ÌâÄ¿ÄÚÈÝ

»¯Ñ§À´Ô´ÓÚÉú»î£¬²¢ÎªÉú»î·þÎñ¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

(1)³¤ÆÚÒûÓÃӲˮ¶ÔÈËÌ彡¿µ²»Àû¡£Éú»îÖÐÎÒÃÇ¿ÉÒÔÓÃ__________À´Çø±ðӲˮºÍÈíË®¡£

(2)´¿µç¶¯Æû³µÊ¹ÓõÄï®µç³Ø·Åµçʱ¿É½«___________ת»¯ÎªµçÄÜ¡£

(3)½ñÄê4Ô£¬ÁÉÄþ½¢º½Ä¸±à¶ÓÊ״ξ«²ÊÁÁÏàÄϺ£Ôıø¡£ÁÉÄþ½¢´óÁ¿Ê¹ÓÃÁ˺Ͻð£¬Ò»°ãÀ´Ëµ£¬ºÏ½ðÓë×é³ÉËüµÄ´¿½ðÊôÏà±È£¬Ó²¶È¸ü__________¡£

(4)ʹÓÃÍÑÁòúÄܼõÉÙ__________µÄÅÅ·Å£¬¼õÇáËáÓêµÄΣº¦£¬µ«ÈÔ»á²úÉú½Ï¶àµÄ_________£¬²»Äܼõ»ºÈÕÒæ¼Ó¾çµÄ¡°ÎÂÊÒЧӦ¡±¡£

(5)ÈÜÒºÔÚÉú²úºÍÉú»îÖÐÓй㷺µÄÓ¦Óá£Ä³Í¬Ñ§ÅäÖÆÈÜÒº£¬ÓÃÁ¿Í²Á¿È¡ËùÐèË®µÄ¹ý³ÌÖи©ÊÓ¶ÁÊý£¬ÕâÑùÅäÖÆµÃµ½µÄÈÜÒºÈÜÖÊÖÊÁ¿·ÖÊý»á____(Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±»ò¡°ÎÞÓ°Ï족)¡£

(6)ÎÒ¹úÑÐÖÆ³ö±ÈƯ°×·Û¸ü¸ßЧµÄÒûÓÃË®Ïû¶¾¼Á¡°ClO2¡±£¬Íê³ÉÖÆÈ¡ClO2·´Ó¦µÄ»¯Ñ§·½³Ìʽ£ºCl2 + 2NaClO2 === 2ClO2 + ________£¬ClO2ÖÐClµÄ»¯ºÏ¼ÛΪ_________¡£

·ÊÔíË® »¯Ñ§ÄÜ ´ó SO2»ò¶þÑõ»¯Áò CO2»ò¶þÑõ»¯Ì¼ Æ«´ó 2NaCl +4 ¡¾½âÎö¡¿(1)Éú»îÖÐÎÒÃÇ¿ÉÒÔÓ÷ÊÔíË®À´Çø±ðӲˮºÍÈíË®£¬Ó²Ë®ÅÝÄ­ÉÙ£¬ÈíË®ÅÝÄ­¶à¡£(2)´¿µç¶¯Æû³µÊ¹ÓõÄï®µç³Ø·Åµçʱ¿É½«»¯Ñ§ÄÜת»¯ÎªµçÄÜ¡£(3)Ò»°ãÀ´Ëµ£¬ºÏ½ðÓë×é³ÉËüµÄ´¿½ðÊôÏà±È£¬Ó²¶È¸ü´ó£¬ºÏ½ð±È½ðÊô¾ßÓиü´óµÄÓÅÔ½ÐÔ¡£(4)ʹÓÃÍÑÁòúÄܼõÉÙSO2»ò¶þÑõ»¯ÁòµÄÅÅ·Å£¬¼õÇáËáÓêµÄΣº¦£¬µ«ÈÔ»á²úÉú½Ï¶àµÄCO2»ò¶þÑõ»¯Ì¼£¬¡°ÎÂ...
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

Ŀǰ¼ÒÓÃÆû³µÈ¼ÁÏÖ÷ÒªÊÇÆûÓͺÍÌìÈ»Æø¡£ÎªÓ¦¶ÔÑϾþµÄ´óÆøÎÛȾÐÎʽ£¬ÌáÉý¿ÕÆøÖÊÁ¿£¬È«¹úÒѾ­È«Ã湩Ӧ·ûºÏµÚÎå½×¶Î¹ú¼Ò±ê×¼³µÓÃÆûÓÍ¡£Ê¹ÓùúÎå±ê×¼ÆûÓÍ£¬Ê¹µÃÆû³µÎ²ÆøÖеĶþÑõ»¯Áòº¬Á¿´ó·ù¶È½µµÍ¡£

(1)¢ÙµÄÒÇÆ÷Ãû³ÆÊÇ ________________

(2)ÓÃÉÏͼËùʾװÖÃÊÕ¼¯¶þÑõ»¯Áòʱ£¬ _________(Ìî¡°a¡±»ò¡°b¡±)¶Ë¹Ü¿ÚÓëc¹ÜÏàÁ¬,ÉÕ±­ÖÐÇâÑõ»¯ÄÆÈÜÒº¿ÉÒÔÎüÊÕ¶àÓàµÄ¶þÑõ»¯Áò¡£

(3)ÌìÈ»ÆøµÄÖ÷Òª³É·ÖΪ¼×Í飬ʵÑéÊÒ²ÉÓüÓÈÈÎÞË®´×ËáÄÆ¹ÌÌåÓë¼îʯ»Ò¹ÌÌåÖÆÈ¡¼×Í飬ÇëÑ¡ÔñÖÆÈ¡¼×ÍéµÄ×°ÖÃ__________(Ìî×ÖĸÐòºÅ).ʵÑéÊÒ¿ÉÓÃ________ ·½·¨ÊÕ¼¯¼×Íé¡£ÓøÃ×°Öû¹¿ÉÒÔÖÆÈ¡ÑõÆø,д³ö·´Ó¦µÄ»¯Ñ§·½³Ìʽ ___________________________

(4)³ä·ÖȼÉÕ1Kg²»Í¬»¯Ê¯È¼ÁÏËù²úÉúCO2ºÍSO2ÆøÌåµÄÖÊÁ¿Èçͼ Ëùʾ,Ôò________ȼÉÕ×îÒ×µ¼ÖÂËáÓ꣬______ȼÉÕ¶Ô»·¾³Ó°Ïì×îС.

·ÖҺ©¶· a B ÅÅË®·¨ 2KClO3 2KCl + 3O2 ¡ü ʯÓÍ ÌìÈ»Æø ¡¾½âÎö¡¿(1)¸ù¾Ý³£¼ûÒÇÆ÷½â´ð£»(2)¸ù¾Ý¶þÑõ»¯ÁòÃÜ¶È±È¿ÕÆø´ó½â´ð£»(3)¸ù¾Ý·´Ó¦ÎïµÄ״̬¡¢·´Ó¦Ìõ¼þ£¬¼×ÍéµÄÃܶȺÍÈܽâÐԵȷÖÎö½â´ð£»¸ù¾ÝÓùÌÌå¼ÓÈÈ·¨ÖÆÈ¡ÑõÆøÇÒÊԹܿÚûÓÐÃÞ»¨½â´ð£»(4)¸ù¾Ýͼʾ·ÖÎö½â´ð¡£(1)¢ÙµÄÒÇÆ÷Ãû³ÆÊÇ·ÖҺ©¶·£»(2)ÓÉÓÚ¶þÑõ»¯ÁòµÄÃÜ¶È±È¿ÕÆø´ó£¬ÓÃC×°ÖÃÊÕ¼¯¶þÑõ»¯ÁòʱӦ¡°³¤½ø¶Ì³ö¡±£¬¹ÊÑ¡a£»(...

ÔÚСӢ¼ÒµÄ²Ö¿âÀ¶Ñ·Å×ÅÒ»´ü´ü»¯·Êһһ̼ËáÇâï§( NH4HCO3).¹ýÁËÒ»¸öÏÄÌ죬СӢ·¢ÏÖÕâÖÖ»¯·ÊËù³ÖÓеĴ̼¤ÐÔÆøÎ¶±äµÃ¸üŨÁÒÁË£¬ÓÐЩ»¯·Ê´üÀï̼ËáÇâï§±äÉÙÁË£¬¼ì²é·¢ÏÖ±äÉٵϝ·Ê°ü×°´üûÓÐÃܷ⣬»¯·ÊҲûÓÐÈöÂäÔÚµØÉÏ£¬¸üûÓÐÈ˽ø¹ý²Ö¿â¿ª´üʹÓá£

ΪÁË̽¾¿ÕâЩ»¯·Ê¼õÉÙµÄÔ­Òò£¬Ð¡Ó¢ÔÚʵÑéÊÒÈ¡ÁËһЩ̼ËáÇâï§·ÛÄ©£¬·ÅÔÚÕô·¢ÃóÖмÓÈÈ£¬¹ýÒ»»á¶ù¹Û²ìµ½·ÛÄ©ÍêÈ«Ïûʧ£¬Í¬Ê±Ò²Îŵ½ÁËÕâÖִ̼¤ÐÔÆøÎ¶.·ÛĩΪʲô»áÏûÊ§ÄØ?

(1)£¨Ìá³öÎÊÌ⣩̼ËáÇâï§·ÛÄ©ÏûʧµÄÔ­ÒòÊÇʲô?

(2)£¨²ÂÏ룩¢Ù̼ËáÇâï§·ÛÄ©ÔÚ²»¼ÓÈÈ»ò¼ÓÈÈÌõ¼þÏÂÓɹÌ̬±ä³ÉËüµÄÆøÌ¬¡£¢Ú̼ËáÇâï§ÔÚ²»¼ÓÈÈ»ò¼ÓÈÈÌõ¼þÏ·¢Éú·Ö½â·´Ó¦£¬¿ÉÄܲúÉúµÄÎïÖÊÓа±ÆøºÍһЩÑõ»¯Îï¡£

(3)£¨²éÔÄ×ÊÁÏ£©¢Ù̼ËáÇâï§ÊôÓÚ°±·Ê£¬²»¾ßÓÐÉý»ªµÄÐÔÖÊ£¬ËµÃ÷ÉÏÊö²ÂÏë__________ (ÌîÐòºÅ)²»³ÉÁ¢£»¢Ú°±Æø(»¯Ñ§Ê½NH3)¾ßÓÐÌØÊâµÄ´Ì¼¤ÐÔÆøÎ¶£¬¼«Ò×ÈÜÓÚË®£¬ÆäË®ÈÜÒºÊǼîÐÔ£¬µ«¸ÉÔïµÄ°±Æø²»ÄÜʹ¸ÉÔïµÄºìɫʯÈïÊÔ¼Á±äÀ¶£»¢ÛNO2Ϊºì×ØÉ«ÆøÌå¡£NOΪÎÞÉ«ÆøÌ壬ÔÚ¿ÕÆøÖÐÒ×·¢Éú·´Ó¦£º2NO+O2=2NO2

(4)£¨ÊµÑé²Ù×÷£¬ÏÖÏóÓë½áÂÛ£©

ʵÑé²Ù×÷

ʵÑéÏÖÏó

ʵÑé½áÂÛ

¢ÙÈ¡ÊÊÁ¿Ì¼ËáÇâï§ÓÚÊÔ¹ÜÖмÓÈÈ£¬ÈçͼһËùʾ£¬½«¸ÉÔïµÄºìɫʯÈïÊÔÖ½½Ó½üµ¼¹Ü¿Ú

²úÉúÇ¿ÁҵĴ̼¤ÐÔÆøÎ¶£¬ÊԹܱÚÉÏÓÐÎÞɫҺµÎÇÒÊÔÖ½±äÀ¶£¬µ«Î´¼ûºì×ØÉ«ÆøÌå

·Ö½â²úÎïÖÐÓÐ______£¬Ã»ÓÐ__________

¢Ú°´Èçͼ¶þËùʾװÖüÌÐøÊµÑ飬ֱµ½·´Ó¦ÍêÈ«

³ÎÇåʯ»ÒË®±ä»ë×Ç

·Ö½â²úÎïÖÐÓÐ__________

(5)£¨Ó¦Óã©¢ÙÀûÓð±ÆøµÄ»¯Ñ§ÐÔÖÊ£¬Çëд³öʵÑéÊÒ¼ìÑé°±ÆøµÄ·½·¨(д³öʵÑé²Ù×÷¡¢ÏÖÏó¡¢½áÂÛ)£º________________________________________£»

¢ÚÈç¹ûÄã¼ÒÀïÓÐ̼ËáÇâï§»¯·Ê£¬ÄãÓ¦¸ÃÈçºÎ±£´æ? ______________________________¡£

¢Ù NH3ºÍH2O NO¡¢NO2 CO2 ÓÃʪÈóµÄºìɫʯÈïÊÔÖ½·ÅÓÚÊԹܿڣ¬ÈôÊÔÖ½±äÀ¶£¬Ôò˵Ã÷´æÔÚ°±Æø Ãܷ⣬µÍα£´æ ¡¾½âÎö¡¿±¾Ìâͨ¹ýʵÑé̽¾¿Ì¼ËáÇâï§·Ö½â²úβéÁ˳£¼ûÆøÌåµÄ¼ìÑé·½·¨¡£½â´ðʱҪ¸ù¾ÝÒÑÓÐ֪ʶºÍÌâÖÐÌṩµÄÐÅÏ¢½øÐзÖÎö¡¢ÅжϵóöÕýÈ·µÄ½áÂÛ¡£ (3) ̼ËáÇâï§·ÛÄ©ÔÚ²»¼ÓÈÈ»ò¼ÓÈÈÌõ¼þÏÂÓɹÌ̬±ä³ÉËüµÄÆøÌ¬³ÆÎªÌ¼ËáÇâï§µÄÉý»ª£¬Ì¼ËáÇâï§ÊôÓÚ°±·Ê£¬²»¾ßÓÐÉý»ªµÄÐÔÖÊ£¬ËµÃ÷ÉÏÊö²ÂÏë¢Ù²»³ÉÁ¢£»...

ѧϰÁË¡¶º£Ë®¡°ÖƼ¡·Ò»¿Îºó£¬Í¬Ñ§ÃÇÄ£Äâ¡°ºîÊÏÖÆ¼î·¨¡±½øÐÐʵÑ飮

£¨²éÔÄ×ÊÁÏ£©°±Æø(NH3)ÊÇÒ»ÖÖÓд̼¤ÐÔÆøÎ¶µÄÆøÌ壬¼«Ò×ÈÜÓÚË®£¬ÆäË®ÈÜÒºÏÔ¼îÐÔ£®

£¨Ä£ÄâʵÑ飩ÑéÖ¤ºîÊÏÖÆ¼î·¨ÄÜ»ñµÃ´¿¼î

²½Öè

ʵÑé×°ÖÃ

ʵÑé²Ù×÷¼°ÏÖÏó

½âÊÍÓë½áÂÛ

(1)

ÏÈ´Óa¹ÜͨÈë____(Ìî ¡°CO2¡±»ò¡°NH3¡±)£¬Ò»¶Îʱ¼äºó£¬ÔÙ´Ób¹ÜͨÈë___(Ìî¡°CO2¡±»ò¡°NH3¡±)£¬ÈÜÒºÖÐÓйÌÌåÎö³ö

ÕºÓÐÏ¡ÁòËáµÄÃÞ»¨¿ÉÎüÊÕ¹ýÁ¿µÄ°±Æø£¬ÒÔÃâ»Ó·¢µ½¿ÕÆøÖС£

(2)

³ä·Ö¼ÓÈÈ£¬¹Û²ìµ½µÄÏÖÏóÊÇ£ºÊԹܿÚÓÐË®Ö飬ÉÕ±­Öеĵ¼¹Ü¿ÚÓÐÆøÅÝð³ö£¬____¡£

²½Öè1Îö³öµÄ¹ÌÌåÊÇ̼ËáÇâÄÆ

(3)

µÎ¼Ó×ãÁ¿Ï¡ÑÎËᣬÓÐÆøÅݲúÉú£¬¹ÌÌåÖð½¥Ïûʧ

²½Öè2¼ÓÈȺóµÃµ½µÄ¹ÌÌåÊÇ´¿¼î

²½Öè1·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ ______________¡£

£¨ÊµÑ鷴˼£©(1)ͬѧÃÇÈÏΪ²½Öè3µÄʵÑ黹²»×ãÒԵóö¡°²½Öè2¼ÓÈȺóµÃµ½µÄ¹ÌÌåÊÇ´¿¼î¡±µÄ½áÂÛ£¬¾­¹ýÌÖÂÛ£¬´ó¼ÒÒ»ÖÂÈÏΪÐèÒª²¹³äµÄʵÑéÊÇ____________¡£

(2)Ò²ÓÐͬѧÈÏÎªÖÆµÃµÄ´¿¼îÖпÉÄܺ¬ÓÐÂÈ»¯ÄÆÔÓÖÊ£¬ ÓÚÊÇÓÖÉè¼ÆÁËÒÔÏÂʵÑé²¢µÃ³ö½áÂÛ£º

ʵÑé²Ù×÷

ʵÑéÏÖÏó

ʵÑé½áÂÛ

_________

____________

ÖÆµÃµÄ´¿¼îÖк¬ÓÐÂÈ»¯ÄÆ

NH3 CO2 ³ÎÇåʯ»ÒË®±ä»ë×Ç NaCl + H2O + NH3 + CO2 = NaHCO3 + NH4Cl ÏȼÓÈë¹ýÁ¿ÂÈ»¯¸Æ£¬¹Û²ìÊÇ·ñ»ë×Ç£¬È¡³ÎÇåÈÜÒº£¬ÔٵμÓÑÎËá¹Û²ìÊÇ·ñÓÐÆøÅÝ¡£ È¡´¿¼îÑùÆ·ÓÚÊÔ¹ÜÖмÓË®Èܽ⣬¼ÓÈë¹ýÁ¿Ï¡ÏõËᣬºóµÎ¼¸µÎÏõËáÒøÈÜÒº ÏÈÓÐÆøÅݲúÉú£¬ºó²úÉú°×É«³Áµí¡£ ¡¾½âÎö¡¿¡¾Ä£ÄâʵÑé¡¿£¨1£©°±Æø¼«Ò×ÈÜÓÚË®£¬ËùÒÔaÊÔ¹ÜÓ¦¸ÃÌí¼Ó°±Æø£¬bÊÔ¹ÜÌí¼Ó¶þÑõ»¯Ì¼£¬·´Ö®¶þÑõ»¯Ì¼ÎÞ·¨³ä·ÖÈܽâºÍ...

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

¾«Ó¢¼Ò½ÌÍø