ÌâÄ¿ÄÚÈÝ

ijУ»¯Ñ§ÐËȤС×éͬѧ¶ÔÖÊÁ¿Êغ㶨ÂɽøÐÐʵÑé̽¾¿£¬¼××éͬѧÓÃÔÚ¿ÕÆøÖÐȼÉÕþ´øµÄʵÑéÀ´Ì½¾¿£¬ÒÒ×éͬѧÓÃͼ¶þ½øÐÐ̽¾¿¡£»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©¼××éͬѧ´ýþÌõȼÉպ󣬳ÆÁ¿ÁôÔÚʯÃÞÍøÉϹÌÌåµÄÖÊÁ¿¡£·¢Ïֱȷ´Ó¦Ç°Ã¾ÌõµÄÖÊÁ¿ÒªÇá¡£ÇëÄã·ÖÎöÆäÖеÄÔ­Òò¿ÉÄÜÊÇ                                             ¡£¾­¹ý·ÖÎöºó£¬¸Ã×éͬѧÔÚȼÉÕµÄþÌõÉÏ·½ÕÖÉÏÕÖ£¬Ê¹¼Ð³ÖµÄþÌõÍêȫȼÉÕ²¢Ê¹Éú³ÉÎïÈ«²¿ÊÕ¼¯ÆðÀ´³ÆÁ¿£¬Ôò³ÆµÃµÄÖÊÁ¿ºÍԭþÌõÖÊÁ¿±È              £¨Ìî¡°ÏàµÈ¡±»ò¡°±ä´ó¡±»ò¡°±äС¡±£©£¬ÇëÄã½âÊÍÔ­Òò£º                                              ¡£
£¨2£©Ì¼ËáÄÆÓëÑÎËá·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                                      ¡£ÒÒ×éͬѧʵÑéÏÖÏóÊÇ£ºÌìÆ½                   £¨Ìî¡°±£³Öƽºâ¡±»ò¡°Ïò×óƫת¡±»ò¡°ÏòÓÒÆ«×ª¡±£© 
£¨3£©Í¨¹ýÒÔÉÏÁ½¸öʵÑ飬Äã»ñµÃµÄÆôʾÊÇ                                   ¡£
£¨1£©Ã¾ÌõÍêȫȼÉÕʱ¿ÉÄܲúÉúÁËÑÌÒݵ½¿ÕÆøÖУ¬ÖÊÁ¿±È·´Ó¦Ç°ÉÙ£¨»òÓÃÓÚ¼ÐȡþÌõ·´Ó¦ÎïµÄÛáÛöǯÉÏÕ´ÓÐÉÙÁ¿·´Ó¦ÎʹʣÓàµÄÖÊÁ¿¼õÉٵȺÏÀí´ð°¸Òà¿É£©   ´ó    ²Î¼Ó·´Ó¦µÄþµÄÖÊÁ¿¼ÓÑõÆøÖÊÁ¿µÈÓÚÉú³ÉµÄÑõ»¯Ã¾µÄÖÊÁ¿£¬ËùÒÔÉú³ÉµÄÑõ»¯Ã¾µÄÖÊÁ¿´óÓÚþµÄÖÊÁ¿ 
£¨2£©Na2CO3+2HCl=2NaCl+H2O+CO2¡ü    ÏòÓÒÆ«×ª
£¨3£©ÑéÖ¤ÖÊÁ¿Êغ㶨ÂÉʱ£¬¶ÔÓÚÓÐÆøÌå²Î¼Ó»òÓÐÆøÌåÉú³ÉµÄ·´Ó¦Ó¦ÔÚÃÜ·âµÄÈÝÆ÷ÖнøÐУ¨»ò½øÐпÆÑ§Ì½¾¿Ê±£¬ÒªÉÆÓÚ͸¹ýÏÖÏó¿´µ½±¾ÖʵȺÏÀí´ð°¸Òà¿É£©

ÊÔÌâ·ÖÎö£ºÖÊÁ¿Êغ㶨ÂÉ£ºÔÚ»¯Ñ§·´Ó¦ÖУ¬²Î¼Ó·´Ó¦µÄ¸÷ÎïÖÊÖÊÁ¿×ܺͣ¬µÈÓÚ·´Ó¦Éú³ÉµÄ¸÷ÎïÖÊÖÊÁ¿×ܺ͡£ÔÚÑéÖ¤ÖÊÁ¿Êغ㶨ÂÉʱ£¬ÓÐÆøÌå²ÎÓë»òÉú³ÉµÄ·´Ó¦ÐèÔÚÃܱÕÈÝÆ÷ÖнøÐС£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨5·Ö£©Ä³»¯Ñ§ÐËȤС×éµÄͬѧ¶ÔºôÎüÃæ¾ßµÄÖÆÑõÔ­Àí²úÉúºÃÆæÐÄ£¬Í¨¹ý²éÔÄ×ÊÁϵÃÖª£¬ºôÎüÃæ¾ßÖÐÖÆÈ¡ÑõÆøµÄÖ÷ÒªÔ­ÁÏÊǹÌÌå¹ýÑõ»¯ÄÆ£¨Na2O2£©¡£¹ýÑõ»¯ÄÆ·Ö±ðÄܺͶþÑõ»¯Ì¼¡¢Ë®·´Ó¦¶¼Éú³ÉÑõÆø£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º
2Na2O2£«2H2O4NaOH£«O2¡ü          2Na2O2£«2CO22Na2CO3£«O2
ΪÁ˲ⶨºôÎüÃæ¾ßÖйýÑõ»¯ÄƵÄÖÊÁ¿·ÖÊý£¬¸ÃС×éµÄͬѧÔÚÀÏʦµÄÖ¸µ¼Ï£¬ÀûÓÃÏÂͼËùʾװÖ㨹̶¨×°ÖÃÒ»ÂÔÈ¥£©¿ªÕ¹Ì½¾¿£¬²¢µÃµ½ÕýÈ·µÄ½áÂÛ¡£
ÒÑÖª£¬×°ÖÃBÖÐÊ¢Óб¥ºÍNaHCO3ÈÜÒº£¨NaHCO3²»ÓëCO2·´Ó¦£©£¬¼îʯ»ÒÊÇÓɹÌÌåNaOHºÍCaO×é³ÉµÄ»ìºÏÎï¡£ÕûÌ××°ÖÃÆøÃÜÐÔÁ¼ºÃ£¬·´Ó¦ËùÐèÊÔ¼Á¾ù×ãÁ¿¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺   
£¨1£©ÒÇÆ÷aµÄÃû³ÆÊÇ           £¬×°ÖÃAµÄ×÷ÓÃÊÇ                                  £»
£¨2£©²»ÓÃ×°ÖÃE´úÌæ×°ÖÃCµÄÀíÓÉÊÇ                                                ¡£
£¨3£©×°ÖÃDÖйýÑõ»¯ÄÆÒ©Æ·µÄÖÊÁ¿Îªmg£¬×°ÖÃDÔÚ·´Ó¦Ç°ºóµÄ×ÜÖÊÁ¿n1gºÍn2g¡£ÈôÒ©Æ·ÖÐËùº¬ÔÓÖʼȲ»ÈÜÓÚˮҲ²»²Î¼Ó»¯Ñ§·´Ó¦£¬ÔòÒ©Æ·ÖйýÑõ»¯ÄƵÄÖÊÁ¿·ÖÊý =                   ¡£
£¨6·Ö£©Þ­´ºÊµÑéÖÐѧ»¯Ñ§ÐËȤС×飬ÔÚѧϰ¼îµÄ»¯Ñ§ÐÔÖÊʱ£¬½øÐÐÁËÈçͼ1ËùʾµÄʵÑé¡£

£¨1£©µÚÒ»×éͬѧÔÚÏòAÖмÓÈëÒ»¶¨Á¿µÄÏ¡ÑÎËáºó£¬·¢ÏÖAÖÐÈÜÒºÈÔÈ»³ÊºìÉ«£¬Ôò·´Ó¦ºóÈÜÒºÖеÄÈÜÖÊÊÇ£¨·Ó̪³ýÍ⣬Óû¯Ñ§Ê½±íʾ£©¡¡                    £»µÚ¶þ×éͬѧ×öÍêAʵÑéºóÈÜÒº³ÊÎÞÉ«£¬ÓÃ΢¹ÛÀíÂÛ½âÊÍAÖÐÈÜÒº±äΪÎÞÉ«µÄ·´Ó¦µÄʵÖÊÊÇ¡¡                                    ¡£
£¨2£©µÚÈý×éͬѧ×öBʵÑéÓÃÓÚÈ·È϶þÑõ»¯Ì¼ÓëÇâÑõ»¯ÄÆÄÜ·¢Éú»¯Ñ§·´Ó¦£¨ÊµÑéǰK1¡¢K2¾ù´¦ÓڹرÕ״̬£©¡£BʵÑéÈ«¹ý³ÌµÄ²Ù×÷¼°ÏÖÏóÊÇ£º
²½Öè
ÏÖÏó
µÚÒ»²½£º                        £»
ÆøÇòÕÍ´ó
µÚ¶þ²½£º                        ¡£
ÆøÇòÓÖËõСÖÁÔ­À´×´Ì¬
 
£¨3£©Çëд³öÆøÇòËõСÖÁÔ­À´×´Ì¬µÄ»¯Ñ§·½³Ìʽ¡¡                            ¡£
£¨4£©ÊµÑé½áÊøºó£¬Ð¡¾ü½«ÒÔÉÏͬѧ×öµÄA¡¢BÁ½¸öʵÑéµÄ·ÏÒº¾ùµ¹Èëͬһ¸ö´óÉÕ±­ÖУ¬¹Û²ìµ½»ìºÏºóµÄ·ÏÒº³ÊºìÉ«£¬Óɴ˲úÉúÒÉÎÊ£º
[Ìá³öÎÊÌâ]´óÉÕ±­µÄ·ÏÒºÖк¬ÓÐÄÄЩÈÜÖÊ£¿£¨·Ó̪³ýÍ⣩
[²éÔÄ×ÊÁÏ]ÂÈ»¯¸ÆÈÜÒº³ÊÖÐÐÔ£¬CaCl2+Na2CO3=CaCO3¡ý+2NaCl
[ʵÑéÑéÖ¤]ȡһ¶¨Á¿µÄ·ÏÒº£¬ÖðµÎ¼ÓÈëÂÈ»¯¸ÆÈÜÒº£¬Èçͼ2ΪͬѧÃǸù¾Ý²¿·ÖʵÑéÏÖÏó»æÖƵĹØÏµÇúÏß¡£
[ʵÑé½áÂÛ]Èç¹û¹Û²ìµ½µÄÏÖÏóÖ»ÓгÁµí²úÉú£¬ÈÜÒºÈÔȻΪºìÉ«£¬Ôò·ÏÒºÖеÄÈÜÖÊΪ¡¡        £»Èç¹û¹Û²ìµ½ÓгÁµí²úÉú£¬ÈÜÒºÈÔÓɺìÉ«±äΪÎÞÉ«£¬Ôò·ÏÒºÖеÄÈÜÖÊΪ¡¡           ¡£
(9·Ö) Ò»´ÎȤζ»¯Ñ§»î¶¯ÖУ¬ÍõÀÏʦÏòͬѧÃÇչʾÁËһƿ±êÇ©ÊÜËðµÄÎÞÉ«ÈÜÒº£¬ÈçͼËùʾ¡£ÒªÇóͬѧÃǽøÐÐ̽¾¿£ºÈ·ÈÏÕâÆ¿ÈÜÒº¾¿¾¹ÊÇʲôÈÜÒº£¿

¡¾Ìá³ö²ÂÏë¡¿ ÍõÀÏʦÌáʾ£ºÕâÆ¿ÎÞÉ«ÈÜÒºÖ»ÄÜÊÇÏÂÁÐËÄÖÖÈÜÒºÖеÄÒ»ÖÖ£º¢ÙÁòËáþÈÜÒº ¢ÚÁòËáÄÆÈÜÒº ¢ÛÁòËáÈÜÒº¢ÜÁòËáï§ÈÜÒº
¡¾²éÔÄ×ÊÁÏ¡¿
¢Ù³£ÎÂÏ£¬Ïà¹ØÎïÖʵÄÈܽâ¶ÈÈçÏ£º
ÎïÖÊ
MgSO4
Na2SO4
(NH4)2SO4
H2SO4
Èܽâ¶È
35£®1g
19£®5g
75£®4g
ÓëË®ÈÎÒâ±È»¥ÈÜ
¢Ú(NH4)2SO4µÄË®ÈÜÒºÏÔËáÐÔ
¡¾ÊµÑé̽¾¿¡¿£¨1£©Ð¡Ã÷¶ÔÍõÀÏʦµÄÌáʾ½øÐÐÁËÆÀ¼Û                          £¬Ô­ÒòÊÇ             ¡£
£¨2£©ÎªÈ·¶¨ÆäËü¼¸ÖÖ²ÂÏëÊÇ·ñÕýÈ·£¬Ð¡Ã÷ͬѧ¼ÌÐø½øÐÐ̽¾¿£º
ʵÑé²Ù×÷
ʵÑéÏÖÏó
ʵÑé½áÂÛ
¢ÙÈ¡¸ÃÈÜÒºÉÙÐíÓÚÊÔ¹ÜÖУ¬ÏòÆäÖеμӼ¸µÎ     ÈÜÒº
ÈÜÒºÖÐÓа×É«³ÁµíÉú³É
²ÂÏë¢Ù³ÉÁ¢
¢ÚÓò£Á§°ôպȡÉÙÐíÔ­ÈÜÒºµÎÔÚpHÊÔÖ½ÉÏ£¬²¢¸ú±ê×¼±ÈÉ«¿¨¶ÔÕÕ
 
ÈÜÒºpHСÓÚ7
 
²ÂÏë¢Û³ÉÁ¢
СÑŶÔСÃ÷ʵÑé²Ù×÷¢ÚµÄ½áÂÛÆÀ¼ÛÊÇ         £¬ÀíÓÉÊÇ                      £»
£¨3£©ÇëÄãÉè¼ÆÊµÑé·½°¸£¬È·ÈϸÃÈÜÒºÊÇÁòËáï§ÈÜÒº²¢Íê³ÉʵÑ鱨¸æ£º
ʵÑé²Ù×÷
ʵÑéÏÖÏó
ʵÑé½áÂÛ
È¡¸ÃÈÜÒºÉÙÐíÓÚÊÔ¹ÜÖУ¬
                      
                      
                    
                       
²ÂÏë¢Ü³ÉÁ¢£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ               
                     

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø