èç´ËÖØ¸´²Ù×÷£¬ÆøÇò¿ÉÒÔ·´¸´ÅòÕͺÍËõС¡£Ôò£º·ÖҺ©¶·AÖпÉÄÜÊÇ           ÈÜÒº£»·ÖҺ©¶·BÖпÉÄÜÊÇ             ÈÜÒº¡£

д³öÉÏÊöÓйط´Ó¦ÖеÄÒ»¸ö»¯Ñ§·½³Ìʽ£º                                    ¡£

 

¢ÅÆøÇò»áÅòÕÍ£»¢ÆÇâÑõ»¯ÄÆ£¨»ò¼îÐԵȣ©£¬ÑÎËᣨ»òËáÐԵȣ©£¬CO2+2NaOH=Na2CO3+H2O £¨»òNa2CO3+2HCl=2NaCl+CO2¡ü+H2OµÈ£¨·½³Ìʽ2·Ö£¬ÆäÓàÿ¿Õ1·Ö£©

 

21£®£¨6·Ö£©ÎªÁ˶ÔÒ»°ü·ÅÖúܾõĸÉÔï¼Á(Ö÷Òª³É·ÖÊÇÉúʯ»Ò)½øÐÐ̽¾¿£¬Ð¡Ã÷Ìá³öÁËÁ½Ïî̽¾¿ÈÎÎñ£º

  (1)̽¾¿¸ÉÔï¼ÁÊÇ·ñʧЧ£»

  (2)̽¾¿Ê§Ð§µÄ¸ÉÔï¼ÁµÄÖ÷Òª³É·ÖÊÇ·ñÊÇÇâÑõ»¯¸Æ¡£ËûÉè¼ÆÁËÒÔϵÄ̽¾¿·½°¸£º

ÎÊÌâÓë²ÂÏë

ʵÑé²½Öè

ʵÑéÏÖÏó

ʵÑé½áÂÛ

(1)¸ÉÔï¼ÁÊÇ·ñʧЧ

È¡ÑùÆ··ÅÈëÊÔ¹ÜÖÐ ¼ÓÊÊÁ¿µÄË®£¬´¥ÃþÊԹܱÚ

ûÓÐÈȸÐ

 

(2)ʧЧµÄ¸ÉÔï¼ÁÖÐ º¬ÓÐÇâÑõ»¯¸Æ

¼ÓË®³ä·Ö½Á°è¡¢¹ýÂ˺ó£¬ÓÃpHÊÔÖ½²âÆäÈÜÒºµÄpHÖµ

 

ÓÐÇâÑõ»¯¸Æ

 

ÄãÈÏΪ̽¾¿(2)ÊÇ·ñÑÏÃÜ?       

Çë˵Ã÷Ô­Òò        ¡£

ÄãÈÏΪÒÔCaOΪÖ÷Òª³É·ÖµÄ¸ÉÔï¼ÁÔÚ¿ÕÆøÖзÅÖþÃÁË£¬³ýÄÜת»¯ÎªCa(OH)2Í⣬»¹¿ÉÄÜת»¯Îª        £¬Çë¼òµ¥ËµÃ÷ÑéÖ¤µÄ˼·               

£¨1£©Ã»ÓÐÑõ»¯¸Æ£¬£¨2£©pHÖµ´óÓÚ7¡£

²»ÑÏÃÜ¡£Éúʯ»ÒÓöË®»áÉú³ÉÇâÑõ»¯¸Æ¡£  CaCO3 ¡£ 

 ¼ÓÈë¹ýÁ¿HCl£¬¹Û²ìÓÐÎÞÆøÌå·Å³ö¡£ÓÐÆøÌå˵Ã÷ÓÐCaCO3´æÔÚ¡££¨Ã¿¿Õ1·Ö£©

 

 

20£®£¨12·Ö£©ÏÖÓÐÒ»°ü´Óº£Ë®»ñµÃµÄ´ÖÑΣ¬ÒѾ­¾­¹ý³õ²½µÄÌá´¿¡£ÄÏÃÅÖÐѧ³õÈý£¨5£©°à¿ÎÍâ»î¶¯Ð¡×é¶ÔËüµÄ³É·Ö½øÐÐ̽¾¿£¬²¢½«´ÖÑνøÒ»²½Ìá´¿¡£

̽¾¿Ò»£ºÕâ°ü´ÖÑÎÖл¹ÓжàÉÙÔÓÖÊ?

(1)¡¢¸ù¾Ýº£Ë®µÄ³É·ÖºÍ³õ²½Ìá´¿µÄʵÑé²Ù×÷£¬¹À¼Æ¸Ã´ÖÑοÉÄÜ»¹º¬ÓеÄÔÓÖÊÊÇCaCl2ºÍMgCl2£¬ÏÖÓÐʵÑéÑéÖ¤ÕâÖÖÍÆ²â£ºÈ¡Ñù²¢ÖƳÉÈÜÒº£¬¼ÓÈëÊýµÎNaOHÈÜÒº£¬Ä¿µÄÊǼìÑéÓÐûÓÐ______________________(Ìѧʽ)½Ó×ÅÔÙ¼ÓÈëÊýµÎNa2CO3Òº£¬Ä¿µÄÊǼìÑéÓÐûÓÐ___________________(Ìѧʽ)

ʵÑéÖ¤Ã÷£¬Õâ°ü´ÖÑκ¬ÓеÄÔÓÖÊÊÇ CaCl2

̽¾¿¶þ£ºÕâ°ü´ÖÑÎÖÐNaClµÄÖÊÁ¿·ÖÊýÊǶàÉÙ?

  °´ÏÂÃæ²½Öè¼ÌÐø½øÐÐʵÑ飺

¢Ù ³ÆÈ¡Ò»¶¨ÖÊÁ¿µÄÑùÆ·£»

¢Ú ½«ÑùÆ·¼ÓË®ÈÜ½â£¬ÖÆ³É´ÖÑÎÈÜÒº£»

¢Û Ïò´ÖÑÎÈÜÒº¼ÓÈë¹ýÁ¿µÄijÖÖÊÔ¼Á£¬¹ýÂË£»

¢Ü ³ÁµíÏ´µÓºóСÐĺæ¸É£¬µÃµ½´¿¾»¹ÌÌåA£»

¢Ý ÂËÒºÔÚ½øÐÐijһ²Ù×÷ºó£¬ÒÆÈëÕô·¢ÃóÕô·¢£¬µÃµ½´¿¾»¹ÌÌåB£»

¢Þ ³ÆÁ¿ÊµÑéÖеõ½µÄijÖÖ¹ÌÌå¡£

(2)¡¢ÔڢڢۢݵIJ½ÖèÖУ¬¶¼Ê¹Óõ½Í¬Ò»ÒÇÆ÷(ÌîÃû³Æ) ______________£»ËüÔÚ²½Öè¢ÚºÍ¢ÝµÄ²Ù×÷·½·¨Ïàͬ£¬µ«Ä¿µÄ²»Í¬£¬ÔÚ²½Öè¢ÚµÄÄ¿µÄÊÇ__________________£»ÔÚ²½Öè¢ÝµÄÄ¿µÄÊÇ____________________________________¡£

(3)¡¢²½Öè¢ÛÖмÓÈëµÄÊÔ¼ÁÊÇ(д»¯Ñ§Ê½) _________£»²½Öè¢Ý½øÐеġ°Ä³Ò»²Ù×÷¡±ÊÇ___________________________________£¬Ä¿µÄÊÇ_______________________

(4)¡¢²½Öè¢ÞÖУ¬ÄãÈÏΪÐèÒª³ÆÁ¿µÄ¹ÌÌåÊǹÌÌåA»¹ÊǹÌÌåB___________(ÌîA»òB)£¬Ä㲻ѡÔñ³ÆÁ¿ÁíÒ»ÖÖ¹ÌÌåµÄÀíÓÉÊÇ______________________________________

________________________________________________________________

20¡¢(1)MgCl£» CaCl2         £¨2£©²£Á§°ô£»¼ÓËÙÈܽ⣻¼ÓÈÈʱ£¬Ê¹ÒºÌåÊÜÈȾùÔÈ

£¨3£©Na2CO3£»¼ÓÈëÊÊÁ¿µÄÏ¡ÁòËá°ÑÈÜÒºµÄpHÖµµ÷½Úµ½7£»³ýÈ¥¹ýÁ¿µÄ̼ËáÄÆ£»

£¨4£©A£¬BµÄÖÊÁ¿²»ÊÇÔ­À´´ÖÑÎÖеÄÂÈ»¯ÄÆÖÊÁ¿£¬»¹°üÀ¨·´Ó¦¹ý³ÌÖÐÉú³ÉµÄÂÈ»¯ÄƵÄÖÊÁ¿¡£

(6·Ö)ÏÖÓû̽¾¿Fe2O3ÄÜ·ñ¼Ó¿ìH2O2µÄ·Ö½â£¬²¢ÓëMnO2µÄ´ß»¯Ð§¹û½øÐбȽϣ¬¼×¡¢ÒÒ¡¢±ûÈýλͬѧͬʱ½øÐÐH2O2µÄ·Ö½âÓëÆøÌåµÄÊÕ¼¯ÊµÑ顣ʵÑéʱ¼ä¾ùÒÔ30ÃëΪ׼(H2O2¾ùδȫ²¿·Ö½â)£¬ÆäËû¿ÉÄÜÓ°ÏìʵÑéµÄÒòËØ¾ùÒѺöÂÔ¡£Ïà¹ØÊý¾ÝÈçÏ£º

ͬѧ

ÐòºÅ

ÖÊÁ¿·ÖÊýÏàͬµÄH2O2ÈÜÒº

ÈÜÒºÖмÓÈëµÄÎïÖÊ

ÊÕ¼¯µ½µÄÆøÌåÌå»ý

¼×

    100 ml

δ¼ÓÆäËûÎïÖÊ

    Aml

  ÒÒ

    100 ml

Fe203 0.5 g

    B ml

±û

    100 ml

 MnO2   O£®5 g

    c ml

    (1)ÈôÒª¼ìÑéÒÒͬѧÊÕ¼¯µÄÆøÌåÊÇ·ñΪO2£¬¿ÉÔÚÒÒͬѧʢÓÐÆøÌåµÄÈÝÆ÷ÖР      

  (2)Óû̽¾¿Fe203ÔÚʵÑéÖÐÊÇ·ñÆð´ß»¯×÷Óã¬Ê×ÏÈÓ¦±È½Ï        ºÍ         (Ñ¡Ìîa¡¢b¡¢c)µÄ´óС£¬Æä´ÎÐè²¹³ä×öÈçÏÂʵÑé(ÎÞÐèд³ö¾ßÌå²Ù×÷)£»¢Ù        £»

    ¢ÚEe203µÄ»¯Ñ§ÐÔÖÊÓÐûÓиı䡣

  (3)ÒÒͬѧµÄʵÑé±íÃ÷£ºFe203ÄܼӿìH2O2µÄ·Ö½â¡£¾Ý´Ë£¬ËûÔ¤²âÌúÐâÒ²¿ÉÒÔ¼Ó¿ìH2O2µÄ·Ö½â¡£ËûµÄÔ¤²âÊÇ·ñÕýÈ·?²¢¼òҪ˵Ã÷ÀíÓÉ¡£

¡¡¡¡±¾ÍøÕ¾²¿·ÖÄÚÈÝÀ´Ô´ÓÚ»¥ÁªÍø£¬Èç¹ûÇÖ·¸ÁËÄúµÄºÏ·¨È¨Ò棬Ç뼰ʱÓë±¾Õ¾ÁªÏµ£¬ÎÒÃǽ«ÔÚµÚһʱ¼äÄÚɾ³ý£¡
ÎÄÕ¼È룺edit    ÔðÈα༭£ºedit 
  • ÉÏһƪÎÄÕ£º

  • ÏÂһƪÎÄÕ£º
  • ¡¾·¢±íÆÀÂÛ¡¿¡¾¼ÓÈëÊղء¿¡¾¸æËߺÃÓÑ¡¿¡¾´òÓ¡´ËÎÄ¡¿¡¾¹Ø±Õ´°¿Ú¡¿
      µ÷²é
    ¡¡¡¡ÍøÓÑÆÀÂÛ£º£¨Ö»ÏÔʾ×îÐÂ10Ìõ¡£ÆÀÂÛÄÚÈÝÖ»´ú±íÍøÓѹ۵㣬Óë±¾Õ¾Á¢³¡Î޹أ¡£©
    Ïà¹ØÎÄÕÂ