2£®Ñо¿ÐÔѧϰС×éµÄͬѧ£¬Îª²â¶¨Ä³º¬Ã¾3%¡«5%µÄÂÁþºÏ½ð£¨²»º¬ÆäËüÔªËØ£©ÖÐþµÄÖÊÁ¿·ÖÊý£¬Éè¼ÆÏÂÁÐÁ½ÖÖ²»Í¬ÊµÑé·½°¸½øÐÐ̽¾¿£®ÌîдÏÂÁпհף®
¡¾·½°¸Ò»¡¿½«ÂÁþºÏ½ðÓë×ãÁ¿NaOHÈÜÒº·´Ó¦£¬²â¶¨Ê£Óà¹ÌÌåÖÊÁ¿£®
ʵÑéÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ2Al+2NaOH+2H2O=2NaAlO2+3H2¡ü
£¨1£©³ÆÈ¡5.4gÂÁþºÏ½ð·ÛÄ©ÑùÆ·£¬ÈÜÓÚV mL 2.0mol/L NaOHÈÜÒºÖУ¬³ä·Ö·´Ó¦£®ÔòNaOHÈÜÒºµÄÌå»ýV¡Ý97mL
£¨2£©¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔï¡¢³ÆÁ¿¹ÌÌ壮¸Ã²½ÖèÖÐÈôδϴµÓ¹ÌÌ壬²âµÃþµÄÖÊÁ¿·ÖÊý½«Æ«¸ß£¨Ìî¡°Æ«¸ß¡±¡¢¡°Æ«µÍ¡±»ò¡°ÎÞÓ°Ï족£©£®
¡¾·½°¸¶þ¡¿½«ÂÁþºÏ½ðÓë×ãÁ¿Ï¡ÁòËáÈÜÒº·´Ó¦£¬²â¶¨Éú³ÉÆøÌåÔÚͨ³£×´¿ö£¨Ô¼20¡æ£¬1.01¡Á105Pa£©µÄÌå»ý£®
£¨1£©Í¬Ñ§ÃÇÄâÑ¡ÓÃÏÂÁÐʵÑé×°ÖÃͼ1Íê³ÉʵÑ飺
¢ÙÄãÈÏΪ×î¼òÒ×µÄ×°ÖÃÆäÁ¬½Ó˳ÐòÊÇ£ºA½ÓE½ÓD½ÓG£¨Ìî½Ó¿Ú×Öĸ£¬¿É²»ÌîÂú£®£©
¢ÚʵÑé½áÊøʱ£¬ÔÚ¶ÁÈ¡²âÁ¿ÊµÑéÖÐÉú³ÉÇâÆøµÄÌå»ýʱ£¬ÄãÈÏΪºÏÀíµÄÊÇACD
A£®´ýʵÑé×°ÖÃÀäÈ´ºóÔÙ¶ÁÊý
B£®ÉÏÏÂÒƶ¯Á¿Í²F£¬Ê¹ÆäÖÐÒºÃæÓë¹ã¿ÚÆ¿ÖÐÒºÃæÏàƽ
C£®ÉÏÏÂÒƶ¯Á¿Í²G£¬Ê¹ÆäÖÐÒºÃæÓë¹ã¿ÚÆ¿ÖÐÒºÃæÏàƽ
D£®ÊÓÏßÓë°¼ÒºÃæ×îµÍµãˮƽ£¬¶ÁÈ¡Á¿Í²ÖÐË®µÄÌå»ý
£¨2£©×Ðϸ·ÖÎöʵÑé×°Öúó£¬Í¬Ñ§ÃǾ­ÌÖÂÛÈÏΪÒÔÏÂÁ½µã»áÒýÆð½Ï´óÎó²î£ºÏ¡ÁòËáµÎÈë׶ÐÎÆ¿ÖУ¬¼´Ê¹²»Éú³ÉÇâÆø£¬Ò²»á½«Æ¿ÄÚ¿ÕÆøÅųö£¬Ê¹Ëù²âÇâÆøÌå»ýÆ«´ó£»ÊµÑé½áÊøʱ£¬Á¬½Ó¹ã¿ÚÆ¿ºÍÁ¿Í²µÄµ¼¹ÜÖÐÓÐÉÙÁ¿Ë®´æÔÚ£¬Ê¹Ëù²âÇâÆøÌå»ýƫС£®ÓÚÊÇËûÃÇÉè¼ÆÁËͼ2ËùʾµÄʵÑé×°Öã®
¢Ù×°ÖÃÖе¼¹ÜaµÄ×÷ÓÃÓУºÊ¹·ÖҺ©¶·ÄÚÆøÌåѹǿÓë׶ÐÎÆ¿ÄÚÆøÌåѹǿÏàµÈ£¬´ò¿ª·ÖҺ©¶·»îÈûʱϡÁòËáÄÜ˳ÀûµÎϺ͵ÎÈë׶ÐÎÆ¿µÄÏ¡ÁòËáÌå»ýµÈÓÚ½øÈë·ÖҺ©¶·µÄÆøÌåÌå»ý£¬´Ó¶øÏû³ýÓÉÓÚ¼ÓÈëÏ¡ÁòËáÒýÆðµÄÇâÆøÌå»ýÎó²î
¢ÚʵÑéÇ°ºó¼îʽµÎ¶¨¹ÜÖÐÒºÃæ¶ÁÊý·Ö±ðΪV1 mL¡¢V2 mL£®Ôò²úÉúÇâÆøµÄÌå»ýΪV1-V2 mL

·ÖÎö ·½°¸Ò»£ºÂÁÓëÇâÑõ»¯ÄÆÈÜÒº·´Ó¦Éú³ÉÆ«ÂÁËáÄÆÓëÇâÆø£»
£¨1£©Ã¾µÄÖÊÁ¿·ÖÊý×îСʱ£¬½ðÊôÂÁµÄÖÊÁ¿×î´ó£¬ÐèÒªµÄÇâÑõ»¯ÄÆÈÜÒº×î¶à£¬Êµ¼ÊÐèÒªÇâÑõ»¯ÄÆÈÜÒºµÄÌå»ýÓ¦´óÓÚ»òµÈÓÚ×î´óÖµ£¬¾Ý´Ë¼ÆË㣻
£¨2£©Ã¾Éϻḽ×ÅÆ«ÂÁËáÄƵÈÎïÖÊ£¬Î´Ï´µÓµ¼Ö²ⶨµÄþµÄÖÊÁ¿Æ«´ó£»
·½°¸¶þ£º£¨1£©¢Ù×°ÖõÄ×éװ˳Ðò£ººÏ½ðÓëË®·´Ó¦£¬ÓÃÅÅË®Á¿Æø·¨²â¶¨ÇâÆøµÄÌå»ý£¬ÆäÖÐʢˮµÄÊÔ¼ÁÆ¿µ¼¹ÜÒ»¶¨Òª¶Ì½ø³¤³ö£¬Ôö´óѹǿԭÀí½«Ë®Åųö£¬Á¿Í²ÖÐË®µÄÌå»ý¾ÍÊÇÉú³ÉÇâÆøµÄÌå»ý£¬Á¿Í²ÄÚµ¼¹ÜÓ¦ÉìÈëÁ¿Í²µ×²¿£»
¢Ú·´Ó¦·ÅÈȵ¼ÖÂÇâÆøµÄζÈÆ«¸ß£¬¹ÊÓ¦ÀäÈ´ºóÔÙ½øÐжÁÈ¡ÇâÆøµÄÌå»ý£¬¶ÁȡʵÑéÖÐÉú³ÉÇâÆøµÄÌå»ýʱÉÏÏÂÒƶ¯Á¿Í²£¬Ê¹ÆäÖÐÒºÃæÓë¹ã¿ÚÆ¿ÖÐÒºÃæÏàƽ£¬ÊÓÏßÓë°¼ÒºÃæµÄ×îµÍµãˮƽ¶ÁÈ¡ÇâÆøµÄÌå»ý£»
£¨2£©¢Ù±£³Ö·ÖҺ©¶·ÄÚÆøÌåѹǿÓë׶ÐÎÆ¿ÄÚÆøÌåѹǿÏàµÈ£¬´ò¿ª·ÖҺ©¶·»îÈûʱϡÁòËáÄÜ˳ÀûµÎÏ£¬µÎÈë׶ÐÎÆ¿µÄÏ¡ÁòËáÌå»ýµÈÓÚ½øÈë·ÖҺ©¶·µÄÆøÌåÌå»ý£¬´Ó¶øÏû³ýÓÉÓÚ¼ÓÈëÏ¡ÁòËáÒýÆðµÄÇâÆøÌå»ýÎó²î£»
¢ÚµÎ¶¨¹ÜµÄÊýÖµÁã¿Ì¶ÈÔÚÉÏ·½£¬Á½´ÎµÄÌå»ýÖ®²îΪ²â¶¨µÄÇâÆøµÄÌå»ý£¨×¢ÒâÓ¦±£³Ö¸ÉÔï¹ÜÓëµÎ¶¨¹ÜÄÚÒºÃæµÈ¸ß£©£¬ÊÕ¼¯ÇâÆøºóµÎ¶¨¹ÜÄÚÒºÃæÉÏÉý£¬¶ÁÊý¼õС£®

½â´ð ½â£º·½°¸Ò»£ºÂÁÓëÇâÑõ»¯ÄÆÈÜÒº·´Ó¦Éú³ÉÆ«ÂÁËáÄÆÓëÇâÆø£¬·´Ó¦·½³ÌʽΪ2Al+2NaOH+2H2O=2NaAlO2+3H2¡ü£¬
¹Ê´ð°¸Îª£º2Al+2NaOH+2H2O=2NaAlO2+3H2¡ü£»
£¨1£©º¬Ã¾Îª3%ʱ£¬½ðÊôÂÁµÄº¬Á¿×î¸ß£¬5.4gºÏ½ðÖÐÂÁµÄÖÊÁ¿Îª£¬5.4g¡Á£¨1-3%£©=5.4¡Á97%g£¬Ôò£º
       2Al+2NaOH+2H2O=2NaAlO2+3H2¡ü
      54g   2mol 
5.4g¡Á97%g   V¡Á10-3L¡Á2.0mol/L
ËùÒÔ54g£º£¨5.4g¡Á97%g£©=2mol£º£¨V¡Á10-3L¡Á2.0mol/L£©£¬½âµÃ£ºV=97£¬¹ÊV£¨NaOHÈÜÒº£©¡Ý97mL£»
¹Ê´ð°¸Îª£º97mL£»
£¨2£©Ã¾Éϻḽ×ÅÆ«ÂÁËáÄƵÈÎïÖÊ£¬Î´Ï´µÓµ¼Ö²ⶨµÄþµÄÖÊÁ¿Æ«´ó£¬Ã¾µÄÖÊÁ¿·ÖÊýÆ«¸ß£¬¹Ê´ð°¸Îª£ºÆ«¸ß£»
·½°¸¶þ£º£¨1£©¢Ù×°ÖõÄ×éװ˳Ðò£ººÏ½ðÓëË®·´Ó¦£¬ÓÃÅÅË®Á¿Æø·¨²â¶¨ÇâÆøµÄÌå»ý£¬ÆäÖÐʢˮµÄÊÔ¼ÁÆ¿µ¼¹ÜÒ»¶¨Òª¶Ì½ø³¤³ö£¬ÀûÓÃÔö´óѹǿԭÀí½«Ë®Åųö£¬Á¿Í²ÖÐË®µÄÌå»ý¾ÍÊÇÉú³ÉÇâÆøµÄÌå»ý£¬Á¿Í²ÄÚµ¼¹ÜÓ¦ÉìÈëÁ¿Í²µ×²¿£¬¹ÊÁ¬½Ó˳ÐòΪ£º£¨A£©½Ó£¨E£©£¨D£©½Ó£¨G£©£»
¹Ê´ð°¸Îª£ºE£»D£»G£»
¢Ú·´Ó¦·ÅÈȵ¼ÖÂÇâÆøµÄζÈÆ«¸ß£¬¹ÊÓ¦ÀäÈ´ºóÔÙ½øÐжÁÈ¡ÇâÆøµÄÌå»ý£¬¶ÁȡʵÑéÖÐÉú³ÉÇâÆøµÄÌå»ýʱÉÏÏÂÒƶ¯Á¿Í²£¬Ê¹ÆäÖÐÒºÃæÓë¹ã¿ÚÆ¿ÖÐÒºÃæÏàƽ£¬ÊÓÏßÓë°¼ÒºÃæµÄ×îµÍµãˮƽ¶ÁÈ¡ÇâÆøµÄÌå»ý£»
¹ÊÑ¡ACD£»
£¨2£©¢Ù×°ÖÃÖе¼¹ÜaµÄ×÷ÓÃÊÇ£º±£³Ö·ÖҺ©¶·ÄÚÆøÌåѹǿÓë׶ÐÎÆ¿ÄÚÆøÌåѹǿÏàµÈ£¬´ò¿ª·ÖҺ©¶·»îÈûʱϡÁòËáÄÜ˳ÀûµÎÏ£¬µÎÈë׶ÐÎÆ¿µÄÏ¡ÁòËáÌå»ýµÈÓÚ½øÈë·ÖҺ©¶·µÄÆøÌåÌå»ý£¬´Ó¶øÏû³ýÓÉÓÚ¼ÓÈëÏ¡ÁòËáÒýÆðµÄÇâÆøÌå»ýÎó²î£¬
¹Ê´ð°¸Îª£ºµÎÈë׶ÐÎÆ¿µÄÏ¡ÁòËáÌå»ýµÈÓÚ½øÈë·ÖҺ©¶·µÄÆøÌåÌå»ý£¬´Ó¶øÏû³ýÓÉÓÚ¼ÓÈëÏ¡ÁòËáÒýÆðµÄÇâÆøÌå»ýÎó²î£»
¢ÚµÎ¶¨¹ÜµÄÊýÖµÁã¿Ì¶ÈÔÚÉÏ·½£¬Á½´ÎµÄÌå»ýÖ®²îΪ²â¶¨µÄÇâÆøµÄÌå»ý£¬ÊÕ¼¯ÇâÆøºóµÎ¶¨¹ÜÄÚÒºÃæ¶ÁÊý¼õС£¬ËùÒԲⶨÇâÆøµÄÌå»ýΪV1-V2£¬
¹Ê´ð°¸Îª£ºV1-V2£®

µãÆÀ ±¾Ì⿼²éÎïÖʺ¬Á¿µÄ²â¶¨¡¢¶ÔʵÑéÔ­ÀíÓë×°ÖõÄÀí½â¡¢ÊµÑé·½°¸Éè¼ÆµÈ£¬ÄѶÈÖеȣ¬Àí½âʵÑéÔ­ÀíÊǽâÌâµÄ¹Ø¼ü£¬ÊǶÔ֪ʶµÄ×ۺϿ¼²é£¬ÐèҪѧÉú¾ßÓÐ֪ʶµÄ»ù´¡Óë×ÛºÏÔËÓÃ֪ʶ·ÖÎöÎÊÌâ¡¢½â¾öÎÊÌâµÄÄÜÁ¦£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

12£®ÉèNA´ú±í°¢·ü¼ÓµÂÂÞ³£Êý£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®ÔÚ³£Î³£Ñ¹Ï£¬11.2 L Cl2º¬ÓеķÖ×ÓÊýΪ0.5NA
B£®ÔÚ³£Î³£Ñ¹Ï£¬1 molÇâÆøº¬ÓеķÖ×ÓÊýΪNA
C£®ÔÚ³£Î³£Ñ¹Ï£¬32 gÑõÆøº¬Ô­×ÓÊýΪNA
D£®±ê×¼×´¿öÏ£¬1molË®µÄÌå»ýԼΪ22.4L

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

13£®¸ßÌúËá¼Ø£¨K2FeO4£©¾ßÓиßЧµÄÏû¶¾×÷Óã¬ÎªÒ»ÖÖÐÂÐÍ·ÇÂȸßЧÏû¶¾¼Á£®µç½â·¨ÖƱ¸¸ßÌúËá¼Ø²Ù×÷¼ò±ã£¬²úÂʸߣ¬Ò×ÓÚʵÑéÊÒÖƱ¸£®ÆäÔ­ÀíÈçͼËùʾ£¬Æä×ܵç½â·´Ó¦Îª£º
Fe+2NaOH+2H2O$\frac{\underline{\;µç½â\;}}{\;}$Na2FeO4+3H2¡ü£®
¢ñ£®ÊµÑé¹ý³ÌÖУ¬Xµç¼«ÓÐÆøÌå²úÉú£¬Y¼«ÇøÈÜÒºÖð½¥±ä³É×ϺìÉ«£»Í£Ö¹ÊµÑ飬Ìúµç¼«Ã÷ÏÔ±äϸ£¬µç½âÒºÈÔÈ»³ÎÇ壮²éÔÄ×ÊÁÏ·¢ÏÖ£¬¸ßÌúËá¸ù£¨FeO${\;}_{4}^{2-}$£©ÔÚÈÜÒºÖгÊ×ϺìÉ«£®
£¨1£©µç½â¹ý³ÌÖУ¬X¼«ÊÇÒõ¼«£®
£¨2£©Éú³É¸ßÌúËá¸ù£¨FeO${\;}_{4}^{2-}$£©µÄµç¼«·´Ó¦ÊÇFe+8OH--6e-=FeO42-+4H2O£®
¢ò£®ÈôÓò»Í¬ÖÖµç³Ø×÷ΪÉÏÊöʵÑéµÄµçÔ´£¬Çë·ÖÎöµç³Ø·´Ó¦£®
£¨1£©Ç¦Ðîµç³Ø×ܵĻ¯Ñ§·½³ÌʽΪ£ºPb+PbO2+2H2SO4$\frac{\underline{\;·Åµç\;}}{³äµç}$2H2O+2PbSO4£¬Ôò¸ÃǦÐîµç³ØÔÚ³äµçʱµÄÑô¼«·´Ó¦Îª
PbSO4-2e-+2H2O¨TPbO2+SO42-+4H+£¬ÈôÓô˵ç³ØÏòÍâµç·Êä³ö2molµç×Ó£¬Õý¼«ÔöÖØ64g£¬µç½âÖÊÈÜÒºÏûºÄµÄÁòËá¸ùÀë×ÓΪ1mol£®
£¨2£©ë£¨N2H4£©-¿ÕÆøȼÁϵç³ØÊÇÒ»ÖÖ¼îÐÔȼÁϵç³Ø£¬µç½âÖÊÈÜÒºÊÇ20%¡«30%µÄKOHÈÜÒº£¬·Åµçʱ¸º¼«µÄµç¼«·´Ó¦Ê½ÎªN2H4+4OH--4e-=N2+4H2O£®
£¨3£©ÒÔ±ûÍ飨C3H8£©ÎªÈ¼ÁÏÖÆ×÷ÐÂÐÍȼÁϵç³Ø£¬µç³ØµÄÕý¼«Í¨ÈëO2£¬¸º¼«Í¨Èë±ûÍ飬µ¼µçµç½âÖÊÊÇÈÛÈÚ½ðÊôÑõ»¯Îд³ö¸Ãµç³Ø¸º¼«µÄµç¼«·´Ó¦C3H8-20e-+10O2-=3CO2+4H2O£®
£¨4£©µ±ÖƱ¸ÏàͬÎïÖʵÄÁ¿µÄ¸ßÌúËá¼Øʱ£¬ÀíÂÛÉÏÉÏÊöÈýÖÖµç³ØÖзֱðÏûºÄµÄPb¡¢ë¡¢±ûÍéµÄÎïÖʵÄÁ¿Ö®±ÈÊÇ10£º10£º5£º1£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

10£®ÓÃNA±íʾ°¢·ü¼ÓµÂÂÞ³£Êý£¬ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®10g 46%µÄÒÒ´¼ÈÜÒºËùº¬ÇâÔ­×ÓÊýΪ1.2NA
B£®0.5 molÈÛÈÚµÄNaHSO4Öк¬ÓеÄÀë×ÓÊýĿΪ1.5NA
C£®±ê×¼×´¿öÏ£¬2.24L¼×´¼Öк¬ÓÐC-H¼üµÄÊýĿΪ0.3NA
D£®S2ºÍS8µÄ»ìºÏÎï¹²38.4g£¬ÆäÖÐËùº¬ÁòÔ­×ÓÊýΪ1.4NA

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

17£®ÑÇÁòËáÄÆ£¨Na2SO3£©¡¢Æ¯°×Òº£¨NaClO£©ºÍÃ÷·¯[KAl£¨SO4£©2•12H2O]¶¼ÊÇÖØÒªµÄ»¯¹¤²úÆ·£¬ÔÚÈÕ³£Éú»îºÍ¹¤ÒµÉú²úÖж¼Óй㷺ÓÃ;£¬ÇÒÈýÕ߶¼¿ÉÓÃÔÚÔìÖ½ÒµµÄ²»Í¬Éú²ú¹¤ÐòÖУ®
£¨1£©NaClO×öÏû¶¾ÒºÊÇËüÄÜÓëË®·´Ó¦²úÉúÒ»ÖÖÇ¿Ñõ»¯ÐÔÎïÖÊ£¬Ð´³öNaClOµÄµç×Óʽ£®ÔÚ¹¤ÒµÉÏ£¬ÓÃÂÈ»¯ÄÆΪԭÁÏ£¬ÔÚ¼îÐÔÈÜÒºÖУ¬Í¨¹ýµç½âµÄ·½·¨¿ÉÖƵÃNaClO£¬ÓÃÀë×Ó·½³Ìʽ±íʾÖÆÈ¡NaClOµÄµç½â×Ü·´Ó¦£ºCl-+H2O$\frac{\underline{\;ͨµç\;}}{\;}$ClO-+H2¡ü£®½«µÈŨ¶ÈµÈÌå»ýµÄNaClOÓëNa2SO3ÈÜÒº»ìºÏºó£¬Á½ÕßÇ¡ºÃÍêÈ«·´Ó¦£¬Ð´³ö»ìºÏ¹ý³ÌµÄÀë×Ó·´Ó¦·½³ÌʽClO-+SO32-=Cl-+SO42-£®
£¨2£©½«±¥ºÍNaClOºÍKAl£¨SO4£©2ÈÜÒº»ìºÏ£¬ÖÃÓÚÒ»Ö»´øµ¥¿×ÏðƤÈûµÄ´óÊÔ¹ÜÖУ¬²úÉú´óÁ¿µÄ°×É«½º×´³Áµí£®´Ëʱ·´Ó¦µÄÀë×Ó·½³ÌΪ3ClO-+Al3++3H2O=3HClO+Al£¨OH£©3¡ý£®ÔÙ½«´óÊÔ¹ÜÖÃÓÚÑô¹âÏÂÕÕÉ䣬²»¾ÃÊÔ¹ÜÒºÃæÉÏ·½ÓÐdz»ÆÂÌÉ«ÆøÌå²úÉú£¬½«ÆøÌåͨÈëNaOHÈÜÒº³ä·ÖÎüÊպ󣬻¹ÄÜÊÕ¼¯µ½Ò»ÖÖÎÞÉ«ÎÞζµÄÆøÌ壮д³öÔÚÑô¹âÕÕÉäÏ£¬»ìºÏÒºÖз´Ó¦µÄ»¯Ñ§·½³Ìʽ4HClO$\frac{\underline{\;¹âÕÕ\;}}{\;}$2H2O+2Cl2¡ü+O2¡ü£®Èô½«KAl£¨SO4£©2ÈÜÒº»»³ÉÁòËáÑÇÌúï§ £¨NH4£©2SO4•FeSO4ÈÜÒººó£¬ÔÙÓëNaClOÈÜÒº»ìºÏ£¬¹Û²ìµ½´óÊÔ¹ÜÖÐÓкìºÖÉ«³Áµí²úÉú£¬Í¬Ê±ÈÜÒºÀïÏÖ»ÆÉ«£¬µ«Ã»ÓÐÆøÌåÉú³É£®´ËʱÊÔ¹ÜÖз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ3ClO-+6Fe2++3H2O=2Fe£¨OH£©3¡ý+4Fe3++3Cl-£®
£¨3£©È¡Ò»¶¨Ìå»ýKAl£¨SO4£©2ÈÜÒºÓÚÊÔ¹ÜÖУ¬ÖðµÎ¼ÓÈëÒ»¶¨Å¨¶ÈµÄBa£¨OH£©2ÈÜÒº£¬Ö±ÖÁ²úÉú³ÁµíµÄÖÊÁ¿×î´ó£¬Ð´³ö´ËʱµÄÀë×Ó·´Ó¦·½³ÌʽAl3++2SO42-+2Ba2++4OH-¨T2BaSO4¡ý+AlO2-+2H2O£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

7£®ÃÀ¹ú°¢²¨ÂÞÓîÖæ·É´¬ÉÏʹÓõÄÇâÑõȼÁϵç³ØÊÇÒ»ÖÖÐÂÐ͵Ļ¯Ñ§µç³Ø£¬Æä¹¹ÔìÈçͼËùʾ£»Á½¸öµç¼«¾ùÓɶà¿×ÐÔ̼ÖƳɣ¬Í¨ÈëµÄÆøÌåÓÉ¿×϶ÖÐÒݳö£¬²¢Ôڵ缫±íÃæ·Åµç£®
£¨1£©¢Ùa¼«¸º¼«
¢Úaµç¼«·´Ó¦Ê½Îª2H2-4e-+4OH-=4H2O
£¨2£©ÇâÆøÊÇȼÁϵç³Ø×î¼òµ¥µÄȼÁÏ£¬ËäȻʹÓ÷½±ã£¬È´Êܵ½¼Û¸ñºÍÀ´Ô´µÄÏÞÖÆ£®³£ÓõÄȼÁÏ£¬ÍùÍùÊÇijЩ̼Ç⻯ºÏÎÈç¼×Í飨ÌìÈ»Æø£©¡¢ÆûÓ͵ȣ®Çëд³ö½«Í¼ÖÐH2»»³ÉCH4ʱËù¹¹³ÉµÄ¼×ÍéȼÁϵç³ØÖР¢Ùa¼«µÄµç¼«·´Ó¦Ê½CH4+10OH--8e-=CO32-+7H2O
¢Ú¼×ÍéȼÁϵç³Ø×ܵÄÀë×Ó·´Ó¦·½³ÌʽΪCH4+2O2+2OH-=CO32-+3H2O£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

14£®ÏÂÁÐ˵·¨ÖУ¬´íÎóµÄÊÇ£¨¡¡¡¡£©
A£®ÈËÀàÄ¿Ç°ËùÖ±½ÓÀûÓõÄÄÜÁ¿´ó²¿·ÖÊÇÓÉ»¯Ñ§·´Ó¦²úÉúµÄ
B£®»¯Ñ§±ä»¯ÖеÄÄÜÁ¿±ä»¯Ö÷ÒªÊÇÓÉ»¯Ñ§¼ü±ä»¯ÒýÆðµÄ
C£®»¯Ñ§·´Ó¦ÖÐÄÜÁ¿±ä»¯µÄ´óСÓë·´Ó¦ÎïµÄÖÊÁ¿¶àÉÙÎÞ¹Ø
D£®ÄÜÁ¿±ä»¯ÊÇ»¯Ñ§·´Ó¦µÄ»ù±¾ÌØÕ÷Ö®Ò»

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

11£®ÔÚÈÝ»ý²»±äµÄÃܱÕÈÝÆ÷ÖÐÓз´Ó¦£ºX£¨g£©+Y£¨g£©?Z£¨g£©£¬ÈôZ£¨g£©µÄÎïÖʵÄÁ¿Å¨¶Èc£¨Z£©ÓëζÈTµÄ¹ØϵÈçͼËùʾ£¨ÇúÏßÉϵÄÈÎÒâÒ»µã¶¼±íʾƽºâ״̬£©£®ÔòÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®AµãÓëBµãÏà±È£¬BµãµÄc£¨X£©´óB£®AµãÓëCµãµÄ»¯Ñ§·´Ó¦ËÙÂÊ£ºA£¼C
C£®ÔÚ·´Ó¦½øÐе½Dµãʱ£¬vÕý£¾vÄæD£®¸Ã·´Ó¦µÄÕý·´Ó¦ÊÇÎüÈÈ·´Ó¦

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

12£®±ê×¼×´¿öÏ£¬½«1g H2¡¢11g CO2ºÍ4g O2»ìºÏ£¬¸Ã»ìºÏÆøÌåµÄÌå»ýԼΪ£¨¡¡¡¡£©
A£®16.8LB£®14LC£®19.6LD£®18.4L

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸