ijͬѧÔÚѧϰÈÜÒºµÄËá¼îÐÔʱ·¢ÏÖÑÎÈÜÒºNaCl³ÊÖÐÐÔNa2CO3ÈÜÒº³Ê¼îÐÔ£¬¸ÃͬѧÈÏΪÓпÉÄÜÓеÄÑÎÈÜÒº»á³ÊËáÐÔ£¬ÉÏÍø²é×ÊÁÏ·¢ÏÖ£¨NH4£©2SO4£¬FeCl3ÈÜÒº³ÊËáÐÔ£®¸ÃͬѧÀ´ÐËȤÁËд³öÈçϵÄѧϰС½á£¬ÇëÄã×öÒ»×ö£º
£¨1£©ÑÎÈÜÒºpHµÄ¿ÉÄÜΪ£¨½«ÄãµÄ¼ÙÉèÌîÈë¿Õ¸ñÖУ©¢Ù¡¡   ¡¡¢Ú¡¡   ¡¡¢ÛpH=7
£¨2£©Òª²â¶¨¸ÃÈÜÒºµÄËá¼îÐÔÓá¡     ¡¡£¬ÈçÒªÖªµÀ¸ÃÈÜÒºµÄËá¼î¶È¿ÉÓá¡  ¡¡£®²Ù×÷·½·¨¡¡                                            ¡¡£®
£¨3£©¡°×¯¼ÚÒ»Ö¦»¨£¬È«¿¿·Êµ±¼Ò£®¡±£¨NH4£©2SO4ÊÇÒ»ÖÖµª·ÊÄÜ´Ù½øÅ©×÷ÎïµÄ¾¥¡¢Ò¶Éú³¤Ã¯Ê¢£¬ÄÜʹ×ÏɫʯÈïÊÔÒº±ä¡¡     ¡¡É«£®
£¨4£©ï§Ì¬µª·ÊµÄ¼ì²â·½·¨ÊÇ£º¡¡                                   ¡¡£®
£¨5£©¸Ãͬѧ¼ÓÈÈFeCl3ÈÜÒº·¢ÏÖÓкìºÖÉ«³ÁµíÉú³É£¬Çëд³ö»¯Ñ§·½³Ìʽ£º¡¡               ¡¡£®£¨Ìáʾ£ºCu£¨OH£©2µÈÄÑÈÜÐÔ¼îÔÚpH£¾5ÑÎËáÖв»Èܽ⣩
£¨6£©ÎªÊ²Ã´Na2CO3ÈÜÒº³Ê¼îÐÔ£¬£¨NH4£©2SO4£¬FeCl3ÈÜÒº³ÊËáÐÔ£¬ÉÏÍø²é²»ÊǺÜÃ÷°×£¬ÊǸßÖÐҪѧµÄ£¬µÈÉϸßÖкóÎÒÒ»¶¨Òª¸ãÇå³þ£®

£¨1£©pH£¾7£¬pH£¼7   
£¨2£©Ëá¼îָʾ¼Á£¬pHÊÔÖ½£¬Óò£Á§°ôպȡ´ý²âҺͿÔÚpHÊÔÖ½ÉÏ£¬°ÑÏÔʾµÄÑÕÉ«Óë±ê×¼±ÈÉ«¿¨¶ÔÕÕ¼´¿É
£¨3£©ºì    £¨4£©ÓëÊìʯ»Ò»ìºÏÑÐÄ¥£¬¹Û²ìÊÇ·ñÓд̼¤ÐÔµÄÆøζ²úÉú
£¨5£©FeCl3+3H20Fe£¨OH£©3¡ý+3HCl

½âÎöÊÔÌâ·ÖÎö£º£¨1£©pH£¾7µÄÈÜÒº³Ê¼îÐÔ£¬pH£¼7µÄÈÜÒº³ÊËáÐÔ£¬pH=7µÄÈÜÒº³ÊÖÐÐÔ£¬ÑÎÈÜÒº¿ÉÄܳʼîÐÔ¡¢ËáÐÔ»òÖÐÐÔ£¬¹ÊÌpH£¾7£¬pH£¼7£»
£¨2£©Ëá¼îָʾ¼Á¿ÉÒԲⶨÈÜÒºµÄËá¼îÐÔ£¬²â¶¨ÈÜÒºµÄËá¼î¶ÈʹÓõÄÊÇpHÊÔÖ½£¬²â¶¨Ê±ÒªÓò£Á§°ôպȡ´ý²âҺͿÔÚpHÊÔÖ½ÉÏ£¬°ÑÏÔʾµÄÑÕÉ«Óë±ê×¼±ÈÉ«¿¨¶ÔÕÕ¼´¿É£¬¹ÊÌËá¼îָʾ¼Á£¬pHÊÔÖ½£¬Óò£Á§°ôպȡ´ý²âҺͿÔÚpHÊÔÖ½ÉÏ£¬°ÑÏÔʾµÄÑÕÉ«Óë±ê×¼±ÈÉ«¿¨¶ÔÕÕ¼´¿É£»
£¨3£©£¨NH4£©2SO4ÈÜÒº³ÊËáÐÔ£¬ËáÐÔÈÜÒºÄÜʹʯÈïÊÔÒº±äºì£¬¹ÊÌºì£»
£¨4£©ï§Ì¬µª·ÊÓë¼îÐÔÎïÖÊ»ìºÏ»á²úÉúÓд̼¤ÐÔÆøζµÄ°±Æø£¬¹Ê¿ÉÒÔʹÓÃÓëÊìʯ»Ò»ìºÏÑÐÄ¥µÄ·½·¨½øÐмìÑéï§Ì¬µª·Ê£¬¹ÊÌÓëÊìʯ»Ò»ìºÏÑÐÄ¥£¬¹Û²ìÊÇ·ñÓд̼¤ÐÔµÄÆøζ²úÉú£»
£¨5£©¼ÓÈÈÂÈ»¯ÌúÈÜÒºÉú³ÉºìºÖÉ«³Áµí£¬ÕâÊÇÂÈ»¯ÌúÓëË®·´Ó¦Éú³ÉÁËÇâÑõ»¯ÌúºÍÑÎËáµÄÔµ¹Ê£¬¸ù¾ÝÌâ¸ÉÌṩµÄÐÅÏ¢¿ÉÒÔÖªµÀ£¬ÔÚpH£¾5µÄÑÎËáÈÜÒºÖÐÄÑÈÜÐԼÈܽ⣬¹ÊÌFeCl3+3H20Fe£¨OH£©3¡ý+3HCl£®
¿¼µã£ºÑεĻ¯Ñ§ÐÔÖÊ£»ÈÜÒºµÄËá¼î¶È²â¶¨£»ÈÜÒºµÄËá¼îÐԲⶨ£»³£¼û»¯·ÊµÄÖÖÀàºÍ×÷Óã»ï§Ì¬µª·ÊµÄ¼ìÑ飻Êéд»¯Ñ§·½³Ìʽ¡¢ÎÄ×Ö±í´ïʽ¡¢µçÀë·½³Ìʽ£®
µãÆÀ£º±¾Ì⿼²éÁ˳£¼ûÑεÄÐÔÖÊÒÔ¼°ÓйØÈÜÒºËá¼îÐԵIJⶨ£¬Íê³É´ËÌ⣬¿ÉÒÔÒÀ¾ÝÒÑÓеÄ֪ʶ½øÐÐ

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

Ä¿Ç°Å©´åÕýÔÚÍƹ㡰²âÍÁÅ䷽ʩ·Ê¡±¼¼Êõ£¬Å©¼¼Ô±¶ÔijÍÁµØ¼ì²âºó¸ø³öÁËÊ©·ÊÅä·½£¬Åä·½ÖÐÖ÷ÒªÓÐKNO3¡¢K2SO4¡¢NH4NO3¡¢NH4HCO3µÈÎïÖÊ£¬ÉÏÊö·ÊÁÏÖÐÊôÓÚ¸´ºÏ·ÊÁϵÄÊÇ            ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÎÊ´ðÌâ

С³Çͬѧ½«Ò»¸öÐÂÏʵļ¦µ°·ÅÔÚÊ¢ÓÐ×ãÁ¿Ï¡ÑÎËáµÄ²£Á§±­
ÖУ¬¿É¹Û²ìµ½µÄÏÖÏóÊÇ£¨1£©___________________________________£»
²úÉú´ËÏÖÏóµÄÔ­ÒòÊÇ£¨2£©______________________________________
____________________________________________________________
___________________________________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÎÊ´ðÌâ

ÈçͼÊÇijͬѧ¼ø±ð̼ËáÇâ李¢ÁòËá李¢ÏõËáï§ÈýÖÖ»¯·ÊµÄ¹ý³Ì£¨·´Ó¦Ìõ¼þδ±ê³ö£©£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©²½Öè¢ÙÖÐͨ¹ý¹Û²ìµ½¡¡     ¡¡ÏÖÏ󣬿ÉÒÔ¼ø±ð³ö̼ËáÇâ泥®
£¨2£©²½Öè¢ÚÖÐËùÐèÊÔ¼Á¿ÉÒÔÑ¡ÓÃÁ½ÖÖ²»Í¬Àà±ð£¨°´Ëá¡¢¼î¡¢ÑΡ¢Ñõ»¯Îï½øÐзÖÀࣩµÄÎïÖÊ£¬Æ仯ѧʽ·Ö±ðΪ¡¡      ¡¡¡¢¡¡      ¡¡£®
Çëд³öÁòËá立ֱðÓëÕâÁ½ÖÖÊÔ¼Á·¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º¡¡               ¡¡¡¢¡¡              ¡¡£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÎÊ´ðÌâ

Éúʯ»Ò¿É×÷ʳƷ°ü×°´üÄڵĸÉÔï¼Á£¬Æä¸ÉÔïµÄÔ­ÀíÊÇʲô£¨Óû¯Ñ§·½³Ìʽ±íʾ£©             ÓùýÒ»¶Îʱ¼äºó£¬¸ÉÔï¼ÁÖеÄÎïÖÊ×î¶àÓР  ÖÖ£®ÓÐЩʳƷ°ü×°´üÄڵĸÉÔï¼ÁÊÇÌú·Û£¬Ìú·ÛʧЧºó»á±ä³É       É«£»Ìú·ÛΪʲô±ÈÉúʯ»Ò¸üÄÜÑÓ³¤Ê³Æ·±£ÖÊÆÚ£¿

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÎÊ´ðÌâ

ͼÊÇij»¯¹¤³§Éú²úÉÕ¼îµÄ¹¤ÒµÁ÷³Ìͼ£®

Çë¸ù¾ÝÒÔÉÏÐÅÏ¢»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Çëд³öXÎïÖÊÔÚʵÑéÊÒÖеÄÒ»ÖÖÓÃ;¡¡                         ¡¡£®
£¨2£©·´Ó¦³ØÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ¡¡                                  ¡¡£®
£¨3£©²Ù×÷¢ÙµÄÃû³ÆÊÇ¡¡  ¡¡£¬½á¾§µÃµ½µÄ¹ÌÌåÉÕ¼îÖпÉÄܺ¬ÓÐÉÙÁ¿µÄ¡¡    ¡¡£¨Ð´»¯Ñ§Ê½£©£®
£¨4£©ÂËÒºD¿É¼ÓÈë·´Ó¦³ØÑ­»·ÔÙÀûÓã¬Ä¿µÄÊǽµµÍÉú²ú³É±¾ºÍ·ÀÖ¹¡¡     ¡¡£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌ½¾¿Ìâ

(4·Ö)ʵÑéÊÒÓÐһƿ±êÇ©ÆÆËðµÄÈÜÒº£¬ÒÑÖªÆäÖк¬ÓÐÄÆÔªËØ£¬ÍƲâ¿ÉÄÜÊÇNa2SO4¡¢Na2CO3¡¢NaCl¡¢NaOHÆäÖеÄÒ»ÖÖ»ò¼¸ÖÖ£¬ÎªÌ½¾¿Æä³É·Ö£¬×öÁËÒÔÏÂʵÑ飺
¢Ù²âµÃÈÜÒºpH´óÓÚ7£»
¢Ú¼ÓÈë×ãÁ¿BaCl2ÈÜÒººó£¬Óа×É«³Áµí²úÉú£¬¹ýÂË£»
¢ÛÏòÂ˳öµÄ³ÁµíÖеμÓÏ¡ÏõËᣬ²¿·Ö³ÁµíÈܽâÇÒÓÐÆøÌå²úÉú£»
ÇëÍƲ⣺£¨1£©¸ÃÈÜÒºÖÐÒ»¶¨º¬ÓÐʲôÎïÖÊ£¿Ð´³öÉú³É¢ÛÖÐδÈܽâ³ÁµíµÄ»¯Ñ§·½³Ìʽ¡£
£¨2£©ÓûÖ¤Ã÷ÈÜÒºÖв»º¬NaOH£¬ÐèÒª²¹³äʲôʵÑ飿Çëд³öʵÑé²½Öè¼°ÏÖÏó¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌ½¾¿Ìâ

£¨7·Ö£©ÒÑÖª£º³£ÎÂÏ£¬CO2¡¢Ë®¶¼ÄÜÓë¹ýÑõ»¯ÄÆ£¨»¯Ñ§Ê½ Na2O2£©·´Ó¦²úÉú O2£¬ÆäÖÐCO2 Óë Na2O2 ·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ 2CO2 + 2Na2O2 =2Na2CO3 +O2¡£Ä³Ð£»¯Ñ§»î¶¯Ð¡×éΪ̽¾¿ CO2ÓëNa2O2·´Ó¦µÄ²úÎïµÄÐÔÖÊ£¬Éè¼ÆÁËÈçÏÂͼËùʾµÄʵÑé×°Öá£

¢ÅÒÇÆ÷ a µÄÃû³ÆÊÇ£º                    ¡£
¢Æ×°Öà B µÄ×÷ÓÃÊÇ                            ¡£
¢Ç×°Öà A ÖУ¬·¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ                                        ¡£
¢È×°Öà D ²£Á§¹ÜÖУ¬¿ÉÒԹ۲쵽µÄÏÖÏóÊÇ                                                               £¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                                                                          ¡£
¢É·´Ó¦ºó£¬½«×°Öà C ÖеĹÌÌåÎïÖÊÈÜÓÚË®Åä³ÉÈÜÒº£¬È»ºóÏò¸ÃÈÜÒºÖмÓÈë              £¨Ñ¡Ìî¡°ÑÎËᡱ¡¢¡°CaCl2 ÈÜÒº¡±¡¢¡°Ê¯ÈïÊÔÒº¡±Ö®Ò»£©£¬»á³öÏÖ                       µÄÏÖÏó¡£
¢Ê·´Ó¦Íê±Ïºó£¬²âµÃ×°Öà C µÄ×ÜÖÊÁ¿Ôö¼ÓÁË 14g£¬Ôò²úÉú O2 µÄÖÊÁ¿Îª   g¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌ½¾¿Ìâ

ʵÑéÊÒÓÐһƿ±£¹Ü²»µ±µÄÊÔ¼Á£¨Èçͼ£©£¬Æä²ÐȱµÄ±êÇ©ÖÐֻʣÏ¡°Na¡±ºÍ¡°10%¡±×ÖÑù¡£ÒÑÖªËüÊÇÎÞÉ«ÒºÌ壬ÊdzõÖл¯Ñ§³£ÓõÄÊÔ¼Á¡£

£¨1£©¸ù¾ÝÊÜËð±êÇ©µÄÇé¿öÅжϣ¬ÕâÆ¿ÊÔ¼Á²»¿ÉÄÜÊÇ         
A£®Ëá        B£®¼î       C£®ÑÎ
£¨2£© ÒÑÖª ¢ñ£®³õÖл¯Ñ§³£¼ûµÄº¬ÄÆ»¯ºÏÎïÓÐNaCl¡¢NaOH¡¢Na2CO3¡¢NaHCO3¡£
¢ò£®Na2CO3ºÍNaHCO3ÈÜÒº¶¼³Ê¼îÐÔ¡£
¢ó£®ÊÒΣ¨20¡æ£©Ê±£¬ËÄÖÖÎïÖʵÄÈܽâ¶ÈµÄÊý¾ÝÈçÏÂͼ£º

ÎïÖÊ
NaCl
NaOH
Na2CO3
NaHCO3
Èܽâ¶Èg
36
109
215
9£®6
 
¸ù¾ÝÊÔ¼ÁÆ¿±ê×¢µÄÈÜÖÊÖÊÁ¿·ÖÊý10%ºÍÓÒ±íÖеÄÈܽâ¶ÈµÄÊý¾ÝÅжϣ¬ÕâÆ¿ÊÔ¼Á²»¿ÉÄÜÊÇ            ¡£
£¨3£© ÓÃpHÊÔÖ½ÉϲâµÃ¸ÃÈÜÒºµÄpH£¾7£¬ÕâÆ¿ÊÔ¼Á²»¿ÉÄÜÊÇ          ¡£
£¨4£© ΪÁËÈ·¶¨¸ÃÈÜÒºÊÇÄÄÖÖÈÜÒº£¬ÏÖ½øÐÐÈçϽøÒ»²½µÄʵÑ飺
²Ù×÷²½Öè
ʵÑéÏÖÏó
  ½áÂÛ¼°»¯Ñ§·½³Ìʽ
È¡ÑùÓÚÊÔ¹ÜÖУ¬µÎ¼Ó
            
 ²úÉú´óÁ¿µÄÆøÅÝ
¸ÃÈÜÒºÊÇ           £¬·´Ó¦µÄ»¯Ñ§·½³Ìʽ
                                    
£¨5£©»¹¿ÉÒÔÑ¡ÔñÓ루4£©Öв»Í¬µÄÊÔ¼ÁÈ·¶¨¸ÃÈÜÒº£¬ÄãÑ¡ÔñµÄÊÔ¼ÁÊÇ         £¨ÒªÇóÀà±ð²»Í¬£©¡£
£¨6£© ΪÁË̽¾¿Ò»Æ¿ÂÈ»¯¸ÆÈÜÒºµÄÖÊÁ¿·ÖÊý£¬È¡¸ÃÈÜÒº50g£¬¼ÓÈë50g̼Ëá¼ØÈÜÒº£¬Ç¡ºÃÍêÈ«·´
Ó¦£¬ËùµÃÈÜÒºÖÊÁ¿95g£¬¼ÆËã´ËÂÈ»¯¸ÆÈÜÒºµÄÖÊÁ¿·ÖÊý¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸