¡¾ÌâÄ¿¡¿µâ»¯ÄÆÓÃ×÷¼××´ÏÙÖ×Áö·ÀÖμÁ¡¢ìî̵¼ÁºÍÀûÄò¼ÁµÈ¡£ÊµÑéÊÒÓÃNaOH¡¢µ¥ÖʵâºÍË®ºÏëÂ(N2H4¡¤H2O)ΪÔÁÏ¿ÉÖƱ¸µâ»¯ÄÆ¡£
×ÊÁÏÏÔʾ£ºË®ºÏëÂÓл¹ÔÐÔ£¬ÄÜÏû³ýË®ÖÐÈܽâµÄÑõÆø£»NaIO3ÊÇÒ»ÖÖÑõ»¯¼Á¡£
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ë®ºÏëµÄÖƱ¸£ºÓйط´Ó¦ÔÀíΪ£ºNaClO + 2NH3 = N2H4¡¤H2O + NaCl¡£
¢ÙÓÃÏÂͼװÖÃÖÆÈ¡Ë®ºÏ룬ÆäÁ¬½Ó˳ÐòΪ_________(°´ÆøÁ÷·½Ïò£¬ÓÃСд×Öĸ±íʾ)¡£
¢Ú×°ÖÃAµÄ×÷ÓÃÊÇ_______¡£
¢Û¿ªÊ¼ÊµÑéʱ£¬ÏÈÏòÑõ»¯¸ÆÖеμÓŨ°±Ë®£¬Ò»¶Îʱ¼äºóÔÙÏòBµÄÈý¿ÚÉÕÆ¿ÖеμÓNaClOÈÜÒº¡£µÎ¼ÓNaClOÈÜҺʱ²»Äܹý¿ìµÄÀíÓÉ___________¡£
£¨2£©µâ»¯ÄƵÄÖƱ¸
¢¡.ÏòÈý¿ÚÉÕÆ¿ÖмÓÈë8.4gNaOH¼°30mLË®£¬½Á°è¡¢ÀäÈ´£¬¼ÓÈë25.4gµâµ¥ÖÊ£¬¿ª¶¯´ÅÁ¦½Á°èÆ÷£¬±£³Ö60~70¡æÖÁ·´Ó¦³ä·Ö£»
¢¢£®¼ÌÐø¼ÓÈëÉÔ¹ýÁ¿µÄN2H4¡¤H2O(Ë®ºÏëÂ)£¬»¹ÔNaIOºÍNaIO3£¬µÃNaIÈÜÒº´ÖÆ·£¬Í¬Ê±ÊÍ·ÅÒ»ÖÖ¿ÕÆøÖеÄÆøÌ壻
¢££®ÏòÉÏÊö·´Ó¦ÒºÖмÓÈë1.0g»îÐÔÌ¿£¬Öó·Ð°ëСʱ£¬È»ºó½«ÈÜÒºÓë»îÐÔÌ¿·ÖÀ룻
¢¤£®½«²½Ö袣·ÖÀë³öµÄÈÜÒºÕô·¢Å¨Ëõ¡¢ÀäÈ´½á¾§¡¢¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔµÃ²úÆ·24.0g¡£
¢Ü²½Ö袡ÊÊÒ˲ÉÓõļÓÈÈ·½·¨ÊÇˮԡ¼ÓÈÈ£¬ ¸Ã²½Öè·´Ó¦ÍêÈ«µÄÏÖÏóÊÇ_________¡£²½Ö袢ÖÐIO3£²ÎÓë·´Ó¦µÄÀë×Ó·½³ÌʽΪ________________________________¡£
¢Ý²½Ö袣 ¡°½«ÈÜÒºÓë»îÐÔÌ¿·ÖÀ롱µÄ·½·¨ÊdzÃÈȹýÂË¡£
¢Þ±¾´ÎʵÑé²úÂÊΪ__________¡£ÊµÑé·¢ÏÖ£¬Ë®ºÏëÂʵ¼ÊÓÃÁ¿±ÈÀíÂÛֵƫ¸ß£¬¿ÉÄܵÄÔÒòÊÇ___________¡£
¢ßijͬѧ¼ìÑé²úÆ·NaIÖÐÊÇ·ñ»ìÓÐNaIO3ÔÓÖÊ¡£È¡ÉÙÁ¿¹ÌÌåÑùÆ·ÓÚÊÔ¹ÜÖУ¬¼ÓË®Èܽ⣬µÎ¼ÓÉÙÁ¿µí·ÛÒººóÔٵμÓÊÊÁ¿Ï¡ÁòËᣬƬ¿ÌºóÈÜÒº±äÀ¶¡£µÃ³öNaIÖк¬ÓÐNaIO3ÔÓÖÊ¡£ÇëÆÀ¼Û¸ÃʵÑé½áÂ۵ĺÏÀíÐÔ____________¡££¨ÈôÈÏΪºÏÀíд³öÀë×Ó·½³Ìʽ£¬ÈôÈÏΪ²»ºÏÀí˵Ã÷ÀíÓÉ£©
¡¾´ð°¸¡¿f a b c d e (ab˳Ðò¿É»¥»») ·ÀÖ¹µ¹Îü»ò°²È«Æ¿ ¹ý¿ìµÎ¼ÓNaClOÈÜÒº£¬¹ýÁ¿µÄNaClOÈÜÒºÑõ»¯Ë®ºÏ룬½µµÍ²úÂÊ ÎÞ¹ÌÌå²ÐÁôÇÒÈÜÒº³ÊÎÞÉ« 2IO3£+ 3N2H4¡¤H2O =3N2¡ü+2I-+9H2O 80% Ë®ºÏëÂÄÜÓëË®ÖеÄÈܽâÑõ·´Ó¦ ¿ÉÄÜÊÇI£ÔÚËáÐÔ»·¾³Öб»O2Ñõ»¯³ÉI2¶øʹµí·Û±äÀ¶
¡¾½âÎö¡¿
(1)¢ÙÓÉ·´Ó¦ÔÀíNaClO+2NH3=N2H4H2O+NaCl¿ÉÖª,ÏÈÓÃD×°ÖòúÉú°±Æø,Ϊ·ÀÖ¹²úÉúµ¹Îü,ÆøÌå½øÈë×°ÖÃAÖÐ,È»ºó°±Æøͨ¹ýcµ¼¹Ü½øÈë×°ÖÃBÖÐ,È»ºó´ÎÂÈËáÄÆÓë°±·´Ó¦,¶àÓàµÄ°±×îºó±»C×°ÖÃÎüÊÕ;ÆäÁ¬½Ó˳ÐòΪfabcde;ÕýÈ·´ð°¸:fabcde¡£¢Ú×°ÖÃAµÄ×÷ÓÃÊÇ·ÀÖ¹µ¹Îü»ò°²È«Æ¿;ÕýÈ·´ð°¸:·ÀÖ¹µ¹Îü»ò°²È«Æ¿¡£
¢Û¹ý¿ìµÎ¼ÓNaClOÈÜÒº,µ¼ÖÂNaClO²»ÄÜÍêÈ«·´Ó¦,¹ýÁ¿µÄNaClOÈÜÒºÄܹ»Ñõ»¯·´Ó¦²úÉúµÄË®ºÏëÂ,µ¼Ö½µµÍ²úÂÊ;ÕýÈ·´ð°¸: ¹ý¿ìµÎ¼ÓNaClOÈÜÒº£¬¹ýÁ¿µÄNaClOÈÜÒºÑõ»¯Ë®ºÏ룬½µµÍ²úÂÊ¡£
(2)¢Ü±£³Ö60~70¡æÇé¿öÏ·¢Éú·´Ó¦,Òò´Ë¿ÉÒÔ²ÉÓÃˮԡ¼ÓÈȵķ½·¨;µâ¹ÌÌåÓëÇâÑõ»¯ÄÆÈÜÒº³ä·Ö·´Ó¦Éú³ÉÎÞÉ«µÄNaIOºÍNaIO3,×îºó¹ÌÌåÈ«²¿Ïûʧ;ËùÒԸò½Öè·´Ó¦ÍêÈ«µÄÏÖÏóÊÇÎÞ¹ÌÌå²ÐÁôÇÒÈÜÒº³ÊÎÞÉ«;N2H4H2O¾ßÓл¹ÔÐÔ,Äܹ»°ÑI03-»¹ÔΪµâÀë×Ó,¶ø-2¼ÛµÄµªÔªËر»Ñõ»¯ÎªµªÆø,Àë×Ó·½³ÌʽΪ: 2IO3£+ 3N2H4¡¤H2O =3N2¡ü+2I-+9H2O;ÕýÈ·´ð°¸:¹ÌÌå²ÐÁôÇÒÈÜÒº³ÊÎÞÉ«; 2IO3£+ 3N2H4¡¤H2O =3N2¡ü+2I-+9H2O¡£
¢Þ8.4gNaOHÓë25.4gµ¥ÖʵⷴӦ£¬ÇâÑõ»¯ÄƹýÁ¿,µâµ¥ÖÊ·´Ó¦ÍêÈ«,µâºÍÇâÑõ»¯ÄÆ·¢Éú·´Ó¦Îª3I2+6NaOH=5NaI+NaIO3+3H2O,ÔòÉú³ÉµÄNaIµÄÖÊÁ¿Îª25.4750/762=25g,Éú³ÉµÄNaIOÓëN2H4
H2O·´Ó¦ËùµÃµÄNaI,·´Ó¦Îª3N2H4
H20+2NaIO3=2NaI+3N2
+9H2O,Ôò6I2
2NaIO3
2NaI¸Ã²½Éú³ÉµÄNaIÖÊÁ¿Îª25.4
300/1524=5g,¹ÊÀíÂÛÉÏÉú³ÉµÄNaIΪ25g+5g=30g,ËùÒÔ,±¾´ÎʵÑéµÄ²úÂÊΪ24/30
100%=80%;Ë®ºÏëÂʵ¼ÊÓÃÁ¿±ÈÀíÂÛֵƫ¸ß,Ë®ºÏ뻹ÔÐÔ½ÏÇ¿,¿ÉÄÜÓëË®ÖеÄÈܽâÑõ·´Ó¦;ÕýÈ·´ð°¸:80%;Ë®ºÏëÂÄÜÓëË®ÖеÄÈܽâÑõ·´Ó¦¡£
¢ß¸Ã·½°¸²»ºÏÀí,µâÀë×Ó¾ßÓÐÇ¿»¹ÔÐÔ,ÔÚËáÐÔÌõ¼þϱ»ÑõÑõ»¯Éú³Éµâµ¥ÖÊ,¿ÉÒÔʹµí·ÛÈÜÒº±äÀ¶;ÕýÈ·´ð°¸:¿ÉÄÜÊÇ1-ÔÚËáÐÔ»·¾³Öб»O2Ñõ»¯³ÉI2¶øʹµí·Û±äÀ¶¡£
Ä꼶 | ¸ßÖÐ¿Î³Ì | Ä꼶 | ³õÖÐ¿Î³Ì |
¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ |
¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍƼö£¡ |
¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍƼö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿³£¼ûµÄÌ«ÑôÄܵç³ØÓе¥¾§¹èÌ«ÑôÄܵç³Ø¡¢¶à¾§¹èÌ«ÑôÄܵç³Ø¡¢GaAsÌ«ÑôÄܵç³Ø¼°Íî÷ïØÎøµÈ¶àÖÖÔªËع¹³É±¡Ä¤Ì«ÑôÄܵç³Ø¡£»Ø´ðÏÂÁÐÎÊÌ⣺
(1)»ù̬ÑÇÍÀë×Ó(Cu+)µÄ¼Û²ãµç×ÓÅŲ¼Ê½Îª__________¡£
(2)Éé¡¢ÎøÊǵÚËÄÖÜÆÚµÄÏàÁÚÔªËØ£¬ÒÑÖªÉéµÄµÚÒ»µçÀëÄÜ(947kJ¡¤mo1£1)´óÓÚÎø(941k]¡¤mol£l)¡£Çë´ÓÔ×ӽṹµÄ½Ç¶È¼ÓÒÔ½âÊÍ__________¡£
(3)H2OµÄ·Ðµã¸ßÓÚH2SeµÄ·Ðµã£¬ÆäÔÒòÊÇ__________¡£
(4)GaCl3ºÍAsF3µÄ·Ö×ÓÁ¢Ìå¹¹ÐÍ·Ö±ðÊÇ__________£¬__________¡£
(5)ÅðËá(H3BO3)±¾Éí²»ÄܵçÀë³öH+£¬ÔÚË®ÖÐÒ×½áºÏÒ»¸öOH£Éú³É[B(OH)4]££¬¶øÌåÏÖÈõËáÐÔ¡£
¢Ù[B(OH)4] £ÖÐBÔ×ÓµÄÔÓ»¯ÀàÐÍΪ__________¡£
¢Ú[B(OH)4] £µÄ½á¹¹Ê½Îª__________¡£
(6)½ð¸ÕʯµÄ¾§°ûÈçͼ£¬ÈôÒÔ¹èÔ×Ó´úÌæ½ð¸Õʯ¾§ÌåÖеÄ̼Ô×Ó£¬±ãµÃµ½¾§Ìå¹è£»Èô½«½ð¸Õʯ¾§ÌåÖÐÒ»°ëµÄ̼Ô×Ó»»³É¹èÔ×Ó£¬ÇÒ̼¡¢¹èÔ×Ó½»Ì棬¼´µÃµ½Ì¼»¯¹è¾§Ìå(½ð¸ÕÉ°)¡£
¢Ù½ð¸Õʯ¡¢¾§Ìå¹è¡¢Ì¼»¯¹èµÄÈÛµãÓɸߵ½µÍµÄÅÅÁÐ˳ÐòÊÇ__________(Óû¯Ñ§Ê½±íʾ)
¢Ú½ð¸ÕʯµÄ¾§°û²ÎÊýΪapm(lpm=10£12m)£¬1cm3¾§ÌåµÄƽ¾ùÖÊÁ¿Îª__________g(Ö»ÒªÇóÁÐËãʽ£¬°¢·ü¼ÓµÂÂÞ³£ÊýµÄֵΪNA)¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÏÂÁйØÓÚ¹èµÄ˵·¨²»ÕýÈ·µÄÊÇ£¨ £©
A.¹èÔڵؿÇÖеĺ¬Á¿½ö´ÎÓÚÑõ
B.¹èÔÚ×ÔÈ»½çÖмÈÓл¯ºÏ̬£¬ÓÖÓÐÓÎÀë̬
C.¹èÊÇÈËÀཫ̫ÑôÄÜת»»ÎªµçÄܵij£ÓòÄÁÏ
D.¹èµ¥Öʵĵ¼µçÄÜÁ¦½éÓÚµ¼ÌåºÍ¾øÔµÌåÖ®¼ä
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿°´ÒªÇóÍê³ÉÏÂÁÐÎÊÌâ:
£¨1£©GÔÚͬһֱÏßÉϵÄ̼Ô×ÓÊýΪ___________£»
£¨2£©ÉÏÊöÓлúÎﻥΪͬ·ÖÒì¹¹ÌåµÄÊÇ__________(Ìî×Öĸ)£»
£¨3£©AµÄϵͳÃüÃûΪ______________£¬ËüÓëCl2ÒÔ1:1±ÈÀý·´Ó¦Ê±²úÎïÓÐ_________ÖÖ£»
£¨4£©ÉÏÊöÓлúÎïÄÜÓë½ðÊôÄÆ·´Ó¦µÄÊÇ_________£¨Ìî×Öĸ£©£¬ÄÜÓë̼ËáÇâÄÆ·´Ó¦µÄÊÇ______£¨Ìî×Öĸ£©£¬BÓëËáÐÔ¸ßÃÌËá¼Ø·´Ó¦µÄ²úÎïÊÇ______________(Ìî½á¹¹¼òʽ)£»
£¨5£©CµÄͬ·ÖÒì¹¹ÌåÓÐ________ÖÖ¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿A£¨C2H4£©ÊÇ»ù±¾Óлú»¯¹¤ÔÁÏ¡£ÓÉAÖƱ¸ÖØÒªÖмäÌåG·Ïߣ¨²¿·Ö·´Ó¦Ìõ¼þÂÔÈ¥£©ÈçÏÂËùʾ£º
ÒÑÖª£º£¨1£©
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©BµÄµç×Óʽ_________£¬CÖк¬ÓеĹÙÄÜÍŵÄÃû³ÆÊÇ______________¡£
£¨2£©·´Ó¦£¨1£©ºÍ£¨4£©µÄ·´Ó¦ÀàÐÍ·Ö±ðÊÇ_____________£¬_____________¡£
£¨3£©·´Ó¦£¨2£©µÄ·´Ó¦Ìõ¼þÊÇ____________¡£
£¨4£©ÒÑÖªGµÄ·Ö×ÓʽΪC8H12O3£¬Æä½á¹¹¼òʽΪ_______________¡£
£¨5£©Ð´³ö·´Ó¦(3)µÄ»¯Ñ§·´Ó¦·½³Ìʽ_________________________________¡£
£¨6£©XÓëC»¥ÎªÍ¬ÏµÎïÇÒ±ÈC¶àÒ»¸ö̼£¬ÔòXÓÐ___Öֽṹ£¨ÒÑ֪˫¼üÁ¬ôÇ»ù²»Îȶ¨²»¿¼ÂÇ£©
£¨7£©²ÎÕÕÉÏÊöºÏ³É·Ïߣ¬Éè¼ÆÒ»ÌõÓÉÒÒȲºÍ±ûͪΪÆðʼÔÁÏÖƱ¸2-¼×»ù-1,3¶¡¶þÏ©µÄºÏ³É·Ïß__________¡£Êéд¸ñʽÈçͼ£º
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÒÒÏ©µÄ²úÁ¿ÊǺâÁ¿Ò»¸ö¹ú¼ÒʯÓÍ»¯¹¤·¢Õ¹Ë®Æ½µÄÖØÒª±êÖ¾¡£
I£®½«8.96L (±ê×¼×´¿ö)ÒÒÏ©ºÍÒÒÍéµÄ»ìºÏÆøÌåͨÈë×ãÁ¿µÄäåµÄËÄÂÈ»¯Ì¼ÈÜÒºÖУ¬³ä·Ö·´Ó¦£¬äåµÄËÄÂÈ»¯Ì¼ÈÜÒºµÄÖÊÁ¿Ôö¼Ó8.4g£¬ÔòÔÆøÌå»ìºÏÎïÖÐÒÒÏ©ÓëÒÒÍéµÄÎïÖʵÄÁ¿Ö®±ÈΪ______________¡£
II£®ÒÑÖªÒÒÏ©ÄÜ·¢ÉúÒÔÏÂת»¯£º
¢Åд³öB¡¢D»¯ºÏÎïÖйÙÄÜÍŵÄÃû³Æ£ºB____________________£»D___________________£»
¢Æд³öÏà¹Ø·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
¢Ù_________________________________ ·´Ó¦ÀàÐÍ£º________________
¢Ú__________________________________¡¡·´Ó¦ÀàÐÍ£º_________________
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÏÂÁÐÓйؽðÊô¼°Æ仯ºÏÎïµÄ˵·¨ÕýÈ·µÄÊÇ
A.ÂÁºÍÇâÑõ»¯ÄÆÈÜÒº·´Ó¦Éú³ÉAl(OH)3ºÍH2O
B.ÄÆÔÚ¿ÕÆøÖÐȼÉÕÉú³Éµ»ÆÉ«µÄNa2O
C.ÌúÔÚ¸ßÎÂÏÂÓëË®ÕôÆø·´Ó¦Éú³ÉFe2O3ºÍH2
D.̼ËáÄÆÈÜÒººÍÇâÑõ»¯¸ÆÈÜÒº·´Ó¦Éú³ÉNaOHºÍCaCO3
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÑÇÏõËáÄÆ(NaNO2)ÊÇÒ»ÖÖ¹¤ÒµÑΣ¬ÔÚÉú²ú¡¢Éú»îÖÐÓ¦Óù㷺£¬Íâ¹Û¿áËÆʳÑÎÇÒÓÐÏÌ棬ÊÇÒ»ÖÖ³£Óõķ¢É«¼ÁºÍ·À¸¯¼Á£¬µ«Ê¹ÓùýÁ¿»áʹÈËÖж¾¡£Ä³Ñ§Ï°Ð¡×éͨ¹ýÈçÏÂ×°ÖÃ(¼Ð³Ö¼°¼ÓÈÈ×°ÖÃÂÔÈ¥)ÖÆÈ¡NaNO2ÑùÆ·²¢¶ÔÑùÆ·º¬Á¿½øÐвⶨ
¸ÃС×éÏȲéÔÄ×ÊÁÏÖª£ºNO¡¢NO2Äܱ»ËáÐÔ¸ßÃÌËá¼ØÑõ»¯ÎªNO3£¡£
(1)×°ÖÃAÖÐÓÃÓÚÊ¢·ÅŨÏõËáµÄÒÇÆ÷Ãû³ÆΪ__________£¬×°ÖÃDÖз¢ÉúµÄ»¯Ñ§·´Ó¦·½³ÌʽΪ___________¡£
(2)¢Ù×°ÖÃCÖÐËù×°µÄÒºÌåÐèÂú×ãµÄÌõ¼þÊÇ____________________¡£
¢ÚÆä×÷ÓÃΪ____________________¡£
(3)×°ÖÃEµÄ×÷ÓÃÊÇ____________________¡£
(4)Ϊ²â¶¨ÖƵÃÑùÆ·ÖÐNaNO2µÄº¬Á¿£¬¸ÃʵÑéÖÐÐèÅäÖÆKMnO4±ê×¼ÈÜÒº²¢Ëữ£¬Ó¦Ñ¡Ôñ__________(Ìѧʽ)À´ËữKMnO4ÈÜÒº¡£
(5)ÒÑÖª£ºÔÚËáÐÔÈÜÒºÖУ¬NO2£¿É½«MnO4£»¹ÔΪMn2+¡£Îª²â¶¨ÑùÆ·ÖÐÑÇÏõËáÄƵĺ¬Á¿£¬¸ÃС×é³ÆÈ¡4.0gÑùÆ·ÈÜÓÚË®ÅäÖƳÉ250mLÈÜÒº£¬È¡25.00mLÈÜÒºÓÚ׶ÐÎÆ¿ÖУ¬ÓÃ0.100mol¡¤L£1µÄËáÐÔKMnO4ÈÜÒº½øÐе樣¬ÏûºÄ20.00mLËáÐÔKMnO4ÈÜÒº(ÔÓÖʲ»·´Ó¦)¡£µÎ¶¨¹ý³ÌÖÐËáÐÔKMnO4ÈÜÒºµÄ×÷ÓÃÊÇ____________________£¬ËùµÃÑùÆ·ÖÐNaNO2µÄÖÊÁ¿·ÖÊýΪ__________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÏÂÁÐÊÔ¼Á¿É¼ø±ð±½¡¢1-¼ºÏ©¡¢±½·Ó¡¢ÒÒ´¼µÄÊÇ (¡¡¡¡¡¡)
A.Ë®B.ŨäåË®C.ËÄÂÈ»¯Ì¼D.FeCl3ÈÜÒº
²é¿´´ð°¸ºÍ½âÎö>>
°Ù¶ÈÖÂÐÅ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com