ij»¯Ñ§ÐËȤС×éÓûÔÚʵÑéÊÒ̽¾¿ÂÈÆøµÄÐÔÖʼ°Ä£Ä⹤ҵÖÆȡƯ°×·Û£¬Éè¼ÆÁËÈçÏÂ×°ÖýøÐÐʵÑ飨ʵÑéÊÒÖÆÈ¡ÂÈÆøµÄ·´Ó¦ÎªMnO2 + 4 HCI(Ũ£©MnCl+ C12¡ü+ 2 H2O£©£º

Çë°´ÒªÇó»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©µí·Ûµâ»¯¼ØÈÜÒºÖй۲쵽µÄÏÖÏóÊÇ_________£¬·´Ó¦µÄÀë×Ó·½³Ìʽ____________¡£
£¨2£©Èô¸ÉÔïµÄÓÐÉ«²¼Ìõ²»ÍÊÉ«£¬ÊªÈóµÄÓÐÉ«²¼ÌõÍÊÉ«£¬Ôò¼×ÖÐÊ¢·Å________£¨ÌîÃû³Æ£©¡£
£¨3£©C12Óëʯ»ÒÈé·´Ó¦ÖÆȡƯ°×·ÛµÄ»¯Ñ§·½³ÌʽΪ___________________¡£
£¨4£©¸ÃÐËȤС×éÓÃ8.7g MnO2Óë×ãÁ¿µÄŨÑÎËáÖƱ¸ÂÈÆø£¬ÔòÀíÂÛÉÏ×î¶à¿ÉÖƵñê×¼×´¿öϵÄCl2______________L¡£

£¨1£©ÈÜÒº±äÀ¶É«    Cl2+2I-=2Cl-+I2
£¨2£©Å¨ÁòËá
£¨3£©2Cl2+2Ca(OH)2£½CaCl2+Ca(ClO)2+2H2O
£¨4£©2.24

½âÎöÊÔÌâ·ÖÎö£º£¨1£©MnO2ÓëŨÑÎËá»ìºÏ¼ÓÈÈ·¢Éú·´Ó¦ÖÆÈ¡Cl2,ÓÉÓÚHClÓлӷ¢ÐÔ£¬ÊµÑéÔÚCl2Öк¬ÓÐÔÓÖÊHCl£¬Óñ¥ºÍNaClÈÜÒºÀ´³ýÈ¥HClÔÓÖÊÓпÉÒÔ¼õÉÙCl2µÄÏûºÄ¡£ÓÉÓÚCl2ÓÐÑõ»¯ÐÔ£¬»áÓëKI·¢Éú·´Ó¦£ºCl2+2KI=2KCl+I2¡£I2Óöµí·Û±äÀ¶É«¡£·´Ó¦µÄÀë×Ó·½³ÌʽΪCl2+2I-=2Cl-+I2¡££¨2£©Èô¸ÉÔïµÄÓÐÉ«²¼Ìõ²»ÍÊÉ«£¬ÊªÈóµÄÓÐÉ«²¼ÌõÍÊÉ«£¬Ôò¼×ÖÐÊ¢·ÅŨÁòËá¡££¨3£©C12Óëʯ»ÒÈé·´Ó¦ÖÆȡƯ°×·ÛµÄ»¯Ñ§·½³ÌʽΪ2Cl2+2Ca(OH)2£½CaCl2+Ca(ClO)2+2H2O¡££¨4£©n(MnO2)=8.7g¡Â87g/mol="0.1mol." ÓÉÓÚHCl¹ýÁ¿¡£ËùÒԷųöµÄÆøÌåÓÉMnO2¾ö¶¨¡£¸ù¾Ý·½³ÌʽMnO2 + 4 HCI(Ũ£©MnCl+ C12¡ü+ 2 H2O¿ÉµÃ·Å³öµÄC12µÄÎïÖʵÄÁ¿Îª0.1mol.ÔÚ±ê×¼×´¿öÏ£¬ÆäÌå»ýΪ2.24L¡£
¿¼µã£º¿¼²éC12µÄʵÑéÊÒÖÆ·¨¼°ÓйصÄÐÔÖÊ¡¢ÏÖÏó¼°ÓйØÆøÌåĦ¶ûÌå»ýµÄ¼ÆËãµÄ֪ʶ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

°±Ë®ÊÇÒ»Öֵij£ÓõĻ¯Ñ§ÊÔ¼Á¡£
(1)ÒÔŨ°±Ë®ºÍÉúʯ»ÒΪԭÁÏ£¬ÓÃÈçÏÂ×°ÖÃΪ·¢Éú×°Ö㬿ÉÖÆÈ¡ÉÙ Á¿°±Æø¡£ÒÑÖª£ºNH3¡¤H2O+CaO=Ca(OH)2+NH3¡ü¡£

¢ÙÉúʯ»ÒÓ¦¸Ã×°ÔÚ   (ÌîA»òB)¡£
¢ÚÊÔ´ÓµçÀëƽºâ½Ç¶È·ÖÎö¸ÃʵÑéÖа±ÆøÒݳöµÄÔ­Òò£º                                            ¡£
¢ÛÏÖÒªÓÃÊԹܺͱØҪʵÑéÓÃÆ·ÊÕ¼¯Ò»ÊԹܰ±Æø£¬ÇëÔÚC´¦²¹³äÏàÓ¦ÊÕ¼¯×°Öá£
(2)»¯Ñ§ÐËȤС×éÓð±Ë®ÓëÏõËáÒøÈÜÒºÅäÖƺÃÒø°±ÈÜÒººó£¬½øÐÐÒÒÈ©µÄÒø¾µ·´Ó¦ÊµÑ飬µ«²¿·ÖͬѧʵÑéʱ¼ä³¤£¬ÇÒЧ¹û²»Ã÷ÏÔ¡£Ó°ÏìÒÒÈ©Òø¾µ·´Ó¦ËÙÂʵÄÒòËØÓÐÄÄЩÄØ£¿ÇëÄãÍê³ÉÈçϲÂÏ룺
¢Ù²ÂÏ룺ÒòËØÒ»£ºÒø°±ÈÜÒºµÄpH´óС£»
ÒòËضþ£º                    ¡£
¢ÚÄãÉè¼ÆʵÑéÑéÖ¤ÉÏÊöÓ°ÏìÒòËØÒ»£¬Íê³ÉϱíÖÐÄÚÈÝ¡£
ÌṩÊÔ¼Á£ºÒÒÈ©¡¢2%Ï¡°±Ë®¡¢ 0£®25mol/L NaOHÈÜÒº¡¢ÑÎËá ¡¢ 2% AgNO3ÈÜÒº
ʵÑéÓÃÆ·£ºÉÕ±­(×°ÓÐÈÈË®)¡¢ÊÔ¹Ü

ʵÑé²½Öè
ʵÑé²Ù×÷
Ô¤ÆÚÏÖÏó¼°½áÂÛ(²½Öè1Ö»ÌîÏÖÏó)
1
ÔÚA¡¢BÁ½Ö§½à¾»µÄÊÔ¹ÜÖи÷Èë1mL 2%µÄAgNO3ÈÜÒº£¬È»ºó±ßÕñµ´ÊԹܱßÖðµÎµÎÈë2%Ï¡°±Ë®£¬ÖÁ×î³õ²úÉúµÄ³ÁµíÇ¡ºÃÍêÈ«Èܽ⣬´ËʱÖƵÃpH¾ùԼΪ8µÄÒø°±ÈÜÒº¡£
 
2
 
 
 
(3)ÎüÈ¡20£®00ml ÉÏÊö°±Ë®ÓÚ׶ÐÎÆ¿£¬µÎ¼Ó3µÎ¼×»ù³Èָʾ¼Á£¬ÓÃ0£®0050mol/L±ê×¼ÁòËáµÎ¶¨£¬µÎ¶¨ÖÁ³öÏÖºìɫΪÖյ㣬Öظ´3´ÎʵÑ飬ƽ¾ùÏûºÄ±ê×¼ÁòËá20£®04 ml,¼ÆËãÉÏÊö°±Ë®µÄÎïÖʵÄÁ¿Å¨¶È                  

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

ij»¯Ñ§¿ÎÍâ»î¶¯Ð¡×éÔÚʵÑéÊÒÉè¼ÆÁËÈçͼËùʾµÄʵÑé×°Ö㬽øÐС°°±µÄ´ß»¯Ñõ»¯¡±ÊµÑé¡£

(1)A´¦ÊÇÆøÌå·¢Éú×°Öã¬AÖÐËùÓõÄÊÔ¼ÁÖ»ÄÜ´ÓÏÂÁÐÎïÖÊÖÐÑ¡È¡£º
¢ÙÏõËá泥»¢Ú̼Ëá泥»¢Û̼ËáÇâ泥»¢ÜÂÈ»¯ï§£»¢ÝÊìʯ»Ò£»¢ÞÇâÑõ»¯ÄÆ¡£
ÈôAÖÐÖÆÈ¡ÆøÌåʱֻÓÃÁËÒ»ÖÖÒ©Æ·£¬Ôò¸ÃÒ©Æ·¿ÉÒÔÊÇ       (ÌîÑ¡Ïî±àºÅ)£¬ÔÚÖ»ÓÃÒ»ÖÖÒ©Æ·ÖÆÈ¡°±Æøʱ£¬Í¼Öпհ״¦ËùÐèÒÇÆ÷ӦΪ       (Ñ¡ÌîÏÂÁÐÒÇÆ÷±àºÅ£¬¹Ì¶¨×°ÖÃÊ¡ÂÔ)¡£

(2)¸Ã×°Öò¹³äÍêÕûºó£¬ÈÔÈ»´æÔÚÒ»¶¨µÄȱÏÝ£¬ÊÔ´Ó°²È«Óë»·±£µÄ½Ç¶ÈÀ´¿¼ÂÇ£¬¶Ô¸Ã×°ÖýøÐиĽø£º
¢Ù                                                                  £»
¢Ú                                                                  ¡£
(3)°´ÕոĽøºóµÄ×°ÖýøÐÐʵÑ飬ÇëÍê³ÉÒÔÏÂÎÊÌ⣺
¢Ù×°ÖÃBµÄ×÷ÓÃÊÇ                                                 £»
¢Úд³öCÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º                                    £»
¢ÛÈôA¡¢B´¦ÊÔ¼Á×ãÁ¿£¬Ôò×°ÖÃDÖпÉÒԹ۲쵽µÄʵÑéÏÖÏóÓР              ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

ijУ»¯Ñ§ÊµÑéÐËȤС×éΪÁËÑéÖ¤ÔÚʵÑéÊÒÖƱ¸Cl2µÄ¹ý³ÌÖÐÓÐË®ÕôÆøºÍHCl»Ó·¢³öÀ´£¬Í¬Ê±Ö¤Ã÷Cl2µÄijЩÐÔÖÊ£¬¼×ͬѧÉè¼ÆÁËÈçͼËùʾµÄʵÑé×°ÖÃ(Ö§³ÅÓõÄÌú¼Ų̈ʡÂÔ)£¬Çë°´ÒªÇó»Ø´ðÏÂÁÐÎÊÌâ¡£

(1)ÏÂÁз½·¨ÖУ¬¿ÉÖƵÃCl2µÄÕýÈ·×éºÏÊÇ       ¡£
¢ÙMnO2ºÍŨÑÎËá»ìºÏ¹²ÈÈ£»¢ÚMnO2¡¢NaClºÍŨÁòËá»ìºÏ¹²ÈÈ£»¢ÛNaClOºÍŨÑÎËá»ìºÏ£»¢ÜK2Cr2O7ºÍŨÑÎËá»ìºÏ£»¢ÝKClO3ºÍŨÑÎËá»ìºÏ¹²ÈÈ£»¢ÞKMnO4ºÍŨÑÎËá»ìºÏ¡£
A£®¢Ù¢Ú¢Þ  B£®¢Ú¢Ü¢Þ
C£®¢Ù¢Ü¢Þ  D£®È«²¿¿ÉÒÔ
(2)д³öʵÑéÊÒÖÆÈ¡Cl2µÄÀë×Ó·½³Ìʽ                              ¡£
(3)¢Ù×°ÖÃBµÄ×÷ÓÃÊÇ                         £»
¢Ú×°ÖÃCºÍD³öÏֵIJ»Í¬ÏÖÏó˵Ã÷µÄÎÊÌâÊÇ                        £»
¢Û×°ÖÃEµÄ×÷ÓÃÊÇ                       ¡£
(4)ÒÒͬѧÈÏΪ¼×ͬѧµÄʵÑéÓÐȱÏÝ£¬²»ÄÜÈ·±£×îÖÕͨÈëAgNO3ÈÜÒºÖеÄÆøÌåÖ»ÓÐÒ»ÖÖ¡£ÎªÁËÈ·±£ÊµÑé½áÂ۵Ŀɿ¿ÐÔ£¬Ö¤Ã÷×îÖÕͨÈëAgNO3ÈÜÒºÖеÄÆøÌåÖ»ÓÐÒ»ÖÖ£¬ÒÒͬѧÌá³öÓ¦¸ÃÔÚ×°Öà      Óë       Ö®¼ä(Ìî×°ÖÃ×ÖĸÐòºÅ)Ôö¼ÓÒ»¸ö×°Öã¬ËùÔö¼Ó×°ÖÃÀïÃæµÄÊÔ¼Á¿ÉÒÔΪ       (Ìî×ÖĸÐòºÅ)¡£
A£®ÊªÈóµÄµâ»¯¼Øµí·ÛÊÔÖ½      B£®Å¨ÁòËá
C£®ÊªÈóµÄºìÉ«²¼Ìõ            D£®±¥ºÍʳÑÎË®¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

¸ù¾ÝÏÂͼװÖýøÐÐʵÑ飬ÒÑÖª£ºNa2O2ÓëH2OºÍCO2¶¼ÄÜ·´Ó¦²¢Éú³ÉO2,µ«ÓëNH3²»·´Ó¦
»Ø´ðÏÂÁÐÎÊÌ⣺¡£

£¨1£©ÔÚÊÜÈȵÄÊÔ¹ÜAÖÐNH4HCO3·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º                               ¡£
£¨2£©±»¼ÓÈȵIJ¬Ë¿´¦·¢ÉúµÄ»¯Ñ§·½³ÌʽΪ£º___________________________________¡£
£¨3£©BÖгöÏÖµÄÏÖÏóΪ£º___________________________________________________¡£
£¨4£©ÉÕ±­CÖз¢ÉúµÄÏÖÏóΪ________________________________________________¡£
£¨5£©µ¹Ö鶷µÄ×÷Óà                                                    ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

ij»¯Ñ§¿ÎÍâ»î¶¯Ð¡×éͨ¹ýʵÑé̽¾¿NO2µÄÐÔÖÊ¡£ÒÑÖª£º2NO2£«2NaOH=NaNO3£«NaNO2£«H2O¡£
ÈÎÎñ1£ºÀûÓÃÈçͼËùʾװÖÃ̽¾¿NO2ÄÜ·ñ±»NH3»¹Ô­(K1¡¢K2¾ùΪֹˮ¼Ð£¬¼Ð³Ö×°ÖÃÒÑÂÔÈ¥)¡£

(1)E×°ÖÃÖÐÖÆÈ¡NO2µÄ»¯Ñ§·´Ó¦·½³ÌʽÊÇ____________________________
____________________________________________¡£
(2)ÈôNO2Äܹ»±»NH3»¹Ô­£¬Ô¤ÆÚÔÚC×°ÖÃÖй۲쵽µÄÏÖÏóÊÇ________________________________________________________________¡£
(3)ʵÑé¹ý³ÌÖУ¬Î´Äܹ۲쵽C×°ÖÃÖеÄÔ¤ÆÚÏÖÏ󡣸ÃС×éͬѧ´Ó·´Ó¦Ô­ÀíµÄ½Ç¶È·ÖÎöÁËÔ­Òò£¬ÈÏΪÓÐÒÔÏÂÈýÖÖ¿ÉÄÜ£º
¢ÙNH3»¹Ô­ÐÔ½ÏÈõ£¬²»Äܽ«NO2»¹Ô­£»
¢ÚÔÚ´ËÌõ¼þÏ£¬NO2µÄת»¯Âʼ«µÍ£»
¢Û______________________________________________________________¡£
(4)´ËʵÑé×°ÖÃÖÐÒ»¸öÃ÷ÏÔµÄȱÏÝÊÇ__________________________________¡£
ÈÎÎñ2£ºÌ½¾¿NO2ÄÜ·ñÓëNa2O2·¢ÉúÑõ»¯»¹Ô­·´Ó¦¡£
(5)ʵÑéÇ°£¬¸ÃС×éͬѧÌá³öÈýÖÖ¼ÙÉè¡£
¼ÙÉè1£º¶þÕß²»·´Ó¦£»
¼ÙÉè2£ºNO2Äܱ»Na2O2Ñõ»¯£»
¼ÙÉè3£º________________________________________________________¡£
(6)ΪÁËÑéÖ¤¼ÙÉè2£¬¸ÃС×éͬѧѡÓÃÈÎÎñ1ÖеÄB¡¢D¡¢E×°Ö㬽«BÖеÄÒ©Æ·¸ü»»³ÉNa2O2£¬ÁíÑ¡F×°ÖÃ(ÈçͼËùʾ)£¬ÖØÐÂ×é×°½øÐÐʵÑé¡£

¢Ù×°ÖõĺÏÀíÁ¬½Ó˳ÐòÊÇ(ijЩװÖÿÉÒÔÖظ´Ê¹ÓÃ)________¡£
¢ÚʵÑé¹ý³ÌÖУ¬B×°ÖÃÖеķÛÄ©Óɵ­»ÆÉ«Öð½¥±ä³É°×É«¡£¾­¼ìÑ飬¸Ã°×É«ÎïÖÊΪ´¿¾»ÎÇÒÎÞÆäËûÎïÖÊÉú³É¡£ÍƲâB×°ÖÃÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ__________________________________

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

ΪÁË̽¾¿AgNO3µÄÑõ»¯ÐÔºÍÈÈÎȶ¨ÐÔ£¬Ä³»¯Ñ§ÐËȤС×éÉè¼ÆÁËÈçÏÂʵÑé¡£
¢ñ.AgNO3µÄÑõ»¯ÐÔ
½«¹âÁÁµÄÌúË¿ÉìÈëAgNO3ÈÜÒºÖУ¬Ò»¶Îʱ¼äºó½«ÌúË¿È¡³ö¡£Îª¼ìÑéÈÜÒºÖÐFeµÄÑõ»¯²úÎ½«ÈÜÒºÖеÄAg£«³ý¾¡ºó£¬½øÐÐÁËÈçÏÂʵÑ飬¿ÉÑ¡ÓÃÊÔ¼Á£ºKSCNÈÜÒº¡¢K3[Fe(CN)6]ÈÜÒº¡¢ÂÈË®¡£
(1)ÇëÍê³ÉÏÂ±í£º

²Ù×÷
ÏÖÏó
½áÂÛ
È¡ÉÙÁ¿³ý¾¡Ag£«ºóµÄÈÜÒºÓÚÊÔ¹ÜÖУ¬¼ÓÈëKSCNÈÜÒº£¬Õñµ´
 
´æÔÚFe3£«
È¡ÉÙÁ¿³ý¾¡Ag£«ºóµÄÈÜÒºÓÚÊÔ¹ÜÖУ¬¼ÓÈë________£¬Õñµ´
 
´æÔÚFe2£«
 
¡¾ÊµÑé½áÂÛ¡¿ FeµÄÑõ»¯²úÎïΪFe2£«ºÍFe3£«¡£
¢ò.AgNO3µÄÈÈÎȶ¨ÐÔ
ÓÃÏÂͼËùʾµÄʵÑé×°ÖÃA¼ÓÈÈAgNO3¹ÌÌ壬²úÉúºì×ØÉ«ÆøÌ壬ÔÚ×°ÖÃDÖÐÊÕ¼¯µ½ÎÞÉ«ÆøÌå¡£µ±·´Ó¦½áÊøºó£¬ÊÔ¹ÜÖвÐÁô¹ÌÌåΪºÚÉ«¡£

(2)×°ÖÃBµÄ×÷ÓÃÊÇ________¡£
(3)¾­Ð¡×éÌÖÂÛ²¢ÑéÖ¤¸ÃÎÞÉ«ÆøÌåΪO2£¬ÆäÑéÖ¤·½·¨ÊÇ________¡£
(4)¡¾²éÔÄ×ÊÁÏ¡¿ Ag2OºÍ·Ûĩ״µÄAg¾ùΪºÚÉ«£»Ag2O¿ÉÈÜÓÚ°±Ë®¡£
¡¾Ìá³öÉèÏë¡¿ÊÔ¹ÜÖвÐÁôµÄºÚÉ«¹ÌÌå¿ÉÄÜÊÇ£º¢¡.Ag£»¢¢.Ag2O£»¢£.AgºÍAg2O¡£
¡¾ÊµÑéÑéÖ¤¡¿¸ÃС×éΪÑéÖ¤ÉÏÊöÉèÏ룬·Ö±ðÈ¡ÉÙÁ¿ºÚÉ«¹ÌÌå·ÅÈëÊÔ¹ÜÖУ¬½øÐÐÁËÈçÏÂʵÑé¡£
ʵÑé±àºÅ
²Ù×÷
ÏÖÏó
a
¼ÓÈë×ãÁ¿°±Ë®£¬Õñµ´
ºÚÉ«¹ÌÌå²»Èܽâ
b
¼ÓÈë×ãÁ¿Ï¡ÏõËᣬÕñµ´
ºÚÉ«¹ÌÌåÈܽ⣬²¢ÓÐÆøÌå²úÉú
 
¡¾ÊµÑéÆÀ¼Û¡¿¸ù¾ÝÉÏÊöʵÑ飬²»ÄÜÈ·¶¨¹ÌÌå²úÎï³É·ÖµÄʵÑéÊÇ________(ÌîʵÑé±àºÅ)¡£
¡¾ÊµÑé½áÂÛ¡¿¸ù¾ÝÉÏÊöʵÑé½á¹û£¬¸ÃС×éµÃ³öAgNO3¹ÌÌåÈÈ·Ö½âµÄ²úÎïÓÐ________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

ʵÑéÊÒÓÃ×ãÁ¿MnO2ÓëŨÑÎËá·´Ó¦ÖÆÈ¡ÂÈÆø£¬Æä×°ÖÃÈçͼ1Ëùʾ£º
          
ͼ1                         Í¼2
£¨1£©Í¼1ÖÐÒÇÆ÷aµÄÃû³ÆÊÇ£º        £»ÒÇÆ÷bµÄÃû³ÆÊÇ£º       £»bÖмÓÈëËé´ÉƬµÄ×÷ÓÃÊÇ£º       ¡£
£¨2£©Çëд³öÒÇÆ÷bÖз¢ÉúµÄ·´Ó¦µÄÀë×Ó·½³Ìʽ£º                       
£¨3£©¼ì²é×°ÖõÄÆøÃÜÐÔÖ®ºóµÄ²Ù×÷ÒÀ´ÎÊÇ£º       ¡¢       ¡¢        ¡££¨ÌîÐòºÅ£©
A£®ÏòÉÕÆ¿ÖмÓÈëMnO2·ÛÄ©
B£®¼ÓÈÈ
C£®ÏòÉÕÆ¿ÖмÓÈëŨÑÎËá
£¨4£©¸Ã·´Ó¦»áÒòΪÑÎËáŨ¶ÈϽµ¶øÍ£Ö¹¡£ÎªÁ˲ⶨ·´Ó¦²ÐÁôÒºÖÐÑÎËáµÄŨ¶È£¬Ä³Ì½¾¿Ð¡×éÌá³öÏÂÁÐʵÑé·½°¸£º
¢Ù¼×ͬѧµÄ·½°¸Îª£ºÓë×ãÁ¿AgNO3ÈÜÒº·´Ó¦£¬³ÆÁ¿Éú³É³ÁµíµÄÖÊÁ¿¡£
¢ÚÒÒͬѧµÄ·½°¸Îª£ºÓë×ãÁ¿µÄп·´Ó¦£¬²âÁ¿Éú³ÉÆøÌåµÄÌå»ý£¬ÊµÑé×°ÖÃÈçͼ2Ëùʾ£¨¼Ð³Ö×°ÖÃÒÑÂÔÈ¥£©¡£Ê¹YÐιÜÖеIJÐÁôÈÜÒºÓëпÁ£·´Ó¦µÄÕýÈ·²Ù×÷ÊÇ           £¨¡°Ð¿Á£×ªÒƵ½²ÐÁôÈÜÒºÖС±»ò¡°²ÐÁôÈÜҺתÒƵ½Ð¿Á£ÖС±£©¡£ÔÚÕýÈ·¶ÁÈ¡Á¿Æø¹Ü¶ÁÊýʱ£¬ÊÓÏßҪƽÊÓ£¬Òª×¢Òâʹ©¶·ÒºÃæÓëÁ¿Æø¹ÜÖÐÒºÃæÏàƽ£¬³ý´ËÍ⻹Ðë×¢Ò⣺               ¡£
Á½ÖÖ·½°¸ÎÒÈÏΪ    £¨Ìî¼×»òÒÒ£©Í¬Ñ§µÄ·½°¸¿ÉÐС£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸